We have for $k>0$, $n>0$, $m\geqslant0$ $$p_k(n,m)=k(n-1)!\sum\limits_{s=0}^{n-1}\frac{p_{k-1}(s+1,m+1)+p_{k-1}(m+1,s)}{s!}$$ also $$p_0(n,m)=\begin{cases} (n-1)!,&\text{$n>0, m=0$}\\ 0,&\text{$n>0, m>0$} \end{cases}$$ so the closed form $$p_{k}(n,m)=k!\sum\limits_{s=0}^{k-1}\binom{n+\left\lfloor\frac{k+s}{2}\right\rfloor-1}{n+s-1}\binom{m+\left\lfloor\frac{k+s-1}{2}\right\rfloor}{m+s}\frac{(n+s-1)!(m+s)!}{s!}$$ But how can one derive it from recurrence?
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1$\begingroup$ Some reference? $\endgroup$– Konstantinos KanakoglouCommented Sep 26, 2019 at 22:56
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1$\begingroup$ Substituting $q_k(n,m)=\frac{p_k(n,m)}{k!(n-1)!m!}$ results in a simpler recurrence: $$q_k(n,m) = \sum_{s=0}^{n-1} (m+1)q_{k-1}(s+1,m+1) + q_{k-1}(m+1,s).$$ $\endgroup$– Max AlekseyevCommented Sep 27, 2019 at 0:24
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