Solving for f given constraint involving f(x, y) and f(xy, y) 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.
 A: It seems experimentally that
\begin{align*}
 f_{ss} &= 1 & \text{ for } s \geq 0 \\
 f_{s,s+1} &= s-1 & \text{ for } s \geq 1 \\
 f_{s,s+2} &= \left(\begin{array}{c}s-1\\2\end{array}\right) & \text{ for } s \geq 1 \\
 f_{s,s+3} &=
  \left(\begin{array}{c}s-2\\1\end{array}\right) +
  \left(\begin{array}{c}s-2\\2\end{array}\right) +
  \left(\begin{array}{c}s-2\\3\end{array}\right)
 & \text{ for } s \geq 2.
\end{align*}
There does not seem to be a similarly tidy expression for $f_{s,s+4}$.
A: I recently started thinking about this problem again.  Here is the best I have.  It is possible to turn the constraint on $f$ into a continued fraction.  (This may be what Mark was thinking in the comments above.)  Specifically,
$$ f(x, y) = \frac{1}{1 - x y f(xy, y)} $$
so
$$ f(x, y) = \frac{1}{1 - \frac{x y}{1 - \frac{x y^2}{1 - \dots}}} $$
There is an explicit closed-form expression for the series arising from such a continued fraction, provided by P. Flajolet in this 1980 paper.  This expression is not as simple as I would have liked, but it makes the problem feasible again.
