# Task
 let F(N) be the sum square of digits of N. So:
 `F(1) = 1, F(3) = 9, F(123) = 14`
 Choose a number A, the sequence {A0, A1, ...} is defined as followed:
  ```
  A0 = A
  A1 = F(A0)
  A2 = F(A1) ...
  ```
  if A = 123, we have:
  ```
  123 → 14(1 x 1 + 2 x 2 + 3 x 3)
      → 17(1 x 1 + 4 x 4)
      → 50(1 x 1 + 7 x 7)
      → 25(5 x 5 + 0 x 0)
      → 29(2 x 2 + 5 x 5)
      → 85(2 x 2 + 9 x 9)
      → 89(8 x 8 + 5 x 5)             ---
      → 145(8 x 8 + 9 x 9)             |r
      → 42(1 x 1 + 4 x 4 + 5 x 5)      |e
      → 20(4 x 4 + 2 x 2)              |p
      → 4(2 x 2 + 0 x 0)               |e
      → 16(4 x 4)                      |a
      → 37(1 x 1 + 6 x 6)              |t
      → 58(3 x 3 + 7 x 7)              |
      → 89(5 x 5 + 8 x 8)             --- 
      → ......
  ```
 As you can see, the sequence repeats itself. Interestingly, whatever A is, there's an index such that from it, the sequence repeats again and again.
 Let `G(A)` be the minimum length of the repeat sequence with A0 = A. 
 So `G(85) = 8` (8 number : `89,145,42, 20,4,16,37,58`)
 Your task is to find G(A) and return it.
# Input/Output
 - `[input]` integer `a0`
  the A0 number
 - `[output]` an integer
  the length of the repeat sequence
def repeat_sequence_len(n):
    s = []

    while True:
        n = sum(int(i) ** 2 for i in str(n))
        if n not in s:
            s.append(n)
        else:
            return len(s[s.index(n):])
def repeat_sequence_len(n):
    s = []

    while True:
        n = sum(int(i) ** 2 for i in str(n))
        if n not in s:
            s.append(n)
        else:
            return len(s[s.index(n):])