Is this a new Fibonacci Identity?












5












$begingroup$


I have found the following Fibonacci Identity (and proved it).



If $F_n$ denotes the nth Fibonacci Number, we have the following identity
begin{equation}
F_{n-r+h}F_{n+k+g+1} - F_{n-r+g}F_{n+k+h+1} = (-1)^{n+r+h+1} F_{g-h}F_{k+r+1}
end{equation}
where $F_1 = F_2 = 1$, $r leq n$, $h leq g$, and $n, g, k in mathbb{N}$.



It is not too hard to show that this identity subsumes Cassini's Identity, Catalan's Identity, Vajda's Idenity, and d'Ocagne's identity to name a few.



I have done a pretty thorough literature review, and I have not found anything like this, but I am still wondering if anyone has seen this identity before? I found this by accident after noticing some patterns in some analysis work I was doing, so if this is already known I would be curious to see what the connections are. Thanks for your patience and input!






nt.number-theory co.combinatorics







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    This can be simplified to $F_{a - r}F_{b + k + 1} - F_{b - r} F_{a + k + 1} = (-1)^{a + r + 1} F_{b - a} F_{k + r + 1}$, using the substitution $a = n + h, b = n + g$, reducing to 5 variables instead of 6.
    $endgroup$
    – user44191
    4 hours ago










  • $begingroup$
    It doesn't have 6 variables? It has 5: $n, r, k, h, g$.
    $endgroup$
    – Grassi
    4 hours ago






  • 1




    $begingroup$
    Sorry, I meant 4 variables instead of 5. I'm pretty sure this can be reduced by changes of variables (with no Fibonacci arithmetic) to Vajda's identity.
    $endgroup$
    – user44191
    4 hours ago








  • 1




    $begingroup$
    This is a disguised version of Vajda's identity (with minor amounts of arithmetic for powers of $-1$); try doing variable substitutions to see for yourself.
    $endgroup$
    – user44191
    4 hours ago












  • $begingroup$
    Yes, but from what I can tell this is no different than saying Vajda's identity is a disguised version of Catalan's identity. You can't go the other way without adding a variable.
    $endgroup$
    – Grassi
    3 hours ago
















5












$begingroup$


I have found the following Fibonacci Identity (and proved it).



If $F_n$ denotes the nth Fibonacci Number, we have the following identity
begin{equation}
F_{n-r+h}F_{n+k+g+1} - F_{n-r+g}F_{n+k+h+1} = (-1)^{n+r+h+1} F_{g-h}F_{k+r+1}
end{equation}
where $F_1 = F_2 = 1$, $r leq n$, $h leq g$, and $n, g, k in mathbb{N}$.



It is not too hard to show that this identity subsumes Cassini's Identity, Catalan's Identity, Vajda's Idenity, and d'Ocagne's identity to name a few.



I have done a pretty thorough literature review, and I have not found anything like this, but I am still wondering if anyone has seen this identity before? I found this by accident after noticing some patterns in some analysis work I was doing, so if this is already known I would be curious to see what the connections are. Thanks for your patience and input!






nt.number-theory co.combinatorics







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    This can be simplified to $F_{a - r}F_{b + k + 1} - F_{b - r} F_{a + k + 1} = (-1)^{a + r + 1} F_{b - a} F_{k + r + 1}$, using the substitution $a = n + h, b = n + g$, reducing to 5 variables instead of 6.
    $endgroup$
    – user44191
    4 hours ago










  • $begingroup$
    It doesn't have 6 variables? It has 5: $n, r, k, h, g$.
    $endgroup$
    – Grassi
    4 hours ago






  • 1




    $begingroup$
    Sorry, I meant 4 variables instead of 5. I'm pretty sure this can be reduced by changes of variables (with no Fibonacci arithmetic) to Vajda's identity.
    $endgroup$
    – user44191
    4 hours ago








  • 1




    $begingroup$
    This is a disguised version of Vajda's identity (with minor amounts of arithmetic for powers of $-1$); try doing variable substitutions to see for yourself.
    $endgroup$
    – user44191
    4 hours ago












  • $begingroup$
    Yes, but from what I can tell this is no different than saying Vajda's identity is a disguised version of Catalan's identity. You can't go the other way without adding a variable.
    $endgroup$
    – Grassi
    3 hours ago














5












5








5


1



$begingroup$


I have found the following Fibonacci Identity (and proved it).



