Digital Sumorial












20














Given an input n, write a program or function that outputs/returns the sum of the digital sums of n for all the bases 1 to n.



$$n + sum_{b=2}^n sum_{i=0}^infty leftlfloor frac{n}{b^i} rightrfloor bmod b$$



Example:



n = 5





Create the range [1...n]: [1,2,3,4,5]





For each element x, get an array of the base-x digits of n: [[1,1,1,1,1],[1,0,1],[1,2],[1,1],[1,0]]



bijective base-1 of 5 is [1,1,1,1,1]



base-2 (binary) of 5 is [1,0,1]



base-3 of 5 is [1,2]



base-4 of 5 is [1,1]



base-5 of 5 is [1,0]





Sum the digits: 13





Test cases:



1    1
2 3
3 6
4 8
5 13
6 16
7 23
8 25
9 30
10 35

36 297
37 334

64 883
65 932


The sequence can be found on OEIS: A131383



Scoring:



code-golf: The submission with the lowest score wins.










share|improve this question


















  • 4




    A fun one: 227 -> 9999. And also: 1383 -> 345678.
    – Arnauld
    Dec 3 at 18:54












  • Any feedback for the downvote would be appreciated.
    – Oliver
    Dec 14 at 18:29
















20














Given an input n, write a program or function that outputs/returns the sum of the digital sums of n for all the bases 1 to n.



$$n + sum_{b=2}^n sum_{i=0}^infty leftlfloor frac{n}{b^i} rightrfloor bmod b$$



Example:



n = 5





Create the range [1...n]: [1,2,3,4,5]





For each element x, get an array of the base-x digits of n: [[1,1,1,1,1],[1,0,1],[1,2],[1,1],[1,0]]



bijective base-1 of 5 is [1,1,1,1,1]



base-2 (binary) of 5 is [1,0,1]



base-3 of 5 is [1,2]



base-4 of 5 is [1,1]



base-5 of 5 is [1,0]





Sum the digits: 13





Test cases:



1    1
2 3
3 6
4 8
5 13
6 16
7 23
8 25
9 30
10 35

36 297
37 334

64 883
65 932


The sequence can be found on OEIS: A131383



Scoring:



code-golf: The submission with the lowest score wins.










share|improve this question


















  • 4




    A fun one: 227 -> 9999. And also: 1383 -> 345678.
    – Arnauld
    Dec 3 at 18:54












  • Any feedback for the downvote would be appreciated.
    – Oliver
    Dec 14 at 18:29














20












20








20







Given an input n, write a program or function that outputs/returns the sum of the digital sums of n for all the bases 1 to n.



$$n + sum_{b=2}^n sum_{i=0}^infty leftlfloor frac{n}{b^i} rightrfloor bmod b$$



Example:



n = 5





Create the range [1...n]: [1,2,3,4,5]





For each element x, get an array of the base-x digits of n: [[1,1,1,1,1],[1,0,1],[1,2],[1,1],[1,0]]



bijective base-1 of 5 is [1,1,1,1,1]



base-2 (binary) of 5 is [1,0,1]



base-3 of 5 is [1,2]



base-4 of 5 is [1,1]



base-5 of 5 is [1,0]





Sum the digits: 13





Test cases:



1    1
2 3
3 6
4 8
5 13
6 16
7 23
8 25
9 30
10 35

36 297
37 334

64 883
65 932


The sequence can be found on OEIS: A131383



Scoring:



code-golf: The submission with the lowest score wins.










share|improve this question













Given an input n, write a program or function that outputs/returns the sum of the digital sums of n for all the bases 1 to n.



$$n + sum_{b=2}^n sum_{i=0}^infty leftlfloor frac{n}{b^i} rightrfloor bmod b$$



Example:



n = 5





Create the range [1...n]: [1,2,3,4,5]





For each element x, get an array of the base-x digits of n: [[1,1,1,1,1],[1,0,1],[1,2],[1,1],[1,0]]



bijective base-1 of 5 is [1,1,1,1,1]



base-2 (binary) of 5 is [1,0,1]



base-3 of 5 is [1,2]



base-4 of 5 is [1,1]



base-5 of 5 is [1,0]





Sum the digits: 13





Test cases:



1    1
2 3
3 6
4 8
5 13
6 16
7 23
8 25
9 30
10 35

36 297
37 334

64 883
65 932


The sequence can be found on OEIS: A131383



Scoring:



code-golf: The submission with the lowest score wins.







code-golf math sequence integer base-conversion






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Dec 3 at 16:31









Oliver

4,7351831




4,7351831








  • 4




    A fun one: 227 -> 9999. And also: 1383 -> 345678.
    – Arnauld
    Dec 3 at 18:54












  • Any feedback for the downvote would be appreciated.
    – Oliver
    Dec 14 at 18:29














  • 4




    A fun one: 227 -> 9999. And also: 1383 -> 345678.
    – Arnauld
    Dec 3 at 18:54












  • Any feedback for the downvote would be appreciated.
    – Oliver
    Dec 14 at 18:29








4




4




A fun one: 227 -> 9999. And also: 1383 -> 345678.
– Arnauld
Dec 3 at 18:54






A fun one: 227 -> 9999. And also: 1383 -> 345678.
– Arnauld
Dec 3 at 18:54














Any feedback for the downvote would be appreciated.
– Oliver
Dec 14 at 18:29




Any feedback for the downvote would be appreciated.
– Oliver
Dec 14 at 18:29










25 Answers
25






active

oldest

votes


















7















Canvas, 3 bytes



┬]∑


Try it here!



Canvas beating Jelly?



{ ]   map over 1..input (implicit "{")
┬ decode input from that base
∑ sum the resulting (nested) array





share|improve this answer































    7















    Haskell, 46 bytes





    f n=sum$do a<-[1..n];mapM(pure[0..a])[1..n]!!n


    Try it online!



    Explanation



    The function b n -> mapM(pure[0..b])[1..n], generates all strings $[0 dotsc b]^n$ in lexicographic order. For example:



    mapM(pure[0..2])[0..1] == [[0,0],[0,1],[0,2],[1,0],[1,1],[1,2],[2,0],[2,1],[2,2]]


    By indexing into it with (!!n) this can be used to convert n to base b+1, however this won't work for unary (base-$1$), but we're summing the results.. We can even save some bytes with a <- [1..n] and using base-$(n+1)$ rather than a work-around for base-$1$ since we're missing $[underset{ntext{ times}}{underbrace{1,dotsc,1}}]$ which is the same as $[n]$ when summing.



    Using do-notation just concatenates all the lists instead of nesting them:



    λ [ mapM (pure [0..a]) [1..5] !! n | a <- [1..5] ]
    [[0,0,1,0,1],[0,0,0,1,2],[0,0,0,1,1],[0,0,0,1,0],[0,0,0,0,5]]

    λ do a <- [1..5]; mapM (pure [0..a]) [1..5] !! n
    [0,0,1,0,1,0,0,0,1,2,0,0,0,1,1,0,0,0,1,0,0,0,0,0,5]





    share|improve this answer































      6















      Ruby, 39 37



      ->n{(2..n).sum{|b|n.digits(b).sum}+n}


      The only golfing here is removing some whitespace. Try it online






      share|improve this answer























      • @Giuseppe The OP starts with: " Given an input n,(...)" . I haven' t been around here for a long time. Is there a list somewhere of requirements to a solution?
        – steenslag
        Dec 4 at 0:46






      • 2




        Usually it must be a full program or a function (named or unnamed), here's more information: loopholes and i/o defaults. I don't know ruby, but this seems the shortest fix.
        – BMO
        Dec 4 at 0:52












      • @BMO Thanks. For someone who doesn't know Ruby, you are creating and calling a lambda with the greatest of ease!
        – steenslag
        Dec 4 at 1:04






      • 1




        you can remove the parentheses around n (37b): ->n{(2..n).sum{|b|n.digits(b).sum}+n}
        – Conor O'Brien
        Dec 4 at 1:06






      • 1




        Well, google "ruby lambda" did the trick ;P But I stand corrected in that you were able to save two bytes.
        – BMO
        Dec 4 at 12:42



















      5















      Python 2, 57 bytes





      lambda n:sum(n/(k/n+2)**(k%n)%(k/n+2)for k in range(n*n))


      This uses the formula $$a(n)=sum_{b=2}^{n+1}sum_{i=0}^{n-1}leftlfloorfrac{n}{b^i}rightrfloorbmod b,$$ which works since $leftlfloorfrac{n}{(n+1)^0}rightrfloorbmod(n+1)=n$.



      Try it online!






      share|improve this answer































        5















        APL (Dyalog Unicode), 14 bytes





        +/∘∊⊢,⊢(⍴⊤⊣)¨⍳


        Try it online!



        Explanation



        Some parentheses are implied and can be added (lighter than the "official" parenthesizing):



        +/∘∊(⊢,(⊢(⍴⊤⊣)¨⍳))


        This is a monadic atop. Given an argument Y, this function behaves like:



        +/∘∊(⊢,(⊢(⍴⊤⊣)¨⍳))Y


        The two functions are applied in order. We'll start from the right one:



        ⊢,(⊢(⍴⊤⊣)¨⍳)


        The are three functions in this train, so this is a fork. Given an argument Y, it acts as:



        (⊢Y),((⊢Y)(⍴⊤⊣)¨⍳Y)


        We can easily reduce to this (monadic returns its argument, hence called identity):



        Y,Y(⍴⊤⊣)¨⍳Y


        Now, we know that Y is an integer (simple scalar, i.e. number or character), since we're given one. Therefore ⍳Y, with ⎕IO=1, returns 1 2 ... Y. ⍳Y actually returns an array with shape Y (Y must be a vector), where every scalar is the index of itself in the array (that's why monadic is called the index generator). These indices are vectors, except for the case where 1≡⍴Y, where they are scalars (this is our case).



        Let's parse the middle function, (⍴⊤⊣)¨, next. ⍴⊤⊣ is the operand of ¨ (each), and the function is dyadic, so the ¨ operator will first reshape each length-1 argument to the shape of the other (that is, take the element and use it to replace every scalar in the other argument), and then apply the function to each pair of the two arguments. In this case, ⍳Y is a vector and Y is a scalar, so, if n≡⍴⍳Y, then Y will be converted to n⍴Y ( represents the shape (monadic) and reshape (dyadic) functions). That is, in simpler terms, Y will be converted to an array containing Y times Y.



        Now, for each pair, let's call the left argument X and the right Z (so that we don't conflict with the input Y). ⍴⊤⊣ is a dyadic fork, so it will expand to:



        (X⍴Z)⊤X⊣Z


        Let's make the easy first step of reducing X⊣Z to X (dyadic is the left function):



        (X⍴Z)⊤X


        The in X⍴Z is, again, the reshape function, so X⍴Z, in our case, is simply X times Z. is the encode function. Given two arrays of numbers, where the left array is the base of each digit in the result (doesn't need to be integer or positive), i.e. the encoding, and the right is an array of numbers, returns the transposed array of those numbers in the specified encoding (transposition is the reversal of an array's dimensions relative to its elements). The representation of a digit is based on the quotient of the division of the number and the product of the less significant bases. If any base is 0, it acts as base +∞. The arguments' scalars are all simple. Since X is a positive integer, and X⍴Z is a vector of equal elements, this is really just a case of converting X to base Z and reshaping to X digits. For $X,Zinmathbb N$, $X_Z$ ($X$ in base $Z$) can't have more than $X$ digits, since $X_1$ has $X$ digits. Therefore, X⍴Z is enough for our purposes.



        The result of Y(⍴⊤⊣)¨⍳Y is, therefore, Y converted to each base from 1 to Y, possibly with leading zeroes. However, there is one issue: in APL, base 1 isn't special-cased, while this challenge does special-case it, so we have to include the sum of the base-1 digits of Y ourselves. Fortunately, this sum is just Y, since $Y_1=underbrace{[1,1,...,1]}_Y$, so the sum is simply $Ytimes1=Y$. It follows that we have to add Y somewhere into the array. This is how we do it:



        Y,Y(⍴⊤⊣)¨⍳Y


        I have already included this part here. Dyadic , is the catenate function, it concatenates its arguments on their last axes, and errors if that's not possible. Here, we simply concatenate the scalar Y to the vector Y(⍴⊤⊣)¨⍳Y, so that we increment the sum we're going to calculate by Y, as explained above.



        The final part is the left function of our atop, +/∘∊:



        +/∘∊Y,Y(⍴⊤⊣)¨⍳Y


        is the compose operator. f∘g Y is the same as f g Y. However, we're using it here so that our train doesn't fork on the . So, we can reduce:



        +/∊Y,Y(⍴⊤⊣)¨⍳Y


        Now, it's time for the sum, but wait... there's a problem. The array isn't flat, so we can't just sum its elements before flattening it first. The enlist function flattens an array. Now that the array has been flattened, we finally use +/ to sum it. / is the reduce operator, it applies a dyadic function between an array's elements on its second-to-last axis, with right-to-left priority. If the rank (number of dimensions, i.e. length of shape) of the array doesn't decrease, the array is then enclosed, although this is not the case here. The function that's applied here is +, which is the plus function that adds the pairs on the last axes of two arrays (and errors if the arrays can't be added like that). Here, it simply adds two numbers a number of times so that the reduce is completed.



        Lo and behold, our train:



        +/∘∊⊢,⊢(⍴⊤⊣)¨⍳





        share|improve this answer























        • +1 Impressive explanation. Not shorter, but no parens: +/∘∊⊢,⊢⊤¨⍨⊢⍴¨⍳
          – Adám
          Dec 4 at 9:19












        • @Adám Thanks! I was almost sleeping when I wrote it. :-P
          – Erik the Outgolfer
          Dec 4 at 11:13










        • @Adám Regarding your version, it looks like it's a tad more difficult to understand. ;-)
          – Erik the Outgolfer
          Dec 4 at 11:26



















        5














        Java 8, 76 65 bytes





        n->{int r=n,b=n,N;for(;b>1;b--)for(N=n;N>0;N/=b)r+=N%b;return r;}


        -11 bytes thanks to @OlivierGrégoire.



        Try it online.



        Explanation:



        n->{           // Method with integer as both parameter and return-type
        int r=n, // Result-sum, starting at the input to account for Base-1
        b=n, // Base, starting at the input
        N; // Temp integer
        for(;b>1 // Loop the Base `b` in the range [n, 1):
        ;b--) // After every iteration: Go to the next Base (downwards)
        for(N=n; // Set `N` to the input `n`
        N>0; // Loop as long as `N` isn't 0 yet:
        N/=b) // After every iteration: Divide `N` by the Base `b`
        r+=N%b; // Increase the result `r` by `N` modulo the Base `b`
        return r;} // Return the result-sum `r`





        share|improve this answer



















        • 1




          There is no need for i, so... 65 bytes.
          – Olivier Grégoire
          Dec 5 at 11:01












        • @OlivierGrégoire Damn, I'm an idiot. Thanks a lot! Hmm, now I also need to golf my derived C and Whitespace answers.. ;)
          – Kevin Cruijssen
          Dec 5 at 11:03



















        4














        SAS, 81 74 bytes



        data;input n;s=n;do b=2to n;do i=0to n;s+mod(int(n/b**i),b);end;end;cards;


        Input is entered after the cards; statement, on newlines, like so:



        data;input n;s=n;do b=2to n;do i=0to n;s+mod(int(n/b**i),b);end;end;cards;
        1
        2
        3
        4
        5
        6
        7
        8
        9
        10
        36
        37
        64
        65


        Outputs a dataset containing the answer s (along with helper variables), with a row for each input value



        enter image description here



        Ungolfed:



        data; /* Implicit dataset creation */
        input n; /* Read a line of input */

        s=n; /* Sum = n to start with (the sum of base 1 digits of n is equal to n) */
        do b=2to n; /* For base = 2 to n */
        do i=0to n; /* For each place in the output base (output base must always have less than or equal to n digits) */

        /* Decimal value of current place in the output base = b^i */
        /* Remainder = int(b / decimal_value) */
        /* Current place digit = mod(remainder, base) */
        s+mod(int(n/b**i),b);
        end;
        end;

        cards;
        1
        2
        3





        share|improve this answer































          3














          Japt -x, 6 bytes



          ÆìXÄ x


          Try it



          õ!ìU c


          Try it






          share|improve this answer





























            3















            J, 24 23 bytes



            +1#.1#.]#.inv"0~2+i.@<:


            Try it online!






            share|improve this answer































              3















              05AB1E (legacy), 5 bytes



              LвOO+


              Try it online!



              Explanation:



              LвOO+  //Full program
              L //push [1 .. input]
              в //for each element b, push digits of input converted to base b
              O //sum each element
              O //sum each sum
              + //add input


              In 05AB1E (legacy), base 1 of 5 is [0,0,0,0,0], not [1,1,1,1,1]. Therefore after summing the range, add the input to account for the missing base 1.



              I am using 05AB1E (legacy) because in current 05AB1E, base 1 of 5 is [1]. In order to account for this, I would either need to decrement the result by 1 or remove the first element of the range, both of which would cost 1 byte.






              share|improve this answer





























                3














                Desmos, 127 bytes



                $$fleft(nright)=sum_{b=2}^{n+1}sum_{i=0}^{n-1}operatorname{mod}left(operatorname{floor}left(frac{n}{b^i}right),bright)$$



                fleft(nright)=sum_{b=2}^{n+1}sum_{i=0}^{n-1}operatorname{mod}left(operatorname{floor}left(frac{n}{b^i}right),bright)


                Try it here! Defines a function $f$, called as $fleft(nright)$. Uses the formula given in Dennis's answer.



                Here is the resulting graph (point at $(65, 932)$):



                graph generated on desmos.com



                Desmos, 56 bytes



                This might not work on all browsers, but it works on mine. Just copy and paste the formula. This is $f_2(n)$ in the above link.



                f(n)=sum_{b=2}^{n+1}sum_{i=0}^{n-1}mod(floor(n/b^i),b)





                share|improve this answer





























                  3















                  Perl 6, 45 41 bytes



                  {$^a+sum map {|polymod $a: $_ xx*},2..$a}


                  -4 bytes thanks to Jo King



                  Try it online!






                  share|improve this answer























                  • 36 bytes
                    – nwellnhof
                    Dec 6 at 14:39



















                  3















                  Whitespace, 153 bytes



                  [S S S N
                  _Push_0][S N
                  S _Duplicate_0][T N
                  T T _Read_STDIN_as_integer][T T T _Retrieve_input][S N
                  S _Duplicate_input][N
                  S S N
                  _Create_Label_OUTER_LOOP][S N
                  S _Duplicate_top][S S S T N
                  _Push_1][T S S T _Subtract][N
                  T S S N
                  _If_0_Jump_to_Label_PRINT][S S S N
                  _Push_0][S N
                  S _Duplicate_0][T T T _Retrieve_input][N
                  S S T N
                  _Create_Label_INNER_LOOP][S N
                  S _Duplicate][N
                  T S S S N
                  _If_0_Jump_to_Label_AFTER_INNER_LOOP][S N
                  T _Swap_top_two][S T S S T N
                  _Copy_1st_item_to_top][S T S S T T N
                  _Copy_3rd_item_to_top][T S T T _Modulo][T S S S _Add][S N
                  T _Swap_top_two][S T S S T S N
                  _Copy_2nd_item_to_top][T S T S _Integer_divide][N
                  S N
                  T N
                  _Jump_to_Label_INNER_LOOP][N
                  S S S S N
                  _Create_Label_AFTER_INNER_LOOP][S N
                  N
                  _Discard_top][S T S S T S N
                  _Copy_2nd_item_to_top][T S S S _Add][S N
                  T _Swap_top_two][S S S T N
                  _Push_1][T S S T _Subtract][N
                  S N
                  N
                  _Jump_to_Label_OUTER_LOOP][N
                  S S S N
                  _Create_Label_PRINT][S N
                  N
                  _Discard_top][T N
                  S T _Print_as_integer]


                  Letters S (space), T (tab), and N (new-line) added as highlighting only.
                  [..._some_action] added as explanation only.



                  Try it online (with raw spaces, tabs and new-lines only).



                  Port of my Java 8 answer, because Whitespace has no Base conversion builtins at all.



                  Example run: input = 3



                  Command    Explanation                     Stack            Heap   STDIN  STDOUT  STDERR

                  SSSN Push 0 [0]
                  SNS Duplicate 0 [0,0]
                  TNTT Read STDIN as integer [0] {0:3} 3
                  TTT Retrieve input from heap 0 [3] {0:3}
                  SNS Duplicate 3 [3,3] {0:3}
                  NSSN Create Label_OUTER_LOOP [3,3] {0:3}
                  SNS Duplicate 3 [3,3,3] {0:3}
                  SSSTN Push 1 [3,3,3,1] {0:3}
                  TSST Subtract (3-1) [3,3,2] {0:3}
                  NTSSN If 0: Jump to Label_PRINT [3,3] {0:3}
                  SSSN Push 0 [3,3,0] {0:3}
                  SNS Duplicate 0 [3,3,0,0] {0:3}
                  TTT Retrieve input from heap 0 [3,3,0,3] {0:3}
                  NSSTN Create Label_INNER_LOOP [3,3,0,3] {0:3}
                  SNS Duplicate 3 [3,3,0,3,3] {0:3}
                  NTSSSN If 0: Jump to Label_AFTER_IL [3,3,0,3] {0:3}
                  SNT Swap top two [3,3,3,0] {0:3}
                  STSSTN Copy (0-indexed) 1st [3,3,3,0,3] {0:3}
                  STSSTTN Copy (0-indexed) 3rd [3,3,3,0,3,3] {0:3}
                  TSTT Modulo (3%3) [3,3,3,0,0] {0:3}
                  TSSS Add (0+0) [3,3,3,0] {0:3}
                  SNT Swap top two [3,3,0,3] {0:3}
                  STSSTSN Copy (0-indexed) 2nd [3,3,0,3,3] {0:3}
                  TSTS Integer divide (3/3) [3,3,0,1] {0:3}
                  NSNTN Jump to LABEL_INNER_LOOP [3,3,0,1] {0:3}

                  SNS Duplicate 1 [3,3,0,1,1] {0:3}
                  NTSSSN If 0: Jump to Label_AFTER_IL [3,3,0,1] {0:3}
                  SNT Swap top two [3,3,1,0] {0:3}
                  STSSTN Copy (0-indexed) 1st [3,3,1,0,1] {0:3}
                  STSSTTN Copy (0-indexed) 3rd [3,3,1,0,1,3] {0:3}
                  TSTT Modulo (1%3) [3,3,1,0,1] {0:3}
                  TSSS Add (0+1) [3,3,1,1] {0:3}
                  SNT Swap top two [3,3,1,1] {0:3}
                  STSSTSN Copy (0-indexed) 2nd [3,3,1,1,3] {0:3}
                  TSTS Integer divide (1/3) [3,3,1,0] {0:3}
                  NSNTN Jump to LABEL_INNER_LOOP [3,3,1,0] {0:3}

                  SNS Duplicate 0 [3,3,1,0,0] {0:3}
                  NTSSSN If 0: Jump to Label_AFTER_IL [3,3,1,0] {0:3}
                  NSSSSN Create Label_AFTER_IL [3,3,1,0] {0:3}
                  SNN Discard top [3,3,1] {0:3}
                  STSSTSN Copy (0-indexed) 2nd [3,3,1,3] {0:3}
                  TSSS Add (1+3) [3,3,4] {0:3}
                  SNT Swap top two [3,4,3] {0:3}
                  SSSTN Push 1 [3,4,3,1] {0:3}
                  TSST Subtract (3-1) [3,4,2] {0:3}
                  NSNN Jump to LABEL_OUTER_LOOP [3,4,2] {0:3}

                  SNS Duplicate 2 [3,4,2,2] {0:3}
                  SSSTN Push 1 [3,4,2,2,1] {0:3}
                  TSST Subtract (2-1) [3,4,2,1] {0:3}
                  NTSSN If 0: Jump to Label_PRINT [3,4,2] {0:3}
                  SSSN Push 0 [3,4,2,0] {0:3}
                  SNS Duplicate 0 [3,4,2,0,0] {0:3}
                  TTT Retrieve input from heap 0 [3,4,2,0,3] {0:3}
                  NSSTN Create Label_INNER_LOOP [3,4,2,0,3] {0:3}
                  SNS Duplicate 3 [3,4,2,0,3,3] {0:3}
                  NTSSSN If 0: Jump to Label_AFTER_IL [3,4,2,0,3] {0:3}
                  SNT Swap top two [3,4,2,3,0] {0:3}
                  STSSTN Copy (0-indexed) 1st [3,4,2,3,0,3] {0:3}
                  STSSTTN Copy (0-indexed) 3rd [3,4,2,3,0,3,2] {0:3}
                  TSTT Modulo (3%2) [3,4,2,3,0,1] {0:3}
                  TSSS Add (0+1) [3,4,2,3,1] {0:3}
                  SNT Swap top two [3,4,2,1,3] {0:3}
                  STSSTSN Copy (0-indexed) 2nd [3,4,2,1,3,2] {0:3}
                  TSTS Integer divide (3/2) [3,4,2,1,1] {0:3}
                  NSNTN Jump to LABEL_INNER_LOOP [3,4,2,1,1] {0:3}

                  SNS Duplicate 1 [3,4,2,1,1,1] {0:3}
                  NTSSSN If 0: Jump to Label_AFTER_IL [3,4,2,1,1] {0:3}
                  SNT Swap top two [3,4,2,1,1] {0:3}
                  STSSTN Copy (0-indexed) 1st [3,4,2,1,1,1] {0:3}
                  STSSTTN Copy (0-indexed) 3rd [3,4,2,1,1,1,2] {0:3}
                  TSTT Modulo (1%2) [3,4,2,1,1,1] {0:3}
                  TSSS Add (1+1) [3,4,2,1,2] {0:3}
                  SNT Swap top two [3,4,2,2,1] {0:3}
                  STSSTSN Copy (0-indexed) 2nd [3,4,2,2,1,2] {0:3}
                  TSTS Integer divide (1/2) [3,4,2,2,0] {0:3}
                  NSNTN Jump to LABEL_INNER_LOOP [3,4,2,2,0] {0:3}

                  SNS Duplicate 0 [3,4,2,2,0,0] {0:3}
                  NTSSSN If 0: Jump to Label_AFTER_IL [3,4,2,2,0] {0:3}
                  NSSSSN Create Label_AFTER_IL [3,4,2,2,0] {0:3}
                  SNN Discard top [3,4,2,2] {0:3}
                  STSSTSN Copy (0-indexed) 2nd [3,4,2,2,4] {0:3}
                  TSSS Add (2+4) [3,4,2,6] {0:3}
                  SNT Swap top two [3,4,6,2] {0:3}
                  SSSTN Push 1 [3,4,6,2,1] {0:3}
                  TSST Subtract (2-1) [3,4,6,1] {0:3}
                  NSNN Jump to LABEL_OUTER_LOOP [3,4,6,1] {0:3}

                  SNS Duplicate 1 [3,4,6,1,1] {0:3}
                  SSSTN Push 1 [3,4,6,1,1,1] {0:3}
                  TSST Subtract (1-1) [3,4,6,1,0] {0:3}
                  NTSSN If 0: Jump to Label_PRINT [3,4,6,1] {0:3}
                  NSSSSN Create Label_PRINT [3,4,6,1] {0:3}
                  SNN Discard top [3,4,6] {0:3}
                  TNST Print top (6) to STDOUT as int [3,4] {0:3} 6
                  error


                  Program stops with an error: No exit found. (Although I could add three trailing newlines NNN to get rid of that error.)






                  share|improve this answer





























                    3















                    R, 60 bytes





                    function(n){if(n-1)for(i in 2:n)F=F+sum(n%/%i^(0:n)%%i)
                    n+F}


                    Try it online!



