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.