TITLE: A polynomial sequence
QUESTION [10 upvotes]: I have a sequence of polynomials $Q_k(x, y)$, $k\geq 1$ defined recursively as follows:

$Q_1=x$.
There is a sequence of polynomials $p_j(y)$ of degree $j$ such that $Q_{2m}$ is of the form
\begin{eqnarray}\frac{p_{0}(y)}{(2m)!}x^{2m}+\frac{p_{1}(y)}{(2m-2)!}x^{2m-2}+\cdots+\frac{p_{m-1}(y)}{2!}x^2+p_{m}(y)\end{eqnarray}
and $Q_{2m+1}$
\begin{eqnarray}
\frac{p_{0}(y)}{(2m+1)!}x^{2m+1}+\frac{p_{1}(y)}{(2m-1)!}x^{2m-1}+\cdots+\frac{p_{m-1}(y)}{6}x^3+p_{m}(y)x
\end{eqnarray}
$Q_k(k+1-2i, k+1)$ as a polynomial in $i$ has roots $1, 2, \cdots, k$. 

Find a general formula of $Q_k(x, y)$. The following are the first 4 polynomials in the sequence:
\begin{align*}
Q_1&=x\\
Q_2&=\frac{x^2}{2}-\frac{y}{6}\\
Q_3&=\frac{x^3-xy}{6}\\
Q_4&=\frac{x^4-2x^2y}{24}+\frac{y(5y+2)}{360}
\end{align*}
In fact, it suffices to find the sequence of polynomials $p_j(y)$. Here's what I've got so far. By condition (2) and (3), we have
\begin{eqnarray}
\sum_{j=0}^m \frac{p_j(2m+1)}{(2m-2j)!}(2i-2m-1)^{2m-2j}=\frac{2^{2m}}{(2m)!}(i-1)(i-2)\cdots(i-2m)
\end{eqnarray}
If we let $m=j+k$, differentiate both sides with respect to $i$ $2k$ times and put $\displaystyle i=\frac{2j+2k+1}{2}$, we have
\begin{eqnarray}
p_j(2j+2k+1)=\left.\frac{d^{2k}}{di^{2k}}\right|_{i=\frac{2j+2k+1}{2}}\frac{2^{2j}}{(2(j+k))!}(i-1)(i-2)\cdots(i-2(j+k))
\end{eqnarray}
Letting $k=0, 1, \cdots, j$, we get the $j+1$ values taken by $p_j$ at $2j+1, 2j+3, \cdots, 4j+1$. $p_j$ then can be computed using Lagrangian interpolation. 
It seems to me that, though the above algorithm can be implemented on a computer to get a few polynomials in the sequence, it does not yield directly a general formula I want. Is there another better way to go about getting a general formula?
Edit: The first 6 members in the polynomial sequence $p_j(y)$ are the following:
\begin{align*}
p_0(y)&=1\\
p_1(y)&=-\frac{y}{6}\\
p_2(y)&=\frac{y(5y+2)}{360}\\
p_3(y)&=-\frac{y(35y^2+42y+16)}{45360}\\
p_4(y)&=\frac{y(5y+4)(35y^2+56y+36)}{5443200}\\
p_5(y)&=-\frac{y(385y^4+1540y^3+2684y^2+2288y+768)}{359251200}
\end{align*}
It is noteworthy that the denominators happen to be the first 6 numbers in this sequence, as pointed out by Solomonoff's Secret.

REPLY [2 votes]: 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}