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I'm trying to solve the following partial difference equation with variable coefficients: $$a_{i,j,k}=-\frac{j+1}{i}a_{i-1,j+1,k-1}- \frac{k+1}{i}a_{i-1,j-1,k+1}, $$ defined on the grid $(i,j,k) \in \mathbb{N}_0^3$ (we take $a_{i,j,k}=0$ if any of the subscripts is negative), with the initial conditions $$a_{0,j,k}:=\alpha_{j,k}. $$ I was surprised to be able to solve this explicitly as follows

$$a_{i,j,k}=\sum_{m+n=j+k} \frac{\alpha_{m,n}}{2^{j+k}} \sum_{r=0}^{j+k} \frac{ \left(2r-j-k \right)^i}{i!} \sum _{p=0}^r (-1)^p \binom{m}{p} \binom{n}{r-p} \sum _{q =0}^j (-1)^{q } \binom{r}{q } \binom{j+k-r}{j-q}. $$

I was even luckier to stumble across this result, which expresses these alternating sums in terms of the hypergeometric function $_2F_1$. Using that fact I've managed to "sum up" the two innermost $\Sigma$s:

$$a_{i,j,k}=\sum_{m+n=j+k} \frac{\alpha_{m,n}}{2^{j+k}} \sum_{r=0}^{j+k} \frac{ \left(2r-j-k \right)^i}{i!} \binom{j+k-r}{j} \, _2F_1(-j,-r;k-r+1;-1) \binom{n}{r} \, _2F_1(-m,-r;n+1-r;-1). $$

Notice however, that for some values of $r$ (e.g. $r=n+1$) we get the indeterminate form $0 \times \infty$. In such cases the formula should be interpreted as taking the limit $r$ approaches that problematic value.

I'm wondering if this is the best one can do for this problem, and I would love to see a neater closed form if such exists.

Thank you!


P.S.

I've been looking for some identities of the $_2F_1$ function, and found this one, which involves a sum of products of a power a binomial coefficient and a hypergeometric function. Although it doesn't exactly fit the formula of the solution, it makes me hope that a useful identity exists.

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