Solving a bivariate recursion $f(k, n) = n^m f(k-n, m)$

Question

I have encountered an unknown object that seems to obey the following recursion, for any integers $k \geq n, m \geq 1$,

$$ f(k, n) = n^m f(k-n, m) $$

I think that $f(k, n) = 0$ is a solution to this (or is it?), but I am wondering if there are any other non-trivial solutions to this recursion, given the value $f(k, 1)$. (Or if it is possible to prove this has no non-trivial solution).

Thanks.

Answer 1

OK, so now with $m=1$, $f(k,n) = n f(k-n,1)$. In particular for $n=1$, $f(k,1) = f(k-1,1)$, i.e. $f(\cdot,1)$ is constant. If $c = f(1,1)$, $f(k,n) = n c$, and your equation then says $ n c = n^m m c$. Taking e.g. $m=2$ and $n > 0$, we see that the only possibility is $c=0$.