How many distinct non-isomorphic bipartite graphs with parts of size $m$ and $n$ exist? How many distinct non-isomorphic  bipartite graphs with parts of size $m$ and $n$  exist? (Two bipartite graphs are distinct if there is no way to just rearrange the vertices within a part set of one of them to become the other.) It's not necessary that every vertex has an edge. Is there a closed form formula in terms of $m$ and $n$? If not, is there a recursive way to count it? If there is a good recursive algorithm to count it, what would its pseudocode look like?
Thanks!
 A: On your question whether or not there exists a closed-form formula for the number of bipartite graphs with parts of size $m$ and $n$ (denoted by $|B_u(m,n)|$ below), my coauthor and I proved the following formulas for $m = 2$ and $m = 3$ in an upcoming paper. $|B_u(2,n)|$ corresponds to the integer sequence A002623, i.e.,  1, 3, 7, 13, 22, 34, 50, 70, 95,..., and $|B_u(3,n)|$ corresponds to the integer sequence A002727, i.e., 1, 4, 13, 36, 87, 190, 386, 734, 1324, 2284,... in Sloane's classification of integer sequences. Generalizing these closed-form formulas for $m = 4,5,6,...$ remains an open problem to the best of my knowledge.
$$|B_u(2,n)|\!=\! \frac{2n^{3}+15n^{2} + 34n + 22.5 + 1.5\left ( -1 \right )^{n}}{24}, n=0,1,2,...$$
$\,$
$\!\!\!\!\!\!\!|B_{u}\left( 3,n \right)|\! = 
\left\{\begin{matrix}
\frac{1}{6}\left [ \binom{n+7}{7}\! + \frac{3\left ( n+4 \right )\left ( 2n^{4}+32n^{3}+172n^{2} + 352n
+ 15\left ( -1 \right )^{n} +225 \right )}{960} +\! \frac{2(n^{3}+12n^{2}+45n+54)}{54} \right ] &\!\!\!\!\!\!\text{if}\, n  \bmod\!\!
\text{ } 3 = 0, \\
\\
\frac{1}{6}\left [ \binom{n+7}{7}\! + \frac{3\left ( n+4 \right )\left ( 2n^{4}+32n^{3}+172n^{2} + 352n
+ 15\left ( -1 \right )^{n} +225 \right )}{960} +\! \frac{2(n^{3}+12n^{2}+45n+50)}{54} \right ] &\!\!\!\!\!\!\!\text{ if}\, n  \bmod\!\!
\text{ } 3 = 1, \\
\\
\frac{1}{6}\left [ \binom{n+7}{7}\! + \frac{3\left ( n+4 \right )\left ( 2n^{4}+32n^{3}+172n^{2} + 352n
+ 15\left ( -1 \right )^{n} +225 \right )}{960} +\! \frac{2(n^{3}+12n^{2}+39n+28)}{54} \right ] &\!\!\!\!\!\!\!\!\! \text{if}\, n  \bmod\!\!
\text{ } 3 = 2. \!\!\!\! \end{matrix}\right.
$
Ref: Abdullah Atmaca and A. Yavuz Oruc. "On The Size Of Two Families Of Unlabeled Bipartite Graphs." AKCE International Journal of Graphs and Combinatorics. To appear.
A: In a more recent article by A. Atmaca and A. Yavuz Oruc, the following more general result for the number of unlabeled graphs with $n$ left vertices and $r$ right vertices (denoted by $|B_u(n,r)|$) has been proven:
$$\frac{{r+2^n-1\choose r}}{n!}\le |B_u(n,r)| \le 2\frac{{r+2^n-1\choose r}}{n!}, n < r. $$
Given that $|B_u(n,r)|= |B_u(r,n)|$, the following inequality holds as well when $n > r:$
$$\frac{{n+2^r-1\choose n}}{r!}\le |B_u(n,r)| \le 2\frac{{n+2^r-1\choose n}}{r!}, n > r. $$
Note that the upper bound is twice as large as the lower bound. Tightening the constant factors, i.e., 1 and 2 on the lefthand and righthand side of the inequality remains open.