                    Fails for n>143 since 144^144 is larger than a double can get. Thanks to Josh Eller for suggesting replacing log(n,i) with simply n.



                    The below will work for n>143; not sure at what point it will stop working.




                    R, 67 bytes





                    function(n){if(n-1)for(i in 2:n)F=F+sum(n%/%i^(0:log(n,i))%%i)
                    n+F}


                    Try it online!



                    Uses the classic n%/%i^(0:log(n,i))%%i method to extract the base-i digits for n for each base b>1, then sums them and accumulates the sum in F, which is initialized to 0, then adding n (the base 1 representation of n) to F and returning the result. For n=1, it skips the bases and simply adds n to F.






                    share|improve this answer



















                    • 1




                      I don't know any R, but instead of using 0:log(n,i), couldn't you use 0:n? There's always going to be at most n digits in any base representation of n, and everything after the initial log(n,i) digits should be 0, so it won't affect the sum.
                      – Josh Eller
                      Dec 4 at 20:18










                    • @JoshEller I suppose I could. It would start to fail for n=144, since 143^143 is around 1.6e308 and 144^144 evaluates to Inf. Thanks!
                      – Giuseppe
                      Dec 4 at 20:23





















                    3














                    C (gcc), 67 56 bytes





                    b,a,s;e(n){for(b=a=n;a>1;a--)for(s=n;s;s/=a)b+=s%a;n=b;}


                    Port of my Java 8 answer.

                    -11 bytes thanks to @OlivierGrégoire's golf on my Java answer.



                    Try it online.



                    Explanation:



                    b,             // Result integer
                    a, // Base integer
                    s; // Temp integer
                    e(n){ // Method with integer as parameter
                    for(b=a=n; // Set the result `b` to input `n` to account for Base-1
                    // And set Base `a` to input `n` as well
                    a>1 // Loop the Base `a` in the range [input, 1):
                    ;a--) // After every iteration: Go to the next Base (downwards)
                    for(s=n; // Set `s` to input `n`
                    s; // Inner loop as long as `s` is not 0 yet:
                    s/=a) // After every iteration: Divide `s` by Base `a`
                    b+=s%a; // Add `s` modulo Base `a` to the result `b`
                    n=b;} // Set input `n` to result `b` to 'return it'





                    share|improve this answer































                      2














                      JavaScript (ES6), 42 bytes



                      This version is almost identical to my main answer but relies on arithmetic underflow to stop the recursion. The highest supported value depends on the size of the call stack.





                      f=(n,b=1,x)=>b>n?n:~~x%b+f(n,b+!x,x?x/b:n)


                      Try it online!





                      JavaScript (ES6),  51 48  44 bytes





                      f=(n,b=2,x=n)=>b>n?n:x%b+f(n,b+!x,x?x/b|0:n)


                      Try it online!



                      Commented



                      f = (             // f = recursive function taking:
                      n, // - n = input
                      b = 2, // - b = current base, initialized to 2
                      x = n // - x = current value being converted in base b
                      ) => //
                      b > n ? // if b is greater than n:
                      n // stop recursion and return n, which is the sum of the digits of n
                      // once converted to unary
                      : // else:
                      x % b + // add x modulo b to the final result
                      f( // and add the result of a recursive call:
                      n, // pass n unchanged
                      b + !x, // increment b if x = 0
                      x ? // if x is not equal to 0:
                      x / b | 0 // update x to floor(x / b)
                      : // else:
                      n // reset x to n
                      ) // end of recursive call





                      share|improve this answer































                        2















                        APL (Dyalog Unicode), 22 bytes





                        {1⊥⍵,∊(1+⍳⍵-1)⊥⍣¯1¨⊂⍵}


                        Try it online!






                        share|improve this answer





























                          2















                          Husk, 6 bytes



                          I really wish that there was something like M for cmap :(



                          Σ§ṁ`Bḣ


                          Try it online or test all!



                          Explanation



                          Σ§ṁ`Bḣ  -- example input: 5
                          § -- fork the argument..
                          ḣ -- | range: [1,2,3,4,5]
                          `B -- | convert argument to base: (`asBase` 5)
                          ṁ -- | and concatMap: cmap (`asBase` 5) [1,2,3,4,5]
                          -- : [1,1,1,1,1,1,0,1,1,2,1,1,1,0]
                          Σ -- sum: 13


                          Alternatively, 6 bytes



                          ΣΣṠMBḣ


                          Try it online or test all!



                          Explanation



                          ΣΣṠMBḣ  -- example input: 5
                          Ṡ -- apply ..
                          ḣ -- | range: [1,2,3,4,5]
                          .. to function applied to itself
                          MB -- | map over left argument of base
                          -- : [[1,1,1,1,1],[1,0,1],[1,2],[1,1],[1,0]]
                          Σ -- flatten: [1,1,1,1,1,1,0,1,1,2,1,1,1,0]
                          Σ -- sum: 13





                          share|improve this answer





























                            2















                            Pari/GP, 30 bytes



                            n->n+sum(b=2,n,sumdigits(n,b))


                            Try it online!






                            share|improve this answer





























                              2















                              Python 2, 61 bytes





                              f=lambda n,b=2.:n and n%b+f(n//b,b)+(n//b>1/n==0and f(n,b+1))


                              Try it online!



                              Though this is longer the Dennis's solution off which it's based, I find the method too amusing to not share.



                              The goal is to recurse both on lopping off the last digit n->n/b and incrementing the base b->b+1, but we want to prevent the base from being increased after one or more digits have been lopped off. This is achieved by make the base b a float, so that after the update n->n//b, the float b infects n with its floatness. In this way, whether n is a float or not is a bit-flag for whether we've removed any digits from n.



                              We require that the condition 1/n==0 is met to recurse into incrementing b, which integers n satisfy because floor division is done, but floats fails. (n=1 also fails but we don't want to recurse on it anyway.) Otherwise, floats work just like integers in the function because we're careful to do floor division n//b, and the output is a whole-number float.






                              share|improve this answer





























                                1















                                Jelly, 4 bytes



                                bRFS


                                Try it online!



                                How it works



                                bRFS  Main link. Argument: n

                                R Range; yield [1, ..., n].
                                b Convert n to back k, for each k to the right.
                                F Flatten the resulting 2D array of digits.
                                S Take the sum.





                                share|improve this answer





























                                  1















                                  Attache, 25 bytes



                                  {Sum!`'^^ToBase[_,2:_]'_}


                                  Try it online!



                                  Explanation



                                  {Sum!`'^^ToBase[_,2:_]'_}
                                  { } anonymous lambda, input: _
                                  example: 5
                                  ToBase[_, ] convert `_`
                                  2:_ ...to each base from 2 to `_`
                                  example: [[1, 0, 1], [1, 2], [1, 1], [1, 0]]
                                  '_ append `_`
                                  example: [[1, 0, 1], [1, 2], [1, 1], [1, 0], 5]
                                  ^^ fold...
                                  `' the append operator over the list
                                  example: [1, 0, 1, 1, 2, 1, 1, 1, 0, 5]
                                  Sum! take the sum
                                  example: 13





                                  share|improve this answer





























                                    1















                                    Charcoal, 12 bytes



                                    IΣEIθΣ↨Iθ⁺²ι


                                    Try it online! Link is to verbose version of code. Charcoal can't convert to base 1, but fortunately the digit sum is the same as the conversion to base $n + 1$. Explanation:



                                        θ           First input
                                    I Cast to integer
                                    E Map over implicit range
                                    θ First input
                                    I Cast to integer
                                    ↨ Converted to base
                                    ι Current index
                                    ⁺ Plus
                                    ² Literal 2
                                    Σ Sum
                                    Σ Grand total
                                    I Cast to string
                                    Implicitly print





                                    share|improve this answer





























                                      1















                                      Retina 0.8.2, 49 bytes



                                      .+
                                      $*
                                      1
                                      11$`;$_¶
                                      +`b(1+);(1)+
                                      $1;$#2$*1,
                                      1+;

                                      1


                                      Try it online! Link includes test cases. Explanation:



                                      .+
                                      $*


                                      Convert to unary.



                                      1
                                      11$`;$_¶


                                      List all the numbers from 2 to $n + 1$ (since that's easier than base 1 conversion).



                                      +`b(1+);(1)+
                                      $1;$#2$*1,


                                      Use repeated divmod to convert the original number to each base.



                                      1+;


                                      Delete the list of bases, leaving just the base conversion digits.



                                      1


                                      Take the sum and convert to decimal.






                                      share|improve this answer





























                                        0














                                        APL(NARS), 29 chars, 58 bytes



                                        {+/+/¨⍵{(⍺⍴⍨⌊1+⍺⍟⍵)⊤⍵}⍨¨1+⍳⍵}


                                        little test on how to use:



                                          h←{+/+/¨⍵{(⍺⍴⍨⌊1+⍺⍟⍵)⊤⍵}⍨¨1+⍳⍵}
                                        h¨1 2 4 10 36 37 64 65
                                        1 3 8 35 297 334 883 932





                                        share|improve this answer























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                                          7















                                          Canvas, 3 bytes



                                          ┬]∑


                                          Try it here!



                                          Canvas beating Jelly?



                                          { ]   map over 1..input (implicit "{")
                                          ┬ decode input from that base
                                          ∑ sum the resulting (nested) array





                                          share|improve this answer




























                                            7















                                            Canvas, 3 bytes



                                            ┬]∑


                                            Try it here!



                                            Canvas beating Jelly?



                                            { ]   map over 1..input (implicit "{")
                                            ┬ decode input from that base
                                            ∑ sum the resulting (nested) array





                                            share|improve this answer


























                                              7












                                              7








                                              7







                                              Canvas, 3 bytes



                                              ┬]∑


                                              Try it here!



                                              Canvas beating Jelly?



                                              { ]   map over 1..input (implicit "{")
                                              ┬ decode input from that base
                                              ∑ sum the resulting (nested) array





                                              share|improve this answer















                                              Canvas, 3 bytes



                                              ┬]∑


                                              Try it here!



                                              Canvas beating Jelly?



                                              { ]   map over 1..input (implicit "{")
                                              ┬ decode input from that base
                                              ∑ sum the resulting (nested) array






                                              share|improve this answer














                                              share|improve this answer



                                              share|improve this answer








                                              edited Dec 3 at 17:08

























                                              answered Dec 3 at 17:03









                                              dzaima

                                              14.4k21754




                                              14.4k21754























                                                  7















                                                  Haskell, 46 bytes





                                                  f n=sum$do a<-[1..n];mapM(pure[0..a])[1..n]!!n


                                                  Try it online!



                                                  Explanation



                                                  The function b n -> mapM(pure[0..b])[1..n], generates all strings $[0 dotsc b]^n$ in lexicographic order. For example:



                                                  mapM(pure[0..2])[0..1] == [[0,0],[0,1],[0,2],[1,0],[1,1],[1,2],[2,0],[2,1],[2,2]]


                                                  By indexing into it with (!!n) this can be used to convert n to base b+1, however this won't work for unary (base-$1$), but we're summing the results.. We can even save some bytes with a <- [1..n] and using base-$(n+1)$ rather than a work-around for base-$1$ since we're missing $[underset{ntext{ times}}{underbrace{1,dotsc,1}}]$ which is the same as $[n]$ when summing.



                                                  Using do-notation just concatenates all the lists instead of nesting them:



                                                  λ [ mapM (pure [0..a]) [1..5] !! n | a <- [1..5] ]
                                                  [[0,0,1,0,1],[0,0,0,1,2],[0,0,0,1,1],[0,0,0,1,0],[0,0,0,0,5]]

                                                  λ do a <- [1..5]; mapM (pure [0..a]) [1..5] !! n
                                                  [0,0,1,0,1,0,0,0,1,2,0,0,0,1,1,0,0,0,1,0,0,0,0,0,5]





                                                  share|improve this answer




























                                                    7















                                                    Haskell, 46 bytes





                                                    f n=sum$do a<-[1..n];mapM(pure[0..a])[1..n]!!n


                                                    Try it online!



                                                    Explanation



                                                    The function b n -> mapM(pure[0..b])[1..n], generates all strings $[0 dotsc b]^n$ in lexicographic order. For example:



                                                    mapM(pure[0..2])[0..1] == [[0,0],[0,1],[0,2],[1,0],[1,1],[1,2],[2,0],[2,1],[2,2]]


                                                    By indexing into it with (!!n) this can be used to convert n to base b+1, however this won't work for unary (base-$1$), but we're summing the results.. We can even save some bytes with a <- [1..n] and using base-$(n+1)$ rather than a work-around for base-$1$ since we're missing $[underset{ntext{ times}}{underbrace{1,dotsc,1}}]$ which is the same as $[n]$ when summing.



                                                    Using do-notation just concatenates all the lists instead of nesting them:



                                                    λ [ mapM (pure [0..a]) [1..5] !! n | a <- [1..5] ]
                                                    [[0,0,1,0,1],[0,0,0,1,2],[0,0,0,1,1],[0,0,0,1,0],[0,0,0,0,5]]

                                                    λ do a <- [1..5]; mapM (pure [0..a]) [1..5] !! n
                                                    [0,0,1,0,1,0,0,0,1,2,0,0,0,1,1,0,0,0,1,0,0,0,0,0,5]





                                                    share|improve this answer


























                                                      7












                                                      7








                                                      7







                                                      Haskell, 46 bytes





                                                      f n=sum$do a<-[1..n];mapM(pure[0..a])[1..n]!!n


                                                      Try it online!



                                                      Explanation



                                                      The function b n -> mapM(pure[0..b])[1..n], generates all strings $[0 dotsc b]^n$ in lexicographic order. For example:



                                                      mapM(pure[0..2])[0..1] == [[0,0],[0,1],[0,2],[1,0],[1,1],[1,2],[2,0],[2,1],[2,2]]


                                                      By indexing into it with (!!n) this can be used to convert n to base b+1, however this won't work for unary (base-$1$), but we're summing the results.. We can even save some bytes with a <- [1..n] and using base-$(n+1)$ rather than a work-around for base-$1$ since we're missing $[underset{ntext{ times}}{underbrace{1,dotsc,1}}]$ which is the same as $[n]$ when summing.



                                                      Using do-notation just concatenates all the lists instead of nesting them:



                                                      λ [ mapM (pure [0..a]) [1..5] !! n | a <- [1..5] ]
                                                      [[0,0,1,0,1],[0,0,0,1,2],[0,0,0,1,1],[0,0,0,1,0],[0,0,0,0,5]]

                                                      λ do a <- [1..5]; mapM (pure [0..a]) [1..5] !! n
                                                      [0,0,1,0,1,0,0,0,1,2,0,0,0,1,1,0,0,0,1,0,0,0,0,0,5]





                                                      share|improve this answer















                                                      Haskell, 46 bytes





                                                      f n=sum$do a<-[1..n];mapM(pure[0..a])[1..n]!!n


                                                      Try it online!



                                                      Explanation



                                                      The function b n -> mapM(pure[0..b])[1..n], generates all strings $[0 dotsc b]^n$ in lexicographic order. For example:



                                                      mapM(pure[0..2])[0..1] == [[0,0],[0,1],[0,2],[1,0],[1,1],[1,2],[2,0],[2,1],[2,2]]


                                                      By indexing into it with (!!n) this can be used to convert n to base b+1, however this won't work for unary (base-$1$), but we're summing the results.. We can even save some bytes with a <- [1..n] and using base-$(n+1)$ rather than a work-around for base-$1$ since we're missing $[underset{ntext{ times}}{underbrace{1,dotsc,1}}]$ which is the same as $[n]$ when summing.



                                                      Using do-notation just concatenates all the lists instead of nesting them:



                                                      λ [ mapM (pure [0..a]) [1..5] !! n | a <- [1..5] ]
                                                      [[0,0,1,0,1],[0,0,0,1,2],[0,0,0,1,1],[0,0,0,1,0],[0,0,0,0,5]]

                                                      λ do a <- [1..5]; mapM (pure [0..a]) [1..5] !! n
                                                      [0,0,1,0,1,0,0,0,1,2,0,0,0,1,1,0,0,0,1,0,0,0,0,0,5]






                                                      share|improve this answer














                                                      share|improve this answer



                                                      share|improve this answer








                                                      edited Dec 4 at 0:41

























                                                      answered Dec 3 at 19:24









                                                      BMO

                                                      11.3k22185




                                                      11.3k22185























                                                          6















                                                          Ruby, 39 37



                                                          ->n{(2..n).sum{|b|n.digits(b).sum}+n}


                                                          The only golfing here is removing some whitespace. Try it online






                                                          share|improve this answer























                                                          • @Giuseppe The OP starts with: " Given an input n,(...)" . I haven' t been around here for a long time. Is there a list somewhere of requirements to a solution?
                                                            – steenslag
                                                            Dec 4 at 0:46






                                                          • 2




                                                            Usually it must be a full program or a function (named or unnamed), here's more information: loopholes and i/o defaults. I don't know ruby, but this seems the shortest fix.
                                                            – BMO
                                                            Dec 4 at 0:52












                                                          • @BMO Thanks. For someone who doesn't know Ruby, you are creating and calling a lambda with the greatest of ease!
                                                            – steenslag
                                                            Dec 4 at 1:04






                                                          • 1




                                                            you can remove the parentheses around n (37b): ->n{(2..n).sum{|b|n.digits(b).sum}+n}
                                                            – Conor O'Brien
                                                            Dec 4 at 1:06






                                                          • 1




                                                            Well, google "ruby lambda" did the trick ;P But I stand corrected in that you were able to save two bytes.
                                                            – BMO
                                                            Dec 4 at 12:42
















                                                          6















                                                          Ruby, 39 37



                                                          ->n{(2..n).sum{|b|n.digits(b).sum}+n}


                                                          The only golfing here is removing some whitespace. Try it online






                                                          share|improve this answer























                                                          • @Giuseppe The OP starts with: " Given an input n,(...)" . I haven' t been around here for a long time. Is there a list somewhere of requirements to a solution?
                                                            – steenslag
                                                            Dec 4 at 0:46






                                                          • 2




                                                            Usually it must be a full program or a function (named or unnamed), here's more information: loopholes and i/o defaults. I don't know ruby, but this seems the shortest fix.
                                                            – BMO
                                                            Dec 4 at 0:52












                                                          • @BMO Thanks. For someone who doesn't know Ruby, you are creating and calling a lambda with the greatest of ease!
                                                            – steenslag
                                                            Dec 4 at 1:04






                                                          • 1




                                                            you can remove the parentheses around n (37b): ->n{(2..n).sum{|b|n.digits(b).sum}+n}
                                                            – Conor O'Brien
                                                            Dec 4 at 1:06






                                                          • 1




                                                            Well, google "ruby lambda" did the trick ;P But I stand corrected in that you were able to save two bytes.
                                                            – BMO
                                                            Dec 4 at 12:42














                                                          6












                                                          6








                                                          6







                                                          Ruby, 39 37



                                                          ->n{(2..n).sum{|b|n.digits(b).sum}+n}


                                                          The only golfing here is removing some whitespace. Try it online






                                                          share|improve this answer















                                                          Ruby, 39 37



                                                          ->n{(2..n).sum{|b|n.digits(b).sum}+n}


                                                          The only golfing here is removing some whitespace. Try it online







                                                          share|improve this answer














                                                          share|improve this answer



                                                          share|improve this answer








                                                          edited Dec 4 at 1:14

























                                                          answered Dec 3 at 23:55









                                                          steenslag

                                                          1,760919




                                                          1,760919












                                                          • @Giuseppe The OP starts with: " Given an input n,(...)" . I haven' t been around here for a long time. Is there a list somewhere of requirements to a solution?
                                                            – steenslag
                                                            Dec 4 at 0:46






                                                          • 2




                                                            Usually it must be a full program or a function (named or unnamed), here's more information: loopholes and i/o defaults. I don't know ruby, but this seems the shortest fix.
                                                            – BMO
                                                            Dec 4 at 0:52












                                                          • @BMO Thanks. For someone who doesn't know Ruby, you are creating and calling a lambda with the greatest of ease!
                                                            – steenslag
                                                            Dec 4 at 1:04






                                                          • 1




                                                            you can remove the parentheses around n (37b): ->n{(2..n).sum{|b|n.digits(b).sum}+n}
                                                            – Conor O'Brien
                                                            Dec 4 at 1:06






                                                          • 1




                                                            Well, google "ruby lambda" did the trick ;P But I stand corrected in that you were able to save two bytes.
                                                            – BMO
                                                            Dec 4 at 12:42


















                                                          • @Giuseppe The OP starts with: " Given an input n,(...)" . I haven' t been around here for a long time. Is there a list somewhere of requirements to a solution?
                                                            – steenslag
                                                            Dec 4 at 0:46






                                                          • 2




                                                            Usually it must be a full program or a function (named or unnamed), here's more information: loopholes and i/o defaults. I don't know ruby, but this seems the shortest fix.
                                                            – BMO
                                                            Dec 4 at 0:52












                                                          • @BMO Thanks. For someone who doesn't know Ruby, you are creating and calling a lambda with the greatest of ease!
                                                            – steenslag
                                                            Dec 4 at 1:04






                                                          • 1




                                                            you can remove the parentheses around n (37b): ->n{(2..n).sum{|b|n.digits(b).sum}+n}
                                                            – Conor O'Brien
                                                            Dec 4 at 1:06






                                                          • 1




                                                            Well, google "ruby lambda" did the trick ;P But I stand corrected in that you were able to save two bytes.
                                                            – BMO
                                                            Dec 4 at 12:42
















                                                          @Giuseppe The OP starts with: " Given an input n,(...)" . I haven' t been around here for a long time. Is there a list somewhere of requirements to a solution?
                                                          – steenslag
                                                          Dec 4 at 0:46




                                                          @Giuseppe The OP starts with: " Given an input n,(...)" . I haven' t been around here for a long time. Is there a list somewhere of requirements to a solution?
                                                          – steenslag
                                                          Dec 4 at 0:46




                                                          2




                                                          2




                                                          Usually it must be a full program or a function (named or unnamed), here's more information: loopholes and i/o defaults. I don't know ruby, but this seems the shortest fix.
                                                          – BMO
                                                          Dec 4 at 0:52






                                                          Usually it must be a full program or a function (named or unnamed), here's more information: loopholes and i/o defaults. I don't know ruby, but this seems the shortest fix.
                                                          – BMO
                                                          Dec 4 at 0:52














                                                          @BMO Thanks. For someone who doesn't know Ruby, you are creating and calling a lambda with the greatest of ease!
                                                          – steenslag
                                                          Dec 4 at 1:04




                                                          @BMO Thanks. For someone who doesn't know Ruby, you are creating and calling a lambda with the greatest of ease!
                                                          – steenslag
                                                          Dec 4 at 1:04




                                                          1




                                                          1




                                                          you can remove the parentheses around n (37b): ->n{(2..n).sum{|b|n.digits(b).sum}+n}
                                                          – Conor O'Brien
                                                          Dec 4 at 1:06




                                                          you can remove the parentheses around n (37b): ->n{(2..n).sum{|b|n.digits(b).sum}+n}
                                                          – Conor O'Brien
                                                          Dec 4 at 1:06




                                                          1




                                                          1




                                                          Well, google "ruby lambda" did the trick ;P But I stand corrected in that you were able to save two bytes.
                                                          – BMO
                                                          Dec 4 at 12:42




                                                          Well, google "ruby lambda" did the trick ;P But I stand corrected in that you were able to save two bytes.
                                                          – BMO
                                                          Dec 4 at 12:42











                                                          5















                                                          Python 2, 57 bytes





                                                          lambda n:sum(n/(k/n+2)**(k%n)%(k/n+2)for k in range(n*n))


                                                          This uses the formula $$a(n)=sum_{b=2}^{n+1}sum_{i=0}^{n-1}leftlfloorfrac{n}{b^i}rightrfloorbmod b,$$ which works since $leftlfloorfrac{n}{(n+1)^0}rightrfloorbmod(n+1)=n$.



