generating function of a sequence $a(m,k)$

Question

i am interested if exist a explicit expresion of sequence given by

$$\sum_{k\geq 0}a(m,k)\frac{x^k}{k!}=\frac{1}{m!}(e^x-x-1)^m$$ I have tried to use binomial theorem but without any success

Answer 1

We have that

$$n! [z^n] \frac{1}{m!} (\exp(z)-z-1)^m \\ = n! [z^n] \frac{1}{m!} \sum_{q=0}^m {m\choose q} (-1)^{m-q} z^{m-q} (\exp(z)-1)^q \\ = n! \frac{1}{m!} \sum_{q=0}^m {m\choose q} (-1)^{m-q} q! [z^n] z^{m-q} \frac{(\exp(z)-1)^q}{q!} \\ = n! \frac{1}{m!} \sum_{q=0}^m {m\choose q} (-1)^{m-q} q! [z^{n-(m-q)}] \frac{(\exp(z)-1)^q}{q!} \\ = n! \frac{1}{m!} \sum_{q=0}^m {m\choose q} (-1)^{m-q} q! \frac{1}{(n-(m-q))!} {n-(m-q)\brace q} \\ = \sum_{q=0}^m {n\choose m-q} (-1)^{m-q} {n-(m-q)\brace q}.$$

If desired we may write this as

$$\sum_{q=0}^m {n\choose q} (-1)^{q} {n-q\brace m-q}.$$

These numbers appeared at the following MSE link.

We have the species

$$\mathfrak{P}_{=m}(\mathfrak{P}_{\ge 2}(\mathcal{Z}))$$

so this equation says in fact

$${n\brace m}_{\ge 2} = \sum_{q=0}^m {n\choose q} (-1)^{q} {n-q\brace m-q}.$$

This is inclusion-exclusion of course where we remove set partitions into $m$ sets containing singletons. We first choose the $q$ singletons and combine them with an arbitrary set partition into $m-q$ sets of the remaining elements. The poset is ordered according to set inclusion of the sets of singletons.