A340023 Number of graphs with n integer labeled vertices covering an initial interval of positive integers.
1, 1, 4, 24, 263, 5566, 239428, 21074412, 3779440490, 1372163701412, 1003687569555456, 1474604145003923000, 4343524388729516494384, 25623424478746329214500144, 302549202766446393276528844768, 7147753721248229224770005386691680
Offset: 0
Keywords
Examples
a(2) = 4 because there are 2 graphs on 2 vertices and each of these can either have both vertices labeled 1 or one vertex labeled 1 and the other 2.
Links
- Andrew Howroyd, Table of n, a(n) for n = 0..50
Programs
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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[GCD[v[[i]], v[[j]]], {i, 2, Length[v]}, {j, 1, i - 1}] + Total[Quotient[v, 2]]; G[n_, k_] := Module[{s = 0}, Do[s += permcount[p]*2^edges[p]*k^Length[p], {p, IntegerPartitions[n]}]; s/n!]; a[n_] := Module[{p = G[n, x]}, Sum[(p /. x -> k)*Sum[Binomial[r, k]*(-1)^(r - k), {r, k, n}], {k, 0, n}]]; a /@ Range[0, 15] (* Jean-François Alcover, Jan 06 2021, 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} edges(v) = {sum(i=2, #v, sum(j=1, i-1, gcd(v[i], v[j]))) + sum(i=1, #v, v[i]\2)} G(n,k)={my(s=0); forpart(p=n, s+=permcount(p)*2^edges(p)*k^#p); s/n!} a(n)={my(p=G(n,x)); sum(k=0, n, subst(p,x,k)*sum(r=k, n, binomial(r, k)*(-1)^(r-k)))}
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