                                                          Try it online!






                                                          share|improve this answer




























                                                            5















                                                            Python 2, 57 bytes





                                                            lambda n:sum(n/(k/n+2)**(k%n)%(k/n+2)for k in range(n*n))


                                                            This uses the formula $$a(n)=sum_{b=2}^{n+1}sum_{i=0}^{n-1}leftlfloorfrac{n}{b^i}rightrfloorbmod b,$$ which works since $leftlfloorfrac{n}{(n+1)^0}rightrfloorbmod(n+1)=n$.



                                                            Try it online!






                                                            share|improve this answer


























                                                              5












                                                              5








                                                              5







                                                              Python 2, 57 bytes





                                                              lambda n:sum(n/(k/n+2)**(k%n)%(k/n+2)for k in range(n*n))


                                                              This uses the formula $$a(n)=sum_{b=2}^{n+1}sum_{i=0}^{n-1}leftlfloorfrac{n}{b^i}rightrfloorbmod b,$$ which works since $leftlfloorfrac{n}{(n+1)^0}rightrfloorbmod(n+1)=n$.



                                                              Try it online!






                                                              share|improve this answer















                                                              Python 2, 57 bytes





                                                              lambda n:sum(n/(k/n+2)**(k%n)%(k/n+2)for k in range(n*n))


                                                              This uses the formula $$a(n)=sum_{b=2}^{n+1}sum_{i=0}^{n-1}leftlfloorfrac{n}{b^i}rightrfloorbmod b,$$ which works since $leftlfloorfrac{n}{(n+1)^0}rightrfloorbmod(n+1)=n$.



                                                              Try it online!







                                                              share|improve this answer














                                                              share|improve this answer



                                                              share|improve this answer








                                                              edited Dec 3 at 16:55

























                                                              answered Dec 3 at 16:45









                                                              Dennis

                                                              186k32295735




                                                              186k32295735























                                                                  5















                                                                  APL (Dyalog Unicode), 14 bytes





                                                                  +/∘∊⊢,⊢(⍴⊤⊣)¨⍳


                                                                  Try it online!



                                                                  Explanation



                                                                  Some parentheses are implied and can be added (lighter than the "official" parenthesizing):



                                                                  +/∘∊(⊢,(⊢(⍴⊤⊣)¨⍳))


                                                                  This is a monadic atop. Given an argument Y, this function behaves like:



                                                                  +/∘∊(⊢,(⊢(⍴⊤⊣)¨⍳))Y


                                                                  The two functions are applied in order. We'll start from the right one:



                                                                  ⊢,(⊢(⍴⊤⊣)¨⍳)


                                                                  The are three functions in this train, so this is a fork. Given an argument Y, it acts as:



                                                                  (⊢Y),((⊢Y)(⍴⊤⊣)¨⍳Y)


                                                                  We can easily reduce to this (monadic returns its argument, hence called identity):



                                                                  Y,Y(⍴⊤⊣)¨⍳Y


                                                                  Now, we know that Y is an integer (simple scalar, i.e. number or character), since we're given one. Therefore ⍳Y, with ⎕IO=1, returns 1 2 ... Y. ⍳Y actually returns an array with shape Y (Y must be a vector), where every scalar is the index of itself in the array (that's why monadic is called the index generator). These indices are vectors, except for the case where 1≡⍴Y, where they are scalars (this is our case).



                                                                  Let's parse the middle function, (⍴⊤⊣)¨, next. ⍴⊤⊣ is the operand of ¨ (each), and the function is dyadic, so the ¨ operator will first reshape each length-1 argument to the shape of the other (that is, take the element and use it to replace every scalar in the other argument), and then apply the function to each pair of the two arguments. In this case, ⍳Y is a vector and Y is a scalar, so, if n≡⍴⍳Y, then Y will be converted to n⍴Y ( represents the shape (monadic) and reshape (dyadic) functions). That is, in simpler terms, Y will be converted to an array containing Y times Y.



                                                                  Now, for each pair, let's call the left argument X and the right Z (so that we don't conflict with the input Y). ⍴⊤⊣ is a dyadic fork, so it will expand to:



                                                                  (X⍴Z)⊤X⊣Z


                                                                  Let's make the easy first step of reducing X⊣Z to X (dyadic is the left function):



                                                                  (X⍴Z)⊤X


                                                                  The in X⍴Z is, again, the reshape function, so X⍴Z, in our case, is simply X times Z. is the encode function. Given two arrays of numbers, where the left array is the base of each digit in the result (doesn't need to be integer or positive), i.e. the encoding, and the right is an array of numbers, returns the transposed array of those numbers in the specified encoding (transposition is the reversal of an array's dimensions relative to its elements). The representation of a digit is based on the quotient of the division of the number and the product of the less significant bases. If any base is 0, it acts as base +∞. The arguments' scalars are all simple. Since X is a positive integer, and X⍴Z is a vector of equal elements, this is really just a case of converting X to base Z and reshaping to X digits. For $X,Zinmathbb N$, $X_Z$ ($X$ in base $Z$) can't have more than $X$ digits, since $X_1$ has $X$ digits. Therefore, X⍴Z is enough for our purposes.



                                                                  The result of Y(⍴⊤⊣)¨⍳Y is, therefore, Y converted to each base from 1 to Y, possibly with leading zeroes. However, there is one issue: in APL, base 1 isn't special-cased, while this challenge does special-case it, so we have to include the sum of the base-1 digits of Y ourselves. Fortunately, this sum is just Y, since $Y_1=underbrace{[1,1,...,1]}_Y$, so the sum is simply $Ytimes1=Y$. It follows that we have to add Y somewhere into the array. This is how we do it:



                                                                  Y,Y(⍴⊤⊣)¨⍳Y


                                                                  I have already included this part here. Dyadic , is the catenate function, it concatenates its arguments on their last axes, and errors if that's not possible. Here, we simply concatenate the scalar Y to the vector Y(⍴⊤⊣)¨⍳Y, so that we increment the sum we're going to calculate by Y, as explained above.



                                                                  The final part is the left function of our atop, +/∘∊:



                                                                  +/∘∊Y,Y(⍴⊤⊣)¨⍳Y


                                                                  is the compose operator. f∘g Y is the same as f g Y. However, we're using it here so that our train doesn't fork on the . So, we can reduce:



                                                                  +/∊Y,Y(⍴⊤⊣)¨⍳Y


                                                                  Now, it's time for the sum, but wait... there's a problem. The array isn't flat, so we can't just sum its elements before flattening it first. The enlist function flattens an array. Now that the array has been flattened, we finally use +/ to sum it. / is the reduce operator, it applies a dyadic function between an array's elements on its second-to-last axis, with right-to-left priority. If the rank (number of dimensions, i.e. length of shape) of the array doesn't decrease, the array is then enclosed, although this is not the case here. The function that's applied here is +, which is the plus function that adds the pairs on the last axes of two arrays (and errors if the arrays can't be added like that). Here, it simply adds two numbers a number of times so that the reduce is completed.



                                                                  Lo and behold, our train:



                                                                  +/∘∊⊢,⊢(⍴⊤⊣)¨⍳





                                                                  share|improve this answer























                                                                  • +1 Impressive explanation. Not shorter, but no parens: +/∘∊⊢,⊢⊤¨⍨⊢⍴¨⍳
                                                                    – Adám
                                                                    Dec 4 at 9:19












                                                                  • @Adám Thanks! I was almost sleeping when I wrote it. :-P
                                                                    – Erik the Outgolfer
                                                                    Dec 4 at 11:13










                                                                  • @Adám Regarding your version, it looks like it's a tad more difficult to understand. ;-)
                                                                    – Erik the Outgolfer
                                                                    Dec 4 at 11:26
















                                                                  5















                                                                  APL (Dyalog Unicode), 14 bytes





                                                                  +/∘∊⊢,⊢(⍴⊤⊣)¨⍳


                                                                  Try it online!



                                                                  Explanation



                                                                  Some parentheses are implied and can be added (lighter than the "official" parenthesizing):



                                                                  +/∘∊(⊢,(⊢(⍴⊤⊣)¨⍳))


                                                                  This is a monadic atop. Given an argument Y, this function behaves like:



                                                                  +/∘∊(⊢,(⊢(⍴⊤⊣)¨⍳))Y


                                                                  The two functions are applied in order. We'll start from the right one:



                                                                  ⊢,(⊢(⍴⊤⊣)¨⍳)


                                                                  The are three functions in this train, so this is a fork. Given an argument Y, it acts as:



                                                                  (⊢Y),((⊢Y)(⍴⊤⊣)¨⍳Y)


                                                                  We can easily reduce to this (monadic returns its argument, hence called identity):



                                                                  Y,Y(⍴⊤⊣)¨⍳Y


                                                                  Now, we know that Y is an integer (simple scalar, i.e. number or character), since we're given one. Therefore ⍳Y, with ⎕IO=1, returns 1 2 ... Y. ⍳Y actually returns an array with shape Y (Y must be a vector), where every scalar is the index of itself in the array (that's why monadic is called the index generator). These indices are vectors, except for the case where 1≡⍴Y, where they are scalars (this is our case).



                                                                  Let's parse the middle function, (⍴⊤⊣)¨, next. ⍴⊤⊣ is the operand of ¨ (each), and the function is dyadic, so the ¨ operator will first reshape each length-1 argument to the shape of the other (that is, take the element and use it to replace every scalar in the other argument), and then apply the function to each pair of the two arguments. In this case, ⍳Y is a vector and Y is a scalar, so, if n≡⍴⍳Y, then Y will be converted to n⍴Y ( represents the shape (monadic) and reshape (dyadic) functions). That is, in simpler terms, Y will be converted to an array containing Y times Y.



                                                                  Now, for each pair, let's call the left argument X and the right Z (so that we don't conflict with the input Y). ⍴⊤⊣ is a dyadic fork, so it will expand to:



                                                                  (X⍴Z)⊤X⊣Z


                                                                  Let's make the easy first step of reducing X⊣Z to X (dyadic is the left function):



                                                                  (X⍴Z)⊤X


                                                                  The in X⍴Z is, again, the reshape function, so X⍴Z, in our case, is simply X times Z. is the encode function. Given two arrays of numbers, where the left array is the base of each digit in the result (doesn't need to be integer or positive), i.e. the encoding, and the right is an array of numbers, returns the transposed array of those numbers in the specified encoding (transposition is the reversal of an array's dimensions relative to its elements). The representation of a digit is based on the quotient of the division of the number and the product of the less significant bases. If any base is 0, it acts as base +∞. The arguments' scalars are all simple. Since X is a positive integer, and X⍴Z is a vector of equal elements, this is really just a case of converting X to base Z and reshaping to X digits. For $X,Zinmathbb N$, $X_Z$ ($X$ in base $Z$) can't have more than $X$ digits, since $X_1$ has $X$ digits. Therefore, X⍴Z is enough for our purposes.



                                                                  The result of Y(⍴⊤⊣)¨⍳Y is, therefore, Y converted to each base from 1 to Y, possibly with leading zeroes. However, there is one issue: in APL, base 1 isn't special-cased, while this challenge does special-case it, so we have to include the sum of the base-1 digits of Y ourselves. Fortunately, this sum is just Y, since $Y_1=underbrace{[1,1,...,1]}_Y$, so the sum is simply $Ytimes1=Y$. It follows that we have to add Y somewhere into the array. This is how we do it:



                                                                  Y,Y(⍴⊤⊣)¨⍳Y


                                                                  I have already included this part here. Dyadic , is the catenate function, it concatenates its arguments on their last axes, and errors if that's not possible. Here, we simply concatenate the scalar Y to the vector Y(⍴⊤⊣)¨⍳Y, so that we increment the sum we're going to calculate by Y, as explained above.



                                                                  The final part is the left function of our atop, +/∘∊:



                                                                  +/∘∊Y,Y(⍴⊤⊣)¨⍳Y


                                                                  is the compose operator. f∘g Y is the same as f g Y. However, we're using it here so that our train doesn't fork on the . So, we can reduce:



                                                                  +/∊Y,Y(⍴⊤⊣)¨⍳Y


                                                                  Now, it's time for the sum, but wait... there's a problem. The array isn't flat, so we can't just sum its elements before flattening it first. The enlist function flattens an array. Now that the array has been flattened, we finally use +/ to sum it. / is the reduce operator, it applies a dyadic function between an array's elements on its second-to-last axis, with right-to-left priority. If the rank (number of dimensions, i.e. length of shape) of the array doesn't decrease, the array is then enclosed, although this is not the case here. The function that's applied here is +, which is the plus function that adds the pairs on the last axes of two arrays (and errors if the arrays can't be added like that). Here, it simply adds two numbers a number of times so that the reduce is completed.



                                                                  Lo and behold, our train:



                                                                  +/∘∊⊢,⊢(⍴⊤⊣)¨⍳





                                                                  share|improve this answer























                                                                  • +1 Impressive explanation. Not shorter, but no parens: +/∘∊⊢,⊢⊤¨⍨⊢⍴¨⍳
                                                                    – Adám
                                                                    Dec 4 at 9:19












                                                                  • @Adám Thanks! I was almost sleeping when I wrote it. :-P
                                                                    – Erik the Outgolfer
                                                                    Dec 4 at 11:13










                                                                  • @Adám Regarding your version, it looks like it's a tad more difficult to understand. ;-)
                                                                    – Erik the Outgolfer
                                                                    Dec 4 at 11:26














                                                                  5












                                                                  5








                                                                  5







                                                                  APL (Dyalog Unicode), 14 bytes





                                                                  +/∘∊⊢,⊢(⍴⊤⊣)¨⍳


                                                                  Try it online!



                                                                  Explanation



                                                                  Some parentheses are implied and can be added (lighter than the "official" parenthesizing):



                                                                  +/∘∊(⊢,(⊢(⍴⊤⊣)¨⍳))


                                                                  This is a monadic atop. Given an argument Y, this function behaves like:



                                                                  +/∘∊(⊢,(⊢(⍴⊤⊣)¨⍳))Y


                                                                  The two functions are applied in order. We'll start from the right one:



                                                                  ⊢,(⊢(⍴⊤⊣)¨⍳)


                                                                  The are three functions in this train, so this is a fork. Given an argument Y, it acts as:



                                                                  (⊢Y),((⊢Y)(⍴⊤⊣)¨⍳Y)


                                                                  We can easily reduce to this (monadic returns its argument, hence called identity):



                                                                  Y,Y(⍴⊤⊣)¨⍳Y


                                                                  Now, we know that Y is an integer (simple scalar, i.e. number or character), since we're given one. Therefore ⍳Y, with ⎕IO=1, returns 1 2 ... Y. ⍳Y actually returns an array with shape Y (Y must be a vector), where every scalar is the index of itself in the array (that's why monadic is called the index generator). These indices are vectors, except for the case where 1≡⍴Y, where they are scalars (this is our case).



                                                                  Let's parse the middle function, (⍴⊤⊣)¨, next. ⍴⊤⊣ is the operand of ¨ (each), and the function is dyadic, so the ¨ operator will first reshape each length-1 argument to the shape of the other (that is, take the element and use it to replace every scalar in the other argument), and then apply the function to each pair of the two arguments. In this case, ⍳Y is a vector and Y is a scalar, so, if n≡⍴⍳Y, then Y will be converted to n⍴Y ( represents the shape (monadic) and reshape (dyadic) functions). That is, in simpler terms, Y will be converted to an array containing Y times Y.



                                                                  Now, for each pair, let's call the left argument X and the right Z (so that we don't conflict with the input Y). ⍴⊤⊣ is a dyadic fork, so it will expand to:



                                                                  (X⍴Z)⊤X⊣Z


                                                                  Let's make the easy first step of reducing X⊣Z to X (dyadic is the left function):



                                                                  (X⍴Z)⊤X


                                                                  The in X⍴Z is, again, the reshape function, so X⍴Z, in our case, is simply X times Z. is the encode function. Given two arrays of numbers, where the left array is the base of each digit in the result (doesn't need to be integer or positive), i.e. the encoding, and the right is an array of numbers, returns the transposed array of those numbers in the specified encoding (transposition is the reversal of an array's dimensions relative to its elements). The representation of a digit is based on the quotient of the division of the number and the product of the less significant bases. If any base is 0, it acts as base +∞. The arguments' scalars are all simple. Since X is a positive integer, and X⍴Z is a vector of equal elements, this is really just a case of converting X to base Z and reshaping to X digits. For $X,Zinmathbb N$, $X_Z$ ($X$ in base $Z$) can't have more than $X$ digits, since $X_1$ has $X$ digits. Therefore, X⍴Z is enough for our purposes.



                                                                  The result of Y(⍴⊤⊣)¨⍳Y is, therefore, Y converted to each base from 1 to Y, possibly with leading zeroes. However, there is one issue: in APL, base 1 isn't special-cased, while this challenge does special-case it, so we have to include the sum of the base-1 digits of Y ourselves. Fortunately, this sum is just Y, since $Y_1=underbrace{[1,1,...,1]}_Y$, so the sum is simply $Ytimes1=Y$. It follows that we have to add Y somewhere into the array. This is how we do it:



                                                                  Y,Y(⍴⊤⊣)¨⍳Y


                                                                  I have already included this part here. Dyadic , is the catenate function, it concatenates its arguments on their last axes, and errors if that's not possible. Here, we simply concatenate the scalar Y to the vector Y(⍴⊤⊣)¨⍳Y, so that we increment the sum we're going to calculate by Y, as explained above.



                                                                  The final part is the left function of our atop, +/∘∊:



                                                                  +/∘∊Y,Y(⍴⊤⊣)¨⍳Y


                                                                  is the compose operator. f∘g Y is the same as f g Y. However, we're using it here so that our train doesn't fork on the . So, we can reduce:



                                                                  +/∊Y,Y(⍴⊤⊣)¨⍳Y


                                                                  Now, it's time for the sum, but wait... there's a problem. The array isn't flat, so we can't just sum its elements before flattening it first. The enlist function flattens an array. Now that the array has been flattened, we finally use +/ to sum it. / is the reduce operator, it applies a dyadic function between an array's elements on its second-to-last axis, with right-to-left priority. If the rank (number of dimensions, i.e. length of shape) of the array doesn't decrease, the array is then enclosed, although this is not the case here. The function that's applied here is +, which is the plus function that adds the pairs on the last axes of two arrays (and errors if the arrays can't be added like that). Here, it simply adds two numbers a number of times so that the reduce is completed.



                                                                  Lo and behold, our train:



                                                                  +/∘∊⊢,⊢(⍴⊤⊣)¨⍳





                                                                  share|improve this answer















                                                                  APL (Dyalog Unicode), 14 bytes





                                                                  +/∘∊⊢,⊢(⍴⊤⊣)¨⍳


                                                                  Try it online!



                                                                  Explanation



                                                                  Some parentheses are implied and can be added (lighter than the "official" parenthesizing):



                                                                  +/∘∊(⊢,(⊢(⍴⊤⊣)¨⍳))


                                                                  This is a monadic atop. Given an argument Y, this function behaves like:



                                                                  +/∘∊(⊢,(⊢(⍴⊤⊣)¨⍳))Y


                                                                  The two functions are applied in order. We'll start from the right one:



                                                                  ⊢,(⊢(⍴⊤⊣)¨⍳)


                                                                  The are three functions in this train, so this is a fork. Given an argument Y, it acts as:



                                                                  (⊢Y),((⊢Y)(⍴⊤⊣)¨⍳Y)


                                                                  We can easily reduce to this (monadic returns its argument, hence called identity):



                                                                  Y,Y(⍴⊤⊣)¨⍳Y


                                                                  Now, we know that Y is an integer (simple scalar, i.e. number or character), since we're given one. Therefore ⍳Y, with ⎕IO=1, returns 1 2 ... Y. ⍳Y actually returns an array with shape Y (Y must be a vector), where every scalar is the index of itself in the array (that's why monadic is called the index generator). These indices are vectors, except for the case where 1≡⍴Y, where they are scalars (this is our case).



                                                                  Let's parse the middle function, (⍴⊤⊣)¨, next. ⍴⊤⊣ is the operand of ¨ (each), and the function is dyadic, so the ¨ operator will first reshape each length-1 argument to the shape of the other (that is, take the element and use it to replace every scalar in the other argument), and then apply the function to each pair of the two arguments. In this case, ⍳Y is a vector and Y is a scalar, so, if n≡⍴⍳Y, then Y will be converted to n⍴Y ( represents the shape (monadic) and reshape (dyadic) functions). That is, in simpler terms, Y will be converted to an array containing Y times Y.



                                                                  Now, for each pair, let's call the left argument X and the right Z (so that we don't conflict with the input Y). ⍴⊤⊣ is a dyadic fork, so it will expand to:



                                                                  (X⍴Z)⊤X⊣Z


                                                                  Let's make the easy first step of reducing X⊣Z to X (dyadic is the left function):



                                                                  (X⍴Z)⊤X


                                                                  The in X⍴Z is, again, the reshape function, so X⍴Z, in our case, is simply X times Z. is the encode function. Given two arrays of numbers, where the left array is the base of each digit in the result (doesn't need to be integer or positive), i.e. the encoding, and the right is an array of numbers, returns the transposed array of those numbers in the specified encoding (transposition is the reversal of an array's dimensions relative to its elements). The representation of a digit is based on the quotient of the division of the number and the product of the less significant bases. If any base is 0, it acts as base +∞. The arguments' scalars are all simple. Since X is a positive integer, and X⍴Z is a vector of equal elements, this is really just a case of converting X to base Z and reshaping to X digits. For $X,Zinmathbb N$, $X_Z$ ($X$ in base $Z$) can't have more than $X$ digits, since $X_1$ has $X$ digits. Therefore, X⍴Z is enough for our purposes.



                                                                  The result of Y(⍴⊤⊣)¨⍳Y is, therefore, Y converted to each base from 1 to Y, possibly with leading zeroes. However, there is one issue: in APL, base 1 isn't special-cased, while this challenge does special-case it, so we have to include the sum of the base-1 digits of Y ourselves. Fortunately, this sum is just Y, since $Y_1=underbrace{[1,1,...,1]}_Y$, so the sum is simply $Ytimes1=Y$. It follows that we have to add Y somewhere into the array. This is how we do it:



                                                                  Y,Y(⍴⊤⊣)¨⍳Y


                                                                  I have already included this part here. Dyadic , is the catenate function, it concatenates its arguments on their last axes, and errors if that's not possible. Here, we simply concatenate the scalar Y to the vector Y(⍴⊤⊣)¨⍳Y, so that we increment the sum we're going to calculate by Y, as explained above.



                                                                  The final part is the left function of our atop, +/∘∊:



                                                                  +/∘∊Y,Y(⍴⊤⊣)¨⍳Y


                                                                  is the compose operator. f∘g Y is the same as f g Y. However, we're using it here so that our train doesn't fork on the . So, we can reduce:



                                                                  +/∊Y,Y(⍴⊤⊣)¨⍳Y


                                                                  Now, it's time for the sum, but wait... there's a problem. The array isn't flat, so we can't just sum its elements before flattening it first. The enlist function flattens an array. Now that the array has been flattened, we finally use +/ to sum it. / is the reduce operator, it applies a dyadic function between an array's elements on its second-to-last axis, with right-to-left priority. If the rank (number of dimensions, i.e. length of shape) of the array doesn't decrease, the array is then enclosed, although this is not the case here. The function that's applied here is +, which is the plus function that adds the pairs on the last axes of two arrays (and errors if the arrays can't be added like that). Here, it simply adds two numbers a number of times so that the reduce is completed.



