So here is the tree. For given $\left \{ m,n \right \}$. {m,n} will transform to give these elements which I will represent using a summation operator;
$$\sum^{n}_{k=1}\left \{ m-k,k \right \}$$
Take {9,3}. So In our example it becomes, $\left \{ 8,1 \right \},\left \{ 7,2 \right \},\left \{ 6,3 \right \}$. And with this we can continue on to look like this;
The Rules here are that, when either the $m$ or the $n$ values becomes $0$ or $1$, end the transformation. For example here, it ends at {8,1},{6,1},{4,1},..{0,2} etc. and there are $12$ of them.
The problem here is that, given $m,n$ where $n≤m$. Find the number of times the transformation "tree" stops.

Just a comment, but it won't fit in a comment box. I calculated the values for $S(m,3),\ 0\leq m<100$. Note that I did get $S(9,3)=12$ as shown in the diagram.
EDIT
I've noticed that we always have $S(m,3)=m+S(m-6),\ m\geq6$ so there will definitely be recurrence relations though there may be $6$ different ones.