- Let $a(n)$ be A208832. Here $$ \frac{1}{1-x} = \sum\limits_{n=0}^{\infty}a(n)x^n\prod\limits_{k=1}^{n}\frac{1-kx}{1+kx}. $$
- Start with vector $\nu$ of fixed length $m$ with elements $\nu_i = 1$ (that is, $\nu = \{1, 1, \dotsc, 1\}$), reserve $t$ as an empty vector of fixed length $m$, set $t:=\nu$ and for $i$ from $1$ to $m-1$, for $j$ from $1$ to $m-i$ consecutively apply $$ \nu_{j+1} := i\nu_j + \nu_{j+1}, \\ \nu_{j} := i\nu_j + \nu_{j+1}. $$ We also need to apply $t_{i+1} = \nu_1$ (after ending each cycle for $j$).
I conjecture that after the whole transform we have $$ t_n = a(n). $$
Here is the PARI/GP program to generate $t$:
upto1(n) = my(v1); v1 = vector(n, i, 1); v2 = v1; for(i=1, n-1, for(j=1, n-i, v1[j+1] += i*v1[j]; v1[j] = i*v1[j] + v1[j+1]); v2[i+1] = v1[1]); v2
In addition, this question can be rephrased as follows:
- Let $$ R(n, q) = \begin{cases} 1 & \textrm{if } n = 0 \\ R(n-1, q+1) + 2\sum\limits_{j=0}^{q} n^{q-j+1}R(n-1,j) & \textrm{otherwise} \end{cases} $$
I conjecture that $$ R(n,0)=a(n+1). $$
Is there a way to prove it?
