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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?

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    Very similar question: http://math.stackexchange.com/questions/9548/general-expression-of-fa-b-if-fa-b-fa-1-b-fa-b-1-fa-1-b-12011-06-04

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