OFFSET
1,2
COMMENTS
Also the number of colored digraphs on n unlabeled nodes with nodes of exactly two colors. (Understand "(x,x) in the relation" for several nodes x as a special color!)
LINKS
Andrew Howroyd, Table of n, a(n) for n = 1..50
EXAMPLE
n=2: We label node 1 with (1,1) in the relation and node 2 with (2,2) not in the relation, and we have two differently labeled nodes and so a(2) = A053763(2) = 4.
MATHEMATICA
permcount[v_] := 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];
edges[v_] := Sum[2*GCD[v[[i]], v[[j]]], {i, 2, Length[v]}, {j, 1, i - 1}] + Total[v];
a[n_] := Module[{s = 0}, Do[t = 2^edges[p]; s += t*(1 - 2^(1 - Length[p]))* permcount[p], {p, IntegerPartitions[n]}]; s/n!];
Array[a, 14] (* Jean-François Alcover, Jan 08 2021, after Andrew Howroyd *)
PROG
(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}
edges(v) = {sum(i=2, #v, sum(j=1, i-1, 2*gcd(v[i], v[j]))) + sum(i=1, #v, v[i])}
a(n) = {my(s=0); forpart(p=n, my(t=2^edges(p)); s+=t*(1 - 2^(1-#p))*permcount(p)); s/n!} \\ Andrew Howroyd, Nov 02 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
Peter Dolland, Nov 02 2019
STATUS
approved