If $F_n$ denotes the nth Fibonacci Number, we have the following identity
begin{equation}
F_{n-r+h}F_{n+k+g+1} - F_{n-r+g}F_{n+k+h+1} = (-1)^{n+r+h+1} F_{g-h}F_{k+r+1}
end{equation}
where $F_1 = F_2 = 1$, $r leq n$, $h leq g$, and $n, g, k in mathbb{N}$.



It is not too hard to show that this identity subsumes Cassini's Identity, Catalan's Identity, Vajda's Idenity, and d'Ocagne's identity to name a few.



I have done a pretty thorough literature review, and I have not found anything like this, but I am still wondering if anyone has seen this identity before? I found this by accident after noticing some patterns in some analysis work I was doing, so if this is already known I would be curious to see what the connections are. Thanks for your patience and input!






nt.number-theory co.combinatorics







share|cite|improve this question









$endgroup$




I have found the following Fibonacci Identity (and proved it).



If $F_n$ denotes the nth Fibonacci Number, we have the following identity
begin{equation}
F_{n-r+h}F_{n+k+g+1} - F_{n-r+g}F_{n+k+h+1} = (-1)^{n+r+h+1} F_{g-h}F_{k+r+1}
end{equation}
where $F_1 = F_2 = 1$, $r leq n$, $h leq g$, and $n, g, k in mathbb{N}$.



It is not too hard to show that this identity subsumes Cassini's Identity, Catalan's Identity, Vajda's Idenity, and d'Ocagne's identity to name a few.



I have done a pretty thorough literature review, and I have not found anything like this, but I am still wondering if anyone has seen this identity before? I found this by accident after noticing some patterns in some analysis work I was doing, so if this is already known I would be curious to see what the connections are. Thanks for your patience and input!






nt.number-theory co.combinatorics




nt.number-theory co.combinatorics






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 4 hours ago









GrassiGrassi

10626




10626








  • 1




    $begingroup$
    This can be simplified to $F_{a - r}F_{b + k + 1} - F_{b - r} F_{a + k + 1} = (-1)^{a + r + 1} F_{b - a} F_{k + r + 1}$, using the substitution $a = n + h, b = n + g$, reducing to 5 variables instead of 6.
    $endgroup$
    – user44191
    4 hours ago










  • $begingroup$
    It doesn't have 6 variables? It has 5: $n, r, k, h, g$.
    $endgroup$
    – Grassi
    4 hours ago






  • 1




    $begingroup$
    Sorry, I meant 4 variables instead of 5. I'm pretty sure this can be reduced by changes of variables (with no Fibonacci arithmetic) to Vajda's identity.
    $endgroup$
    – user44191
    4 hours ago








  • 1




    $begingroup$
    This is a disguised version of Vajda's identity (with minor amounts of arithmetic for powers of $-1$); try doing variable substitutions to see for yourself.
    $endgroup$
    – user44191
    4 hours ago












  • $begingroup$
    Yes, but from what I can tell this is no different than saying Vajda's identity is a disguised version of Catalan's identity. You can't go the other way without adding a variable.
    $endgroup$
    – Grassi
    3 hours ago














  • 1




    $begingroup$
    This can be simplified to $F_{a - r}F_{b + k + 1} - F_{b - r} F_{a + k + 1} = (-1)^{a + r + 1} F_{b - a} F_{k + r + 1}$, using the substitution $a = n + h, b = n + g$, reducing to 5 variables instead of 6.
    $endgroup$
    – user44191
    4 hours ago










  • $begingroup$
    It doesn't have 6 variables? It has 5: $n, r, k, h, g$.
    $endgroup$
    – Grassi
    4 hours ago






  • 1




    $begingroup$
    Sorry, I meant 4 variables instead of 5. I'm pretty sure this can be reduced by changes of variables (with no Fibonacci arithmetic) to Vajda's identity.
    $endgroup$
    – user44191
    4 hours ago








  • 1




    $begingroup$
    This is a disguised version of Vajda's identity (with minor amounts of arithmetic for powers of $-1$); try doing variable substitutions to see for yourself.
    $endgroup$
    – user44191
    4 hours ago












  • $begingroup$
    Yes, but from what I can tell this is no different than saying Vajda's identity is a disguised version of Catalan's identity. You can't go the other way without adding a variable.
    $endgroup$
    – Grassi
    3 hours ago