                                                                  Lo and behold, our train:



                                                                  +/∘∊⊢,⊢(⍴⊤⊣)¨⍳






                                                                  share|improve this answer














                                                                  share|improve this answer



                                                                  share|improve this answer








                                                                  edited Dec 3 at 23:23

























                                                                  answered Dec 3 at 20:42









                                                                  Erik the Outgolfer

                                                                  31.3k429102




                                                                  31.3k429102












                                                                  • +1 Impressive explanation. Not shorter, but no parens: +/∘∊⊢,⊢⊤¨⍨⊢⍴¨⍳
                                                                    – Adám
                                                                    Dec 4 at 9:19












                                                                  • @Adám Thanks! I was almost sleeping when I wrote it. :-P
                                                                    – Erik the Outgolfer
                                                                    Dec 4 at 11:13










                                                                  • @Adám Regarding your version, it looks like it's a tad more difficult to understand. ;-)
                                                                    – Erik the Outgolfer
                                                                    Dec 4 at 11:26


















                                                                  • +1 Impressive explanation. Not shorter, but no parens: +/∘∊⊢,⊢⊤¨⍨⊢⍴¨⍳
                                                                    – Adám
                                                                    Dec 4 at 9:19












                                                                  • @Adám Thanks! I was almost sleeping when I wrote it. :-P
                                                                    – Erik the Outgolfer
                                                                    Dec 4 at 11:13










                                                                  • @Adám Regarding your version, it looks like it's a tad more difficult to understand. ;-)
                                                                    – Erik the Outgolfer
                                                                    Dec 4 at 11:26
















                                                                  +1 Impressive explanation. Not shorter, but no parens: +/∘∊⊢,⊢⊤¨⍨⊢⍴¨⍳
                                                                  – Adám
                                                                  Dec 4 at 9:19






                                                                  +1 Impressive explanation. Not shorter, but no parens: +/∘∊⊢,⊢⊤¨⍨⊢⍴¨⍳
                                                                  – Adám
                                                                  Dec 4 at 9:19














                                                                  @Adám Thanks! I was almost sleeping when I wrote it. :-P
                                                                  – Erik the Outgolfer
                                                                  Dec 4 at 11:13




                                                                  @Adám Thanks! I was almost sleeping when I wrote it. :-P
                                                                  – Erik the Outgolfer
                                                                  Dec 4 at 11:13












                                                                  @Adám Regarding your version, it looks like it's a tad more difficult to understand. ;-)
                                                                  – Erik the Outgolfer
                                                                  Dec 4 at 11:26




                                                                  @Adám Regarding your version, it looks like it's a tad more difficult to understand. ;-)
                                                                  – Erik the Outgolfer
                                                                  Dec 4 at 11:26











                                                                  5














                                                                  Java 8, 76 65 bytes





                                                                  n->{int r=n,b=n,N;for(;b>1;b--)for(N=n;N>0;N/=b)r+=N%b;return r;}


                                                                  -11 bytes thanks to @OlivierGrégoire.



                                                                  Try it online.



                                                                  Explanation:



                                                                  n->{           // Method with integer as both parameter and return-type
                                                                  int r=n, // Result-sum, starting at the input to account for Base-1
                                                                  b=n, // Base, starting at the input
                                                                  N; // Temp integer
                                                                  for(;b>1 // Loop the Base `b` in the range [n, 1):
                                                                  ;b--) // After every iteration: Go to the next Base (downwards)
                                                                  for(N=n; // Set `N` to the input `n`
                                                                  N>0; // Loop as long as `N` isn't 0 yet:
                                                                  N/=b) // After every iteration: Divide `N` by the Base `b`
                                                                  r+=N%b; // Increase the result `r` by `N` modulo the Base `b`
                                                                  return r;} // Return the result-sum `r`





                                                                  share|improve this answer



















                                                                  • 1




                                                                    There is no need for i, so... 65 bytes.
                                                                    – Olivier Grégoire
                                                                    Dec 5 at 11:01












                                                                  • @OlivierGrégoire Damn, I'm an idiot. Thanks a lot! Hmm, now I also need to golf my derived C and Whitespace answers.. ;)
                                                                    – Kevin Cruijssen
                                                                    Dec 5 at 11:03
















                                                                  5














                                                                  Java 8, 76 65 bytes





                                                                  n->{int r=n,b=n,N;for(;b>1;b--)for(N=n;N>0;N/=b)r+=N%b;return r;}


                                                                  -11 bytes thanks to @OlivierGrégoire.



                                                                  Try it online.



                                                                  Explanation:



                                                                  n->{           // Method with integer as both parameter and return-type
                                                                  int r=n, // Result-sum, starting at the input to account for Base-1
                                                                  b=n, // Base, starting at the input
                                                                  N; // Temp integer
                                                                  for(;b>1 // Loop the Base `b` in the range [n, 1):
                                                                  ;b--) // After every iteration: Go to the next Base (downwards)
                                                                  for(N=n; // Set `N` to the input `n`
                                                                  N>0; // Loop as long as `N` isn't 0 yet:
                                                                  N/=b) // After every iteration: Divide `N` by the Base `b`
                                                                  r+=N%b; // Increase the result `r` by `N` modulo the Base `b`
                                                                  return r;} // Return the result-sum `r`





                                                                  share|improve this answer



















                                                                  • 1




                                                                    There is no need for i, so... 65 bytes.
                                                                    – Olivier Grégoire
                                                                    Dec 5 at 11:01












                                                                  • @OlivierGrégoire Damn, I'm an idiot. Thanks a lot! Hmm, now I also need to golf my derived C and Whitespace answers.. ;)
                                                                    – Kevin Cruijssen
                                                                    Dec 5 at 11:03














                                                                  5












                                                                  5








                                                                  5






                                                                  Java 8, 76 65 bytes





                                                                  n->{int r=n,b=n,N;for(;b>1;b--)for(N=n;N>0;N/=b)r+=N%b;return r;}


                                                                  -11 bytes thanks to @OlivierGrégoire.



                                                                  Try it online.



                                                                  Explanation:



                                                                  n->{           // Method with integer as both parameter and return-type
                                                                  int r=n, // Result-sum, starting at the input to account for Base-1
                                                                  b=n, // Base, starting at the input
                                                                  N; // Temp integer
                                                                  for(;b>1 // Loop the Base `b` in the range [n, 1):
                                                                  ;b--) // After every iteration: Go to the next Base (downwards)
                                                                  for(N=n; // Set `N` to the input `n`
                                                                  N>0; // Loop as long as `N` isn't 0 yet:
                                                                  N/=b) // After every iteration: Divide `N` by the Base `b`
                                                                  r+=N%b; // Increase the result `r` by `N` modulo the Base `b`
                                                                  return r;} // Return the result-sum `r`





                                                                  share|improve this answer














                                                                  Java 8, 76 65 bytes





                                                                  n->{int r=n,b=n,N;for(;b>1;b--)for(N=n;N>0;N/=b)r+=N%b;return r;}


                                                                  -11 bytes thanks to @OlivierGrégoire.



                                                                  Try it online.



                                                                  Explanation:



                                                                  n->{           // Method with integer as both parameter and return-type
                                                                  int r=n, // Result-sum, starting at the input to account for Base-1
                                                                  b=n, // Base, starting at the input
                                                                  N; // Temp integer
                                                                  for(;b>1 // Loop the Base `b` in the range [n, 1):
                                                                  ;b--) // After every iteration: Go to the next Base (downwards)
                                                                  for(N=n; // Set `N` to the input `n`
                                                                  N>0; // Loop as long as `N` isn't 0 yet:
                                                                  N/=b) // After every iteration: Divide `N` by the Base `b`
                                                                  r+=N%b; // Increase the result `r` by `N` modulo the Base `b`
                                                                  return r;} // Return the result-sum `r`






                                                                  share|improve this answer














                                                                  share|improve this answer



                                                                  share|improve this answer








                                                                  edited Dec 5 at 11:03

























                                                                  answered Dec 4 at 8:38









                                                                  Kevin Cruijssen

                                                                  35.6k554186




                                                                  35.6k554186








                                                                  • 1




                                                                    There is no need for i, so... 65 bytes.
                                                                    – Olivier Grégoire
                                                                    Dec 5 at 11:01












                                                                  • @OlivierGrégoire Damn, I'm an idiot. Thanks a lot! Hmm, now I also need to golf my derived C and Whitespace answers.. ;)
                                                                    – Kevin Cruijssen
                                                                    Dec 5 at 11:03














                                                                  • 1




                                                                    There is no need for i, so... 65 bytes.
                                                                    – Olivier Grégoire
                                                                    Dec 5 at 11:01












                                                                  • @OlivierGrégoire Damn, I'm an idiot. Thanks a lot! Hmm, now I also need to golf my derived C and Whitespace answers.. ;)
                                                                    – Kevin Cruijssen
                                                                    Dec 5 at 11:03








                                                                  1




                                                                  1




                                                                  There is no need for i, so... 65 bytes.
                                                                  – Olivier Grégoire
                                                                  Dec 5 at 11:01






                                                                  There is no need for i, so... 65 bytes.
                                                                  – Olivier Grégoire
                                                                  Dec 5 at 11:01














                                                                  @OlivierGrégoire Damn, I'm an idiot. Thanks a lot! Hmm, now I also need to golf my derived C and Whitespace answers.. ;)
                                                                  – Kevin Cruijssen
                                                                  Dec 5 at 11:03




                                                                  @OlivierGrégoire Damn, I'm an idiot. Thanks a lot! Hmm, now I also need to golf my derived C and Whitespace answers.. ;)
                                                                  – Kevin Cruijssen
                                                                  Dec 5 at 11:03











                                                                  4














                                                                  SAS, 81 74 bytes



                                                                  data;input n;s=n;do b=2to n;do i=0to n;s+mod(int(n/b**i),b);end;end;cards;


                                                                  Input is entered after the cards; statement, on newlines, like so:



                                                                  data;input n;s=n;do b=2to n;do i=0to n;s+mod(int(n/b**i),b);end;end;cards;
                                                                  1
                                                                  2
                                                                  3
                                                                  4
                                                                  5
                                                                  6
                                                                  7
                                                                  8
                                                                  9
                                                                  10
                                                                  36
                                                                  37
                                                                  64
                                                                  65


                                                                  Outputs a dataset containing the answer s (along with helper variables), with a row for each input value



                                                                  enter image description here



                                                                  Ungolfed:



                                                                  data; /* Implicit dataset creation */
                                                                  input n; /* Read a line of input */

                                                                  s=n; /* Sum = n to start with (the sum of base 1 digits of n is equal to n) */
                                                                  do b=2to n; /* For base = 2 to n */
                                                                  do i=0to n; /* For each place in the output base (output base must always have less than or equal to n digits) */

                                                                  /* Decimal value of current place in the output base = b^i */
                                                                  /* Remainder = int(b / decimal_value) */
                                                                  /* Current place digit = mod(remainder, base) */
                                                                  s+mod(int(n/b**i),b);
                                                                  end;
                                                                  end;

                                                                  cards;
                                                                  1
                                                                  2
                                                                  3





                                                                  share|improve this answer




























                                                                    4














                                                                    SAS, 81 74 bytes



                                                                    data;input n;s=n;do b=2to n;do i=0to n;s+mod(int(n/b**i),b);end;end;cards;


                                                                    Input is entered after the cards; statement, on newlines, like so:



                                                                    data;input n;s=n;do b=2to n;do i=0to n;s+mod(int(n/b**i),b);end;end;cards;
                                                                    1
                                                                    2
                                                                    3
                                                                    4
                                                                    5
                                                                    6
                                                                    7
                                                                    8
                                                                    9
                                                                    10
                                                                    36
                                                                    37
                                                                    64
                                                                    65


                                                                    Outputs a dataset containing the answer s (along with helper variables), with a row for each input value



                                                                    enter image description here



                                                                    Ungolfed:



                                                                    data; /* Implicit dataset creation */
                                                                    input n; /* Read a line of input */

                                                                    s=n; /* Sum = n to start with (the sum of base 1 digits of n is equal to n) */
                                                                    do b=2to n; /* For base = 2 to n */
                                                                    do i=0to n; /* For each place in the output base (output base must always have less than or equal to n digits) */

                                                                    /* Decimal value of current place in the output base = b^i */
                                                                    /* Remainder = int(b / decimal_value) */
                                                                    /* Current place digit = mod(remainder, base) */
                                                                    s+mod(int(n/b**i),b);
                                                                    end;
                                                                    end;

                                                                    cards;
                                                                    1
                                                                    2
                                                                    3





                                                                    share|improve this answer


























                                                                      4












                                                                      4








                                                                      4






                                                                      SAS, 81 74 bytes



                                                                      data;input n;s=n;do b=2to n;do i=0to n;s+mod(int(n/b**i),b);end;end;cards;


                                                                      Input is entered after the cards; statement, on newlines, like so:



                                                                      data;input n;s=n;do b=2to n;do i=0to n;s+mod(int(n/b**i),b);end;end;cards;
                                                                      1
                                                                      2
                                                                      3
                                                                      4
                                                                      5
                                                                      6
                                                                      7
                                                                      8
                                                                      9
                                                                      10
                                                                      36
                                                                      37
                                                                      64
                                                                      65


                                                                      Outputs a dataset containing the answer s (along with helper variables), with a row for each input value



                                                                      enter image description here



                                                                      Ungolfed:



                                                                      data; /* Implicit dataset creation */
                                                                      input n; /* Read a line of input */

                                                                      s=n; /* Sum = n to start with (the sum of base 1 digits of n is equal to n) */
                                                                      do b=2to n; /* For base = 2 to n */
                                                                      do i=0to n; /* For each place in the output base (output base must always have less than or equal to n digits) */

                                                                      /* Decimal value of current place in the output base = b^i */
                                                                      /* Remainder = int(b / decimal_value) */
                                                                      /* Current place digit = mod(remainder, base) */
                                                                      s+mod(int(n/b**i),b);
                                                                      end;
                                                                      end;

                                                                      cards;
                                                                      1
                                                                      2
                                                                      3





                                                                      share|improve this answer














                                                                      SAS, 81 74 bytes



                                                                      data;input n;s=n;do b=2to n;do i=0to n;s+mod(int(n/b**i),b);end;end;cards;


                                                                      Input is entered after the cards; statement, on newlines, like so:



                                                                      data;input n;s=n;do b=2to n;do i=0to n;s+mod(int(n/b**i),b);end;end;cards;
                                                                      1
                                                                      2
                                                                      3
                                                                      4
                                                                      5
                                                                      6
                                                                      7
                                                                      8
                                                                      9
                                                                      10
                                                                      36
                                                                      37
                                                                      64
                                                                      65


                                                                      Outputs a dataset containing the answer s (along with helper variables), with a row for each input value



                                                                      enter image description here



                                                                      Ungolfed:



                                                                      data; /* Implicit dataset creation */
                                                                      input n; /* Read a line of input */

                                                                      s=n; /* Sum = n to start with (the sum of base 1 digits of n is equal to n) */
                                                                      do b=2to n; /* For base = 2 to n */
                                                                      do i=0to n; /* For each place in the output base (output base must always have less than or equal to n digits) */

                                                                      /* Decimal value of current place in the output base = b^i */
                                                                      /* Remainder = int(b / decimal_value) */
                                                                      /* Current place digit = mod(remainder, base) */
                                                                      s+mod(int(n/b**i),b);
                                                                      end;
                                                                      end;

                                                                      cards;
                                                                      1
                                                                      2
                                                                      3






                                                                      share|improve this answer














                                                                      share|improve this answer



                                                                      share|improve this answer








                                                                      edited Dec 4 at 20:25

























                                                                      answered Dec 4 at 19:41









                                                                      Josh Eller

                                                                      2113




                                                                      2113























                                                                          3














                                                                          Japt -x, 6 bytes



                                                                          ÆìXÄ x


                                                                          Try it



                                                                          õ!ìU c


                                                                          Try it






                                                                          share|improve this answer


























                                                                            3














                                                                            Japt -x, 6 bytes



                                                                            ÆìXÄ x


                                                                            Try it



                                                                            õ!ìU c


                                                                            Try it






                                                                            share|improve this answer
























                                                                              3












                                                                              3








                                                                              3






                                                                              Japt -x, 6 bytes



                                                                              ÆìXÄ x


                                                                              Try it



                                                                              õ!ìU c


                                                                              Try it






                                                                              share|improve this answer












                                                                              Japt -x, 6 bytes



                                                                              ÆìXÄ x


                                                                              Try it



                                                                              õ!ìU c


                                                                              Try it







                                                                              share|improve this answer












                                                                              share|improve this answer



                                                                              share|improve this answer










                                                                              answered Dec 3 at 17:20









                                                                              Shaggy

                                                                              18.9k21666




                                                                              18.9k21666























                                                                                  3















                                                                                  J, 24 23 bytes



                                                                                  +1#.1#.]#.inv"0~2+i.@<:


                                                                                  Try it online!






                                                                                  share|improve this answer




























                                                                                    3















                                                                                    J, 24 23 bytes



                                                                                    +1#.1#.]#.inv"0~2+i.@<:


                                                                                    Try it online!






                                                                                    share|improve this answer


























                                                                                      3












                                                                                      3








                                                                                      3







                                                                                      J, 24 23 bytes



                                                                                      +1#.1#.]#.inv"0~2+i.@<:


                                                                                      Try it online!






                                                                                      share|improve this answer















                                                                                      J, 24 23 bytes



                                                                                      +1#.1#.]#.inv"0~2+i.@<:


                                                                                      Try it online!







                                                                                      share|improve this answer














                                                                                      share|improve this answer



                                                                                      share|improve this answer








                                                                                      edited Dec 3 at 18:40

























                                                                                      answered Dec 3 at 17:29









                                                                                      Jonah

                                                                                      2,011816




                                                                                      2,011816























                                                                                          3















                                                                                          05AB1E (legacy), 5 bytes



                                                                                          LвOO+


                                                                                          Try it online!



                                                                                          Explanation:



                                                                                          LвOO+  //Full program
                                                                                          L //push [1 .. input]
                                                                                          в //for each element b, push digits of input converted to base b
                                                                                          O //sum each element
                                                                                          O //sum each sum
                                                                                          + //add input


                                                                                          In 05AB1E (legacy), base 1 of 5 is [0,0,0,0,0], not [1,1,1,1,1]. Therefore after summing the range, add the input to account for the missing base 1.



                                                                                          I am using 05AB1E (legacy) because in current 05AB1E, base 1 of 5 is [1]. In order to account for this, I would either need to decrement the result by 1 or remove the first element of the range, both of which would cost 1 byte.






                                                                                          share|improve this answer


























                                                                                            3















                                                                                            05AB1E (legacy), 5 bytes



                                                                                            LвOO+


                                                                                            Try it online!



                                                                                            Explanation:



                                                                                            LвOO+  //Full program
                                                                                            L //push [1 .. input]
                                                                                            в //for each element b, push digits of input converted to base b
                                                                                            O //sum each element
                                                                                            O //sum each sum
                                                                                            + //add input


                                                                                            In 05AB1E (legacy), base 1 of 5 is [0,0,0,0,0], not [1,1,1,1,1]. Therefore after summing the range, add the input to account for the missing base 1.



                                                                                            I am using 05AB1E (legacy) because in current 05AB1E, base 1 of 5 is [1]. In order to account for this, I would either need to decrement the result by 1 or remove the first element of the range, both of which would cost 1 byte.






                                                                                            share|improve this answer
























                                                                                              3












                                                                                              3








                                                                                              3







                                                                                              05AB1E (legacy), 5 bytes



                                                                                              LвOO+


                                                                                              Try it online!



                                                                                              Explanation:



                                                                                              LвOO+  //Full program
                                                                                              L //push [1 .. input]
                                                                                              в //for each element b, push digits of input converted to base b
                                                                                              O //sum each element
                                                                                              O //sum each sum
                                                                                              + //add input


                                                                                              In 05AB1E (legacy), base 1 of 5 is [0,0,0,0,0], not [1,1,1,1,1]. Therefore after summing the range, add the input to account for the missing base 1.



                                                                                              I am using 05AB1E (legacy) because in current 05AB1E, base 1 of 5 is [1]. In order to account for this, I would either need to decrement the result by 1 or remove the first element of the range, both of which would cost 1 byte.






                                                                                              share|improve this answer













                                                                                              05AB1E (legacy), 5 bytes



                                                                                              LвOO+


                                                                                              Try it online!



                                                                                              Explanation:



                                                                                              LвOO+  //Full program
                                                                                              L //push [1 .. input]
                                                                                              в //for each element b, push digits of input converted to base b
                                                                                              O //sum each element
                                                                                              O //sum each sum
                                                                                              + //add input


                                                                                              In 05AB1E (legacy), base 1 of 5 is [0,0,0,0,0], not [1,1,1,1,1]. Therefore after summing the range, add the input to account for the missing base 1.



                                                                                              I am using 05AB1E (legacy) because in current 05AB1E, base 1 of 5 is [1]. In order to account for this, I would either need to decrement the result by 1 or remove the first element of the range, both of which would cost 1 byte.







                                                                                              share|improve this answer












                                                                                              share|improve this answer



                                                                                              share|improve this answer










                                                                                              answered Dec 3 at 20:25









                                                                                              Cowabunghole

                                                                                              1,065418




                                                                                              1,065418























                                                                                                  3














                                                                                                  Desmos, 127 bytes



                                                                                                  $$fleft(nright)=sum_{b=2}^{n+1}sum_{i=0}^{n-1}operatorname{mod}left(operatorname{floor}left(frac{n}{b^i}right),bright)$$



                                                                                                  fleft(nright)=sum_{b=2}^{n+1}sum_{i=0}^{n-1}operatorname{mod}left(operatorname{floor}left(frac{n}{b^i}right),bright)


                                                                                                  Try it here! Defines a function $f$, called as $fleft(nright)$. Uses the formula given in Dennis's answer.



                                                                                                  Here is the resulting graph (point at $(65, 932)$):



                                                                                                  graph generated on desmos.com



                                                                                                  Desmos, 56 bytes



                                                                                                  This might not work on all browsers, but it works on mine. Just copy and paste the formula. This is $f_2(n)$ in the above link.



                                                                                                  f(n)=sum_{b=2}^{n+1}sum_{i=0}^{n-1}mod(floor(n/b^i),b)





                                                                                                  share|improve this answer


























                                                                                                    3














                                                                                                    Desmos, 127 bytes



                                                                                                    $$fleft(nright)=sum_{b=2}^{n+1}sum_{i=0}^{n-1}operatorname{mod}left(operatorname{floor}left(frac{n}{b^i}right),bright)$$



                                                                                                    fleft(nright)=sum_{b=2}^{n+1}sum_{i=0}^{n-1}operatorname{mod}left(operatorname{floor}left(frac{n}{b^i}right),bright)


                                                                                                    Try it here! Defines a function $f$, called as $fleft(nright)$. Uses the formula given in Dennis's answer.



                                                                                                    Here is the resulting graph (point at $(65, 932)$):



                                                                                                    graph generated on desmos.com



                                                                                                    Desmos, 56 bytes



                                                                                                    This might not work on all browsers, but it works on mine. Just copy and paste the formula. This is $f_2(n)$ in the above link.



                                                                                                    f(n)=sum_{b=2}^{n+1}sum_{i=0}^{n-1}mod(floor(n/b^i),b)





                                                                                                    share|improve this answer
























                                                                                                      3












                                                                                                      3








                                                                                                      3






                                                                                                      Desmos, 127 bytes



                                                                                                      $$fleft(nright)=sum_{b=2}^{n+1}sum_{i=0}^{n-1}operatorname{mod}left(operatorname{floor}left(frac{n}{b^i}right),bright)$$



                                                                                                      fleft(nright)=sum_{b=2}^{n+1}sum_{i=0}^{n-1}operatorname{mod}left(operatorname{floor}left(frac{n}{b^i}right),bright)


                                                                                                      Try it here! Defines a function $f$, called as $fleft(nright)$. Uses the formula given in Dennis's answer.



                                                                                                      Here is the resulting graph (point at $(65, 932)$):



                                                                                                      graph generated on desmos.com



                                                                                                      Desmos, 56 bytes



                                                                                                      This might not work on all browsers, but it works on mine. Just copy and paste the formula. This is $f_2(n)$ in the above link.



                                                                                                      f(n)=sum_{b=2}^{n+1}sum_{i=0}^{n-1}mod(floor(n/b^i),b)





                                                                                                      share|improve this answer












                                                                                                      Desmos, 127 bytes



                                                                                                      $$fleft(nright)=sum_{b=2}^{n+1}sum_{i=0}^{n-1}operatorname{mod}left(operatorname{floor}left(frac{n}{b^i}right),bright)$$



                                                                                                      fleft(nright)=sum_{b=2}^{n+1}sum_{i=0}^{n-1}operatorname{mod}left(operatorname{floor}left(frac{n}{b^i}right),bright)


                                                                                                      Try it here! Defines a function $f$, called as $fleft(nright)$. Uses the formula given in Dennis's answer.



                                                                                                      Here is the resulting graph (point at $(65, 932)$):



                                                                                                      graph generated on desmos.com



                                                                                                      Desmos, 56 bytes



                                                                                                      This might not work on all browsers, but it works on mine. Just copy and paste the formula. This is $f_2(n)$ in the above link.



                                                                                                      f(n)=sum_{b=2}^{n+1}sum_{i=0}^{n-1}mod(floor(n/b^i),b)






                                                                                                      share|improve this answer












                                                                                                      share|improve this answer



                                                                                                      share|improve this answer










                                                                                                      answered Dec 4 at 1:14









                                                                                                      Conor O'Brien

                                                                                                      29.1k263162




                                                                                                      29.1k263162























                                                                                                          3















                                                                                                          Perl 6, 45 41 bytes



                                                                                                          {$^a+sum map {|polymod $a: $_ xx*},2..$a}


                                                                                                          -4 bytes thanks to Jo King



                                                                                                          Try it online!






                                                                                                          share|improve this answer























                                                                                                          • 36 bytes
                                                                                                            – nwellnhof
                                                                                                            Dec 6 at 14:39
















                                                                                                          3















                                                                                                          Perl 6, 45 41 bytes



                                                                                                          {$^a+sum map {|polymod $a: $_ xx*},2..$a}


                                                                                                          -4 bytes thanks to Jo King



                                                                                                          Try it online!






