TITLE: How many $n\times m$ binary matrices are there, up to row and column permutations?
QUESTION [17 upvotes]: I'm interested in the number of binary matrices of a given size that are distinct with regard to row and column permutations.
If $\sim$ is the equivalence relation on $n\times m$ binary matrices such that $A \sim B$ iff one can obtain B from applying a permutation matrix to A, I'm interested in the number of $\sim$-equivalence classes over all $n\times m$ binary matrices.
I know there are $2^{nm}$ binary matrices of size $n\times m$, and $n!m!$ possible permutations, but somehow I fail to get an intuition on what this implies for the equivalence classes.

REPLY [10 votes]: Here is a computational contribution that  treats the case of a square
matrix.  As  pointed out this  problem can  be solved using  the Polya
Enumeration Theorem.  In fact  if we are  interested only  in counting
these matrices, then the Burnside lemma  will suffice. We just need to
compute  the cycle  index of  the  group acting  on the  slots of  the
matrix.

These cycle indices are easy to  compute and we do not need to iterate
over all $(n!)^2$  pairs of permutations (acting on  rows and columns)
but instead it  is sufficient to iterate over pairs  of terms from the
cycle index $Z(S_n)$  of the symmetric group $S_n$  according to their
multiplicities  to obtain  the cycle  index $Z(Q_n)$  of  the combined
action  on rows  and columns.  The number  of terms  here is  the much
better count of the number of partitions of $n$ squared (upper bound).

Now for a pair  of cycles, one of length $l_1$  from a row permutation
$\alpha$ and another of length $l_2$ from a column permutation $\beta$
their  contribution to  the disjoint  cycle decomposition  product for
$(\alpha,\beta)$  in  the  cycle   index  $Z(Q_n)$  is  by  inspection
$$a_{\mathrm{lcm}(l_1,  l_2)}^{l_1  l_2  / \mathrm{lcm}(l_1,  l_2)}  =
a_{\mathrm{lcm}(l_1, l_2)}^{\gcd(l_1, l_2)}.$$

The algorithm now  becomes very simple -- iterate over  pairs of terms
as described above, collect the  contribution from each pair of cycles
and add it to the cycle index being computed.

