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!
 A: 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 $t_0=1$.
A: With this question we run into the problem of determining exactly what
the notation is supposed to mean  and which family of trees from among
the  many possibilities is  being  referenced. I  would  like to  help
clarify this. 
It appears  that these at most  binary trees are not  ordered trees or
plane trees so that two trees  that have the same pair of subtrees but
in  opposite order are  considered to  be the  same.  This  yields the
unlabeled species
$$\mathcal{T} = \mathcal{Z}
+ \mathcal{Z}\mathfrak{M}_{=1}(\mathcal{T})
+ \mathcal{Z}\mathfrak{M}_{=2}(\mathcal{T}).$$
Here   we   have   the    convention   that   the   species   operator
$\mathfrak{M}_{=q}$ represents  the symmetric group acting  on the $q$
slots, so that  the cycle index $Z(S_q)$ is  used. These are multisets
as  opposed to  sets (operator  $\mathfrak{P}_{=q}$.)   Translating to
generating functions we thus obtain
$$T(z) = z + z T(z) + \frac{1}{2} z (T(z)^2 + T(z^2))$$
where we have used the fact that $Z(S_2) = \frac{1}{2} (a_1^2 + a_2).$
Now we introduce the function $$A(z) = z + z T(z)$$ which is a shifted
version  of $T(z)$  with  a  singleton added  and  get the  functional
equation
$$(A(z)-z)/z = z + A(z) - z +
\frac{1}{2} z ((A(z)-z)^2/z^2 + (A(z^2)-z^2)/z^2)
\\ = A(z) +
\frac{1}{2} z (A(z)^2/z^2-2zA(z)/z^2+1 + A(z^2)/z^2-1)
\\ = A(z) +
\frac{1}{2} z (A(z)^2/z^2-2zA(z)/z^2 + A(z^2)/z^2).$$
Multiply by $z$ to get
$$A(z)-z = z A(z) + \frac{1}{2} (A(z)^2 - 2z A(z) + A(z^2))
= \frac{1}{2} (A(z)^2 + A(z^2)).$$
This finally yields
$$\bbox[5px,border:2px solid #00A000]{
A(z) = z + \frac{1}{2} (A(z)^2 + A(z^2)).}$$
We have the functional equation of the ordinary generating function of
the  Wedderburn-Etherington  numbers  and  may continue  as  in  OEIS
A001190.
Remark.   Using  the  terminology   from  Harary   and  Palmer,
Graphical Enumeration  we see  that taking $z+zT(z)$  corresponds to
planting the  trees from $T(z)$  and adding the single  node planted
tree  on  one  node.   Hence  $A(z)$ counts  the  species  of  planted
unordered at most binary trees.