                                                                                                          share|improve this answer























                                                                                                          • 36 bytes
                                                                                                            – nwellnhof
                                                                                                            Dec 6 at 14:39














                                                                                                          3












                                                                                                          3








                                                                                                          3







                                                                                                          Perl 6, 45 41 bytes



                                                                                                          {$^a+sum map {|polymod $a: $_ xx*},2..$a}


                                                                                                          -4 bytes thanks to Jo King



                                                                                                          Try it online!






                                                                                                          share|improve this answer















                                                                                                          Perl 6, 45 41 bytes



                                                                                                          {$^a+sum map {|polymod $a: $_ xx*},2..$a}


                                                                                                          -4 bytes thanks to Jo King



                                                                                                          Try it online!







                                                                                                          share|improve this answer














                                                                                                          share|improve this answer



                                                                                                          share|improve this answer








                                                                                                          edited Dec 4 at 1:27

























                                                                                                          answered Dec 3 at 18:46









                                                                                                          Sean

                                                                                                          3,32636




                                                                                                          3,32636












                                                                                                          • 36 bytes
                                                                                                            – nwellnhof
                                                                                                            Dec 6 at 14:39


















                                                                                                          • 36 bytes
                                                                                                            – nwellnhof
                                                                                                            Dec 6 at 14:39
















                                                                                                          36 bytes
                                                                                                          – nwellnhof
                                                                                                          Dec 6 at 14:39




                                                                                                          36 bytes
                                                                                                          – nwellnhof
                                                                                                          Dec 6 at 14:39











                                                                                                          3















                                                                                                          Whitespace, 153 bytes



                                                                                                          [S S S N
                                                                                                          _Push_0][S N
                                                                                                          S _Duplicate_0][T N
                                                                                                          T T _Read_STDIN_as_integer][T T T _Retrieve_input][S N
                                                                                                          S _Duplicate_input][N
                                                                                                          S S N
                                                                                                          _Create_Label_OUTER_LOOP][S N
                                                                                                          S _Duplicate_top][S S S T N
                                                                                                          _Push_1][T S S T _Subtract][N
                                                                                                          T S S N
                                                                                                          _If_0_Jump_to_Label_PRINT][S S S N
                                                                                                          _Push_0][S N
                                                                                                          S _Duplicate_0][T T T _Retrieve_input][N
                                                                                                          S S T N
                                                                                                          _Create_Label_INNER_LOOP][S N
                                                                                                          S _Duplicate][N
                                                                                                          T S S S N
                                                                                                          _If_0_Jump_to_Label_AFTER_INNER_LOOP][S N
                                                                                                          T _Swap_top_two][S T S S T N
                                                                                                          _Copy_1st_item_to_top][S T S S T T N
                                                                                                          _Copy_3rd_item_to_top][T S T T _Modulo][T S S S _Add][S N
                                                                                                          T _Swap_top_two][S T S S T S N
                                                                                                          _Copy_2nd_item_to_top][T S T S _Integer_divide][N
                                                                                                          S N
                                                                                                          T N
                                                                                                          _Jump_to_Label_INNER_LOOP][N
                                                                                                          S S S S N
                                                                                                          _Create_Label_AFTER_INNER_LOOP][S N
                                                                                                          N
                                                                                                          _Discard_top][S T S S T S N
                                                                                                          _Copy_2nd_item_to_top][T S S S _Add][S N
                                                                                                          T _Swap_top_two][S S S T N
                                                                                                          _Push_1][T S S T _Subtract][N
                                                                                                          S N
                                                                                                          N
                                                                                                          _Jump_to_Label_OUTER_LOOP][N
                                                                                                          S S S N
                                                                                                          _Create_Label_PRINT][S N
                                                                                                          N
                                                                                                          _Discard_top][T N
                                                                                                          S T _Print_as_integer]


                                                                                                          Letters S (space), T (tab), and N (new-line) added as highlighting only.
                                                                                                          [..._some_action] added as explanation only.



                                                                                                          Try it online (with raw spaces, tabs and new-lines only).



                                                                                                          Port of my Java 8 answer, because Whitespace has no Base conversion builtins at all.



                                                                                                          Example run: input = 3



                                                                                                          Command    Explanation                     Stack            Heap   STDIN  STDOUT  STDERR

                                                                                                          SSSN Push 0 [0]
                                                                                                          SNS Duplicate 0 [0,0]
                                                                                                          TNTT Read STDIN as integer [0] {0:3} 3
                                                                                                          TTT Retrieve input from heap 0 [3] {0:3}
                                                                                                          SNS Duplicate 3 [3,3] {0:3}
                                                                                                          NSSN Create Label_OUTER_LOOP [3,3] {0:3}
                                                                                                          SNS Duplicate 3 [3,3,3] {0:3}
                                                                                                          SSSTN Push 1 [3,3,3,1] {0:3}
                                                                                                          TSST Subtract (3-1) [3,3,2] {0:3}
                                                                                                          NTSSN If 0: Jump to Label_PRINT [3,3] {0:3}
                                                                                                          SSSN Push 0 [3,3,0] {0:3}
                                                                                                          SNS Duplicate 0 [3,3,0,0] {0:3}
                                                                                                          TTT Retrieve input from heap 0 [3,3,0,3] {0:3}
                                                                                                          NSSTN Create Label_INNER_LOOP [3,3,0,3] {0:3}
                                                                                                          SNS Duplicate 3 [3,3,0,3,3] {0:3}
                                                                                                          NTSSSN If 0: Jump to Label_AFTER_IL [3,3,0,3] {0:3}
                                                                                                          SNT Swap top two [3,3,3,0] {0:3}
                                                                                                          STSSTN Copy (0-indexed) 1st [3,3,3,0,3] {0:3}
                                                                                                          STSSTTN Copy (0-indexed) 3rd [3,3,3,0,3,3] {0:3}
                                                                                                          TSTT Modulo (3%3) [3,3,3,0,0] {0:3}
                                                                                                          TSSS Add (0+0) [3,3,3,0] {0:3}
                                                                                                          SNT Swap top two [3,3,0,3] {0:3}
                                                                                                          STSSTSN Copy (0-indexed) 2nd [3,3,0,3,3] {0:3}
                                                                                                          TSTS Integer divide (3/3) [3,3,0,1] {0:3}
                                                                                                          NSNTN Jump to LABEL_INNER_LOOP [3,3,0,1] {0:3}

                                                                                                          SNS Duplicate 1 [3,3,0,1,1] {0:3}
                                                                                                          NTSSSN If 0: Jump to Label_AFTER_IL [3,3,0,1] {0:3}
                                                                                                          SNT Swap top two [3,3,1,0] {0:3}
                                                                                                          STSSTN Copy (0-indexed) 1st [3,3,1,0,1] {0:3}
                                                                                                          STSSTTN Copy (0-indexed) 3rd [3,3,1,0,1,3] {0:3}
                                                                                                          TSTT Modulo (1%3) [3,3,1,0,1] {0:3}
                                                                                                          TSSS Add (0+1) [3,3,1,1] {0:3}
                                                                                                          SNT Swap top two [3,3,1,1] {0:3}
                                                                                                          STSSTSN Copy (0-indexed) 2nd [3,3,1,1,3] {0:3}
                                                                                                          TSTS Integer divide (1/3) [3,3,1,0] {0:3}
                                                                                                          NSNTN Jump to LABEL_INNER_LOOP [3,3,1,0] {0:3}

                                                                                                          SNS Duplicate 0 [3,3,1,0,0] {0:3}
                                                                                                          NTSSSN If 0: Jump to Label_AFTER_IL [3,3,1,0] {0:3}
                                                                                                          NSSSSN Create Label_AFTER_IL [3,3,1,0] {0:3}
                                                                                                          SNN Discard top [3,3,1] {0:3}
                                                                                                          STSSTSN Copy (0-indexed) 2nd [3,3,1,3] {0:3}
                                                                                                          TSSS Add (1+3) [3,3,4] {0:3}
                                                                                                          SNT Swap top two [3,4,3] {0:3}
                                                                                                          SSSTN Push 1 [3,4,3,1] {0:3}
                                                                                                          TSST Subtract (3-1) [3,4,2] {0:3}
                                                                                                          NSNN Jump to LABEL_OUTER_LOOP [3,4,2] {0:3}

                                                                                                          SNS Duplicate 2 [3,4,2,2] {0:3}
                                                                                                          SSSTN Push 1 [3,4,2,2,1] {0:3}
                                                                                                          TSST Subtract (2-1) [3,4,2,1] {0:3}
                                                                                                          NTSSN If 0: Jump to Label_PRINT [3,4,2] {0:3}
                                                                                                          SSSN Push 0 [3,4,2,0] {0:3}
                                                                                                          SNS Duplicate 0 [3,4,2,0,0] {0:3}
                                                                                                          TTT Retrieve input from heap 0 [3,4,2,0,3] {0:3}
                                                                                                          NSSTN Create Label_INNER_LOOP [3,4,2,0,3] {0:3}
                                                                                                          SNS Duplicate 3 [3,4,2,0,3,3] {0:3}
                                                                                                          NTSSSN If 0: Jump to Label_AFTER_IL [3,4,2,0,3] {0:3}
                                                                                                          SNT Swap top two [3,4,2,3,0] {0:3}
                                                                                                          STSSTN Copy (0-indexed) 1st [3,4,2,3,0,3] {0:3}
                                                                                                          STSSTTN Copy (0-indexed) 3rd [3,4,2,3,0,3,2] {0:3}
                                                                                                          TSTT Modulo (3%2) [3,4,2,3,0,1] {0:3}
                                                                                                          TSSS Add (0+1) [3,4,2,3,1] {0:3}
                                                                                                          SNT Swap top two [3,4,2,1,3] {0:3}
                                                                                                          STSSTSN Copy (0-indexed) 2nd [3,4,2,1,3,2] {0:3}
                                                                                                          TSTS Integer divide (3/2) [3,4,2,1,1] {0:3}
                                                                                                          NSNTN Jump to LABEL_INNER_LOOP [3,4,2,1,1] {0:3}

                                                                                                          SNS Duplicate 1 [3,4,2,1,1,1] {0:3}
                                                                                                          NTSSSN If 0: Jump to Label_AFTER_IL [3,4,2,1,1] {0:3}
                                                                                                          SNT Swap top two [3,4,2,1,1] {0:3}
                                                                                                          STSSTN Copy (0-indexed) 1st [3,4,2,1,1,1] {0:3}
                                                                                                          STSSTTN Copy (0-indexed) 3rd [3,4,2,1,1,1,2] {0:3}
                                                                                                          TSTT Modulo (1%2) [3,4,2,1,1,1] {0:3}
                                                                                                          TSSS Add (1+1) [3,4,2,1,2] {0:3}
                                                                                                          SNT Swap top two [3,4,2,2,1] {0:3}
                                                                                                          STSSTSN Copy (0-indexed) 2nd [3,4,2,2,1,2] {0:3}
                                                                                                          TSTS Integer divide (1/2) [3,4,2,2,0] {0:3}
                                                                                                          NSNTN Jump to LABEL_INNER_LOOP [3,4,2,2,0] {0:3}

                                                                                                          SNS Duplicate 0 [3,4,2,2,0,0] {0:3}
                                                                                                          NTSSSN If 0: Jump to Label_AFTER_IL [3,4,2,2,0] {0:3}
                                                                                                          NSSSSN Create Label_AFTER_IL [3,4,2,2,0] {0:3}
                                                                                                          SNN Discard top [3,4,2,2] {0:3}
                                                                                                          STSSTSN Copy (0-indexed) 2nd [3,4,2,2,4] {0:3}
                                                                                                          TSSS Add (2+4) [3,4,2,6] {0:3}
                                                                                                          SNT Swap top two [3,4,6,2] {0:3}
                                                                                                          SSSTN Push 1 [3,4,6,2,1] {0:3}
                                                                                                          TSST Subtract (2-1) [3,4,6,1] {0:3}
                                                                                                          NSNN Jump to LABEL_OUTER_LOOP [3,4,6,1] {0:3}

                                                                                                          SNS Duplicate 1 [3,4,6,1,1] {0:3}
                                                                                                          SSSTN Push 1 [3,4,6,1,1,1] {0:3}
                                                                                                          TSST Subtract (1-1) [3,4,6,1,0] {0:3}
                                                                                                          NTSSN If 0: Jump to Label_PRINT [3,4,6,1] {0:3}
                                                                                                          NSSSSN Create Label_PRINT [3,4,6,1] {0:3}
                                                                                                          SNN Discard top [3,4,6] {0:3}
                                                                                                          TNST Print top (6) to STDOUT as int [3,4] {0:3} 6
                                                                                                          error


                                                                                                          Program stops with an error: No exit found. (Although I could add three trailing newlines NNN to get rid of that error.)






                                                                                                          share|improve this answer


























                                                                                                            3















                                                                                                            Whitespace, 153 bytes



                                                                                                            [S S S N
                                                                                                            _Push_0][S N
                                                                                                            S _Duplicate_0][T N
                                                                                                            T T _Read_STDIN_as_integer][T T T _Retrieve_input][S N
                                                                                                            S _Duplicate_input][N
                                                                                                            S S N
                                                                                                            _Create_Label_OUTER_LOOP][S N
                                                                                                            S _Duplicate_top][S S S T N
                                                                                                            _Push_1][T S S T _Subtract][N
                                                                                                            T S S N
                                                                                                            _If_0_Jump_to_Label_PRINT][S S S N
                                                                                                            _Push_0][S N
                                                                                                            S _Duplicate_0][T T T _Retrieve_input][N
                                                                                                            S S T N
                                                                                                            _Create_Label_INNER_LOOP][S N
                                                                                                            S _Duplicate][N
                                                                                                            T S S S N
                                                                                                            _If_0_Jump_to_Label_AFTER_INNER_LOOP][S N
                                                                                                            T _Swap_top_two][S T S S T N
                                                                                                            _Copy_1st_item_to_top][S T S S T T N
                                                                                                            _Copy_3rd_item_to_top][T S T T _Modulo][T S S S _Add][S N
                                                                                                            T _Swap_top_two][S T S S T S N
                                                                                                            _Copy_2nd_item_to_top][T S T S _Integer_divide][N
                                                                                                            S N
                                                                                                            T N
                                                                                                            _Jump_to_Label_INNER_LOOP][N
                                                                                                            S S S S N
                                                                                                            _Create_Label_AFTER_INNER_LOOP][S N
                                                                                                            N
                                                                                                            _Discard_top][S T S S T S N
                                                                                                            _Copy_2nd_item_to_top][T S S S _Add][S N
                                                                                                            T _Swap_top_two][S S S T N
                                                                                                            _Push_1][T S S T _Subtract][N
                                                                                                            S N
                                                                                                            N
                                                                                                            _Jump_to_Label_OUTER_LOOP][N
                                                                                                            S S S N
                                                                                                            _Create_Label_PRINT][S N
                                                                                                            N
                                                                                                            _Discard_top][T N
                                                                                                            S T _Print_as_integer]


                                                                                                            Letters S (space), T (tab), and N (new-line) added as highlighting only.
                                                                                                            [..._some_action] added as explanation only.



                                                                                                            Try it online (with raw spaces, tabs and new-lines only).



                                                                                                            Port of my Java 8 answer, because Whitespace has no Base conversion builtins at all.



                                                                                                            Example run: input = 3



                                                                                                            Command    Explanation                     Stack            Heap   STDIN  STDOUT  STDERR

                                                                                                            SSSN Push 0 [0]
                                                                                                            SNS Duplicate 0 [0,0]
                                                                                                            TNTT Read STDIN as integer [0] {0:3} 3
                                                                                                            TTT Retrieve input from heap 0 [3] {0:3}
                                                                                                            SNS Duplicate 3 [3,3] {0:3}
                                                                                                            NSSN Create Label_OUTER_LOOP [3,3] {0:3}
                                                                                                            SNS Duplicate 3 [3,3,3] {0:3}
                                                                                                            SSSTN Push 1 [3,3,3,1] {0:3}
                                                                                                            TSST Subtract (3-1) [3,3,2] {0:3}
                                                                                                            NTSSN If 0: Jump to Label_PRINT [3,3] {0:3}
                                                                                                            SSSN Push 0 [3,3,0] {0:3}
                                                                                                            SNS Duplicate 0 [3,3,0,0] {0:3}
                                                                                                            TTT Retrieve input from heap 0 [3,3,0,3] {0:3}
                                                                                                            NSSTN Create Label_INNER_LOOP [3,3,0,3] {0:3}
                                                                                                            SNS Duplicate 3 [3,3,0,3,3] {0:3}
                                                                                                            NTSSSN If 0: Jump to Label_AFTER_IL [3,3,0,3] {0:3}
                                                                                                            SNT Swap top two [3,3,3,0] {0:3}
                                                                                                            STSSTN Copy (0-indexed) 1st [3,3,3,0,3] {0:3}
                                                                                                            STSSTTN Copy (0-indexed) 3rd [3,3,3,0,3,3] {0:3}
                                                                                                            TSTT Modulo (3%3) [3,3,3,0,0] {0:3}
                                                                                                            TSSS Add (0+0) [3,3,3,0] {0:3}
                                                                                                            SNT Swap top two [3,3,0,3] {0:3}
                                                                                                            STSSTSN Copy (0-indexed) 2nd [3,3,0,3,3] {0:3}
                                                                                                            TSTS Integer divide (3/3) [3,3,0,1] {0:3}
                                                                                                            NSNTN Jump to LABEL_INNER_LOOP [3,3,0,1] {0:3}

                                                                                                            SNS Duplicate 1 [3,3,0,1,1] {0:3}
                                                                                                            NTSSSN If 0: Jump to Label_AFTER_IL [3,3,0,1] {0:3}
                                                                                                            SNT Swap top two [3,3,1,0] {0:3}
                                                                                                            STSSTN Copy (0-indexed) 1st [3,3,1,0,1] {0:3}
                                                                                                            STSSTTN Copy (0-indexed) 3rd [3,3,1,0,1,3] {0:3}
                                                                                                            TSTT Modulo (1%3) [3,3,1,0,1] {0:3}
                                                                                                            TSSS Add (0+1) [3,3,1,1] {0:3}
                                                                                                            SNT Swap top two [3,3,1,1] {0:3}
                                                                                                            STSSTSN Copy (0-indexed) 2nd [3,3,1,1,3] {0:3}
                                                                                                            TSTS Integer divide (1/3) [3,3,1,0] {0:3}
                                                                                                            NSNTN Jump to LABEL_INNER_LOOP [3,3,1,0] {0:3}

                                                                                                            SNS Duplicate 0 [3,3,1,0,0] {0:3}
                                                                                                            NTSSSN If 0: Jump to Label_AFTER_IL [3,3,1,0] {0:3}
                                                                                                            NSSSSN Create Label_AFTER_IL [3,3,1,0] {0:3}
                                                                                                            SNN Discard top [3,3,1] {0:3}
                                                                                                            STSSTSN Copy (0-indexed) 2nd [3,3,1,3] {0:3}
                                                                                                            TSSS Add (1+3) [3,3,4] {0:3}
                                                                                                            SNT Swap top two [3,4,3] {0:3}
                                                                                                            SSSTN Push 1 [3,4,3,1] {0:3}
                                                                                                            TSST Subtract (3-1) [3,4,2] {0:3}
                                                                                                            NSNN Jump to LABEL_OUTER_LOOP [3,4,2] {0:3}

                                                                                                            SNS Duplicate 2 [3,4,2,2] {0:3}
                                                                                                            SSSTN Push 1 [3,4,2,2,1] {0:3}
                                                                                                            TSST Subtract (2-1) [3,4,2,1] {0:3}
                                                                                                            NTSSN If 0: Jump to Label_PRINT [3,4,2] {0:3}
                                                                                                            SSSN Push 0 [3,4,2,0] {0:3}
                                                                                                            SNS Duplicate 0 [3,4,2,0,0] {0:3}
                                                                                                            TTT Retrieve input from heap 0 [3,4,2,0,3] {0:3}
                                                                                                            NSSTN Create Label_INNER_LOOP [3,4,2,0,3] {0:3}
                                                                                                            SNS Duplicate 3 [3,4,2,0,3,3] {0:3}
                                                                                                            NTSSSN If 0: Jump to Label_AFTER_IL [3,4,2,0,3] {0:3}
                                                                                                            SNT Swap top two [3,4,2,3,0] {0:3}
                                                                                                            STSSTN Copy (0-indexed) 1st [3,4,2,3,0,3] {0:3}
                                                                                                            STSSTTN Copy (0-indexed) 3rd [3,4,2,3,0,3,2] {0:3}
                                                                                                            TSTT Modulo (3%2) [3,4,2,3,0,1] {0:3}
                                                                                                            TSSS Add (0+1) [3,4,2,3,1] {0:3}
                                                                                                            SNT Swap top two [3,4,2,1,3] {0:3}
                                                                                                            STSSTSN Copy (0-indexed) 2nd [3,4,2,1,3,2] {0:3}
                                                                                                            TSTS Integer divide (3/2) [3,4,2,1,1] {0:3}
                                                                                                            NSNTN Jump to LABEL_INNER_LOOP [3,4,2,1,1] {0:3}

                                                                                                            SNS Duplicate 1 [3,4,2,1,1,1] {0:3}
                                                                                                            NTSSSN If 0: Jump to Label_AFTER_IL [3,4,2,1,1] {0:3}
                                                                                                            SNT Swap top two [3,4,2,1,1] {0:3}
                                                                                                            STSSTN Copy (0-indexed) 1st [3,4,2,1,1,1] {0:3}
                                                                                                            STSSTTN Copy (0-indexed) 3rd [3,4,2,1,1,1,2] {0:3}
                                                                                                            TSTT Modulo (1%2) [3,4,2,1,1,1] {0:3}
                                                                                                            TSSS Add (1+1) [3,4,2,1,2] {0:3}
                                                                                                            SNT Swap top two [3,4,2,2,1] {0:3}
                                                                                                            STSSTSN Copy (0-indexed) 2nd [3,4,2,2,1,2] {0:3}
                                                                                                            TSTS Integer divide (1/2) [3,4,2,2,0] {0:3}
                                                                                                            NSNTN Jump to LABEL_INNER_LOOP [3,4,2,2,0] {0:3}

                                                                                                            SNS Duplicate 0 [3,4,2,2,0,0] {0:3}
                                                                                                            NTSSSN If 0: Jump to Label_AFTER_IL [3,4,2,2,0] {0:3}
                                                                                                            NSSSSN Create Label_AFTER_IL [3,4,2,2,0] {0:3}
                                                                                                            SNN Discard top [3,4,2,2] {0:3}
                                                                                                            STSSTSN Copy (0-indexed) 2nd [3,4,2,2,4] {0:3}
                                                                                                            TSSS Add (2+4) [3,4,2,6] {0:3}
                                                                                                            SNT Swap top two [3,4,6,2] {0:3}
                                                                                                            SSSTN Push 1 [3,4,6,2,1] {0:3}
                                                                                                            TSST Subtract (2-1) [3,4,6,1] {0:3}
                                                                                                            NSNN Jump to LABEL_OUTER_LOOP [3,4,6,1] {0:3}

                                                                                                            SNS Duplicate 1 [3,4,6,1,1] {0:3}
                                                                                                            SSSTN Push 1 [3,4,6,1,1,1] {0:3}
                                                                                                            TSST Subtract (1-1) [3,4,6,1,0] {0:3}
                                                                                                            NTSSN If 0: Jump to Label_PRINT [3,4,6,1] {0:3}
                                                                                                            NSSSSN Create Label_PRINT [3,4,6,1] {0:3}
                                                                                                            SNN Discard top [3,4,6] {0:3}
                                                                                                            TNST Print top (6) to STDOUT as int [3,4] {0:3} 6
                                                                                                            error


                                                                                                            Program stops with an error: No exit found. (Although I could add three trailing newlines NNN to get rid of that error.)






                                                                                                            share|improve this answer
























                                                                                                              3












                                                                                                              3








                                                                                                              3







                                                                                                              Whitespace, 153 bytes



                                                                                                              [S S S N
                                                                                                              _Push_0][S N
                                                                                                              S _Duplicate_0][T N
                                                                                                              T T _Read_STDIN_as_integer][T T T _Retrieve_input][S N
                                                                                                              S _Duplicate_input][N
                                                                                                              S S N
                                                                                                              _Create_Label_OUTER_LOOP][S N
                                                                                                              S _Duplicate_top][S S S T N
                                                                                                              _Push_1][T S S T _Subtract][N
                                                                                                              T S S N
                                                                                                              _If_0_Jump_to_Label_PRINT][S S S N
                                                                                                              _Push_0][S N
                                                                                                              S _Duplicate_0][T T T _Retrieve_input][N
                                                                                                              S S T N
                                                                                                              _Create_Label_INNER_LOOP][S N
                                                                                                              S _Duplicate][N
                                                                                                              T S S S N
                                                                                                              _If_0_Jump_to_Label_AFTER_INNER_LOOP][S N
                                                                                                              T _Swap_top_two][S T S S T N
                                                                                                              _Copy_1st_item_to_top][S T S S T T N
                                                                                                              _Copy_3rd_item_to_top][T S T T _Modulo][T S S S _Add][S N
                                                                                                              T _Swap_top_two][S T S S T S N
                                                                                                              _Copy_2nd_item_to_top][T S T S _Integer_divide][N
                                                                                                              S N
                                                                                                              T N
                                                                                                              _Jump_to_Label_INNER_LOOP][N
                                                                                                              S S S S N
                                                                                                              _Create_Label_AFTER_INNER_LOOP][S N
                                                                                                              N
                                                                                                              _Discard_top][S T S S T S N
                                                                                                              _Copy_2nd_item_to_top][T S S S _Add][S N
                                                                                                              T _Swap_top_two][S S S T N
                                                                                                              _Push_1][T S S T _Subtract][N
                                                                                                              S N
                                                                                                              N
                                                                                                              _Jump_to_Label_OUTER_LOOP][N
                                                                                                              S S S N
                                                                                                              _Create_Label_PRINT][S N
                                                                                                              N
                                                                                                              _Discard_top][T N
                                                                                                              S T _Print_as_integer]


                                                                                                              Letters S (space), T (tab), and N (new-line) added as highlighting only.
                                                                                                              [..._some_action] added as explanation only.



