Considering the success of a [previous question](https://mathoverflow.net/questions/29118/need-help-proving-that-sum-limits-j0k-1-1j1k-j2k-2-binom2k1) 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:


1. $\sum\_{m=0}^{n}
 (-1)^mm!(2n-m-2)!\left\langle\left\langle
 n\atop
 m\right\rangle\right\rangle=0$ for
all $n$;
2. $\sum\_{m=0}^{n}
(-1)^m(m)m!(2n-m-2)!\left\langle\left\langle
n\atop
m\right\rangle\right\rangle=0$ for
odd $n$;
3. $\sum\_{m=0}^{n}
(-1)^m(m+1)!(2n-m)!\left\langle\left\langle
n\atop
m\right\rangle\right\rangle=0$ 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 $n$.



Expressions involving Eulerian numbers of the second kind: trying to 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 $n$. 
Following Mike Spivey's comment above I will consider $B(n)$ in (3). It turns out that your conjecture is true, because for odd $n$ the sum $B(n)$ is a certain weighted $L^2$ norm of the Eulerian polynomial of the second kind of order $\frac{n+1}{2},$ with the sign of $(-1)^{\frac{n-1}{2}}\, . $


The product of the two factorials in $B(n)$ may be expressed in terms of the Eulerian Beta integral
$$(m+1)!(2n−m)!=(2n+2)!\int\_0^1 t^{m+1 }(1-t)^{2n-m}dt,$$
so that dividing it by $(2n+2)!
$ we have
$$ \frac {B(n)}{(2n+2)!}= \int\_0^1 \sum\_{m\ge0}(-1)^m\left\langle\!\!\left\langle n\atop m\right\rangle\!\!\right\rangle t^{m+1 }(1-t)^{2n-m}dt = $$
$$= \int\_0^1 t(1-t)^{2n}\sum\_{m\ge0} \left\langle\!\! \left\langle n\atop m\right\rangle\!\! \right\rangle \Big(\frac{t}{t-1}\Big)^{ m} dt= \int\_0^1 t(1-t)^{2n} E\_n \Big(\frac{t}{t-1}\Big) dt. $$
Changing variable with $x:=\frac{t}{t-1}$ this becomes:
$$ \int\_{-\infty}^0 x(x-1)^{-2n-3} E\_n(x) dx, $$


where $E\_n$ denotes the Eulerian polynomial of the second kind
$$E\_n(x):=\sum\_{m\ge0} \left\langle\!\!\left\langle n\atop m\right\rangle\!\! \right\rangle x^m,$$
and satisfies the recursive relation (corresponding to the relation for the coefficient that you gave in your question):
$$(x-1)^{-2n-2}E\_{n+1}(x)=\left( -x(x-1)^{-2n-1}E\_n(x) \right)^{\prime}.$$
By the above formula it is now easy to show, integrating by parts repeatedly, that $B(n)=0$ for even $n$ while for odd $n=2p-1$ 
$$B(2p-1)=(-1) ^ { p + 1 } (4p)! \int\_0^{+\infty} E\_p (-x)^2 (x+1)^{-4p-1} x dx . $$ 


(To check this, it is convenient to introduce the sequence of rational functions $U\_ n(x):= (x-1)^{-2n}E\_n(x)$ that satisfy the recurrence $U\_ {n+1}= \big (\frac{x}{1-x} U\_ n \big) ^ {\prime} $ with initial condition $U\_0=1.$ Hence for all $n+m>0$ we have $\int\_{-\infty}^0 U\_{n+1}U\_{m}\frac{x}{1-x}dx=-\int\_{-\infty}^0 U\_ {n}U\_ {m+1}\frac{x}{1-x}dx\, .$ The integral found above for $B(n) / (2n+2)!\, $ was $-\int\_{-\infty}^0 U\_{n}U\_{1}\frac{x}{1-x}dx$ ). 


