Find two-index Inverse relations Solve the following system of $ mn$ equations
$$
a_{p,q}=\sum_{i=0}^m \sum_{j=0}^n i^p j^q b_{i,j}, p=0 \ldots m, q=0 \ldots n. 
$$
where $b_{i,j}$ are unknowns.
Of course, for small $m,n$, it is possible to solve it by hand by using Gauss elimination, but what about the general case? I hope there exists a nice close expression for $b_{p,q}$ something like to
$$
b_{p,q}=\sum_{i=0}^n \sum_{j=0}^m X^{(p,q)}_{i,j} \, a_{i,j}, p=0 \ldots m, q=0 \ldots n. 
$$
Numerical experiments for small $n,m$ show that it is very likely that the Stirling numbers of the first kind should be involved in the expression for $X^{(p,q)}_{i,j}$
EDIT.  I checked for some small $n,m$  that the following holds
$$
b_{0,0}=- \frac{1}{n! m! } \sum_{i=0}^m \sum_{j=0}^n s(m+1,i+1) s(n+1,j+1) a_{i,j}, 
$$
where $s(n,i)$ is the signed Stirling numbers of the first kind.
For other cases it seems that we are dealing with some generalizations of the Stirling numbers.
 A: This got too long for a comment.
Certainly the $X_{i, j}^{(p, q)}$ do exist. They're just the entries of the inverse of the $mn \times mn$ coefficient matrix $(i^p j^q)_{(p, q), (i, j)}$. This matrix is indeed invertible since it's the Kronecker product of $(i^p)_{p, i}$ and $(j^q)_{q, j}$, both of which are invertible by the usual Vandermonde determinant formula.
In fact, the inverse of the Kronecker product is the Kronecker product of the inverses, so you just need to know the inverse of the Vandermonde specialized at $x_i=i$. It seems likely that the most convenient explicit approach is this one, which expresses the inverse Vandermonde as a product of upper and lower triangular matrices. In the "Properties of Factor Matrices" section they specialize to your case, where it turns out the upper triangular factor's coefficients indeed involve Stirling numbers of the first kind. Surely your conjectured formula is a special case.
So, you can indeed get a very explicit general formula, though if your goal is computation I'd expect a general purpose matrix inversion routine will be faster. (If you write out the formula, feel free to edit this with it for the benefit of future searchers.)
