How to solve the recursion:
$$p[n,m] = p[n-1,m-1] + p[n+1,m-1] + p[n,m-1]$$
Ideally in general, but if you need base cases:
$$p[n,0] = 0 \text{ (for } n \neq 0),$$ $$p[0,0] = 1$$
I've asked a similar question previously and believe $n^{m-2}$ is somehow involved.
Would turning this into a DFQ help?