Closed form for $q_k(0,0)$ from recurrence

Question

We have $$p_0(n,m)=\begin{cases} 0,&\text{$n=m=0$}\\ (n-1)!,&\text{$n>0, m=0$}\\ 0,&\text{$n\geqslant0, m>0$} \end{cases}$$ $$q_0(n,m)=0, n\geqslant0, m\geqslant0$$ and $$p_k(n,m)=\begin{cases} 0,&\text{$n=0, m\geqslant0, k>0$}\\ \sum\limits_{d=0}^{n-1}\frac{(n-1)!}{d!}q_k(d,m),&\text{$n>0, m\geqslant0, k>0$} \end{cases}$$ $$q_k(n,m)=p_{k-1}(n+1,m+1)+p_{k-1}(m+1,n), n\geqslant0, m\geqslant0, k>0$$ so when $n\ne0$ for $p_k(n,m)$ and $q_1(n,m)$, and when $n\geqslant0$ for $q_k(n,m)$ with $k>0$, first three cases are $$p_1(n,m)=(n-1)!m!, q_1(n,m)=0$$ $$p_2(n,m)=2n!m!(m+2), q_2(n,m)=n!m!(m+2)$$ $$p_3(n,m)=3n!(m+1)!((n+1)(m+4)+2), q_3(n,m)=2n!(m+1)!((n+1)(m+4)+1)$$ also $$q_1(0,m)=m!, m\geqslant0$$ General interest is to find $q_k(0,0)$ for which we need $$q_k(0,0)=p_{k-1}(1,1)+p_{k-1}(1,0)=(k-1)(q_{k-1}(0,1)+q_{k-1}(0,0))$$ Is there closed form for it?