Here is the problem:
Let $f$ be a function taking the nonnegative integers to the positive integers such that $f(0) = f(1) = 1$ and $\frac{f(n) f(m + 1)}{f(m)} + \frac{f(n) f(n - m)}{f(n - m - 1)} = f(n + 1)$ for all nonnegative integers $n$ and $m,$ where $n \ge m + 1.$
Find the smallest nonnegative integer $n$ such that $f(n) > 10^6.$
Plugging in $m=0,$ I got $$f(n) + \frac{f(n)^2}{f(n-1)} = f(n+1).$$
Usually in these types of problems a few substitutions do the trick, but I can't see any other substitutions for $m$ or $n$ that could reduce the problem.
Thanks in advance.
If $g(n) =f(n+1)/f(n)$ then $g(0)=f(1)/f(0)=1$ and
$$g(m) +g(n-m-1)= g(n)$$
so for $n=m+1$ we have $g(m)+1 =g(m+1)$ so $g(n) = n$ for all $n$. (use induction for the prove)
So $f(n+1)=n\cdot f(n)$ and thus $f(n) =(n-1)!$ for all $n\geq 1$ again by induction and $f(0)=1$ .