This  gives the  following  cycle  indices (only  the  first four  are
shown):
$$Z(Q_2) = 1/4\,{a_{{1}}}^{4}+3/4\,{a_{{2}}}^{2},$$
$$Z(Q_3) = 
1/36\,{a_{{1}}}^{9}+1/6\,{a_{{1}}}^{3}{a_{{2}}}^{3}+1/4\,a_{{
1}}{a_{{2}}}^{4}+2/9\,{a_{{3}}}^{3}+1/3\,a_{{3}}a_{{6}},$$
$$Z(Q_4) =
{\frac {{a_{{1}}}^{16}}{576}}+1/48\,{a_{{1}}}^{8}{a_{{2}}}^{4
}+1/16\,{a_{{1}}}^{4}{a_{{2}}}^{6}+1/36\,{a_{{1}}}^{4}{a_{{3}
}}^{4}+{\frac {17\,{a_{{2}}}^{8}}{192}}\\+1/6\,{a_{{1}}}^{2}a_{
{2}}{a_{{3}}}^{2}a_{{6}}+1/9\,a_{{1}}{a_{{3}}}^{5}+1/12\,{a_{
{2}}}^{2}{a_{{6}}}^{2}+{\frac {13\,{a_{{4}}}^{4}}{48}}+1/6\,a
_{{4}}a_{{12}},$$
and
$$Z(Q_5) = 
{\frac {{a_{{1}}}^{25}}{14400}}+{\frac {{a_{{1}}}^{15}{a_{{2}
}}^{5}}{720}}+{\frac {{a_{{1}}}^{9}{a_{{2}}}^{8}}{144}}+{
\frac {{a_{{1}}}^{10}{a_{{3}}}^{5}}{360}}+{\frac {{a_{{1}}}^{
5}{a_{{2}}}^{10}}{480}}\\+1/48\,{a_{{1}}}^{3}{a_{{2}}}^{11}+{
\frac {a_{{1}}{a_{{2}}}^{12}}{64}}+1/36\,{a_{{1}}}^{6}{a_{{2}
}}^{2}{a_{{3}}}^{3}a_{{6}}+1/36\,{a_{{1}}}^{4}{a_{{3}}}^{7}+{
\frac {{a_{{1}}}^{5}{a_{{4}}}^{5}}{240}}\\+{\frac {{a_{{2}}}^{5
}{a_{{3}}}^{5}}{360}}+1/24\,{a_{{1}}}^{3}a_{{2}}{a_{{4}}}^{5}
+1/24\,{a_{{1}}}^{2}{a_{{2}}}^{4}a_{{3}}{a_{{6}}}^{2}+1/36\,{
a_{{2}}}^{5}{a_{{3}}}^{3}a_{{6}}\\+1/16\,a_{{1}}{a_{{2}}}^{2}{a
_{{4}}}^{5}+1/24\,{a_{{2}}}^{5}a_{{3}}{a_{{6}}}^{2}+1/18\,{a_
{{2}}}^{2}{a_{{3}}}^{5}a_{{6}}+1/16\,a_{{1}}{a_{{4}}}^{6}\\+1/
36\,{a_{{2}}}^{2}{a_{{3}}}^{3}{a_{{6}}}^{2}+1/12\,{a_{{1}}}^{
2}a_{{3}}{a_{{4}}}^{2}a_{{12}}+1/12\,a_{{2}}a_{{3}}{a_{{4}}}^
{2}a_{{12}}+{\frac {13\,{a_{{5}}}^{5}}{300}}\\+1/30\,{a_{{5}}}^
{3}a_{{10}}+1/15\,{a_{{5}}}^{2}a_{{15}}+1/20\,a_{{5}}{a_{{10}
}}^{2}+1/10\,a_{{5}}a_{{20}}+1/15\,a_{{10}}a_{{15}}.$$
Evaluating  these cycle  indices  with  the variables  set  to two  we
quickly obtain the sequence
$$2, 7, 36, 317, 5624, 251610, 33642660, 14685630688,\\
21467043671008, 105735224248507784,1764356230257807614296,\\
100455994644460412263071692,19674097197480928600253198363072,\\
13363679231028322645152300040033513414,\\
31735555932041230032311939400670284689732948,\ldots$$
which is indeed OEIS A002724.
Note  that   the  cycle   indices  make   it  possible   to  enumerate
configurations with  more than  two possible  entries or  with entries
having different  weights. For  example, with a  3x3 square  and three
colors $A,B$ and $C$ we get the generating function
$$Z(Q_3)(A+B+C) = 1/36\, \left( A+B+C \right) ^{9}+1/6\, 
\left( A+B+C \right) ^{3} \left( {A}^{2}+{B}^{2}+{C}^{2} 
\right) ^{3}\\+2/9\, \left( {A}^{3}+{B}^{3}+{C}^{3} \right)^{3}
+1/4\, \left( A+B+C \right)  \left( {A}^{2}+{B}^{2}+{C}^{2
} \right) ^{4}\\+1/3\, \left( {A}^{3}+{B}^{3}+{C}^{3}
 \right)  \left( {A}^{6}+{B}^{6}+{C}^{6} \right)$$
which expands to
$${A}^{9}+{A}^{8}B+{A}^{8}C+3\,{A}^{7}{B}^{2}+3\,{A}^{7}B
C+3\,{A}^{7}{C}^{2}+6\,{A}^{6}{B}^{3}+10\,{A}^{6}{B}^{2
}C\\+10\,{A}^{6}B{C}^{2}+6\,{A}^{6}{C}^{3}+7\,{A}^{5}{B}^
{4}+17\,{A}^{5}{B}^{3}C+28\,{A}^{5}{B}^{2}{C}^{2}\\+17\,{
A}^{5}B{C}^{3}+7\,{A}^{5}{C}^{4}+7\,{A}^{4}{B}^{5}+22\,
{A}^{4}{B}^{4}C+43\,{A}^{4}{B}^{3}{C}^{2}+43\,{A}^{4}{B
}^{2}{C}^{3}\\+22\,{A}^{4}B{C}^{4}+7\,{A}^{4}{C}^{5}+6\,{
A}^{3}{B}^{6}+17\,{A}^{3}{B}^{5}C+43\,{A}^{3}{B}^{4}{C}
^{2}+54\,{A}^{3}{B}^{3}{C}^{3}\\+43\,{A}^{3}{B}^{2}{C}^{4
}+17\,{A}^{3}B{C}^{5}+6\,{A}^{3}{C}^{6}+3\,{A}^{2}{B}^{
7}+10\,{A}^{2}{B}^{6}C+28\,{A}^{2}{B}^{5}{C}^{2}\\+43\,{A
}^{2}{B}^{4}{C}^{3}+43\,{A}^{2}{B}^{3}{C}^{4}+28\,{A}^{
2}{B}^{2}{C}^{5}+10\,{A}^{2}B{C}^{6}+3\,{A}^{2}{C}^{7}\\+
A{B}^{8}+3\,A{B}^{7}C+10\,A{B}^{6}{C}^{2}+17\,A{B}^{5}{
C}^{3}+22\,A{B}^{4}{C}^{4}+17\,A{B}^{3}{C}^{5}\\+10\,A{B}
^{2}{C}^{6}+3\,AB{C}^{7}+A{C}^{8}+{B}^{9}+{B}^{8}C+3\,{
B}^{7}{C}^{2}+6\,{B}^{6}{C}^{3}+7\,{B}^{5}{C}^{4}\\+7\,{B
}^{4}{C}^{5}+6\,{B}^{3}{C}^{6}+3\,{B}^{2}{C}^{7}+B{C}^{
8}+{C}^{9}.$$
This is the Maple code for this computation. Here we have two slightly
different ways of evaluating the count, the first by substituting into
the cycle index and the second  by skipping the cycle index altogether
and  evaluating all  variables at  two during  processing. The  latter
should be used when  we are interested ony in the  count as opposed to
classifying configurations  according to  the number  of each  color /
value that are present.

