By generating functions, we can find the number of unlabeled plane rooted trees.
I am thinking about is there a way to find the number of unlabeled plane rooted trees with $n$ vertices and $m$ leaves. I define $F(x,y)=\sum C(m,n)x^my^n$, where $C(m,n)$ is the number of unlabeled plane rooted trees with $m$ vertices and $n$ leaves.
I think $F(x,y)=x(y+F(x,y)+F(x,y)^2+F(x,y)^3+\cdots)=x(y+\frac{1}{1-F(x,y)}-1)$. But it seems that it doesn't have a simple representation of $F(x,y)$.
So I want to know if it is on the right track?
I think you just have the $x$ and $y$ in the wrong place. The leading factor $y$ represents the root node. The general term $F(x,y)^k$ represents the case of $k$ first-level children. The $x$ by itself represents the trivial no-children case (i.e. a single node tree).
Also, we can re-arrange the expression as a quadratic in $F(x,y)$ and solve as follows:
\begin{align} F(x,y) &= y(x+F(x,y)+F(x,y)^2+F(x,y)^3+\cdots) \\ &= y(x+\frac{1}{1-F(x,y)}-1) \\ F(x,y)(1-F(x,y)) &= (xy-y)(1-F(x,y)) + y \\ F(x,y)^2 + (y-xy-1)F(x,y) + xy &= 0 \\ F(x,y) &= \dfrac{1+xy-y - \sqrt{y^2+x^2y^2+1-2xy-2xy^2-2y}}{2}. \end{align}
Note that the other root of the quadratic doesn't work as the generating function so it is discarded.