                                                                                                              Try it online (with raw spaces, tabs and new-lines only).



                                                                                                              Port of my Java 8 answer, because Whitespace has no Base conversion builtins at all.



                                                                                                              Example run: input = 3



                                                                                                              Command    Explanation                     Stack            Heap   STDIN  STDOUT  STDERR

                                                                                                              SSSN Push 0 [0]
                                                                                                              SNS Duplicate 0 [0,0]
                                                                                                              TNTT Read STDIN as integer [0] {0:3} 3
                                                                                                              TTT Retrieve input from heap 0 [3] {0:3}
                                                                                                              SNS Duplicate 3 [3,3] {0:3}
                                                                                                              NSSN Create Label_OUTER_LOOP [3,3] {0:3}
                                                                                                              SNS Duplicate 3 [3,3,3] {0:3}
                                                                                                              SSSTN Push 1 [3,3,3,1] {0:3}
                                                                                                              TSST Subtract (3-1) [3,3,2] {0:3}
                                                                                                              NTSSN If 0: Jump to Label_PRINT [3,3] {0:3}
                                                                                                              SSSN Push 0 [3,3,0] {0:3}
                                                                                                              SNS Duplicate 0 [3,3,0,0] {0:3}
                                                                                                              TTT Retrieve input from heap 0 [3,3,0,3] {0:3}
                                                                                                              NSSTN Create Label_INNER_LOOP [3,3,0,3] {0:3}
                                                                                                              SNS Duplicate 3 [3,3,0,3,3] {0:3}
                                                                                                              NTSSSN If 0: Jump to Label_AFTER_IL [3,3,0,3] {0:3}
                                                                                                              SNT Swap top two [3,3,3,0] {0:3}
                                                                                                              STSSTN Copy (0-indexed) 1st [3,3,3,0,3] {0:3}
                                                                                                              STSSTTN Copy (0-indexed) 3rd [3,3,3,0,3,3] {0:3}
                                                                                                              TSTT Modulo (3%3) [3,3,3,0,0] {0:3}
                                                                                                              TSSS Add (0+0) [3,3,3,0] {0:3}
                                                                                                              SNT Swap top two [3,3,0,3] {0:3}
                                                                                                              STSSTSN Copy (0-indexed) 2nd [3,3,0,3,3] {0:3}
                                                                                                              TSTS Integer divide (3/3) [3,3,0,1] {0:3}
                                                                                                              NSNTN Jump to LABEL_INNER_LOOP [3,3,0,1] {0:3}

                                                                                                              SNS Duplicate 1 [3,3,0,1,1] {0:3}
                                                                                                              NTSSSN If 0: Jump to Label_AFTER_IL [3,3,0,1] {0:3}
                                                                                                              SNT Swap top two [3,3,1,0] {0:3}
                                                                                                              STSSTN Copy (0-indexed) 1st [3,3,1,0,1] {0:3}
                                                                                                              STSSTTN Copy (0-indexed) 3rd [3,3,1,0,1,3] {0:3}
                                                                                                              TSTT Modulo (1%3) [3,3,1,0,1] {0:3}
                                                                                                              TSSS Add (0+1) [3,3,1,1] {0:3}
                                                                                                              SNT Swap top two [3,3,1,1] {0:3}
                                                                                                              STSSTSN Copy (0-indexed) 2nd [3,3,1,1,3] {0:3}
                                                                                                              TSTS Integer divide (1/3) [3,3,1,0] {0:3}
                                                                                                              NSNTN Jump to LABEL_INNER_LOOP [3,3,1,0] {0:3}

                                                                                                              SNS Duplicate 0 [3,3,1,0,0] {0:3}
                                                                                                              NTSSSN If 0: Jump to Label_AFTER_IL [3,3,1,0] {0:3}
                                                                                                              NSSSSN Create Label_AFTER_IL [3,3,1,0] {0:3}
                                                                                                              SNN Discard top [3,3,1] {0:3}
                                                                                                              STSSTSN Copy (0-indexed) 2nd [3,3,1,3] {0:3}
                                                                                                              TSSS Add (1+3) [3,3,4] {0:3}
                                                                                                              SNT Swap top two [3,4,3] {0:3}
                                                                                                              SSSTN Push 1 [3,4,3,1] {0:3}
                                                                                                              TSST Subtract (3-1) [3,4,2] {0:3}
                                                                                                              NSNN Jump to LABEL_OUTER_LOOP [3,4,2] {0:3}

                                                                                                              SNS Duplicate 2 [3,4,2,2] {0:3}
                                                                                                              SSSTN Push 1 [3,4,2,2,1] {0:3}
                                                                                                              TSST Subtract (2-1) [3,4,2,1] {0:3}
                                                                                                              NTSSN If 0: Jump to Label_PRINT [3,4,2] {0:3}
                                                                                                              SSSN Push 0 [3,4,2,0] {0:3}
                                                                                                              SNS Duplicate 0 [3,4,2,0,0] {0:3}
                                                                                                              TTT Retrieve input from heap 0 [3,4,2,0,3] {0:3}
                                                                                                              NSSTN Create Label_INNER_LOOP [3,4,2,0,3] {0:3}
                                                                                                              SNS Duplicate 3 [3,4,2,0,3,3] {0:3}
                                                                                                              NTSSSN If 0: Jump to Label_AFTER_IL [3,4,2,0,3] {0:3}
                                                                                                              SNT Swap top two [3,4,2,3,0] {0:3}
                                                                                                              STSSTN Copy (0-indexed) 1st [3,4,2,3,0,3] {0:3}
                                                                                                              STSSTTN Copy (0-indexed) 3rd [3,4,2,3,0,3,2] {0:3}
                                                                                                              TSTT Modulo (3%2) [3,4,2,3,0,1] {0:3}
                                                                                                              TSSS Add (0+1) [3,4,2,3,1] {0:3}
                                                                                                              SNT Swap top two [3,4,2,1,3] {0:3}
                                                                                                              STSSTSN Copy (0-indexed) 2nd [3,4,2,1,3,2] {0:3}
                                                                                                              TSTS Integer divide (3/2) [3,4,2,1,1] {0:3}
                                                                                                              NSNTN Jump to LABEL_INNER_LOOP [3,4,2,1,1] {0:3}

                                                                                                              SNS Duplicate 1 [3,4,2,1,1,1] {0:3}
                                                                                                              NTSSSN If 0: Jump to Label_AFTER_IL [3,4,2,1,1] {0:3}
                                                                                                              SNT Swap top two [3,4,2,1,1] {0:3}
                                                                                                              STSSTN Copy (0-indexed) 1st [3,4,2,1,1,1] {0:3}
                                                                                                              STSSTTN Copy (0-indexed) 3rd [3,4,2,1,1,1,2] {0:3}
                                                                                                              TSTT Modulo (1%2) [3,4,2,1,1,1] {0:3}
                                                                                                              TSSS Add (1+1) [3,4,2,1,2] {0:3}
                                                                                                              SNT Swap top two [3,4,2,2,1] {0:3}
                                                                                                              STSSTSN Copy (0-indexed) 2nd [3,4,2,2,1,2] {0:3}
                                                                                                              TSTS Integer divide (1/2) [3,4,2,2,0] {0:3}
                                                                                                              NSNTN Jump to LABEL_INNER_LOOP [3,4,2,2,0] {0:3}

                                                                                                              SNS Duplicate 0 [3,4,2,2,0,0] {0:3}
                                                                                                              NTSSSN If 0: Jump to Label_AFTER_IL [3,4,2,2,0] {0:3}
                                                                                                              NSSSSN Create Label_AFTER_IL [3,4,2,2,0] {0:3}
                                                                                                              SNN Discard top [3,4,2,2] {0:3}
                                                                                                              STSSTSN Copy (0-indexed) 2nd [3,4,2,2,4] {0:3}
                                                                                                              TSSS Add (2+4) [3,4,2,6] {0:3}
                                                                                                              SNT Swap top two [3,4,6,2] {0:3}
                                                                                                              SSSTN Push 1 [3,4,6,2,1] {0:3}
                                                                                                              TSST Subtract (2-1) [3,4,6,1] {0:3}
                                                                                                              NSNN Jump to LABEL_OUTER_LOOP [3,4,6,1] {0:3}

                                                                                                              SNS Duplicate 1 [3,4,6,1,1] {0:3}
                                                                                                              SSSTN Push 1 [3,4,6,1,1,1] {0:3}
                                                                                                              TSST Subtract (1-1) [3,4,6,1,0] {0:3}
                                                                                                              NTSSN If 0: Jump to Label_PRINT [3,4,6,1] {0:3}
                                                                                                              NSSSSN Create Label_PRINT [3,4,6,1] {0:3}
                                                                                                              SNN Discard top [3,4,6] {0:3}
                                                                                                              TNST Print top (6) to STDOUT as int [3,4] {0:3} 6
                                                                                                              error


                                                                                                              Program stops with an error: No exit found. (Although I could add three trailing newlines NNN to get rid of that error.)






                                                                                                              share|improve this answer













                                                                                                              Whitespace, 153 bytes



                                                                                                              [S S S N
                                                                                                              _Push_0][S N
                                                                                                              S _Duplicate_0][T N
                                                                                                              T T _Read_STDIN_as_integer][T T T _Retrieve_input][S N
                                                                                                              S _Duplicate_input][N
                                                                                                              S S N
                                                                                                              _Create_Label_OUTER_LOOP][S N
                                                                                                              S _Duplicate_top][S S S T N
                                                                                                              _Push_1][T S S T _Subtract][N
                                                                                                              T S S N
                                                                                                              _If_0_Jump_to_Label_PRINT][S S S N
                                                                                                              _Push_0][S N
                                                                                                              S _Duplicate_0][T T T _Retrieve_input][N
                                                                                                              S S T N
                                                                                                              _Create_Label_INNER_LOOP][S N
                                                                                                              S _Duplicate][N
                                                                                                              T S S S N
                                                                                                              _If_0_Jump_to_Label_AFTER_INNER_LOOP][S N
                                                                                                              T _Swap_top_two][S T S S T N
                                                                                                              _Copy_1st_item_to_top][S T S S T T N
                                                                                                              _Copy_3rd_item_to_top][T S T T _Modulo][T S S S _Add][S N
                                                                                                              T _Swap_top_two][S T S S T S N
                                                                                                              _Copy_2nd_item_to_top][T S T S _Integer_divide][N
                                                                                                              S N
                                                                                                              T N
                                                                                                              _Jump_to_Label_INNER_LOOP][N
                                                                                                              S S S S N
                                                                                                              _Create_Label_AFTER_INNER_LOOP][S N
                                                                                                              N
                                                                                                              _Discard_top][S T S S T S N
                                                                                                              _Copy_2nd_item_to_top][T S S S _Add][S N
                                                                                                              T _Swap_top_two][S S S T N
                                                                                                              _Push_1][T S S T _Subtract][N
                                                                                                              S N
                                                                                                              N
                                                                                                              _Jump_to_Label_OUTER_LOOP][N
                                                                                                              S S S N
                                                                                                              _Create_Label_PRINT][S N
                                                                                                              N
                                                                                                              _Discard_top][T N
                                                                                                              S T _Print_as_integer]


                                                                                                              Letters S (space), T (tab), and N (new-line) added as highlighting only.
                                                                                                              [..._some_action] added as explanation only.



                                                                                                              Try it online (with raw spaces, tabs and new-lines only).



                                                                                                              Port of my Java 8 answer, because Whitespace has no Base conversion builtins at all.



                                                                                                              Example run: input = 3



                                                                                                              Command    Explanation                     Stack            Heap   STDIN  STDOUT  STDERR

                                                                                                              SSSN Push 0 [0]
                                                                                                              SNS Duplicate 0 [0,0]
                                                                                                              TNTT Read STDIN as integer [0] {0:3} 3
                                                                                                              TTT Retrieve input from heap 0 [3] {0:3}
                                                                                                              SNS Duplicate 3 [3,3] {0:3}
                                                                                                              NSSN Create Label_OUTER_LOOP [3,3] {0:3}
                                                                                                              SNS Duplicate 3 [3,3,3] {0:3}
                                                                                                              SSSTN Push 1 [3,3,3,1] {0:3}
                                                                                                              TSST Subtract (3-1) [3,3,2] {0:3}
                                                                                                              NTSSN If 0: Jump to Label_PRINT [3,3] {0:3}
                                                                                                              SSSN Push 0 [3,3,0] {0:3}
                                                                                                              SNS Duplicate 0 [3,3,0,0] {0:3}
                                                                                                              TTT Retrieve input from heap 0 [3,3,0,3] {0:3}
                                                                                                              NSSTN Create Label_INNER_LOOP [3,3,0,3] {0:3}
                                                                                                              SNS Duplicate 3 [3,3,0,3,3] {0:3}
                                                                                                              NTSSSN If 0: Jump to Label_AFTER_IL [3,3,0,3] {0:3}
                                                                                                              SNT Swap top two [3,3,3,0] {0:3}
                                                                                                              STSSTN Copy (0-indexed) 1st [3,3,3,0,3] {0:3}
                                                                                                              STSSTTN Copy (0-indexed) 3rd [3,3,3,0,3,3] {0:3}
                                                                                                              TSTT Modulo (3%3) [3,3,3,0,0] {0:3}
                                                                                                              TSSS Add (0+0) [3,3,3,0] {0:3}
                                                                                                              SNT Swap top two [3,3,0,3] {0:3}
                                                                                                              STSSTSN Copy (0-indexed) 2nd [3,3,0,3,3] {0:3}
                                                                                                              TSTS Integer divide (3/3) [3,3,0,1] {0:3}
                                                                                                              NSNTN Jump to LABEL_INNER_LOOP [3,3,0,1] {0:3}

                                                                                                              SNS Duplicate 1 [3,3,0,1,1] {0:3}
                                                                                                              NTSSSN If 0: Jump to Label_AFTER_IL [3,3,0,1] {0:3}
                                                                                                              SNT Swap top two [3,3,1,0] {0:3}
                                                                                                              STSSTN Copy (0-indexed) 1st [3,3,1,0,1] {0:3}
                                                                                                              STSSTTN Copy (0-indexed) 3rd [3,3,1,0,1,3] {0:3}
                                                                                                              TSTT Modulo (1%3) [3,3,1,0,1] {0:3}
                                                                                                              TSSS Add (0+1) [3,3,1,1] {0:3}
                                                                                                              SNT Swap top two [3,3,1,1] {0:3}
                                                                                                              STSSTSN Copy (0-indexed) 2nd [3,3,1,1,3] {0:3}
                                                                                                              TSTS Integer divide (1/3) [3,3,1,0] {0:3}
                                                                                                              NSNTN Jump to LABEL_INNER_LOOP [3,3,1,0] {0:3}

                                                                                                              SNS Duplicate 0 [3,3,1,0,0] {0:3}
                                                                                                              NTSSSN If 0: Jump to Label_AFTER_IL [3,3,1,0] {0:3}
                                                                                                              NSSSSN Create Label_AFTER_IL [3,3,1,0] {0:3}
                                                                                                              SNN Discard top [3,3,1] {0:3}
                                                                                                              STSSTSN Copy (0-indexed) 2nd [3,3,1,3] {0:3}
                                                                                                              TSSS Add (1+3) [3,3,4] {0:3}
                                                                                                              SNT Swap top two [3,4,3] {0:3}
                                                                                                              SSSTN Push 1 [3,4,3,1] {0:3}
                                                                                                              TSST Subtract (3-1) [3,4,2] {0:3}
                                                                                                              NSNN Jump to LABEL_OUTER_LOOP [3,4,2] {0:3}

                                                                                                              SNS Duplicate 2 [3,4,2,2] {0:3}
                                                                                                              SSSTN Push 1 [3,4,2,2,1] {0:3}
                                                                                                              TSST Subtract (2-1) [3,4,2,1] {0:3}
                                                                                                              NTSSN If 0: Jump to Label_PRINT [3,4,2] {0:3}
                                                                                                              SSSN Push 0 [3,4,2,0] {0:3}
                                                                                                              SNS Duplicate 0 [3,4,2,0,0] {0:3}
                                                                                                              TTT Retrieve input from heap 0 [3,4,2,0,3] {0:3}
                                                                                                              NSSTN Create Label_INNER_LOOP [3,4,2,0,3] {0:3}
                                                                                                              SNS Duplicate 3 [3,4,2,0,3,3] {0:3}
                                                                                                              NTSSSN If 0: Jump to Label_AFTER_IL [3,4,2,0,3] {0:3}
                                                                                                              SNT Swap top two [3,4,2,3,0] {0:3}
                                                                                                              STSSTN Copy (0-indexed) 1st [3,4,2,3,0,3] {0:3}
                                                                                                              STSSTTN Copy (0-indexed) 3rd [3,4,2,3,0,3,2] {0:3}
                                                                                                              TSTT Modulo (3%2) [3,4,2,3,0,1] {0:3}
                                                                                                              TSSS Add (0+1) [3,4,2,3,1] {0:3}
                                                                                                              SNT Swap top two [3,4,2,1,3] {0:3}
                                                                                                              STSSTSN Copy (0-indexed) 2nd [3,4,2,1,3,2] {0:3}
                                                                                                              TSTS Integer divide (3/2) [3,4,2,1,1] {0:3}
                                                                                                              NSNTN Jump to LABEL_INNER_LOOP [3,4,2,1,1] {0:3}

                                                                                                              SNS Duplicate 1 [3,4,2,1,1,1] {0:3}
                                                                                                              NTSSSN If 0: Jump to Label_AFTER_IL [3,4,2,1,1] {0:3}
                                                                                                              SNT Swap top two [3,4,2,1,1] {0:3}
                                                                                                              STSSTN Copy (0-indexed) 1st [3,4,2,1,1,1] {0:3}
                                                                                                              STSSTTN Copy (0-indexed) 3rd [3,4,2,1,1,1,2] {0:3}
                                                                                                              TSTT Modulo (1%2) [3,4,2,1,1,1] {0:3}
                                                                                                              TSSS Add (1+1) [3,4,2,1,2] {0:3}
                                                                                                              SNT Swap top two [3,4,2,2,1] {0:3}
                                                                                                              STSSTSN Copy (0-indexed) 2nd [3,4,2,2,1,2] {0:3}
                                                                                                              TSTS Integer divide (1/2) [3,4,2,2,0] {0:3}
                                                                                                              NSNTN Jump to LABEL_INNER_LOOP [3,4,2,2,0] {0:3}

                                                                                                              SNS Duplicate 0 [3,4,2,2,0,0] {0:3}
                                                                                                              NTSSSN If 0: Jump to Label_AFTER_IL [3,4,2,2,0] {0:3}
                                                                                                              NSSSSN Create Label_AFTER_IL [3,4,2,2,0] {0:3}
                                                                                                              SNN Discard top [3,4,2,2] {0:3}
                                                                                                              STSSTSN Copy (0-indexed) 2nd [3,4,2,2,4] {0:3}
                                                                                                              TSSS Add (2+4) [3,4,2,6] {0:3}
                                                                                                              SNT Swap top two [3,4,6,2] {0:3}
                                                                                                              SSSTN Push 1 [3,4,6,2,1] {0:3}
                                                                                                              TSST Subtract (2-1) [3,4,6,1] {0:3}
                                                                                                              NSNN Jump to LABEL_OUTER_LOOP [3,4,6,1] {0:3}

                                                                                                              SNS Duplicate 1 [3,4,6,1,1] {0:3}
                                                                                                              SSSTN Push 1 [3,4,6,1,1,1] {0:3}
                                                                                                              TSST Subtract (1-1) [3,4,6,1,0] {0:3}
                                                                                                              NTSSN If 0: Jump to Label_PRINT [3,4,6,1] {0:3}
                                                                                                              NSSSSN Create Label_PRINT [3,4,6,1] {0:3}
                                                                                                              SNN Discard top [3,4,6] {0:3}
                                                                                                              TNST Print top (6) to STDOUT as int [3,4] {0:3} 6
                                                                                                              error


                                                                                                              Program stops with an error: No exit found. (Although I could add three trailing newlines NNN to get rid of that error.)







                                                                                                              share|improve this answer












                                                                                                              share|improve this answer



                                                                                                              share|improve this answer










                                                                                                              answered Dec 4 at 16:13









                                                                                                              Kevin Cruijssen

                                                                                                              35.6k554186




                                                                                                              35.6k554186























                                                                                                                  3















                                                                                                                  R, 60 bytes





                                                                                                                  function(n){if(n-1)for(i in 2:n)F=F+sum(n%/%i^(0:n)%%i)
                                                                                                                  n+F}


                                                                                                                  Try it online!



                                                                                                                  Fails for n>143 since 144^144 is larger than a double can get. Thanks to Josh Eller for suggesting replacing log(n,i) with simply n.



                                                                                                                  The below will work for n>143; not sure at what point it will stop working.




                                                                                                                  R, 67 bytes





                                                                                                                  function(n){if(n-1)for(i in 2:n)F=F+sum(n%/%i^(0:log(n,i))%%i)
                                                                                                                  n+F}


                                                                                                                  Try it online!



                                                                                                                  Uses the classic n%/%i^(0:log(n,i))%%i method to extract the base-i digits for n for each base b>1, then sums them and accumulates the sum in F, which is initialized to 0, then adding n (the base 1 representation of n) to F and returning the result. For n=1, it skips the bases and simply adds n to F.






                                                                                                                  share|improve this answer



















                                                                                                                  • 1




                                                                                                                    I don't know any R, but instead of using 0:log(n,i), couldn't you use 0:n? There's always going to be at most n digits in any base representation of n, and everything after the initial log(n,i) digits should be 0, so it won't affect the sum.
                                                                                                                    – Josh Eller
                                                                                                                    Dec 4 at 20:18










                                                                                                                  • @JoshEller I suppose I could. It would start to fail for n=144, since 143^143 is around 1.6e308 and 144^144 evaluates to Inf. Thanks!
                                                                                                                    – Giuseppe
                                                                                                                    Dec 4 at 20:23


















                                                                                                                  3















                                                                                                                  R, 60 bytes





                                                                                                                  function(n){if(n-1)for(i in 2:n)F=F+sum(n%/%i^(0:n)%%i)
                                                                                                                  n+F}


                                                                                                                  Try it online!



                                                                                                                  Fails for n>143 since 144^144 is larger than a double can get. Thanks to Josh Eller for suggesting replacing log(n,i) with simply n.



                                                                                                                  The below will work for n>143; not sure at what point it will stop working.




                                                                                                                  R, 67 bytes





                                                                                                                  function(n){if(n-1)for(i in 2:n)F=F+sum(n%/%i^(0:log(n,i))%%i)
                                                                                                                  n+F}


                                                                                                                  Try it online!



                                                                                                                  Uses the classic n%/%i^(0:log(n,i))%%i method to extract the base-i digits for n for each base b>1, then sums them and accumulates the sum in F, which is initialized to 0, then adding n (the base 1 representation of n) to F and returning the result. For n=1, it skips the bases and simply adds n to F.






                                                                                                                  share|improve this answer



















                                                                                                                  • 1




                                                                                                                    I don't know any R, but instead of using 0:log(n,i), couldn't you use 0:n? There's always going to be at most n digits in any base representation of n, and everything after the initial log(n,i) digits should be 0, so it won't affect the sum.
                                                                                                                    – Josh Eller
                                                                                                                    Dec 4 at 20:18










                                                                                                                  • @JoshEller I suppose I could. It would start to fail for n=144, since 143^143 is around 1.6e308 and 144^144 evaluates to Inf. Thanks!
                                                                                                                    – Giuseppe
                                                                                                                    Dec 4 at 20:23
















                                                                                                                  3












                                                                                                                  3








                                                                                                                  3







                                                                                                                  R, 60 bytes





                                                                                                                  function(n){if(n-1)for(i in 2:n)F=F+sum(n%/%i^(0:n)%%i)
                                                                                                                  n+F}


                                                                                                                  Try it online!



                                                                                                                  Fails for n>143 since 144^144 is larger than a double can get. Thanks to Josh Eller for suggesting replacing log(n,i) with simply n.



                                                                                                                  The below will work for n>143; not sure at what point it will stop working.




                                                                                                                  R, 67 bytes





                                                                                                                  function(n){if(n-1)for(i in 2:n)F=F+sum(n%/%i^(0:log(n,i))%%i)
                                                                                                                  n+F}


                                                                                                                  Try it online!



                                                                                                                  Uses the classic n%/%i^(0:log(n,i))%%i method to extract the base-i digits for n for each base b>1, then sums them and accumulates the sum in F, which is initialized to 0, then adding n (the base 1 representation of n) to F and returning the result. For n=1, it skips the bases and simply adds n to F.






