Given $k,n\in\mathbf{N}$ with $n\ge k$, define the set $\mathcal{F}(k,n)$ of $(k,n)$-forests of binary rooted trees (where a $(k,n)$-forest is a collection of $k$ rooted trees, which have a totality of $n$ leaves).


My aim is to count the cardinality of $\mathcal{F}(k,n)$.


For example it is well known that $|\mathcal{F}(1,n+1)|$ is given by the $n$-th Catalan number $c\_n=\frac{1}{n+1}{{2n}\choose{n}}$.


Using generating functions (and something like Maple), we have an explicit way to obtain any $f\_{k,n}:=|\mathcal{F}(k,n)|$: if $C(z)$ denotes the generating function of $c\_n$, then the $n-k$-th coefficient of $C(z)^k$ is exactly $f\_{k,n}$.


Isn't there any explicit formula for such numbers?


Is there an asymptotic estimate and is it a good one?



Counting $(n,k)$-forests of binary trees
The number of $(k,n)$-binary forests is the $(n-1,n-k)$ entry of [Catalan's triangle](http://mathworld.wolfram.com/CatalansTriangle.html). Thus the formula is:
$$
f\_{k,n} \:=\: \frac{\:k\:}{n}\binom{2n-k-1}{n-1}.
$$
Given this formula, you can use [Stirling's approximation](http://en.wikipedia.org/wiki/Stirling%27s_approximation) to obtain asymptotic estimates.


