For $n\ge 1$, let $f(n)$ be the number of rooted complete (unordered) binary trees with $n$ leaves labeled from $1$ to $n$ ("complete binary" means that every vertex has either $0$ or $2$ children and "unordered" means that the we do not specify which child is the left child or the right child). Then it is well known (e.g., Example 5.2.6 of Stanley's *Enumerative Combinatorics 2*) that the exponential generating function of $f(n)$ is given by
$$F(x) := \sum\_{n\ge1} f(n)\, {x^n\over n!} = 1 - \sqrt{1-2x}.$$


Now fix a positive integer $r$ and for $n\ge r$, let $f(n,r)$ be the number of rooted complete binary *forests* with $n$ leaves labeled $1$ to $n$, and $r$ roots labeled $n+1$ to $n+r$. By generatingfunctionology (e.g., Proposition 5.1.3 of Stanley's *Enumerative Combinatorics 2*),
$$\sum\_{n\ge r} f(n,r)\, {x^n\over n!} = F(x)^r = (1 - \sqrt{1-2x})^r.$$
This equation yields a formula for $f(n,r)$, and in fact we have:


**Theorem.** $$f(n,r) = {r(2n-r-1)!\over 2^{n-r}(n-r)!}.$$


The current proof I have of the Theorem simply notes that the generating function $F(x)^r$ coincides with a generating function for the [Catalan tree](http://oeis.org/A102625), for which the coefficients are known to obey the above formula. My question is:



> 
> Is there a bijective proof of the Theorem?
> 
> 
> 



Bijective proof of formula for rooted binary forests
I think the following might do the trick:


*Erdös, Péter L.*, [**A new bijection on rooted forests**](http://dx.doi.org/10.1016/0012-365X(93)90154-L), Discrete Math. 111, No.1-3, 179-188 (1993). [ZBL0785.05049](https://zbmath.org/?q=an:0785.05049).