pet_cycleind_symm :=
proc(n)
option remember;

    if n=0 then return 1; fi;

    expand(1/n*add(a[l]*pet_cycleind_symm(n-l), l=1..n));
end;

pet_varinto_cind :=
proc(poly, ind)
local subs1, subs2, polyvars, indvars, v, pot, res;

    res := ind;

    polyvars := indets(poly);
    indvars := indets(ind);

    for v in indvars do
        pot := op(1, v);

        subs1 :=
        [seq(polyvars[k]=polyvars[k]^pot,
             k=1..nops(polyvars))];

        subs2 := [v=subs(subs1, poly)];

        res := subs(subs2, res);
    od;

    res;
end;


pet_cycleind_sqmat :=
proc(n)
option remember;
local sind, cind, term_a, term_b, v_a, v_b,
    len_a, len_b, inst_a, inst_b, p;

    cind := 0;

    if n=1 then
        sind := return a[1];
    else
        sind := pet_cycleind_symm(n);
    fi;

    for term_a in sind do
        for term_b in sind do
            p := 1;
            for v_a in indets(term_a) do
                len_a := op(1, v_a);
                inst_a := degree(term_a, v_a);
                
                for v_b in indets(term_b) do
                    len_b := op(1, v_b);
                    inst_b := degree(term_b, v_b);

                    p := p*a[lcm(len_a, len_b)]
                    ^(gcd(len_a, len_b)*inst_a*inst_b);
                od;
            od;

            cind := cind +
            lcoeff(term_a)*lcoeff(term_b)*p;
        od;
    od;

    cind;
end;

v :=
proc(n)
    option remember;
    local cind, vars, sbl;

    cind := pet_cycleind_sqmat(n);

    vars := indets(cind);
    sbl := [seq(v=2, v in vars)];

    subs(sbl, cind);
end;

w := 
proc(n)
option remember;
local sind, count, term_a, term_b, v_a, v_b,
    len_a, len_b, inst_a, inst_b, p;

    count := 0;

    if n=1 then
        sind := return 2;
    else
        sind := pet_cycleind_symm(n);
    fi;

    for term_a in sind do
        for term_b in sind do
            p := 1;
            for v_a in indets(term_a) do
                len_a := op(1, v_a);
                inst_a := degree(term_a, v_a);
                
                for v_b in indets(term_b) do
                    len_b := op(1, v_b);
                    inst_b := degree(term_b, v_b);

                    p := p*
                    2^(gcd(len_a, len_b)*inst_a*inst_b);
                od;
            od;

            count := count +
            lcoeff(term_a)*lcoeff(term_b)*p;
        od;
    od;

    count;
end;

This MSE Meta Link
has many more PET computations by various users.