A053513 Number of 2-multigraphs with loops on n nodes.
3, 18, 165, 3132, 137268, 15548004, 4679446950, 3771927027864, 8186669639820081, 48184182482857319682, 774912347548961791914585, 34299111628183837790980740312, 4205499936656520106909422649497294
Offset: 1
Links
- Andrew Howroyd, Table of n, a(n) for n = 1..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] + 1]; a[n_] := (s=0; Do[s += permcount[p]*3^edges[p], {p, IntegerPartitions[n]}]; s/n!); Array[a, 15] (* Jean-François Alcover, Jul 08 2018, 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 + 1)} a(n) = {my(s=0); forpart(p=n, s+=permcount(p)*3^edges(p)); s/n!} \\ Andrew Howroyd, Oct 22 2017 -
Python
from itertools import combinations from math import prod, gcd, factorial from fractions import Fraction from sympy.utilities.iterables import partitions def A053513(n): return int(sum(Fraction(3**(sum(p[r]*p[s]*gcd(r,s) for r,s in combinations(p.keys(),2))+sum(((q>>1)+1)*r+(q*r*(r-1)>>1) for q, r in p.items())),prod(q**r*factorial(r) for q, r in p.items())) for p in partitions(n))) # Chai Wah Wu, Jul 09 2024
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