1




1




$begingroup$
This can be simplified to $F_{a - r}F_{b + k + 1} - F_{b - r} F_{a + k + 1} = (-1)^{a + r + 1} F_{b - a} F_{k + r + 1}$, using the substitution $a = n + h, b = n + g$, reducing to 5 variables instead of 6.
$endgroup$
– user44191
4 hours ago




$begingroup$
This can be simplified to $F_{a - r}F_{b + k + 1} - F_{b - r} F_{a + k + 1} = (-1)^{a + r + 1} F_{b - a} F_{k + r + 1}$, using the substitution $a = n + h, b = n + g$, reducing to 5 variables instead of 6.
$endgroup$
– user44191
4 hours ago












$begingroup$
It doesn't have 6 variables? It has 5: $n, r, k, h, g$.
$endgroup$
– Grassi
4 hours ago




$begingroup$
It doesn't have 6 variables? It has 5: $n, r, k, h, g$.
$endgroup$
– Grassi
4 hours ago




1




1




$begingroup$
Sorry, I meant 4 variables instead of 5. I'm pretty sure this can be reduced by changes of variables (with no Fibonacci arithmetic) to Vajda's identity.
$endgroup$
– user44191
4 hours ago






$begingroup$
Sorry, I meant 4 variables instead of 5. I'm pretty sure this can be reduced by changes of variables (with no Fibonacci arithmetic) to Vajda's identity.
$endgroup$
– user44191
4 hours ago






1




1




$begingroup$
This is a disguised version of Vajda's identity (with minor amounts of arithmetic for powers of $-1$); try doing variable substitutions to see for yourself.
$endgroup$
– user44191
4 hours ago






$begingroup$
This is a disguised version of Vajda's identity (with minor amounts of arithmetic for powers of $-1$); try doing variable substitutions to see for yourself.
$endgroup$
– user44191
4 hours ago














$begingroup$
Yes, but from what I can tell this is no different than saying Vajda's identity is a disguised version of Catalan's identity. You can't go the other way without adding a variable.
$endgroup$
– Grassi
3 hours ago




$begingroup$
Yes, but from what I can tell this is no different than saying Vajda's identity is a disguised version of Catalan's identity. You can't go the other way without adding a variable.
$endgroup$
– Grassi
3 hours ago










2 Answers
2






active

oldest

votes


















2












$begingroup$

Here is an expanded comment of user44191. The basic observation is that one can extend $F_n$ to all $nin {mathbb Z}$ by requiring $F_{-n}=(-1)^{n+1}F_n$. Then by Vajda's formula, one has $$F_{n'+a'}F_{n'+b'}-F_{n'}F_{n'+a'+b'}=(-1)^{n'}F_{a'}F_{b'}=(-1)^{n'+a'+1}F_{-a'}F_{b'},$$ where one uses the extension above in the last equality. Now by the following substitutions, the above identity leads to the one given by user44191: $$n'=b-r,a'=a-b,b'=k+r+1,$$ using the fact that $a-r+1$ is congruent to $a+r+1$ mod $2$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    "Vajda's identity" is really Tagiuro's identity: A. Tagiuri, Di alcune successioni ricorrenti a termini interi e positivi, Periodico di Matematica 16 (1900–1901), 1–12.
    See also https://math.stackexchange.com/questions/1356391/is-there-a-name-for-this-fibonacci-identity.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Here is an expanded comment of user44191. The basic observation is that one can extend $F_n$ to all $nin {mathbb Z}$ by requiring $F_{-n}=(-1)^{n+1}F_n$. Then by Vajda's formula, one has $$F_{n'+a'}F_{n'+b'}-F_{n'}F_{n'+a'+b'}=(-1)^{n'}F_{a'}F_{b'}=(-1)^{n'+a'+1}F_{-a'}F_{b'},$$ where one uses the extension above in the last equality. Now by the following substitutions, the above identity leads to the one given by user44191: $$n'=b-r,a'=a-b,b'=k+r+1,$$ using the fact that $a-r+1$ is congruent to $a+r+1$ mod $2$.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Here is an expanded comment of user44191. The basic observation is that one can extend $F_n$ to all $nin {mathbb Z}$ by requiring $F_{-n}=(-1)^{n+1}F_n$. Then by Vajda's formula, one has $$F_{n'+a'}F_{n'+b'}-F_{n'}F_{n'+a'+b'}=(-1)^{n'}F_{a'}F_{b'}=(-1)^{n'+a'+1}F_{-a'}F_{b'},$$ where one uses the extension above in the last equality. Now by the following substitutions, the above identity leads to the one given by user44191: $$n'=b-r,a'=a-b,b'=k+r+1,$$ using the fact that $a-r+1$ is congruent to $a+r+1$ mod $2$.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Here is an expanded comment of user44191. The basic observation is that one can extend $F_n$ to all $nin {mathbb Z}$ by requiring $F_{-n}=(-1)^{n+1}F_n$. Then by Vajda's formula, one has $$F_{n'+a'}F_{n'+b'}-F_{n'}F_{n'+a'+b'}=(-1)^{n'}F_{a'}F_{b'}=(-1)^{n'+a'+1}F_{-a'}F_{b'},$$ where one uses the extension above in the last equality. Now by the following substitutions, the above identity leads to the one given by user44191: $$n'=b-r,a'=a-b,b'=k+r+1,$$ using the fact that $a-r+1$ is congruent to $a+r+1$ mod $2$.






