Gfrktyl

I am solving this Kata - https://www.codewars.com/kata/simple-fun-number-23-square-digits-sequence/python

Consider a sequence of numbers a0, a1, ..., an, in which an element is equal to the sum of squared digits of the previous element. The sequence ends once an element that has already been in the sequence appears again.

Given the first element a0, find the length of the sequence.

Example
For a0 = 16, the output should be 9

Here's how elements of the sequence are constructed:

a0 = 16

a1 = 12 + 62 = 37

a2 = 32 + 72 = 58

a3 = 52 + 82 = 89

a4 = 82 + 92 = 145

a5 = 12 + 42 + 52 = 42

a6 = 42 + 22 = 20

a7 = 22 + 02 = 4

a8 = 42 = 16, which has already occurred before (a0)

Thus, there are 9 elements in the sequence.

For a0 = 103, the output should be 4

The sequence goes as follows: 103 -> 10 -> 1 -> 1, 4 elements altogether.

I took a wrong approach to solve the code (that's what I already know from other solutions) but still do not understand why is it not working. Any help?

def mysquare(digits, previous = , loop = 2):

    sum = 0

    for digit in str(digits):

        sum+=int(digit)**2

    if sum in previous or digits == sum:

        return loop

    else:

        previous.append(sum)

        return mysquare(sum, previous, loop+1)



def square_digits_sequence(n):

    return mysquare(n)