Find the ordinary generating funtion for thenumber of at most binary trees on n unlabelled nodes. An at most binary tree is an unordered rooted tree in whic each node has at most two children.
I want to find the ordinary generating function for the number of at most binary trees on n unlabelled nodes. 
I am trying to solve this problem using the theory of species. I am thinking the following way.:
Let $t_n$ be thenumber of trees with n vertices and let $T(x) = \sum_n t_n x^n$ be their ordinary generating function. We first pick the root of the tree and then any of the remaining branches must be an at most binary unorderedrooted tree. Thus, the generating function fo any of the branches is $(1+T(x) + T(x)^2/2)$. THus, multiplying our choice for root and the remaining branches, we have that $T(x) = x(1+T(x)+T(x)^2/2)$. From here I should just solve the recurrence and find the ordinary generating function.
I'm not sure if this reasoning is right, but in a previous problem I found the exponential generating fucntion for thenumber of at most binary trees on n laelled nodes, but I got a very similar result. 
I am wondering if they way I approached this problem is right and, if not, if anyone could give me a hint for it. Thanks!

Your $t_n$ are the Wedderburn–Etherington numbers, OEIS A001190. Specifically, $t_n=a_{n+1}$, where the $a_n$ are the Wedderburn-Etherington numbers. The OEIS entry has copious references but neither a closed form nor an explicit generating function. It does note that the generating function $A(x)$ for the $a_n$ satisfies
$$A(x)=x+\frac12\left(A(x)^2+A(x^2)\right)$$
and
$$A(x)=1-\sqrt{1-2x-A(x^2)}\;.$$
Since $a_0=0$, $T(x)=\frac{A(x)}x$. For $n\ge 1$ the Wedderburn–Etherington numbers satisfy the recurrences
$$\begin{align*}
a_{2n-1}&=\sum_{k=1}^{n-1}a_ka_{2n-k-1}\\
a_{2n}&=\frac{a_n(a_n+1)}2+\sum_{k=1}^{n-1}a_ka_{2n-k}\;,
\end{align*}$$
With base case $a_1=1$. This yields the recurrences
$$\begin{align*}
t_{2n}&=\sum_{k=0}^{n-1}t_kt_{2n-k-1}\\
t_{2n-1}&=\frac{t_{n-1}(t_{n-1}+1)}2+\sum_{k=1}^{n-1}t_{k-1}t_{2n-1-k}
\end{align*}$$
for $n\ge 1$, with base case $t_0=1$.
