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Let $$P_m(x,z) = \prod_{k=0}^{m-1} (p(x-k)+q(x-k)z)$$ and $$\mathcal{A_n}(x,z) = \sum_{k=0}^{\infty} A(n,x-k) z^k.$$

Then unrolling the given recurrence $m$ times, we get that $A(n,x)$ equals the coefficient of $z^m$ in $$P_m(x,z)\cdot \mathcal{A}_{n-m}(x,z).$$ In particular, for $A(n,x)$ equals the coefficient of $z^n$ in $$P_n(x,z)\cdot \mathcal{A}_{0}(x,z).$$

More could be said if the boundary constraints were given.

Post Undeleted by Max Alekseyev
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Unrolling

Let $$P_m(x,z) = \prod_{k=0}^{m-1} (p(x-k)+q(x-k)z)$$ and $$\mathcal{A_n}(x,z) = \sum_{k=0}^{\infty} A(n,x-k) z^k.$$

Then unrolling the given recurrence $m$ times, we get : that $$A(n,x) = \sum_{k=0}^m A(n,x) equals the coefficient of z^m in$$P_m(x,z)\cdot \binom{m}{k} A(n-m,x-k) p(x)^k q(x)^{m-k}.$$mathcal{A}_{n-m}(x,z).$$ In particular, for $m=n$ we have A(n,x)\$ equals the coefficient of $$A(n,x) = \sum_{k=0}^n z^n in$$P_n(x,z)\cdot \binom{n}{k} A(0,x-k) p(x)^k q(x)^{n-k}.$$mathcal{A}_{0}(x,z).$$

Post Deleted by Max Alekseyev
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