Considering the success of a previous question involving Eulerian numbers, I thought I might throw this question into the mix. It comes from some localization computations in GW theory, but in this form is purely combinatorial.

Eulerian numbers of the second kind are defined by the recursion relation

$$\left\langle\left\langle n\atop m\right\rangle\right\rangle = (m+1)\left\langle\left\langle n-1\atop m\right\rangle\right\rangle+(2n-m-1)\left\langle\left\langle n-1\atop m-1\right\rangle\right\rangle$$

with the initial conditions $\left\langle\left\langle n\atop 0\right\rangle\right\rangle=1$ and $\left\langle\left\langle n\atop m\right\rangle\right\rangle$ = 0 for $m\geq n$. For references, see:

http://en.wikipedia.org/wiki/Eulerian_number and http://oeis.org/classic/A008517.

The following three statements are known:

$\sum_{m=0}^{n} (-1)^mm!(2n-m-2)!\left\langle\left\langle n\atop m\right\rangle\right\rangle=0$ for all $n$;

$\sum_{m=0}^{n} (-1)^m(m)m!(2n-m-2)!\left\langle\left\langle n\atop m\right\rangle\right\rangle=0$ for odd $n$;

$\sum_{m=0}^{n} (-1)^m(m+1)!(2n-m)!\left\langle\left\langle n\atop m\right\rangle\right\rangle=0$ for even $n$;

(2 and 3 are equivalent).

Question: Show that the expression in the second statement is non-zero for even $n$, i.e. show

$\sum_{m=0}^{n} (-1)^m(m)m!(2n-m-2)!\left\langle\left\langle n\atop m\right\rangle\right\rangle\neq0$ for even $n$.

Certainly we've been checking this on a computer for modest values of $n$.