This problem involves an iterative procedure that begins with a circle of $n\ge 3$ integers. At each step every number is simultaneously replaced with the absolute difference of its two neighbours.

For any initial values, the procedure eventually becomes periodic.

Let $S(N)$ be the sum of all possible periods for $3\le n\le N$. For example, $S(6)=6$, because the possible periods for $3\le n\le 6$ are $1,2,3$. Specifically, $n=3$ and $n=4$ can each have period $1$ only, while $n=5$ can have period $1$ or $3$, and $n=6$ can have period $1$ or $2$.

You are also given $S(30)=20381$.

Find $S(100)$.