I've been trying to solve the linear partial difference equation $$a_{i,j,k}=-\frac{j+1}{i} a_{i-1,j+1,k-1}-\lambda \frac{k+1}{i} a_{i-1,j,k+1}, $$ defined for $(i,j,k) \in \mathbb{N}_0^3$ (where $a_{i,j,k}=0$ if any of the subscripts is negative). The numbers $\{a_{0,j,k}\}_{j,k \in \mathbb{N}_0}$ can be thought of as the initial conditions.
I've found a formula involving the multinomial coefficients
$$a_{i,j,k}=(-1)^i \sum_{m=j}^{j+ \left\lfloor \frac{i+k}{2} \right\rfloor} a_{0,m,i+k-2m+2j} \lambda^{i-m+j} \sum_{r=0}^{m-j} \binom{m}{j,m-j-r,r} \binom{i+k-2m+2j}{k-r} 2^{r-m+j}, $$ and I've been wondering if a simpler form exists. Putting this into Mathematica gives
$$a_{i,j,k}=(-1)^i \sum _{m=j}^{\left\lfloor \frac{i+k}{2}\right\rfloor +j} 2^{j-m} \binom{m}{m-j} \lambda ^{i+j-m} \binom{i+2 j+k-2 m}{k} \, _2F_1(-k,j-m;i+2 j-2 m+1;2) a_{0,m,i+k-2m+2j}$$
which involves the hypergeometric function $_2F_1$. Does this simplify at all? Am I missing some binomial/hypergeometric identities here?
P.S. I don't expect to "sum-up'' this final $\Sigma$, since the initial conditions are independent numbers.
Thanks!