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%I #13 Sep 10 2019 09:11:52
%S 1,0,1,0,1,1,0,0,1,1,0,0,3,1,1,0,0,1,3,1,1,0,0,1,6,3,1,1,0,0,0,7,6,3,
%T 1,1,0,0,0,7,16,6,3,1,1,0,0,0,6,21,16,6,3,1,1,0,0,0,3,39,34,16,6,3,1,
%U 1,0,0,0,1,44,69,34,16,6,3,1,1,0,0,0,1,55,130,90,34,16,6,3,1,1
%N Array read by antidiagonals: T(n,k) is the number of inequivalent binary n X n matrices with k ones, under row and column permutations.
%F T(n,k) = T(k,k) for n > k.
%F T(n,k) = 0 for k > n^2.
%e Array begins:
%e ==========================================================
%e n\k| 0 1 2 3 4 5 6 7 8 9 10 11 12
%e ---+------------------------------------------------------
%e 0 | 1 0 0 0 0 0 0 0 0 0 0 0 0 ...
%e 1 | 1 1 0 0 0 0 0 0 0 0 0 0 0 ...
%e 2 | 1 1 3 1 1 0 0 0 0 0 0 0 0 ...
%e 3 | 1 1 3 6 7 7 6 3 1 1 0 0 0 ...
%e 4 | 1 1 3 6 16 21 39 44 55 44 39 21 16 ...
%e 5 | 1 1 3 6 16 34 69 130 234 367 527 669 755 ...
%e 6 | 1 1 3 6 16 34 90 182 425 870 1799 3323 5973 ...
%e 7 | 1 1 3 6 16 34 90 211 515 1229 2960 6893 15753 ...
%e 8 | 1 1 3 6 16 34 90 211 558 1371 3601 9209 24110 ...
%e 9 | 1 1 3 6 16 34 90 211 558 1430 3825 10278 28427 ...
%e ...
%t permcount[v_List] := Module[{m = 1, s = 0, k = 0, t}, For[i = 1, i <= Length[v], i++, t = v[[i]]; k = If[i > 1 && t == v[[i - 1]], k + 1, 1]; m *= t*k; s += t]; s!/m];
%t c[p_List, q_List, k_] := SeriesCoefficient[Product[Product[(1 + O[x]^(k + 1) + x^LCM[p[[i]], q[[j]]])^GCD[p[[i]], q[[j]]], {j, 1, Length[q]}], {i, 1, Length[p]}], {x, 0, k}];
%t M[m_, n_, k_] := Module[{s = 0}, Do[Do[s += permcount[p]*permcount[q]*c[p, q, k], {q, IntegerPartitions[n]}], {p, IntegerPartitions[m]}]; s/(m!*n!)]
%t Table[M[n - k, n - k, k], {n, 0, 12}, {k, n, 0, -1}] // Flatten (* _Jean-François Alcover_, Sep 10 2019, after _Andrew Howroyd_ *)
%o (PARI)
%o permcount(v) = {my(m=1, s=0, k=0, t); for(i=1, #v, t=v[i]; k=if(i>1&&t==v[i-1], k+1, 1); m*=t*k; s+=t); s!/m}
%o c(p, q, k)={polcoef(prod(i=1, #p, prod(j=1, #q, (1 + x^lcm(p[i], q[j]) + O(x*x^k))^gcd(p[i], q[j]))), k)}
%o M(m, n, k)={my(s=0); forpart(p=m, forpart(q=n, s+=permcount(p) * permcount(q) * c(p, q, k))); s/(m!*n!)}
%o for(n=0, 10, for(k=0, 12, print1(M(n, n, k), ", ")); print); \\ _Andrew Howroyd_, Nov 14 2018
%Y Rows n=6..8 are A052370, A053304, A053305.
%Y Main diagonal is A049311.
%Y Row sums are A002724.
%Y Cf. A052371 (as triangle), A057150, A246106, A318795.
%K nonn,tabl
%O 0,13
%A _Andrew Howroyd_, Nov 14 2018
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