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:

. What explains the cycle-length spectrum for a given $n$?Q

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 $n$...