Numer of permutations that are product of exactly two disjoint cycles Let $a_n$ denote the number of those permutations $\sigma$ on $\{1,2,3....,n\}$ such that $\sigma$ is a product of exactly two disjoint cycles. Then 


*

*$a_5 = 50$

*$a_4 = 14$

*$a_5 = 40$

*$a_4  = 11$
I tried specifically for $a_5$  and $a_4$ with a little bit of calculations. But I want to know about a formula for $a_n$ with less calculations 
 A: There's a well-known formula for the number of permutations with $p_i$ cycles of length $i$ for each $i$, namely 
$$ {n! \over \prod_i i^{p_i} (p_i)!}$$
(see for example this post by Mark Jason Dominus).  In the case where you have one cycle of length $k$ and one of length $n-k$, and $k \not = n-k$, then you have $p_k = p_{n-k} = 1$ and this reduces to 
$$n! \times {1 \over k(n-k)}.$$  If $k = n-k$, that is, if $k = n/2$, then you have $p_k = 2$ and the number of such permutations is 
$$n! \times {2 \over n^2}.$$
For example, if $n = 5, k = 2$, the first formula gives you that there are $5!/(2 \times 3) = 20$ permutations of $[5]$ consisting of a 2-cycle and a 3-cycle.  If $n = 4, k = 2$, the second formula gives you that there are $4! \times 2/4^2 = 3$ permutations consisting of two 2-cycles.
From this you can derive a general formula for the number of permutations of $n$ which are products of two disjoint cycles by summing over the possible values of $k$. If $n$ is even you'll need to handle the $k = n/2$ case separately from the rest of the sum.
(It's not clear if your definition counts a cycle of length 1 as a cycle.  For example, is $(3, 5)(1, 4)(2)(6)$ a product of two distinct cycles in $S_6$?  If so your sum will be more complicated but the general idea still holds.)
A: I would  like to present the  connection to Stirling numbers  since it
has  not been  pointed out.   For the  first interpretation  where the
cycles     may     be     singletons    we     get     the     species
$\mathfrak{P}_{=2}(\mathfrak{C}(\mathcal{Z}))$   which    yields   per
generating function
$$n! [z^n] \frac{1}{2!}\left(\log\frac{1}{1-z}\right)^2
= \left[n\atop 2\right]$$
the sequence
$$0, 1, 3, 11, 50, 274, 1764, 13068, 109584, 1026576,\ldots $$
which is OEIS A000254  which looks to be a
match. The second interpretation is when we do not admit singletons as
cycles  and we  get  the species  $\mathfrak{P}_{=2}(\mathfrak{C}_{\ge
2}(\mathcal{Z}))$ which yields per generating function
$$n! [z^n] \frac{1}{2!}\left(-z + \log\frac{1}{1-z}\right)^2$$
the sequence
$$0, 0, 0, 3, 20, 130, 924, 7308, 64224, 623376,\ldots$$
which is  OEIS A000276. For $n\ge  2$ this
simplifies to
$$\frac{1}{2} n! [z^n]
\left(z^2 - 2z \log\frac{1}{1-z} +
\left(\log\frac{1}{1-z}\right)^2\right)
\\ = [[n=2]] - n! [z^{n-1}] \log\frac{1}{1-z} + \left[n\atop 2\right]
\\ = [[n=2]] - n \times (n-2)! + \left[n\atop 2\right].$$
