Suppose I have known polynomials $p_i, i=0\ldots k-1$. I have the following horrific looking recursive equation: $$p_{k}(x)=-\prod_{i=1}^{k}{(x+i)}^{\left\lfloor\frac{k}{i}\right\rfloor-1}\left[p_o(x)+\sum_{j=1}^{k-1}\prod_{i=1}^{j}\left(\frac{x+k+1-i}{(x+i)^{\left\lfloor\frac{j}{i}\right\rfloor}}\right) p_j(x)\right]$$ $$p_{k}(x)=-\prod_{i=1}^{k}{(x+i)}^{\left\lfloor\frac{k}{i}\right\rfloor-1}\left[p_o(x)+\sum_{j=1}^{k-1}\left(\frac{\prod_{i=1}^{j}(x+k+1-i)}{\prod_{i=1}^{j}(x+i)^{\left\lfloor\frac{j}{i}\right\rfloor}}\right)p_j(x)\right]$$
This looks crazy. Is there any way to modify this kind of expression to get something more "usable"?
This arises out of the study of hypergeometric Bernoulli numbers where
$$\frac{x^N/N!}{e^x-1-x-\cdots.-(x^{N-1}/(N-1)!)}=\sum_{k=0}^{\infty}\frac{B_k(N)}{k!}x^k$$ and where the $B_k(N)$ are the hypergeometric Bernoulli numbers. If $N=1$, these are the 1st order Bernoulli numbers, $B_0=1, B_1=-1/2, B_2=1/6,$ etc...
It appears that the numbers have a particular structure, namely
$$B_k(N)=\frac{k!p_k(N)}{\prod_{j=1}^{k}(N+i)^{\left\lfloor\frac{k}{i}\right\rfloor}}$$
The first few $p_k$ are $$p_0=1 \text{ (trivially) }$$ $$p_1=-1$$ $$p_2=1$$ $$p_3=x-1$$ $$p_4=x^3-x^2-6x+2$$ $$p_5=x^4-4x^3-11x^2+16x-2$$
NOTE: It appears that $p_{2m+1}(x)=(x-1)q(x)$ for some $q\in \mathbb{Z}[x]$. This makes sense since the odd indexed Bernoulli number of order 1 are zero.
$$B_{2k+1}(1)=\frac{(2k+1)!p_{2k+1}(1)}{\prod_{i=1}^{2k+1}{(1+i)^{\left\lfloor\frac{2k+1}{i}\right\rfloor}}}=\frac{(2k+1)!(1-1)q_{2k+1}(1)}{\prod_{i=1}^{2k+1}{(1+i)^{\left\lfloor\frac{2k+1}{i}\right\rfloor}}}=0$$
I checked mathematica for the first few and got aforementioned results.
$$p_5=(x-1)(x^3-3x^2-14x+2)$$
$$p_7=(x-1)(x^7-7x^6-81x^5-37x^4+766x^3+1048x^2-390x+12)$$