I'm interested in solving a particular non-linear recurrence in two variables:


$$\lambda\_{j,k} = (j+k) \lambda\_{j,k-1} + \begin{pmatrix} j+k \\ 2 \end{pmatrix} \lambda\_{j-1, k-1}.$$ 


Here $j \geq -1$ and $k \geq 0$, and we have initial conditions:


$\lambda\_{-1,k} = 0$ for all $k$;


$\lambda\_{j,0} = 0$ for all $j>0$;


$\lambda\_{0,0} = 1$.


This is a relation between the leading coefficients of certain polynomials occurring in a problem in modular invariant theory. $\lambda\_{-1,k}$ doesn't really make sense in context, but introducing it simplifies the relation and initial conditions nicely. Because of the context, I happen to know that for any prime $p$, $\lambda\_{j,k} \neq 0$ mod $p$ unless $j+k \geq p$. This suggests to me that there is a particularly simple solution, possibly a product of binomial coefficients.


Some starting points: Obviously $\lambda\_{0,k} = k!$ and it's easy to show that $\lambda\_{j,k} = 0$ if $j>k$ and $\lambda\_{k,k} = \frac{(2k)!}{2^k}$ for $k \geq 0$. I've tried solving it with generating functions in two variables, but that just produces a truly terrifying PDE of order 2 and degree 4.



A non-linear recurrence relation in two variables
Denote $\lambda\_{j,k}=(j+k)! a\_{j,k}$ (unless $j=-1,k=0$). Then we get $a\_{j,k}=a\_{j,k-1}+a\_{j-1,k-1}/2$. Further denoting $a\_{j,k}=2^{-j}b\_{j,k}$ we get $b\_{j,k}=b\_{j,k-1}+b\_{j-1,k-1}$ that looks like a Pascal triangle recurrence. So $b\_{j,k}=\binom{k}j$ (check the initial conditions also) and $\lambda\_{j,k}=2^{-j}(j+k)!\binom{k}j$.


