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 $n$.  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 $n$.
 A: Surprise: your expression is a multiple of a Bernoulli number:
$$A(n)=-\frac{(2n)!}{n}B_n $$
for all $n>1$, and the Bernoulli numbers are indeed vanishing exactly for all odd integers greater than 1. This follows computing the generating function for the $A(n)$, on the lines of the preceding answer and comments. Note that your identity can be written in the form
$$\sum_{m=0}^n (-1)^n\frac{ \left\langle \!\!\left\langle n\atop  m\right\rangle \!\!\right\rangle }{\left( 2n+1\atop  m+1\right)}=2B_{n+1}\, .$$
After that, I had a vague memory of a similar relation about the Eulerian numbers of first kind, that it is the fourth identity reported here:
$$\sum_{m=0}^n (-1)^n\frac{ \left\langle n\atop  m\right\rangle }{\left( n\atop  m\right)}=(n+1)B_n\, .$$
Checking the source of the latter may be useful for you, and will possibly give a reference for your identity (I'm quite confident that it should be written somewhere; hopefully some expert may recognize and provide a reference). In case, I will  add the details of the   (quite standard)  computation of the generating function of yours $A(n)$ .
Details. Here is a generating function for yor sequence as a real function. This should allow a nice proof of the Bernoulli number formula, provided one is able to compute the last integral.
For $t\in\mathbb{R}$ and $x\in [0,\infty)$ let $$\psi(t,x):=W(x e^{x+t})=L^{-1}\big(t+L(x)\big)$$
where $W$ is the Lambert function and $L:\mathbb{R_+}\to\mathbb{R}$ is the invertible function $L(x):=x+\log(x)$. So $\psi$ is the general solution of the Cauchy problem for the autonomous ODE (the flow)
$$\psi_t=\frac \psi {1+\psi} $$
$$\psi(0,x)=x$$ 
Thus it also solves the linear first-order PDE
$$\psi_t- \frac x {1+x} \psi_x =0$$
and  all derivatives w.r.to $t$ have the form
$$\partial_t^n\psi= v_n(\psi)$$
for a recursively defined sequence $v_n(x)$ 
$$v_0(x)=x$$
$$v_{n+1}(x)=\frac x {1+x}v'_n(x)$$
So the $v_n$ are rational functions with poles at $-1$; $v_n(x)=O(x^{-n})$ as $x\to+\infty$ and in fact
$$v_n(x)=-x(1+x)^{-2n+1}E_n(-x)$$
where $E_n$ are the Eulerian polynomial of second kind (these $v_n$ are just a simple modification of to the previously defined sequence $U_n$ ).
For all $t\in\mathbb{R}$, all $r>0$ and $n\ge2$, it's easy to see that
$$\sup_{|t|\le r}\, \big| v_n\big(\psi(t,x)\big) \big|:=g_{r,n}(x)\in L^1(\mathbb{R}_+),$$
which allows (by the dominated convergence theorem) to differentiate under the sign of integral the function
$$h(t):=\int_0^\infty \psi_{tt}(t,x)dx=\int_0^\infty v_2\big(\psi(t,x)\big)dx\, ,$$
so that 
$$h^{(n)}(t)=\int_0^\infty v_{n+2}\big(\psi(t,x)\big)dx,$$
and in particular we have for $n\ge 1$
$$h^{n-1}(0)=\int_0^\infty v_{n+1}(x)dx=-\frac {A(n)}{(2n)!}$$
The relation $$-\frac {A(n)}{(2n)!} = \frac {B_n} n$$ for all $n\ge 1$ now writes:
$$\int_0^\infty \frac{W(xe^{x+t})}{\big(1+W(xe^{x+t})\big)^3}dx=\frac1 t - \frac 1 {e^t - 1}.$$
A: 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  $-\int_{-\infty}^0 U_{n}U_{1}\frac{x}{1-x}dx$  ). 
