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A000666
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Number of symmetric relations on n nodes.
(Formerly M1650 N0646)
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81
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1, 2, 6, 20, 90, 544, 5096, 79264, 2208612, 113743760, 10926227136, 1956363435360, 652335084592096, 405402273420996800, 470568642161119963904, 1023063423471189431054720, 4178849203082023236058229792, 32168008290073542372004082199424
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OFFSET
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0,2
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COMMENTS
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Each node may or may not be related to itself.
Also the number of rooted graphs on n+1 nodes.
The 1-to-1 correspondence is as follows: Given a rooted graph on n+1 nodes, replace each edge joining the root node to another node with a self-loop at that node and erase the root node. The result is an undirected graph on n nodes which is the graph of the symmetric relation.
Also the number of the graphs with n nodes whereby each node is colored or not colored. A loop can be interpreted as a colored node. - Juergen Will, Oct 31 2011
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REFERENCES
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F. Harary and E. M. Palmer, Graphical Enumeration, Academic Press, NY, 1973, pp. 101, 241.
M. D. McIlroy, Calculation of numbers of structures of relations on finite sets, Massachusetts Institute of Technology, Research Laboratory of Electronics, Quarterly Progress Reports, No. 17, Sept. 15, 1955, pp. 14-22.
R. W. Robinson, Numerical implementation of graph counting algorithms, AGRC Grant, Math. Dept., Univ. Newcastle, Australia, 1976.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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Lorenzo Sauras-Altuzarra, Graphs
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FORMULA
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Let G_{n+1,k} be the number of rooted graphs on n+1 nodes with k edges and let G_{n+1}(x) = Sum_{k=0..n(n+1)/2} G_{n+1,k} x^k. Thus a(n) = G_{n+1}(1). Let S_n(x_1, ..., x_n) denote the cycle index for Sym_n. (cf. the link in A000142).
Compute x_1*S_n and regard it as the cycle index of a set of permutations on n+1 points and find the corresponding cycle index for the action on the n(n+1)/2 edges joining those points (the corresponding "pair group"). Finally, by replacing each x_i by 1+x^i gives G_{n+1}(x). [Harary]
Example, n=2. S_2 = (1/2)*(x_1^2+x_2), x_1*S_2 = (1/2)*(x_1^3+x_1*x_2). The pair group is (1/2)*(x_1^2+x_1*x_2) and so G_3(x) = (1/2)*((1+x)^3+(1+x)*(1+x^2)) = 1+2*x+2*x^2+x^3; set x=1 to get a(2) = 6.
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MAPLE
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# see Riedel link above
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MATHEMATICA
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Join[{1, 2}, Table[CycleIndex[Join[PairGroup[SymmetricGroup[n]], Permutations[Range[n*(n-1)/2+1, n*(n+1)/2]], 2], s] /. Table[s[i]->2, {i, 1, n^2-n}], {n, 2, 8}]] (* Geoffrey Critzer, Nov 04 2011 *)
Table[Module[{eds, pms, leq},
eds=Select[Tuples[Range[n], 2], OrderedQ];
pms=Map[Sort, eds/.Table[i->Part[#, i], {i, n}]]&/@Permutations[Range[n]];
leq=Function[seq, PermutationCycles[Ordering[seq], Length]]/@pms;
Total[Thread[Power[2, leq]]]/n!
], {n, 0, 8}] (* This is after Geoffrey Critzer's program but does not use the (deprecated) Combinatorica package. - Gus Wiseman, Jul 21 2016 *)
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[Sum[GCD[v[[i]], v[[j]]], {j, 1, i-1}], {i, 2, Length[v]}] + Sum[Quotient[v[[i]], 2] + 1, {i, 1, Length[v]}];
a[n_] := a[n] = (s = 0; Do[s += permcount[p]*2^edges[p], {p, IntegerPartitions[n]}]; s/n!);
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PROG
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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 + 1)}
a(n) = {my(s=0); forpart(p=n, s+=permcount(p)*2^edges(p)); s/n!} \\ Andrew Howroyd, Oct 22 2017
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CROSSREFS
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KEYWORD
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nonn,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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