functions satisfying Vandermonde identity Suppose $f(m+n,r)=\sum_{k=0}^rf(m,k)f(n,r-k)$.
Then $f(n,k)$ could be $n\choose k$, or $f(n,k)$ could be $0$.
Could $f(n,k)$ be any other functions?
For non-negative integer function argument and function values?
For more general arguments and values?
 A: Well, the RHS is the Cauchy convolution of the generating function over the second parameter of $f$. So just pick your favorite family of functions $f_a(x)$ such that $f_{a+b}(x)=f_a(x)f_b(x).$ For example, every single exponent. So the binomials work because $f_n(x)=(1+x)^n$ and $0$ works because $f_n(x)=0$ but also for example $f_n(x)=e^{nx}$ will give you the sequence $f(n,m)=n^m/m!$.
A: Knuth wrote a paper Convolution Polynomials which explores this exact question (arXiv link). He gives several examples, including infinite families of examples. For example, for any $s\in \mathbb R$, the function $f_s$ satisfies your property, where
$$
f_s(n,k)=\frac{n(n-s)(n-2s)\cdots(n-(k-1)s)}{k!}
$$
Another infinite family is
$$
B_t(n,k)=\frac{n}{n+tk}\binom{n+tk}{k}
$$
There are many surprising examples, like
$$
n(n+k)^{k-1}\over k!
$$
The fact that this satisfies your convolution condition is equivalent to Abel's binomial theorem, which has connection to the combinatorics of trees on a finite set.
He also gives a complete characterization of functions satisfying $f(n+m,r)=\sum_k f(n,k)f(m,n-k)$ for which $f(n,k)$ is a polynomial in $n$ with degree at most $k$. Namely, any such sequence can be realized by taking a formal power series $F(z)=1+F_1z+F_2z^2+\dots$, expanding $F(z)^x$ as a formal power series in $x$ and $z$, letting $f(x,k)$ be the coefficient of $z^k$ in the expansion of $F(z)^x$.
Finally, $f(n,k)$ is a polynomial satisfying your condition if and only if the polynomial sequence $n\mapsto f(n,k)\cdot k!$ is a sequence of binomial type, so you can find more information there.
