I am interested in a weighted version of the Catalan numbers. The generating function for this case, $$ f(x, y) = \sum_s \sum_n f_{s n} x^s y^n $$ (where the $y^n$ term is the weight), obeys the relation $$ f(x, y) = 1 + x y f(x, y) f(xy, y) $$

Main question:

Is there any hope of solving for $f(x, y)$ in closed form? Alternately, I'd be happy to get a closed form for the coefficients of the series expansion in $x$ of $f(x, y)$ at $x=0$ and a particular value of $y$, where $0 < y < 1$. An exact expression is not required; a good approximation will serve just as well.

Comments/thoughts:

-I haven't encountered any methods for solving such a problem before, so I don't know if there is a name for such problems. I apologize if the title and keywords are uninformative.

-Obviously when $y=1$, this reduces to the ordinary Catalan numbers.

-I believe that it may be possible to turn this into a coupled set of PDEs by defining $g(x, y) = f(xy, y)$. However, this seems to make the problem more complicated rather than less so.

-Since $f(xy, y)$ is known in terms of $f(x, y)$, we may be able to learn about the series at $x=0$ by examining the sequence $f(x y^m, y)$ as $m \to \infty$. However, I haven't been able to make this idea work in any concrete way yet.

Any ideas are appreciated. Thanks!

Edit: As Mark noted, this comes from trying to solve a recurrence relation of the form $$ f_{sn} = \sum_{s_1 + s_2 + 1 = s} \sum_{n_1 + n_2 + s_1 + 1 = n} f_{s_1 n_1} f_{s_2 n_2} $$ This arises when considering the first-return times in a one-dimensional discrete random walk where the probability of moving to the right decreases exponentially with position.

numeratorsof each of the fractions, rather than the usual form of expressions as the constants and 1 as all the numerators. I thought it would be easy to go between the two forms, but I don't see how. $\endgroup$