Consider $n$ entities $x_i$ that may have $m$ different binary properties $P_j$. There are $2^{n\cdot m}$ ways the properties may be attributed to the entities. For $n = m = 3$ the attribution table may look like this:
$$\begin{array}{c|c|c|c} & P_1 & P_2 & P_3 \\ \hline x_1 & 1 & 0 & 0 \\ \hline x_2 & 0 & 1 & 1 \\ \hline x_3 & 1 & 1 & 0 \end{array}$$
When we neglect the order of the entities two attribution tables are the same when you get the one from the other by some permutation of the entities (rows). How many equivalence classes of attributions are there?
For $n = m = 2$ I found that the number of non-equivalent attributions is
$$a(2,2) = 1 + 2 + 4 + 2 + 1 = 10$$
with $a_0 = a_4 = 1$ being the number of non-equivalent ways to place 0 or 4 entries in the 2x2 table, $a_1 = a_3 = 2$ the number of placing 1 or 3 entries, and $a_2 = 4$ the number of placing 2 entries.
For $n=m=3$, the sum starts and ends the same: $a_0 = a_9 = 1$, $a_1 = a_8 = 3$, but I am already stuck to find $a_2$, i.e. the number of non-equivalent ways to place 2 entries in the 3x3 table. But I guess, it's $9$.
So one might tend to guess that $a(n,n) = 2\cdot \sum_{k=0}^{\lfloor n^2/2\rfloor} n^k$ (for odd $n$), but this number grows way too fast.
So my questions is how $a(n,n)$ or more generally $a(n,m)$ can be systematically derived? Is it a case for generating functions?
Added: This is the table of $a(n,m)$ for $n,m \leq 5$:
The diagonal can be found here.
Added: I wonder how many bits you need to represent an equivalence class. Consider $n=m=5$ with 376,992 equivalence classes for which to represent you need at least 19 bits (because $2^{18} =262,144$ and $2^{19} = 524,288$). Being more verbose you need to specify up to five rows, and for each row a number $\leq 5$, the numbers summing up to $5$. So you need $5 \cdot (5 + \lceil \log_2(5)\rceil) = 40$ bits in this representation which gives the essential information.
Added: The number $a(n,m)$ is also the number of equivalence classes of bipartite $n \times m$ graphs. I could have phrased the question like this as well which would have been much shorter.
How many different possible rows are possible? $2^m$. Now, apply stars-and-bars, you have $n$ rows which order doesn't matter for that you wish to have each row assigned to one of the possible values for the row. $$\binom{n+2^m-1}{2^m-1}$$ In the case of $m=n=2$ this would be $\binom{2+2^2-1}{2^2-1}=\binom{5}{3}=10$