A063841 Table T(n,k) giving number of k-multigraphs on n nodes (n >= 1, k >= 0) read by antidiagonals.
1, 1, 1, 1, 2, 1, 1, 3, 4, 1, 1, 4, 10, 11, 1, 1, 5, 20, 66, 34, 1, 1, 6, 35, 276, 792, 156, 1, 1, 7, 56, 900, 10688, 25506, 1044, 1, 1, 8, 84, 2451, 90005, 1601952, 2302938, 12346, 1, 1, 9, 120, 5831, 533358, 43571400, 892341888, 591901884, 274668, 1
Offset: 1
Examples
Table begins =========================================================== n\k| 0 1 2 3 4 5 ---|------------------------------------------------------- 1 | 1 1 1 1 1 1 ... 2 | 1 2 3 4 5 6 ... 3 | 1 4 10 20 35 56 ... 4 | 1 11 66 276 900 2451 ... 5 | 1 34 792 10688 90005 533358 ... 6 | 1 156 25506 1601952 43571400 661452084 ... 7 | 1 1044 2302938 892341888 95277592625 4364646955812 ... ... T(3,2)=10 because there are 10 unlabeled graphs with 3 nodes with at most 2 edges connecting any pair. (. . .),(.-. .),(.-.-.),(.-.-.-),(.=. .),(.=.=.),(.=.=.=),(.-.=.),(.-.-.=),(.=.=.-). - _Geoffrey Critzer_, Jan 23 2012
Links
- Chai Wah Wu, Table of n, a(n) for n = 1..1584 (terms 1..820 from Andrew Howroyd)
- Harald Fripertinger, The cycle type of the induced action on 2-subsets
- Vladeta Jovovic, Formulae for the number T(n,k) of n-multigraphs on k nodes
Crossrefs
Programs
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Mathematica
(* This code gives the array T(n,k). *) Needs["Combinatorica`"]; Transpose[Table[Table[PairGroupIndex[SymmetricGroup[n],s]/.Table[s[i]->k+1, {i,0,Binomial[n,2]}], {n,1,7}], {k,0,6}]]//Grid (* Geoffrey Critzer, Jan 23 2012 *) 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]]; T[n_, k_] := (s=0; Do[s += permcount[p]*(k+1)^edges[p], {p, IntegerPartitions[n]}]; s/n!); Table[T[n-k, k], {n, 1, 10}, {k, n-1, 0, -1}] // Flatten (* Jean-François Alcover, Jul 08 2018, after Andrew Howroyd *)
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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} edges(v) = {sum(i=2, #v, sum(j=1, i-1, gcd(v[i],v[j]))) + sum(i=1, #v, v[i]\2)} T(n, k) = {my(s=0); forpart(p=n, s+=permcount(p)*(k+1)^edges(p)); s/n!} \\ Andrew Howroyd, Oct 22 2017
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Python
from itertools import combinations from math import prod, gcd, factorial from fractions import Fraction from sympy.utilities.iterables import partitions def A063841_T(n,k): return int(sum(Fraction((k+1)**(sum(p[r]*p[s]*gcd(r,s) for r,s in combinations(p.keys(),2))+sum((q>>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
Formula
T(n,k) = A327084(n-1,k+1) for n > 1. - Robert A. Russell, Aug 21 2019
Extensions
More terms from Vladeta Jovovic, Sep 03 2001
Comments