A052371 Triangle T(n,k) of n X n binary matrices with k=0...n^2 ones up to row and column permutations.
1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 3, 6, 7, 7, 6, 3, 1, 1, 1, 1, 3, 6, 16, 21, 39, 44, 55, 44, 39, 21, 16, 6, 3, 1, 1, 1, 1, 3, 6, 16, 34, 69, 130, 234, 367, 527, 669, 755, 755, 669, 527, 367, 234, 130, 69, 34, 16, 6, 3, 1, 1
Offset: 0
Examples
Triangle begins: 1; 1, 1; 1, 1, 3, 1, 1; 1, 1, 3, 6, 7, 7, 6, 3, 1, 1; 1, 1, 3, 6, 16, 21, 39, 44, 55, 44, 39, 21, 16, 6, 3, 1, 1; ... (the last block giving the numbers of 4 X 4 binary matrices with k=0..16 ones up to row and column permutations).
Links
- Andrew Howroyd, Table of n, a(n) for n = 0..2890 (rows n=0..20)
Programs
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Mathematica
permcount[v_] := Module[{m = 1, s = 0, t, i, k = 0}, 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_, q_] := Product[(1 + x^LCM[p[[i]], q[[j]]])^GCD[p[[i]], q[[j]]], {i, 1, Length[p]}, {j, 1, Length[q]}]; row[n_] := Module[{s = 0}, Do[Do[s += permcount[p]*permcount[q]*c[p, q], {q, IntegerPartitions[n]}], {p, IntegerPartitions[n]}]; CoefficientList[ s/(n!^2), x]] row /@ Range[0, 5] // Flatten (* Jean-François Alcover, Sep 22 2019, after Andrew Howroyd *) -
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)={prod(i=1, #p, prod(j=1, #q, (1 + x^lcm(p[i], q[j]))^gcd(p[i], q[j])))} row(n)={my(s=0); forpart(p=n, forpart(q=n, s+=permcount(p) * permcount(q) * c(p, q))); Vec(s/(n!^2))} for(n=1, 5, print(row(n))) \\ Andrew Howroyd, Nov 14 2018
Extensions
a(0)=1 prepended by Andrew Howroyd, Nov 14 2018