I have attempted to calculate the number of unlabelled bipartite graphs as follows:



> 
> Let $G = (V\_1, V\_2, E)$ be a bipartite graph on $n$ vertices with $|V\_1| = m$ and $|V\_2| = n-m$. Assume without loss of generality that $|V\_1| \leq |V\_2|$ so $m \leq \left\lfloor \frac{n}{2} \right\rfloor$. If $G$ is complete bipartite then it has $m(n-m)$ edges since each of the vertices in $V\_1$ is connected to each in $V\_2$. Thus, the total number of bipartite graphs with parts of size $m$ and $n-m$ is $2^{m(n-m)}$. In order to find the total number of possible bipartite graphs on $n$ vertices we sum over all possible $m$:
>  \begin{align}
> \sum^{\left\lfloor \frac{n}{2} \right\rfloor}\_{m=1} 2^{m(n-m)}
> \end{align}
> 
> 
> 


However, I notice that I have counted labelled bipartite graphs where I need the number of unlabelled graphs. I'm struggling to see how to account for this.



What is the cardinality of the family of unlabelled bipartite graphs on n vertices?
It seems that what Andrew wants to count are what are called in enumerative contexts
"bicolored graphs". A bicolored graph is a graph in which the vertices have been colored black and white so that every edge joins two vertices of different colors. A bipartite (or bicolorable) graph is a graph that has a bicoloring. A bicolorable graph with $k$ connected components has $2^k$ bicolorings. (In nonenumerative contexts the distinction between bipartite and bicolored is usually unimportant.) In addition, in counting bicolored graphs one might or might not consider switching the two colors to give an equivalent graph. All of the versions of the enumeration problem have been solved. Counting unlabeled bicolored graphs (with no color-switching equivalence) is a straightforward application of Burnside's lemma; counting unlabeled bipartite graphs is tricky. 


It's not too hard to find appropriate references by searching MathSciNet. (Hint: "color" is sometimes spelled "colour".)


Incidentally, the number of labeled bicolored graphs on $n$ vertices is 
$$\sum\_{m=0}^n 2^{m(n-m)}\binom{n}{m}.$$


