Start with a permutation $s_0$ of the numbers $(1,\ldots,n)$, e.g., for $n=10$, $s_0=(8,2,1,6,9,7,10,5,4,3)$. Form $s_1$ by using the numbers in $s_0$ as indices into $s_0$. So $s_1$ is composed of the $8$-th element of $s_0$, followed the $2$-nd element, followed by the $1$-st element, followed by the $6$-th element, etc., yielding $s_1 = (5,2,8,7,4,10,3,9,6,1)$ Continuing in this manner, $s_{i+1}=s_i[s_i]$, we reach a cycle of length $6$: $s_1 = s_7$: \begin{eqnarray} s_0 &=& (8,2,1,6,9,7,10,5,4,3)\\ s_1 &=& (5,2,8,7,4,10,3,9,6,1)\\ s_2 &=& (4,2,9,3,7,1,8,6,10,5)\\ s_3 &=& (3,2,10,9,8,4,6,1,5,7)\\ s_4 &=& (10,2,7,5,1,9,4,3,8,6)\\ s_5 &=& (6,2,4,1,10,8,5,7,3,9)\\ s_6 &=& (8,2,1,6,9,7,10,5,4,3)\\ s_7 &=& (5,2,8,7,4,10,3,9,6,1) \end{eqnarray} For $n=10$, cycles of lengths $1,2,3,4,6$ occur, but (apparently) no other lengths. Call this list the cycle-length spectrum for $n$. My question is:
Q. What explains the cycle-length spectrum for a given $n$?
Here is a bit of empirical data: $$\left[ \begin{array}{cc} n & \textrm{cycle lengths}\\ - & ------ \\ 2 & (1) \\ 3 & (1,2) \\ 4 & (1,2) \\ 5 & (1,2,4) \\ 6 & (1,2,4) \\ 7 & (1,2,3,4) \\ 8 & (1,2,3,4) \\ 9 & (1,2,3,4,6) \\ 10 & (1,2,3,4,6) \\ 11 & (1,2,3,4,6,10) \\ 12 & (1,2,3,4,6,10,12) \\ 13 & (1,2,3,4,6,10,12) \\ 14 & (1,2,3,4,6,10,12) \\ 15 & (1,2,3,4,6,10,12) \\ 16 & (1,2,3,4,6,10,12,20) \\ 17 & (1,2,3,4,6,8,10,12,20) \\ 18 & (1,2,3,4,6,8,10,12,20,30) \\ 19 & (1,2,3,4,6,8,10,12,18,20,3 0) \\ 20 & (1,2,3,4,6,8,10,12,18,20,3 0) \\ 21 & (1,2,3,4,6,8,10,12,18,20,3 0) \\ 22 & (1,2,3,4,6,8,10,12,18,20,3 0) \\ 23 & (1,2,3,4,6,8,10,11,12,18,2 0,30,60) \end{array} \right]$$ The cycle spectrum for $n$ is a subset of the cycle spectrum of $n+1$. Even cycle lengths dominate, but some cycles have odd length, e.g., $n=8$, $(7,4,8,5,1,6,3,2)$ leads to a cycle of length $3$, and $n=23$ includes cycles of length $11$. So perhaps cycles of length $5$ occur for some $n$...