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I'm struggling to get the following recurrence relation into a closed form if possible:

$$f(m,0)=f(0,n)=2$$ $$f(0,0)=0$$ $$f(m,n)=f(m-1,n)\cdot(2n+1)(2n+2) + f(m,n-1)\cdot(2m+1)(2m+2)$$

where $m$ and $n$ are non-negative integers. This recurrence relation came from one combinatorics problem and I 'd like to ask how to derive the explicit formula for $f(m,n)$.

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  • $\begingroup$ check out math.stackexchange.com/questions/206158/… $\endgroup$ – sku May 23 '18 at 5:05
  • $\begingroup$ actually I've got the variable in my coefficients, so kind of different from the above. $\endgroup$ – Mclalalala May 23 '18 at 6:37
  • $\begingroup$ @sku I mean is there a way to solve this? $\endgroup$ – Mclalalala May 23 '18 at 23:54
  • $\begingroup$ In the world of single variable recurrence if the coefficient has $n$ then we use exponential GF. May be something like that can help in this case. $\endgroup$ – sku May 23 '18 at 23:57

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