                                                                                                                  share|improve this answer















                                                                                                                  R, 60 bytes





                                                                                                                  function(n){if(n-1)for(i in 2:n)F=F+sum(n%/%i^(0:n)%%i)
                                                                                                                  n+F}


                                                                                                                  Try it online!



                                                                                                                  Fails for n>143 since 144^144 is larger than a double can get. Thanks to Josh Eller for suggesting replacing log(n,i) with simply n.



                                                                                                                  The below will work for n>143; not sure at what point it will stop working.




                                                                                                                  R, 67 bytes





                                                                                                                  function(n){if(n-1)for(i in 2:n)F=F+sum(n%/%i^(0:log(n,i))%%i)
                                                                                                                  n+F}


                                                                                                                  Try it online!



                                                                                                                  Uses the classic n%/%i^(0:log(n,i))%%i method to extract the base-i digits for n for each base b>1, then sums them and accumulates the sum in F, which is initialized to 0, then adding n (the base 1 representation of n) to F and returning the result. For n=1, it skips the bases and simply adds n to F.







                                                                                                                  share|improve this answer














                                                                                                                  share|improve this answer



                                                                                                                  share|improve this answer








                                                                                                                  edited Dec 4 at 20:26

























                                                                                                                  answered Dec 3 at 18:26









                                                                                                                  Giuseppe

                                                                                                                  16.6k31052




                                                                                                                  16.6k31052








                                                                                                                  • 1




                                                                                                                    I don't know any R, but instead of using 0:log(n,i), couldn't you use 0:n? There's always going to be at most n digits in any base representation of n, and everything after the initial log(n,i) digits should be 0, so it won't affect the sum.
                                                                                                                    – Josh Eller
                                                                                                                    Dec 4 at 20:18










                                                                                                                  • @JoshEller I suppose I could. It would start to fail for n=144, since 143^143 is around 1.6e308 and 144^144 evaluates to Inf. Thanks!
                                                                                                                    – Giuseppe
                                                                                                                    Dec 4 at 20:23
















                                                                                                                  • 1




                                                                                                                    I don't know any R, but instead of using 0:log(n,i), couldn't you use 0:n? There's always going to be at most n digits in any base representation of n, and everything after the initial log(n,i) digits should be 0, so it won't affect the sum.
                                                                                                                    – Josh Eller
                                                                                                                    Dec 4 at 20:18










                                                                                                                  • @JoshEller I suppose I could. It would start to fail for n=144, since 143^143 is around 1.6e308 and 144^144 evaluates to Inf. Thanks!
                                                                                                                    – Giuseppe
                                                                                                                    Dec 4 at 20:23










                                                                                                                  1




                                                                                                                  1




                                                                                                                  I don't know any R, but instead of using 0:log(n,i), couldn't you use 0:n? There's always going to be at most n digits in any base representation of n, and everything after the initial log(n,i) digits should be 0, so it won't affect the sum.
                                                                                                                  – Josh Eller
                                                                                                                  Dec 4 at 20:18




                                                                                                                  I don't know any R, but instead of using 0:log(n,i), couldn't you use 0:n? There's always going to be at most n digits in any base representation of n, and everything after the initial log(n,i) digits should be 0, so it won't affect the sum.
                                                                                                                  – Josh Eller
                                                                                                                  Dec 4 at 20:18












                                                                                                                  @JoshEller I suppose I could. It would start to fail for n=144, since 143^143 is around 1.6e308 and 144^144 evaluates to Inf. Thanks!
                                                                                                                  – Giuseppe
                                                                                                                  Dec 4 at 20:23






                                                                                                                  @JoshEller I suppose I could. It would start to fail for n=144, since 143^143 is around 1.6e308 and 144^144 evaluates to Inf. Thanks!
                                                                                                                  – Giuseppe
                                                                                                                  Dec 4 at 20:23













                                                                                                                  3














                                                                                                                  C (gcc), 67 56 bytes





                                                                                                                  b,a,s;e(n){for(b=a=n;a>1;a--)for(s=n;s;s/=a)b+=s%a;n=b;}


                                                                                                                  Port of my Java 8 answer.

                                                                                                                  -11 bytes thanks to @OlivierGrégoire's golf on my Java answer.



                                                                                                                  Try it online.



                                                                                                                  Explanation:



                                                                                                                  b,             // Result integer
                                                                                                                  a, // Base integer
                                                                                                                  s; // Temp integer
                                                                                                                  e(n){ // Method with integer as parameter
                                                                                                                  for(b=a=n; // Set the result `b` to input `n` to account for Base-1
                                                                                                                  // And set Base `a` to input `n` as well
                                                                                                                  a>1 // Loop the Base `a` in the range [input, 1):
                                                                                                                  ;a--) // After every iteration: Go to the next Base (downwards)
                                                                                                                  for(s=n; // Set `s` to input `n`
                                                                                                                  s; // Inner loop as long as `s` is not 0 yet:
                                                                                                                  s/=a) // After every iteration: Divide `s` by Base `a`
                                                                                                                  b+=s%a; // Add `s` modulo Base `a` to the result `b`
                                                                                                                  n=b;} // Set input `n` to result `b` to 'return it'





                                                                                                                  share|improve this answer




























                                                                                                                    3














                                                                                                                    C (gcc), 67 56 bytes





                                                                                                                    b,a,s;e(n){for(b=a=n;a>1;a--)for(s=n;s;s/=a)b+=s%a;n=b;}


                                                                                                                    Port of my Java 8 answer.

                                                                                                                    -11 bytes thanks to @OlivierGrégoire's golf on my Java answer.



                                                                                                                    Try it online.



                                                                                                                    Explanation:



                                                                                                                    b,             // Result integer
                                                                                                                    a, // Base integer
                                                                                                                    s; // Temp integer
                                                                                                                    e(n){ // Method with integer as parameter
                                                                                                                    for(b=a=n; // Set the result `b` to input `n` to account for Base-1
                                                                                                                    // And set Base `a` to input `n` as well
                                                                                                                    a>1 // Loop the Base `a` in the range [input, 1):
                                                                                                                    ;a--) // After every iteration: Go to the next Base (downwards)
                                                                                                                    for(s=n; // Set `s` to input `n`
                                                                                                                    s; // Inner loop as long as `s` is not 0 yet:
                                                                                                                    s/=a) // After every iteration: Divide `s` by Base `a`
                                                                                                                    b+=s%a; // Add `s` modulo Base `a` to the result `b`
                                                                                                                    n=b;} // Set input `n` to result `b` to 'return it'





                                                                                                                    share|improve this answer


























                                                                                                                      3












                                                                                                                      3








                                                                                                                      3






                                                                                                                      C (gcc), 67 56 bytes





                                                                                                                      b,a,s;e(n){for(b=a=n;a>1;a--)for(s=n;s;s/=a)b+=s%a;n=b;}


                                                                                                                      Port of my Java 8 answer.

                                                                                                                      -11 bytes thanks to @OlivierGrégoire's golf on my Java answer.



                                                                                                                      Try it online.



                                                                                                                      Explanation:



                                                                                                                      b,             // Result integer
                                                                                                                      a, // Base integer
                                                                                                                      s; // Temp integer
                                                                                                                      e(n){ // Method with integer as parameter
                                                                                                                      for(b=a=n; // Set the result `b` to input `n` to account for Base-1
                                                                                                                      // And set Base `a` to input `n` as well
                                                                                                                      a>1 // Loop the Base `a` in the range [input, 1):
                                                                                                                      ;a--) // After every iteration: Go to the next Base (downwards)
                                                                                                                      for(s=n; // Set `s` to input `n`
                                                                                                                      s; // Inner loop as long as `s` is not 0 yet:
                                                                                                                      s/=a) // After every iteration: Divide `s` by Base `a`
                                                                                                                      b+=s%a; // Add `s` modulo Base `a` to the result `b`
                                                                                                                      n=b;} // Set input `n` to result `b` to 'return it'





                                                                                                                      share|improve this answer














                                                                                                                      C (gcc), 67 56 bytes





                                                                                                                      b,a,s;e(n){for(b=a=n;a>1;a--)for(s=n;s;s/=a)b+=s%a;n=b;}


                                                                                                                      Port of my Java 8 answer.

                                                                                                                      -11 bytes thanks to @OlivierGrégoire's golf on my Java answer.



                                                                                                                      Try it online.



                                                                                                                      Explanation:



                                                                                                                      b,             // Result integer
                                                                                                                      a, // Base integer
                                                                                                                      s; // Temp integer
                                                                                                                      e(n){ // Method with integer as parameter
                                                                                                                      for(b=a=n; // Set the result `b` to input `n` to account for Base-1
                                                                                                                      // And set Base `a` to input `n` as well
                                                                                                                      a>1 // Loop the Base `a` in the range [input, 1):
                                                                                                                      ;a--) // After every iteration: Go to the next Base (downwards)
                                                                                                                      for(s=n; // Set `s` to input `n`
                                                                                                                      s; // Inner loop as long as `s` is not 0 yet:
                                                                                                                      s/=a) // After every iteration: Divide `s` by Base `a`
                                                                                                                      b+=s%a; // Add `s` modulo Base `a` to the result `b`
                                                                                                                      n=b;} // Set input `n` to result `b` to 'return it'






                                                                                                                      share|improve this answer














                                                                                                                      share|improve this answer



                                                                                                                      share|improve this answer








                                                                                                                      edited Dec 5 at 11:08

























                                                                                                                      answered Dec 4 at 14:16









                                                                                                                      Kevin Cruijssen

                                                                                                                      35.6k554186




                                                                                                                      35.6k554186























                                                                                                                          2














                                                                                                                          JavaScript (ES6), 42 bytes



                                                                                                                          This version is almost identical to my main answer but relies on arithmetic underflow to stop the recursion. The highest supported value depends on the size of the call stack.





                                                                                                                          f=(n,b=1,x)=>b>n?n:~~x%b+f(n,b+!x,x?x/b:n)


                                                                                                                          Try it online!





                                                                                                                          JavaScript (ES6),  51 48  44 bytes





                                                                                                                          f=(n,b=2,x=n)=>b>n?n:x%b+f(n,b+!x,x?x/b|0:n)


                                                                                                                          Try it online!



                                                                                                                          Commented



                                                                                                                          f = (             // f = recursive function taking:
                                                                                                                          n, // - n = input
                                                                                                                          b = 2, // - b = current base, initialized to 2
                                                                                                                          x = n // - x = current value being converted in base b
                                                                                                                          ) => //
                                                                                                                          b > n ? // if b is greater than n:
                                                                                                                          n // stop recursion and return n, which is the sum of the digits of n
                                                                                                                          // once converted to unary
                                                                                                                          : // else:
                                                                                                                          x % b + // add x modulo b to the final result
                                                                                                                          f( // and add the result of a recursive call:
                                                                                                                          n, // pass n unchanged
                                                                                                                          b + !x, // increment b if x = 0
                                                                                                                          x ? // if x is not equal to 0:
                                                                                                                          x / b | 0 // update x to floor(x / b)
                                                                                                                          : // else:
                                                                                                                          n // reset x to n
                                                                                                                          ) // end of recursive call





                                                                                                                          share|improve this answer




























                                                                                                                            2














                                                                                                                            JavaScript (ES6), 42 bytes



                                                                                                                            This version is almost identical to my main answer but relies on arithmetic underflow to stop the recursion. The highest supported value depends on the size of the call stack.





                                                                                                                            f=(n,b=1,x)=>b>n?n:~~x%b+f(n,b+!x,x?x/b:n)


                                                                                                                            Try it online!





                                                                                                                            JavaScript (ES6),  51 48  44 bytes





                                                                                                                            f=(n,b=2,x=n)=>b>n?n:x%b+f(n,b+!x,x?x/b|0:n)


                                                                                                                            Try it online!



                                                                                                                            Commented



                                                                                                                            f = (             // f = recursive function taking:
                                                                                                                            n, // - n = input
                                                                                                                            b = 2, // - b = current base, initialized to 2
                                                                                                                            x = n // - x = current value being converted in base b
                                                                                                                            ) => //
                                                                                                                            b > n ? // if b is greater than n:
                                                                                                                            n // stop recursion and return n, which is the sum of the digits of n
                                                                                                                            // once converted to unary
                                                                                                                            : // else:
                                                                                                                            x % b + // add x modulo b to the final result
                                                                                                                            f( // and add the result of a recursive call:
                                                                                                                            n, // pass n unchanged
                                                                                                                            b + !x, // increment b if x = 0
                                                                                                                            x ? // if x is not equal to 0:
                                                                                                                            x / b | 0 // update x to floor(x / b)
                                                                                                                            : // else:
                                                                                                                            n // reset x to n
                                                                                                                            ) // end of recursive call





                                                                                                                            share|improve this answer


























                                                                                                                              2












                                                                                                                              2








                                                                                                                              2






                                                                                                                              JavaScript (ES6), 42 bytes



                                                                                                                              This version is almost identical to my main answer but relies on arithmetic underflow to stop the recursion. The highest supported value depends on the size of the call stack.





                                                                                                                              f=(n,b=1,x)=>b>n?n:~~x%b+f(n,b+!x,x?x/b:n)


                                                                                                                              Try it online!





                                                                                                                              JavaScript (ES6),  51 48  44 bytes





                                                                                                                              f=(n,b=2,x=n)=>b>n?n:x%b+f(n,b+!x,x?x/b|0:n)


                                                                                                                              Try it online!



                                                                                                                              Commented



                                                                                                                              f = (             // f = recursive function taking:
                                                                                                                              n, // - n = input
                                                                                                                              b = 2, // - b = current base, initialized to 2
                                                                                                                              x = n // - x = current value being converted in base b
                                                                                                                              ) => //
                                                                                                                              b > n ? // if b is greater than n:
                                                                                                                              n // stop recursion and return n, which is the sum of the digits of n
                                                                                                                              // once converted to unary
                                                                                                                              : // else:
                                                                                                                              x % b + // add x modulo b to the final result
                                                                                                                              f( // and add the result of a recursive call:
                                                                                                                              n, // pass n unchanged
                                                                                                                              b + !x, // increment b if x = 0
                                                                                                                              x ? // if x is not equal to 0:
                                                                                                                              x / b | 0 // update x to floor(x / b)
                                                                                                                              : // else:
                                                                                                                              n // reset x to n
                                                                                                                              ) // end of recursive call





                                                                                                                              share|improve this answer














                                                                                                                              JavaScript (ES6), 42 bytes



                                                                                                                              This version is almost identical to my main answer but relies on arithmetic underflow to stop the recursion. The highest supported value depends on the size of the call stack.





                                                                                                                              f=(n,b=1,x)=>b>n?n:~~x%b+f(n,b+!x,x?x/b:n)


                                                                                                                              Try it online!





                                                                                                                              JavaScript (ES6),  51 48  44 bytes





                                                                                                                              f=(n,b=2,x=n)=>b>n?n:x%b+f(n,b+!x,x?x/b|0:n)


                                                                                                                              Try it online!



                                                                                                                              Commented



                                                                                                                              f = (             // f = recursive function taking:
                                                                                                                              n, // - n = input
                                                                                                                              b = 2, // - b = current base, initialized to 2
                                                                                                                              x = n // - x = current value being converted in base b
                                                                                                                              ) => //
                                                                                                                              b > n ? // if b is greater than n:
                                                                                                                              n // stop recursion and return n, which is the sum of the digits of n
                                                                                                                              // once converted to unary
                                                                                                                              : // else:
                                                                                                                              x % b + // add x modulo b to the final result
                                                                                                                              f( // and add the result of a recursive call:
                                                                                                                              n, // pass n unchanged
                                                                                                                              b + !x, // increment b if x = 0
                                                                                                                              x ? // if x is not equal to 0:
                                                                                                                              x / b | 0 // update x to floor(x / b)
                                                                                                                              : // else:
                                                                                                                              n // reset x to n
                                                                                                                              ) // end of recursive call






                                                                                                                              share|improve this answer














                                                                                                                              share|improve this answer



                                                                                                                              share|improve this answer








                                                                                                                              edited Dec 3 at 17:14

























                                                                                                                              answered Dec 3 at 16:45









                                                                                                                              Arnauld

                                                                                                                              72.3k689303




                                                                                                                              72.3k689303























                                                                                                                                  2















                                                                                                                                  APL (Dyalog Unicode), 22 bytes





                                                                                                                                  {1⊥⍵,∊(1+⍳⍵-1)⊥⍣¯1¨⊂⍵}


                                                                                                                                  Try it online!






                                                                                                                                  share|improve this answer


























                                                                                                                                    2















                                                                                                                                    APL (Dyalog Unicode), 22 bytes





                                                                                                                                    {1⊥⍵,∊(1+⍳⍵-1)⊥⍣¯1¨⊂⍵}


                                                                                                                                    Try it online!






                                                                                                                                    share|improve this answer
























                                                                                                                                      2












                                                                                                                                      2








                                                                                                                                      2







                                                                                                                                      APL (Dyalog Unicode), 22 bytes





                                                                                                                                      {1⊥⍵,∊(1+⍳⍵-1)⊥⍣¯1¨⊂⍵}


                                                                                                                                      Try it online!






                                                                                                                                      share|improve this answer













                                                                                                                                      APL (Dyalog Unicode), 22 bytes





                                                                                                                                      {1⊥⍵,∊(1+⍳⍵-1)⊥⍣¯1¨⊂⍵}


                                                                                                                                      Try it online!







                                                                                                                                      share|improve this answer












                                                                                                                                      share|improve this answer



                                                                                                                                      share|improve this answer










                                                                                                                                      answered Dec 3 at 18:14









                                                                                                                                      J. Sallé

                                                                                                                                      1,903322




                                                                                                                                      1,903322























                                                                                                                                          2















                                                                                                                                          Husk, 6 bytes



                                                                                                                                          I really wish that there was something like M for cmap :(



                                                                                                                                          Σ§ṁ`Bḣ


                                                                                                                                          Try it online or test all!



                                                                                                                                          Explanation



                                                                                                                                          Σ§ṁ`Bḣ  -- example input: 5
                                                                                                                                          § -- fork the argument..
                                                                                                                                          ḣ -- | range: [1,2,3,4,5]
                                                                                                                                          `B -- | convert argument to base: (`asBase` 5)
                                                                                                                                          ṁ -- | and concatMap: cmap (`asBase` 5) [1,2,3,4,5]
                                                                                                                                          -- : [1,1,1,1,1,1,0,1,1,2,1,1,1,0]
                                                                                                                                          Σ -- sum: 13


                                                                                                                                          Alternatively, 6 bytes



                                                                                                                                          ΣΣṠMBḣ


                                                                                                                                          Try it online or test all!



                                                                                                                                          Explanation



                                                                                                                                          ΣΣṠMBḣ  -- example input: 5
                                                                                                                                          Ṡ -- apply ..
                                                                                                                                          ḣ -- | range: [1,2,3,4,5]
                                                                                                                                          .. to function applied to itself
                                                                                                                                          MB -- | map over left argument of base
                                                                                                                                          -- : [[1,1,1,1,1],[1,0,1],[1,2],[1,1],[1,0]]
                                                                                                                                          Σ -- flatten: [1,1,1,1,1,1,0,1,1,2,1,1,1,0]
                                                                                                                                          Σ -- sum: 13





                                                                                                                                          share|improve this answer


























                                                                                                                                            2















                                                                                                                                            Husk, 6 bytes



                                                                                                                                            I really wish that there was something like M for cmap :(



                                                                                                                                            Σ§ṁ`Bḣ


                                                                                                                                            Try it online or test all!



                                                                                                                                            Explanation



                                                                                                                                            Σ§ṁ`Bḣ  -- example input: 5
                                                                                                                                            § -- fork the argument..
                                                                                                                                            ḣ -- | range: [1,2,3,4,5]
                                                                                                                                            `B -- | convert argument to base: (`asBase` 5)
                                                                                                                                            ṁ -- | and concatMap: cmap (`asBase` 5) [1,2,3,4,5]
                                                                                                                                            -- : [1,1,1,1,1,1,0,1,1,2,1,1,1,0]
                                                                                                                                            Σ -- sum: 13


                                                                                                                                            Alternatively, 6 bytes



                                                                                                                                            ΣΣṠMBḣ


                                                                                                                                            Try it online or test all!



                                                                                                                                            Explanation



                                                                                                                                            ΣΣṠMBḣ  -- example input: 5
                                                                                                                                            Ṡ -- apply ..
                                                                                                                                            ḣ -- | range: [1,2,3,4,5]
                                                                                                                                            .. to function applied to itself
                                                                                                                                            MB -- | map over left argument of base
                                                                                                                                            -- : [[1,1,1,1,1],[1,0,1],[1,2],[1,1],[1,0]]
                                                                                                                                            Σ -- flatten: [1,1,1,1,1,1,0,1,1,2,1,1,1,0]
                                                                                                                                            Σ -- sum: 13





                                                                                                                                            share|improve this answer
























                                                                                                                                              2












                                                                                                                                              2








                                                                                                                                              2







                                                                                                                                              Husk, 6 bytes



                                                                                                                                              I really wish that there was something like M for cmap :(



                                                                                                                                              Σ§ṁ`Bḣ


                                                                                                                                              Try it online or test all!



                                                                                                                                              Explanation



                                                                                                                                              Σ§ṁ`Bḣ  -- example input: 5
                                                                                                                                              § -- fork the argument..
                                                                                                                                              ḣ -- | range: [1,2,3,4,5]
                                                                                                                                              `B -- | convert argument to base: (`asBase` 5)
                                                                                                                                              ṁ -- | and concatMap: cmap (`asBase` 5) [1,2,3,4,5]
                                                                                                                                              -- : [1,1,1,1,1,1,0,1,1,2,1,1,1,0]
                                                                                                                                              Σ -- sum: 13


                                                                                                                                              Alternatively, 6 bytes



                                                                                                                                              ΣΣṠMBḣ


                                                                                                                                              Try it online or test all!



                                                                                                                                              Explanation



                                                                                                                                              ΣΣṠMBḣ  -- example input: 5
                                                                                                                                              Ṡ -- apply ..
                                                                                                                                              ḣ -- | range: [1,2,3,4,5]
                                                                                                                                              .. to function applied to itself
                                                                                                                                              MB -- | map over left argument of base
                                                                                                                                              -- : [[1,1,1,1,1],[1,0,1],[1,2],[1,1],[1,0]]
                                                                                                                                              Σ -- flatten: [1,1,1,1,1,1,0,1,1,2,1,1,1,0]
                                                                                                                                              Σ -- sum: 13





                                                                                                                                              share|improve this answer













                                                                                                                                              Husk, 6 bytes



                                                                                                                                              I really wish that there was something like M for cmap :(



                                                                                                                                              Σ§ṁ`Bḣ


                                                                                                                                              Try it online or test all!



                                                                                                                                              Explanation



                                                                                                                                              Σ§ṁ`Bḣ  -- example input: 5
                                                                                                                                              § -- fork the argument..
                                                                                                                                              ḣ -- | range: [1,2,3,4,5]
                                                                                                                                              `B -- | convert argument to base: (`asBase` 5)
                                                                                                                                              ṁ -- | and concatMap: cmap (`asBase` 5) [1,2,3,4,5]
                                                                                                                                              -- : [1,1,1,1,1,1,0,1,1,2,1,1,1,0]
                                                                                                                                              Σ -- sum: 13


                                                                                                                                              Alternatively, 6 bytes



                                                                                                                                              ΣΣṠMBḣ


                                                                                                                                              Try it online or test all!



                                                                                                                                              Explanation



                                                                                                                                              ΣΣṠMBḣ  -- example input: 5
                                                                                                                                              Ṡ -- apply ..
                                                                                                                                              ḣ -- | range: [1,2,3,4,5]
                                                                                                                                              .. to function applied to itself
                                                                                                                                              MB -- | map over left argument of base
                                                                                                                                              -- : [[1,1,1,1,1],[1,0,1],[1,2],[1,1],[1,0]]
                                                                                                                                              Σ -- flatten: [1,1,1,1,1,1,0,1,1,2,1,1,1,0]
                                                                                                                                              Σ -- sum: 13






                                                                                                                                              share|improve this answer












                                                                                                                                              share|improve this answer



                                                                                                                                              share|improve this answer










                                                                                                                                              answered Dec 3 at 19:06









                                                                                                                                              BMO

                                                                                                                                              11.3k22185




                                                                                                                                              11.3k22185























                                                                                                                                                  2















                                                                                                                                                  Pari/GP, 30 bytes



                                                                                                                                                  n->n+sum(b=2,n,sumdigits(n,b))


                                                                                                                                                  Try it online!






                                                                                                                                                  share|improve this answer


























                                                                                                                                                    2















                                                                                                                                                    Pari/GP, 30 bytes



                                                                                                                                                    n->n+sum(b=2,n,sumdigits(n,b))


                                                                                                                                                    Try it online!






                                                                                                                                                    share|improve this answer
























                                                                                                                                                      2












                                                                                                                                                      2








                                                                                                                                                      2







                                                                                                                                                      Pari/GP, 30 bytes



                                                                                                                                                      n->n+sum(b=2,n,sumdigits(n,b))


                                                                                                                                                      Try it online!






                                                                                                                                                      share|improve this answer













                                                                                                                                                      Pari/GP, 30 bytes



                                                                                                                                                      n->n+sum(b=2,n,sumdigits(n,b))


                                                                                                                                                      Try it online!







