Spanning forests of bipartite graphs and distinct row/column sums of binary matrices Let $F_{m,n}$ be the set of spanning forests on the complete bipartite graph $K_{m,n}$.  Let $$S_{m,n} = \{(r(M), c(M)), M \in B_{m,n} \}$$ where $B_{m,n}$ is the set of $m \times n$ binary matrices and $r(M), c(M)$ are the vectors of row sums and column sums of $M$, respecitvely.  That is, $S_{m,n}$ is the set of the distinct row-sum, column-sum pairs of binary matrices. (The term spanning forest here refers to a forest that spans all of the vertices of the given graph; it doesn't have to be a maximal acyclic subgraph.)
Q: Is it true that $|F_{m,n}| = |S_{m,n}|$?  It is true for $m,n \leq 4$.  For $m=n$ this is OEIS A297077.
There is an obvious mapping from $F_{m,n} \rightarrow B_{m,n}$ given by taking the reduced adjacency matrix, so if $U = \{u_1, \ldots, u_m\}$, $V = \{v_1, \ldots, v_m\}$ are the color categories we set $M_{ij} = 1$ if $v_i \sim u_j$ in a forest $F$, else $M_{ij} = 0$.  However, this does not help because multiple forests may have the same row and column sums - and not every row-sum, column-sum pair is represented by a forest under this mapping.
The numbers $|F_{m,n}|$ are given here:
$$\begin{array}{|c|c|}\hline
m\backslash n & 1 & 2 & 3 & 4 & 5 \\ \hline
1 & 2 &  \\\hline
2 & 4 & 15 \\\hline
3 & 8 & 54 & 328  \\\hline
4 & 16 & 189 & 1856 & 16145 \\\hline
5 & 32 & 648 & 9984 & 129000 & 1475856 \\\hline\end{array}$$
For more see this answer (sum each row.)
 A: This conjectured equivalence (if true) is still amazing to me.  However, here is a baby step: The conjecture is indeed true for $m=2$.  Perhaps someone else will find this useful as a base for generalization.
Claim: $|F_{2,n}| = |S_{2,n}|$ for all $n$
First, lets consider $F_{2,n}$.  In your definition, a spanning forest is any subset $T$ of edges of $K_{2,n}$ which is acyclic.  Let $u,v$ be the two nodes on the $m=2$ side, and let $S(u)$ be the neighbors of $u$, i.e. the subset of nodes (on the $n$-node side) which $u$ is connected to in $T$.  Similarly for $S(v)$. 
Next, note that $T$ contains a cycle iff $S = S(u) \cap S(v)$ contains $2$ nodes (or more).  So, $F_{2,n}$ can be counted by counting all cases where $S$ contains $0$ or $1$ node.


*

*$|S|=0$ case: Each of $n$ nodes can be connected to $u$, or to $v$, or to neither.  Total no. $= 3^n$.

*$|S|=1$ case: There are $n$ ways to pick the unique $w \in S$.  After that, each of the remaining $n-1$ nodes can be connected to $u$, or to $v$, or to neither.  Total no. $= n\, 3^{n-1}$

*Summing up, $|F_{2,n}| = 3^n + n \,3^{n-1} = 3^{n-1} (3+n),$ which matches your table.
Second, lets consider $S_{2,n}$.  There are $n$ column sums, each of which can be $0, 1, 2$.  Let $x,y,z$ be the number of columns whose sum $=0, 1, 2$ respectively.  The columns whose sum $=0$ or $2$ dictate their elements.  In the $y$ columns whose sum $=1$, the number of $1$s in the top row can be any integer $\in [0,y]$, and so the top row sum can be any integer $\in [z,z+y]$, and in short there are $(y+1)$ possibilities.  Obviously, once the top row sum is known that determines the bottom row sum $= (y+2z) \,-$ top row sum.
So $S_{2,n}$ can be counted by (i) summing over all possible $y$, and for each $y$: (ii) picking the $y$ columns whose sum $=1$, and (iii) for the remaining $n-y=x+z$ columns assign them $0$ or $2$ arbitrarily.  I.e.:
$$|S_{2,n}| = \sum_{y=0}^n (y+1) {n \choose y} 2^{n-y}$$
This can be evaluated many ways but my favorite is to recognize a slightly transformed sum as an explicit formula for the expected value of a Binomial random variable:
$$
\begin{align}
|S_{2,n}| &= \sum_{y=0}^n (y+1) {n \choose y} 2^{n-y} \\
&= 3^n \times \sum_{y=0}^n (1+y) {n \choose y} (\frac13)^y (\frac23)^{n-y} \\
&= 3^n \times \mathbb{E}[1 + Bin(n, \frac13)] \\
&= 3^n (1 + {n \over 3}) = 3^{n-1} (3 + n) = |F_{2,n}| & QED
\end{align}
$$

As I said, this is a baby step only.  The key step in $F_{2,n}$ is that $T$ has a cycle iff $|S| \ge 2$, and the key step in $S_{2,n}$ is that the $y$ ones can be all in the top row or all in the bottom row or anywhere in between, for $(1+y)$ possibilities.  These key steps make the counting easy.  However, as far as I can see, neither key step has any easy way to generalize to $m=3$, let alone arbitrary $m$.  (E.g. for $m=3$, a cycle in $T$ can involve $2$ or $3$ nodes on the $m=3$ side, and I don't know any good way to count that.)
