You have $n$ stones. You break the stones into some number of groups, and place the stones within each group into a line. You then arrange these lines in a circle. Let $s_n$ be the number of ways to do this. Find a formula for $s_n$.
So I think there are $n(n-1)$ ways to seperate into groups and $n!$ ways to form into lines? Does this give $\sum_{k=0}^{\inf}n(n-1)n!\frac{x^n}{n!}$ ? I'm not sure how to go about this
There is a way to get an answer via generating function trickery, with a combinatorial proof aided by hindsight.
Each line of $k$ stones can be arranged in $k!$ ways, so the exponential generating function for a single line is $\sum_{k\ge 1}k!\frac{x^k}{k!}=\frac{x}{1-x}$.
Suppose there were $m$ lines total, with $m\ge 1$. The EGF for putting these lines in a linear order would be $\left(\frac{x}{1-x}\right)^m$, but since we only want a circular order, we divide by $n$, for $\frac{1}m\left(\frac{x}{1-x}\right)^m$. Summing over all $m$, we get \begin{align} \sum_{m\ge 1}\frac1m\left(\frac{x}{1-x}\right)^m &=-\log \left(1-\frac{x}{1-x}\right) \\&=-\log(1-2x)+\log(1-x) \\&=\sum_{n\ge 1}\frac{2^n-1}{n}x^n \\&=\sum_{n\ge 1}\color{green}{(2^n-1)(n-1)!}\;\frac{x^n}{n!} \end{align} To explain this, note that the circular arrangement of lines can be constructed by first arranging the $n$ stones in a circle in $(n-1)!$ ways, then placing barriers between some of the stones to break them into lines. There must be at least one barrier, so this can be done in $2^n-1$ ways.