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A321609
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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.
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5
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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, 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, 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
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OFFSET
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0,13
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LINKS
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FORMULA
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T(n,k) = T(k,k) for n > k.
T(n,k) = 0 for k > n^2.
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EXAMPLE
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Array begins:
==========================================================
n\k| 0 1 2 3 4 5 6 7 8 9 10 11 12
---+------------------------------------------------------
0 | 1 0 0 0 0 0 0 0 0 0 0 0 0 ...
1 | 1 1 0 0 0 0 0 0 0 0 0 0 0 ...
2 | 1 1 3 1 1 0 0 0 0 0 0 0 0 ...
3 | 1 1 3 6 7 7 6 3 1 1 0 0 0 ...
4 | 1 1 3 6 16 21 39 44 55 44 39 21 16 ...
5 | 1 1 3 6 16 34 69 130 234 367 527 669 755 ...
6 | 1 1 3 6 16 34 90 182 425 870 1799 3323 5973 ...
7 | 1 1 3 6 16 34 90 211 515 1229 2960 6893 15753 ...
8 | 1 1 3 6 16 34 90 211 558 1371 3601 9209 24110 ...
9 | 1 1 3 6 16 34 90 211 558 1430 3825 10278 28427 ...
...
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MATHEMATICA
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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];
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}];
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!)]
Table[M[n - k, n - k, k], {n, 0, 12}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Sep 10 2019, after Andrew Howroyd *)
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PROG
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(PARI)
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}
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)}
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!)}
for(n=0, 10, for(k=0, 12, print1(M(n, n, k), ", ")); print); \\ Andrew Howroyd, Nov 14 2018
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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