                                                                                                                                                      share|improve this answer












                                                                                                                                                      share|improve this answer



                                                                                                                                                      share|improve this answer










                                                                                                                                                      answered Dec 4 at 2:43









                                                                                                                                                      alephalpha

                                                                                                                                                      21.1k32989




                                                                                                                                                      21.1k32989























                                                                                                                                                          2















                                                                                                                                                          Python 2, 61 bytes





                                                                                                                                                          f=lambda n,b=2.:n and n%b+f(n//b,b)+(n//b>1/n==0and f(n,b+1))


                                                                                                                                                          Try it online!



                                                                                                                                                          Though this is longer the Dennis's solution off which it's based, I find the method too amusing to not share.



                                                                                                                                                          The goal is to recurse both on lopping off the last digit n->n/b and incrementing the base b->b+1, but we want to prevent the base from being increased after one or more digits have been lopped off. This is achieved by make the base b a float, so that after the update n->n//b, the float b infects n with its floatness. In this way, whether n is a float or not is a bit-flag for whether we've removed any digits from n.



                                                                                                                                                          We require that the condition 1/n==0 is met to recurse into incrementing b, which integers n satisfy because floor division is done, but floats fails. (n=1 also fails but we don't want to recurse on it anyway.) Otherwise, floats work just like integers in the function because we're careful to do floor division n//b, and the output is a whole-number float.






                                                                                                                                                          share|improve this answer


























                                                                                                                                                            2















                                                                                                                                                            Python 2, 61 bytes





                                                                                                                                                            f=lambda n,b=2.:n and n%b+f(n//b,b)+(n//b>1/n==0and f(n,b+1))


                                                                                                                                                            Try it online!



                                                                                                                                                            Though this is longer the Dennis's solution off which it's based, I find the method too amusing to not share.



                                                                                                                                                            The goal is to recurse both on lopping off the last digit n->n/b and incrementing the base b->b+1, but we want to prevent the base from being increased after one or more digits have been lopped off. This is achieved by make the base b a float, so that after the update n->n//b, the float b infects n with its floatness. In this way, whether n is a float or not is a bit-flag for whether we've removed any digits from n.



                                                                                                                                                            We require that the condition 1/n==0 is met to recurse into incrementing b, which integers n satisfy because floor division is done, but floats fails. (n=1 also fails but we don't want to recurse on it anyway.) Otherwise, floats work just like integers in the function because we're careful to do floor division n//b, and the output is a whole-number float.






                                                                                                                                                            share|improve this answer
























                                                                                                                                                              2












                                                                                                                                                              2








                                                                                                                                                              2







                                                                                                                                                              Python 2, 61 bytes





                                                                                                                                                              f=lambda n,b=2.:n and n%b+f(n//b,b)+(n//b>1/n==0and f(n,b+1))


                                                                                                                                                              Try it online!



                                                                                                                                                              Though this is longer the Dennis's solution off which it's based, I find the method too amusing to not share.



                                                                                                                                                              The goal is to recurse both on lopping off the last digit n->n/b and incrementing the base b->b+1, but we want to prevent the base from being increased after one or more digits have been lopped off. This is achieved by make the base b a float, so that after the update n->n//b, the float b infects n with its floatness. In this way, whether n is a float or not is a bit-flag for whether we've removed any digits from n.



                                                                                                                                                              We require that the condition 1/n==0 is met to recurse into incrementing b, which integers n satisfy because floor division is done, but floats fails. (n=1 also fails but we don't want to recurse on it anyway.) Otherwise, floats work just like integers in the function because we're careful to do floor division n//b, and the output is a whole-number float.






                                                                                                                                                              share|improve this answer













                                                                                                                                                              Python 2, 61 bytes





                                                                                                                                                              f=lambda n,b=2.:n and n%b+f(n//b,b)+(n//b>1/n==0and f(n,b+1))


                                                                                                                                                              Try it online!



                                                                                                                                                              Though this is longer the Dennis's solution off which it's based, I find the method too amusing to not share.



                                                                                                                                                              The goal is to recurse both on lopping off the last digit n->n/b and incrementing the base b->b+1, but we want to prevent the base from being increased after one or more digits have been lopped off. This is achieved by make the base b a float, so that after the update n->n//b, the float b infects n with its floatness. In this way, whether n is a float or not is a bit-flag for whether we've removed any digits from n.



                                                                                                                                                              We require that the condition 1/n==0 is met to recurse into incrementing b, which integers n satisfy because floor division is done, but floats fails. (n=1 also fails but we don't want to recurse on it anyway.) Otherwise, floats work just like integers in the function because we're careful to do floor division n//b, and the output is a whole-number float.







                                                                                                                                                              share|improve this answer












                                                                                                                                                              share|improve this answer



                                                                                                                                                              share|improve this answer










                                                                                                                                                              answered Dec 5 at 5:57









                                                                                                                                                              xnor

                                                                                                                                                              89.7k18184439




                                                                                                                                                              89.7k18184439























                                                                                                                                                                  1















                                                                                                                                                                  Jelly, 4 bytes



                                                                                                                                                                  bRFS


                                                                                                                                                                  Try it online!



                                                                                                                                                                  How it works



                                                                                                                                                                  bRFS  Main link. Argument: n

                                                                                                                                                                  R Range; yield [1, ..., n].
                                                                                                                                                                  b Convert n to back k, for each k to the right.
                                                                                                                                                                  F Flatten the resulting 2D array of digits.
                                                                                                                                                                  S Take the sum.





                                                                                                                                                                  share|improve this answer


























                                                                                                                                                                    1















                                                                                                                                                                    Jelly, 4 bytes



                                                                                                                                                                    bRFS


                                                                                                                                                                    Try it online!



                                                                                                                                                                    How it works



                                                                                                                                                                    bRFS  Main link. Argument: n

                                                                                                                                                                    R Range; yield [1, ..., n].
                                                                                                                                                                    b Convert n to back k, for each k to the right.
                                                                                                                                                                    F Flatten the resulting 2D array of digits.
                                                                                                                                                                    S Take the sum.





                                                                                                                                                                    share|improve this answer
























                                                                                                                                                                      1












                                                                                                                                                                      1








                                                                                                                                                                      1







                                                                                                                                                                      Jelly, 4 bytes



                                                                                                                                                                      bRFS


                                                                                                                                                                      Try it online!



                                                                                                                                                                      How it works



                                                                                                                                                                      bRFS  Main link. Argument: n

                                                                                                                                                                      R Range; yield [1, ..., n].
                                                                                                                                                                      b Convert n to back k, for each k to the right.
                                                                                                                                                                      F Flatten the resulting 2D array of digits.
                                                                                                                                                                      S Take the sum.





                                                                                                                                                                      share|improve this answer













                                                                                                                                                                      Jelly, 4 bytes



                                                                                                                                                                      bRFS


                                                                                                                                                                      Try it online!



                                                                                                                                                                      How it works



                                                                                                                                                                      bRFS  Main link. Argument: n

                                                                                                                                                                      R Range; yield [1, ..., n].
                                                                                                                                                                      b Convert n to back k, for each k to the right.
                                                                                                                                                                      F Flatten the resulting 2D array of digits.
                                                                                                                                                                      S Take the sum.






                                                                                                                                                                      share|improve this answer












                                                                                                                                                                      share|improve this answer



                                                                                                                                                                      share|improve this answer










                                                                                                                                                                      answered Dec 3 at 16:33









                                                                                                                                                                      Dennis

                                                                                                                                                                      186k32295735




                                                                                                                                                                      186k32295735























                                                                                                                                                                          1















                                                                                                                                                                          Attache, 25 bytes



                                                                                                                                                                          {Sum!`'^^ToBase[_,2:_]'_}


                                                                                                                                                                          Try it online!



                                                                                                                                                                          Explanation



                                                                                                                                                                          {Sum!`'^^ToBase[_,2:_]'_}
                                                                                                                                                                          { } anonymous lambda, input: _
                                                                                                                                                                          example: 5
                                                                                                                                                                          ToBase[_, ] convert `_`
                                                                                                                                                                          2:_ ...to each base from 2 to `_`
                                                                                                                                                                          example: [[1, 0, 1], [1, 2], [1, 1], [1, 0]]
                                                                                                                                                                          '_ append `_`
                                                                                                                                                                          example: [[1, 0, 1], [1, 2], [1, 1], [1, 0], 5]
                                                                                                                                                                          ^^ fold...
                                                                                                                                                                          `' the append operator over the list
                                                                                                                                                                          example: [1, 0, 1, 1, 2, 1, 1, 1, 0, 5]
                                                                                                                                                                          Sum! take the sum
                                                                                                                                                                          example: 13





                                                                                                                                                                          share|improve this answer


























                                                                                                                                                                            1















                                                                                                                                                                            Attache, 25 bytes



                                                                                                                                                                            {Sum!`'^^ToBase[_,2:_]'_}


                                                                                                                                                                            Try it online!



                                                                                                                                                                            Explanation



                                                                                                                                                                            {Sum!`'^^ToBase[_,2:_]'_}
                                                                                                                                                                            { } anonymous lambda, input: _
                                                                                                                                                                            example: 5
                                                                                                                                                                            ToBase[_, ] convert `_`
                                                                                                                                                                            2:_ ...to each base from 2 to `_`
                                                                                                                                                                            example: [[1, 0, 1], [1, 2], [1, 1], [1, 0]]
                                                                                                                                                                            '_ append `_`
                                                                                                                                                                            example: [[1, 0, 1], [1, 2], [1, 1], [1, 0], 5]
                                                                                                                                                                            ^^ fold...
                                                                                                                                                                            `' the append operator over the list
                                                                                                                                                                            example: [1, 0, 1, 1, 2, 1, 1, 1, 0, 5]
                                                                                                                                                                            Sum! take the sum
                                                                                                                                                                            example: 13





                                                                                                                                                                            share|improve this answer
























                                                                                                                                                                              1












                                                                                                                                                                              1








                                                                                                                                                                              1







                                                                                                                                                                              Attache, 25 bytes



                                                                                                                                                                              {Sum!`'^^ToBase[_,2:_]'_}


                                                                                                                                                                              Try it online!



                                                                                                                                                                              Explanation



                                                                                                                                                                              {Sum!`'^^ToBase[_,2:_]'_}
                                                                                                                                                                              { } anonymous lambda, input: _
                                                                                                                                                                              example: 5
                                                                                                                                                                              ToBase[_, ] convert `_`
                                                                                                                                                                              2:_ ...to each base from 2 to `_`
                                                                                                                                                                              example: [[1, 0, 1], [1, 2], [1, 1], [1, 0]]
                                                                                                                                                                              '_ append `_`
                                                                                                                                                                              example: [[1, 0, 1], [1, 2], [1, 1], [1, 0], 5]
                                                                                                                                                                              ^^ fold...
                                                                                                                                                                              `' the append operator over the list
                                                                                                                                                                              example: [1, 0, 1, 1, 2, 1, 1, 1, 0, 5]
                                                                                                                                                                              Sum! take the sum
                                                                                                                                                                              example: 13





                                                                                                                                                                              share|improve this answer













                                                                                                                                                                              Attache, 25 bytes



                                                                                                                                                                              {Sum!`'^^ToBase[_,2:_]'_}


                                                                                                                                                                              Try it online!



                                                                                                                                                                              Explanation



                                                                                                                                                                              {Sum!`'^^ToBase[_,2:_]'_}
                                                                                                                                                                              { } anonymous lambda, input: _
                                                                                                                                                                              example: 5
                                                                                                                                                                              ToBase[_, ] convert `_`
                                                                                                                                                                              2:_ ...to each base from 2 to `_`
                                                                                                                                                                              example: [[1, 0, 1], [1, 2], [1, 1], [1, 0]]
                                                                                                                                                                              '_ append `_`
                                                                                                                                                                              example: [[1, 0, 1], [1, 2], [1, 1], [1, 0], 5]
                                                                                                                                                                              ^^ fold...
                                                                                                                                                                              `' the append operator over the list
                                                                                                                                                                              example: [1, 0, 1, 1, 2, 1, 1, 1, 0, 5]
                                                                                                                                                                              Sum! take the sum
                                                                                                                                                                              example: 13






                                                                                                                                                                              share|improve this answer












                                                                                                                                                                              share|improve this answer



                                                                                                                                                                              share|improve this answer










                                                                                                                                                                              answered Dec 3 at 22:09









                                                                                                                                                                              Conor O'Brien

                                                                                                                                                                              29.1k263162




                                                                                                                                                                              29.1k263162























                                                                                                                                                                                  1















                                                                                                                                                                                  Charcoal, 12 bytes



                                                                                                                                                                                  IΣEIθΣ↨Iθ⁺²ι


                                                                                                                                                                                  Try it online! Link is to verbose version of code. Charcoal can't convert to base 1, but fortunately the digit sum is the same as the conversion to base $n + 1$. Explanation:



                                                                                                                                                                                      θ           First input
                                                                                                                                                                                  I Cast to integer
                                                                                                                                                                                  E Map over implicit range
                                                                                                                                                                                  θ First input
                                                                                                                                                                                  I Cast to integer
                                                                                                                                                                                  ↨ Converted to base
                                                                                                                                                                                  ι Current index
                                                                                                                                                                                  ⁺ Plus
                                                                                                                                                                                  ² Literal 2
                                                                                                                                                                                  Σ Sum
                                                                                                                                                                                  Σ Grand total
                                                                                                                                                                                  I Cast to string
                                                                                                                                                                                  Implicitly print





                                                                                                                                                                                  share|improve this answer


























                                                                                                                                                                                    1















                                                                                                                                                                                    Charcoal, 12 bytes



                                                                                                                                                                                    IΣEIθΣ↨Iθ⁺²ι


                                                                                                                                                                                    Try it online! Link is to verbose version of code. Charcoal can't convert to base 1, but fortunately the digit sum is the same as the conversion to base $n + 1$. Explanation:



                                                                                                                                                                                        θ           First input
                                                                                                                                                                                    I Cast to integer
                                                                                                                                                                                    E Map over implicit range
                                                                                                                                                                                    θ First input
                                                                                                                                                                                    I Cast to integer
                                                                                                                                                                                    ↨ Converted to base
                                                                                                                                                                                    ι Current index
                                                                                                                                                                                    ⁺ Plus
                                                                                                                                                                                    ² Literal 2
                                                                                                                                                                                    Σ Sum
                                                                                                                                                                                    Σ Grand total
                                                                                                                                                                                    I Cast to string
                                                                                                                                                                                    Implicitly print





                                                                                                                                                                                    share|improve this answer
























                                                                                                                                                                                      1












                                                                                                                                                                                      1








                                                                                                                                                                                      1







                                                                                                                                                                                      Charcoal, 12 bytes



                                                                                                                                                                                      IΣEIθΣ↨Iθ⁺²ι


                                                                                                                                                                                      Try it online! Link is to verbose version of code. Charcoal can't convert to base 1, but fortunately the digit sum is the same as the conversion to base $n + 1$. Explanation:



                                                                                                                                                                                          θ           First input
                                                                                                                                                                                      I Cast to integer
                                                                                                                                                                                      E Map over implicit range
                                                                                                                                                                                      θ First input
                                                                                                                                                                                      I Cast to integer
                                                                                                                                                                                      ↨ Converted to base
                                                                                                                                                                                      ι Current index
                                                                                                                                                                                      ⁺ Plus
                                                                                                                                                                                      ² Literal 2
                                                                                                                                                                                      Σ Sum
                                                                                                                                                                                      Σ Grand total
                                                                                                                                                                                      I Cast to string
                                                                                                                                                                                      Implicitly print





                                                                                                                                                                                      share|improve this answer













                                                                                                                                                                                      Charcoal, 12 bytes



                                                                                                                                                                                      IΣEIθΣ↨Iθ⁺²ι


                                                                                                                                                                                      Try it online! Link is to verbose version of code. Charcoal can't convert to base 1, but fortunately the digit sum is the same as the conversion to base $n + 1$. Explanation:



                                                                                                                                                                                          θ           First input
                                                                                                                                                                                      I Cast to integer
                                                                                                                                                                                      E Map over implicit range
                                                                                                                                                                                      θ First input
                                                                                                                                                                                      I Cast to integer
                                                                                                                                                                                      ↨ Converted to base
                                                                                                                                                                                      ι Current index
                                                                                                                                                                                      ⁺ Plus
                                                                                                                                                                                      ² Literal 2
                                                                                                                                                                                      Σ Sum
                                                                                                                                                                                      Σ Grand total
                                                                                                                                                                                      I Cast to string
                                                                                                                                                                                      Implicitly print






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                                                                                                                                                                                      answered Dec 3 at 23:39









                                                                                                                                                                                      Neil

                                                                                                                                                                                      79.3k744177




                                                                                                                                                                                      79.3k744177























                                                                                                                                                                                          1















                                                                                                                                                                                          Retina 0.8.2, 49 bytes



                                                                                                                                                                                          .+
                                                                                                                                                                                          $*
                                                                                                                                                                                          1
                                                                                                                                                                                          11$`;$_¶
                                                                                                                                                                                          +`b(1+);(1)+
                                                                                                                                                                                          $1;$#2$*1,
                                                                                                                                                                                          1+;

                                                                                                                                                                                          1


                                                                                                                                                                                          Try it online! Link includes test cases. Explanation:



                                                                                                                                                                                          .+
                                                                                                                                                                                          $*


                                                                                                                                                                                          Convert to unary.



                                                                                                                                                                                          1
                                                                                                                                                                                          11$`;$_¶


                                                                                                                                                                                          List all the numbers from 2 to $n + 1$ (since that's easier than base 1 conversion).



                                                                                                                                                                                          +`b(1+);(1)+
                                                                                                                                                                                          $1;$#2$*1,


                                                                                                                                                                                          Use repeated divmod to convert the original number to each base.



                                                                                                                                                                                          1+;


                                                                                                                                                                                          Delete the list of bases, leaving just the base conversion digits.



                                                                                                                                                                                          1


                                                                                                                                                                                          Take the sum and convert to decimal.






                                                                                                                                                                                          share|improve this answer


























                                                                                                                                                                                            1















                                                                                                                                                                                            Retina 0.8.2, 49 bytes



                                                                                                                                                                                            .+
                                                                                                                                                                                            $*
                                                                                                                                                                                            1
                                                                                                                                                                                            11$`;$_¶
                                                                                                                                                                                            +`b(1+);(1)+
                                                                                                                                                                                            $1;$#2$*1,
                                                                                                                                                                                            1+;

                                                                                                                                                                                            1


                                                                                                                                                                                            Try it online! Link includes test cases. Explanation:



                                                                                                                                                                                            .+
                                                                                                                                                                                            $*


                                                                                                                                                                                            Convert to unary.



                                                                                                                                                                                            1
                                                                                                                                                                                            11$`;$_¶


                                                                                                                                                                                            List all the numbers from 2 to $n + 1$ (since that's easier than base 1 conversion).



                                                                                                                                                                                            +`b(1+);(1)+
                                                                                                                                                                                            $1;$#2$*1,


                                                                                                                                                                                            Use repeated divmod to convert the original number to each base.



                                                                                                                                                                                            1+;


                                                                                                                                                                                            Delete the list of bases, leaving just the base conversion digits.



                                                                                                                                                                                            1


                                                                                                                                                                                            Take the sum and convert to decimal.






                                                                                                                                                                                            share|improve this answer
























                                                                                                                                                                                              1












                                                                                                                                                                                              1








                                                                                                                                                                                              1







                                                                                                                                                                                              Retina 0.8.2, 49 bytes



                                                                                                                                                                                              .+
                                                                                                                                                                                              $*
                                                                                                                                                                                              1
                                                                                                                                                                                              11$`;$_¶
                                                                                                                                                                                              +`b(1+);(1)+
                                                                                                                                                                                              $1;$#2$*1,
                                                                                                                                                                                              1+;

                                                                                                                                                                                              1


                                                                                                                                                                                              Try it online! Link includes test cases. Explanation:



                                                                                                                                                                                              .+
                                                                                                                                                                                              $*


                                                                                                                                                                                              Convert to unary.



                                                                                                                                                                                              1
                                                                                                                                                                                              11$`;$_¶


                                                                                                                                                                                              List all the numbers from 2 to $n + 1$ (since that's easier than base 1 conversion).



                                                                                                                                                                                              +`b(1+);(1)+
                                                                                                                                                                                              $1;$#2$*1,


                                                                                                                                                                                              Use repeated divmod to convert the original number to each base.



                                                                                                                                                                                              1+;


                                                                                                                                                                                              Delete the list of bases, leaving just the base conversion digits.



                                                                                                                                                                                              1


                                                                                                                                                                                              Take the sum and convert to decimal.






                                                                                                                                                                                              share|improve this answer













                                                                                                                                                                                              Retina 0.8.2, 49 bytes



                                                                                                                                                                                              .+
                                                                                                                                                                                              $*
                                                                                                                                                                                              1
                                                                                                                                                                                              11$`;$_¶
                                                                                                                                                                                              +`b(1+);(1)+
                                                                                                                                                                                              $1;$#2$*1,
                                                                                                                                                                                              1+;

                                                                                                                                                                                              1


                                                                                                                                                                                              Try it online! Link includes test cases. Explanation:



                                                                                                                                                                                              .+
                                                                                                                                                                                              $*


                                                                                                                                                                                              Convert to unary.



                                                                                                                                                                                              1
                                                                                                                                                                                              11$`;$_¶


                                                                                                                                                                                              List all the numbers from 2 to $n + 1$ (since that's easier than base 1 conversion).



                                                                                                                                                                                              +`b(1+);(1)+
                                                                                                                                                                                              $1;$#2$*1,


                                                                                                                                                                                              Use repeated divmod to convert the original number to each base.



                                                                                                                                                                                              1+;


                                                                                                                                                                                              Delete the list of bases, leaving just the base conversion digits.



                                                                                                                                                                                              1


                                                                                                                                                                                              Take the sum and convert to decimal.







                                                                                                                                                                                              share|improve this answer












                                                                                                                                                                                              share|improve this answer



                                                                                                                                                                                              share|improve this answer










                                                                                                                                                                                              answered Dec 3 at 23:50









                                                                                                                                                                                              Neil

                                                                                                                                                                                              79.3k744177




                                                                                                                                                                                              79.3k744177























                                                                                                                                                                                                  0














                                                                                                                                                                                                  APL(NARS), 29 chars, 58 bytes



                                                                                                                                                                                                  {+/+/¨⍵{(⍺⍴⍨⌊1+⍺⍟⍵)⊤⍵}⍨¨1+⍳⍵}


                                                                                                                                                                                                  little test on how to use:



                                                                                                                                                                                                    h←{+/+/¨⍵{(⍺⍴⍨⌊1+⍺⍟⍵)⊤⍵}⍨¨1+⍳⍵}
                                                                                                                                                                                                  h¨1 2 4 10 36 37 64 65
                                                                                                                                                                                                  1 3 8 35 297 334 883 932





                                                                                                                                                                                                  share|improve this answer




























                                                                                                                                                                                                    0














                                                                                                                                                                                                    APL(NARS), 29 chars, 58 bytes



                                                                                                                                                                                                    {+/+/¨⍵{(⍺⍴⍨⌊1+⍺⍟⍵)⊤⍵}⍨¨1+⍳⍵}


                                                                                                                                                                                                    little test on how to use:



                                                                                                                                                                                                      h←{+/+/¨⍵{(⍺⍴⍨⌊1+⍺⍟⍵)⊤⍵}⍨¨1+⍳⍵}
                                                                                                                                                                                                    h¨1 2 4 10 36 37 64 65
                                                                                                                                                                                                    1 3 8 35 297 334 883 932





                                                                                                                                                                                                    share|improve this answer


























                                                                                                                                                                                                      0












                                                                                                                                                                                                      0








                                                                                                                                                                                                      0






                                                                                                                                                                                                      APL(NARS), 29 chars, 58 bytes



                                                                                                                                                                                                      {+/+/¨⍵{(⍺⍴⍨⌊1+⍺⍟⍵)⊤⍵}⍨¨1+⍳⍵}


                                                                                                                                                                                                      little test on how to use:



                                                                                                                                                                                                        h←{+/+/¨⍵{(⍺⍴⍨⌊1+⍺⍟⍵)⊤⍵}⍨¨1+⍳⍵}
                                                                                                                                                                                                      h¨1 2 4 10 36 37 64 65
                                                                                                                                                                                                      1 3 8 35 297 334 883 932





                                                                                                                                                                                                      share|improve this answer














                                                                                                                                                                                                      APL(NARS), 29 chars, 58 bytes



                                                                                                                                                                                                      {+/+/¨⍵{(⍺⍴⍨⌊1+⍺⍟⍵)⊤⍵}⍨¨1+⍳⍵}


                                                                                                                                                                                                      little test on how to use:



                                                                                                                                                                                                        h←{+/+/¨⍵{(⍺⍴⍨⌊1+⍺⍟⍵)⊤⍵}⍨¨1+⍳⍵}
                                                                                                                                                                                                      h¨1 2 4 10 36 37 64 65
                                                                                                                                                                                                      1 3 8 35 297 334 883 932






                                                                                                                                                                                                      share|improve this answer














                                                                                                                                                                                                      share|improve this answer



                                                                                                                                                                                                      share|improve this answer








                                                                                                                                                                                                      edited Dec 22 at 19:33

























                                                                                                                                                                                                      answered Dec 22 at 10:28









                                                                                                                                                                                                      RosLuP

                                                                                                                                                                                                      1,806514




                                                                                                                                                                                                      1,806514






























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