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We have the following recurrence relation: \begin{equation} f(n,m) = f(n-1,m) g_{\alpha, \gamma}(n,m) + f(n,m-1) g_{\beta, \gamma}(n,m) \\ g_{\alpha, \gamma}(n,m)= \sum^{n}_{i = 0} \sum^{m}_{j = 0} \tbinom{i+j}{j} \tbinom{n-i+m-j}{m-j} \alpha^i \gamma^j \end{equation}

With the boundary conditions: \begin{equation} f(n,0) = \prod_{i=1}^n \frac{1-\alpha^i}{1-\alpha}, \ \ \ \ \ f(0,m) = \prod_{i=1}^n \frac{1-\beta^i}{1-\beta} \end{equation}

Is it possible to find the explicit solution for $f(n,m)$? If not, can we find the ratio $\frac{f(n+1,m)}{f(n,m)}$? If it still doesn't work, to what extent can we make this problem a solvable one? for example, let $\alpha = \beta$ or some other modifications.

Currently, the best I can do is in the $\alpha = \beta = \gamma$ limit, we can have: $$ f(n,m) = 2^{n+m} \prod^{n+m}_{i=1} \frac{1-\gamma^i}{1-\gamma}$$

in the $\alpha = \beta$ limit, we can have: $$ f(n,m) = 2^{n+m} \prod_{j=0}^{m} \prod_{i=0}^{n} g_{\alpha, \gamma}(i,j) $$

p.s. this problem maybe somehow related to this one: How to solve recurrence relation with 2 variables？

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  • $\begingroup$ The expression for $g_{\alpha,\gamma}(n,m)$ can be simplified into $$ g_{\alpha,\gamma}(n,m) = (n+m+1)\binom{n+m}n \int_0^1 (1+(\alpha-1)t)^n (1+(\gamma-1)t)^m\,{\rm d}t.$$ $\endgroup$
    – Max Alekseyev
    Commented May 10, 2023 at 1:39
  • $\begingroup$ Just to confirm - both terms in the right-hand side of the recurrence have the same coefficient $g_{\alpha,\gamma}(m,n)$, don't they? $\endgroup$
    – Max Alekseyev
    Commented May 10, 2023 at 2:20
  • $\begingroup$ @MaxAlekseyev yes, $g_{\beta, \gamma}(n,m)= \sum^{n}_{i = 0} \sum^{m}_{j = 0} \tbinom{i+j}{j} \tbinom{n-i+m-j}{m-j} \beta^i \gamma^j$ $\endgroup$
    – Lili Si
    Commented May 10, 2023 at 9:06
  • $\begingroup$ @MaxAlekseyev but if this is not solvable, we can also consider a little modified problem, like $ f(n,m) = f(n-1,m) g_{\alpha, \gamma}(n-1,m) + f(n,m-1) g_{\beta, \gamma}(n,m-1) $ $\endgroup$
    – Lili Si
    Commented May 10, 2023 at 9:11

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