This is again a question that I asked at Stack Exchange, but got no answer so far, so I am trying here.
Let:

$$ a_n=\sum_{k\ge0}(k+1) {n+2\brack k+2}(n+2)^kB_k$$ $B_k$ is the Bernoulli number. ${n\brack {k}}\;$ is the unsigned Stirling number of first kind, $\left( {0\brack {0}}\;=1 \text{ and }{{n}\brack {k}}\;=(n-1){{n-1}\brack {k}}\;+{{n-1}\brack {k-1}}\;\right)$.

From $n=0$, the first terms are: $ \ \ 1\ ,\ 0\ ,\ -5\ ,\ 0\ ,\ 238\ ,\ 0\ ,\ -51508\ ,\ 0\ ,\ 35028576\ , ..$

The $a_n$ seem to be all integers, and the odd-index $a_{2n+1}$ seem to vanish. I could not find proofs for these statements though. A generating function should be even, but I could not find it either.

Also, any possible combinatorial interpretation (when removing the sign)?

I would welcome any help or indication on this. Thank you in advance.

EDIT(23/10/17): I have found a proof that they are all integers. But that they would be zero when the index is odd is still waiting for a proof.