          share|cite|improve this answer









          $endgroup$



          Here is an expanded comment of user44191. The basic observation is that one can extend $F_n$ to all $nin {mathbb Z}$ by requiring $F_{-n}=(-1)^{n+1}F_n$. Then by Vajda's formula, one has $$F_{n'+a'}F_{n'+b'}-F_{n'}F_{n'+a'+b'}=(-1)^{n'}F_{a'}F_{b'}=(-1)^{n'+a'+1}F_{-a'}F_{b'},$$ where one uses the extension above in the last equality. Now by the following substitutions, the above identity leads to the one given by user44191: $$n'=b-r,a'=a-b,b'=k+r+1,$$ using the fact that $a-r+1$ is congruent to $a+r+1$ mod $2$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 3 hours ago









          Cherng-tiao PerngCherng-tiao Perng

          805148




          805148























              1












              $begingroup$

              "Vajda's identity" is really Tagiuro's identity: A. Tagiuri, Di alcune successioni ricorrenti a termini interi e positivi, Periodico di Matematica 16 (1900–1901), 1–12.
              See also https://math.stackexchange.com/questions/1356391/is-there-a-name-for-this-fibonacci-identity.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                "Vajda's identity" is really Tagiuro's identity: A. Tagiuri, Di alcune successioni ricorrenti a termini interi e positivi, Periodico di Matematica 16 (1900–1901), 1–12.
                See also https://math.stackexchange.com/questions/1356391/is-there-a-name-for-this-fibonacci-identity.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  "Vajda's identity" is really Tagiuro's identity: A. Tagiuri, Di alcune successioni ricorrenti a termini interi e positivi, Periodico di Matematica 16 (1900–1901), 1–12.
                  See also https://math.stackexchange.com/questions/1356391/is-there-a-name-for-this-fibonacci-identity.






                  share|cite|improve this answer









                  $endgroup$



                  "Vajda's identity" is really Tagiuro's identity: A. Tagiuri, Di alcune successioni ricorrenti a termini interi e positivi, Periodico di Matematica 16 (1900–1901), 1–12.
                  See also https://math.stackexchange.com/questions/1356391/is-there-a-name-for-this-fibonacci-identity.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 hours ago









                  Ira GesselIra Gessel

                  8,3922642




                  8,3922642






























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                      Алькесар

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                      Муниципалитет Алькесар Alquézar Герб 42°10′16″ с. ш. 0°01′27″ в. д. H G Я O L Страна Испания   Испания Автономное сообщество Арагон Провинция Уэска Комарка Сомонтано-де-Барбастро Алькальд Хосе Марияно Альтемир Ласкорс (ИСРП) История и география Площадь 32,36 км² Высота НУМ 660 м Часовой пояс UTC+1, летом UTC+2 Население Население 313 человек ( 2010 ) Плотность 9,92 чел./км² Этнохороним alquezarano/-a Цифровые идентификаторы Телефонный код +34 974 Почтовые индексы 22112 Автомобильный код HU alquezar.org   (исп.) Показать/скрыть карты Алькесар Алькесар Алькесар  Аудио, фото и видео на Викискладе Алькесар (исп.  Alquézar ) — муниципалитет в Испании, входит в провинцию Уэска, в составе автономного сообщества Арагон. Муниципалитет находится в составе района (комарки) Сомонтано-де-Барбастро. Занимает площадь 32,36 км². Население — 321 челове...
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