* Let $a(n)$ be [A208832][1]. 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

Is there a way to prove it?


  [1]: https://oeis.org/A208832