Generating function of a polynomial sequence Using Wolfram Alpha, I find that the first 6 members $p_j(x)$, $0\leq j\leq 5$, of the polynomial sequence happen to be the first 6 non-zero coefficients of the Maclaurin series of 
\begin{eqnarray}\left(\frac{t}{\sinh(t)}\right)^x.\end{eqnarray}
I conjecture that
\begin{eqnarray}\left(\frac{t}{\sinh(t)}\right)^x=\sum_{j=0}^\infty p_j(x)t^{2j}.\end{eqnarray}
Is that true? Why?
 A: Let $f_x(t)=\left(\dfrac{t}{\sinh t}\right)^x$. It is clear that
$$f_x(t)=\exp(xg(t))\quad\hbox{ with}\quad g(t)=\ln\left(\dfrac{t}{\sinh t}\right)\tag1$$
Now, we have the well-known expansion, (see here):
$$g'(t)=\frac{1}{t}-\coth t=-\sum_{k=1}^\infty\frac{2^{2k}b_{2k}}{(2k)!}t^{2k-1}\tag2$$
in the disk ${\mathbb D}=D(0,\pi/2)$. It follows that $f_x$ is analytic in ${\mathbb D}$. Moreover, since $f_x$ is an even function, we conclude that $f_x$ has
in ${\mathbb D}$ a power series expansion of the form
$$
f(t)=\sum_{n=0}^\infty p_nt^{2n}
$$
In particular, we have clearly, $p_0=1$. So, we only need to show that the $p_n$'s are in fact polynomials in $x$.
We will show more. Indeed, from $(1)$ we have
$f_x'(t)=xg'(t)f_x(t)$, and in terms of power series this yields
$$
\sum_{n=0}^\infty 2n p_nt^{2n-1}=-x\left(\sum_{k=1}^\infty\frac{2^{2k}b_{2k}}{(2k)!}t^{2k-1}\right)\left(\sum_{m=0 }^\infty p_mt^{2m}\right)
$$
considering the coefficient of $t^{2n-1}$ on both sides we get
$$
p_{n}=-x\sum_{k+m=n,\,k\ge1}\frac{2^{2k}b_{2k}}{(2k)!} p_m
=\frac{-x}{2n}\sum_{k=1}^n\frac{2^{2k}b_{2k}}{(2k)!} p_{n-k}\tag3
$$
The importance of $(3)$ is that it defines $p_n$ as a linear combination of
$p_0,\ldots,p_{n-1}$ multiplied by $x$. In particular, we can prove by a simple inductive argument that $p_n$ is a polynomial of degree $n$ with leading term equal to $(-1)^n\dfrac{x^n}{n!\cdot 6^n}$. 
Next let us prove that the obtained sequence $(p_n)_n$ satisfies the desired conditions here. Indeed, note that 
$Q_{2m}(x,y)$ is the coefficient of $t^{2m}$ in the power series expansion of $t\mapsto f_y(t)\cosh(xt)$, and
$Q_{2m+1}(x,y)$ is the coefficient of $t^{2m+1}$ in the power series expansion of  $t\mapsto f_y(t)\sinh(xt)$.
In particular, we have the following integral representations
$$\eqalign{
Q_{2m}(x,y)&=\frac{1}{2\pi i}\int_{C^+(0,1)}\frac{f_y(z)\cosh(xz)}{z^{2m+1}}dz\cr
Q_{2m+1}(x,y)&=\frac{1}{2\pi i}\int_{C^+(0,1)}\frac{f_y(z)\sinh(xz)}{z^{2m+2}}dz}
$$
In particular, 
$$\eqalign{
Q_{2m}(x,2m+1)&=\frac{1}{2\pi i}\int_{C^+(0,1)}\frac{\cosh(xz)}{(\sinh z)^{2m+1}}dz\cr
Q_{2m+1}(x,2m+2)&=\frac{1}{2\pi i}\int_{C^+(0,1)}\frac{\sinh(xz)}{(\sinh z)^{2m+2}}dz}
$$
Now, to finish the argument, note that if $\ell\in\{0,1,\ldots,m-1\}$ then $$\cosh((2\ell+1) z)=T_{2\ell+1}(\cosh z)$$ where
$T_{2\ell+1}(X)=\sum_{j=0}^{\ell}\mu_jX^{2j+1}$ is an odd polynomial (it is fact the Chebyshev polynomial of the first kind). So
$$\cosh((2\ell+1) z)=(\cosh z) \sum_{j=0}^{\ell}\mu_j(1+\sinh^2z)^j
=(\cosh z) \sum_{j=0}^{\ell}\lambda_j\sinh^{2j}z
$$
Thus $$\int_{C^+(0,1)}\frac{\cosh((2\ell+1)z)}{(\sinh z)^{2m+1}}dz=\sum_{j=0}^{\ell}\lambda_j\int_{C^+(0,1)}\frac{\cosh z}{(\sinh z)^{2(m-j)+1}}dz$$
But these last integrals are all zeros because the integrands have simple primitives.
We conclude that $Q_{2m}(2m+1-2i,2m+1)=0$ for $i=0,1,\ldots,2m$.
Similarly, since $\dfrac{\sinh((2\ell+2) z)}{\sinh z}=2\sum_{j=0}^\ell
\cosh((2j+1)z)$ we conclude using the previous result that
$Q_{2m+1}(2m+2-2i,2m+2)=0$ for $i=0,1,\ldots,2m+1$. This proves that the sequence $(Q_k(x,y))_k$ obtained from the $p_n$'s satisfies the conditions of
the previous  question.
