<s>[INST] Find closed formula for the exponential generating function of sequence counting permutations with 2 cycles. Find a closed formula for the exponential generating function of $\{c(n,2)\}_{n\geq2}$, that is the sequence counting permutations of length $n$ that consist of $2$ cycles.
My approach: The exponential generating function for Stirling Numbers of the First Kind,
$$
C(n,k) = \sum_{n=k}^{+\infty}{c(n,k)} \, \frac{x^n}{n!} 
= \frac{\log^k(1/(1-x))}{k!}
$$
Replacing $k$ with $2$,
$$
\sum_{n=2}^{+\infty} c(n,2) \, \frac{x^n}{n!} 
= \frac{1}{2!} \biggl( \log \frac{1}{1-x} \biggr)^2 
$$
Using
$$
\log \frac{1}{1-x} = x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots{}
$$
we have
$$
\biggl( \log \frac{1}{1-x} \biggr)^2 
= \biggl( x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots{} \biggr) 
\biggl( x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots{} \biggr)
$$
Now $c(n,2)$ is given by the coefficient of $\dfrac{x^n}{n!}$ in $C(x)$
Hence,
$$
c(n,2) = \frac{1}{2!} \sum_{k\geq 1} \frac{n!}{k(n-k)}
$$
I may be dead wrong about this. Any help is appreciated.
 [/INST] Your approach is correct, to finish notice that $\frac{1}{k(n-k)}=\frac{1}{n}\left (\frac{1}{k}+\frac{1}{n-k}\right )$ and adding over $k$ you get twice $H_{n-1}.$
For a combinatorial way, choose the elements that you want in the cycle containing one, say of size $k$ in $\binom{n-1}{k}$ ways. Notice that when you pick the elements there are $k!$ possible cycles (by fixing $1$ as the minimal element and permuting the other elements in all ways), the rest $n-k$ elements are in the other cycle and there are $(n-k-1)!$ to arrange it. Adding over $k$ you get
$$\sum _{k=1}^{n-1}\binom{n-1}{k}k!(n-k-1)!=\sum _{k=1}^{n-1}\frac{(n-1)!}{(n-k)}=(n-1)!H_{n-1}.$$
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