I'm trying to reinvent the wheel here by deriving the formula for Dyck Words of length p+q, that is, p left parens and q right parens. The answer of course is $\binom{p+q}{q} - \binom{p+q}{q-1}$.
Using an OGF, if I'm right, starting from the recurrence $c_{p,q} = c_{p-1,q} + c_{p,q-1}, \quad q \leq p$ and letting $c(x,y) = \sum_{p=0} \sum_{q\leq p}c_{p,q}x^p y^q$ I should get $c(x,y)-1 = x \times c(x,y) + y \times f(x,y)$.
It is this $f(x,y)$ that is troubling me. Reverse engineering the answer it seems to me that I need $c(x,y)(1-x-y)=1-y/x$ which would have come from $c(x,y)-1 = x \times c(x,y) + y \times (c(x,y)-1/x)$. I don't see how this could be.