A104533 E.g.f.: exp(2x/(1-2x)).
1, 2, 12, 104, 1168, 16032, 259264, 4817024, 100954368, 2353435136, 60355677184, 1687701792768, 51077784506368, 1662782678736896, 57917727119818752, 2148722382829027328, 84569896954751942656, 3518839711497761980416, 154306731918073225019392
Offset: 0
Keywords
Examples
Let "a_i" and "b_j" be elements situated in the classes A and B with _i and _j as labels. Let : denote a separator among levels (ranks). Let | denote a separator among groups. E.g., a_1:b_2|b_1 is a hierarchy composed of two groups which contain three elements in total. a(2) = 12 from b_2:b_1, b_2:a_1, b_2|b_1, a_1:a_2, b_2:a_1, a_1|a_2, a_1:b_2, a_2:a_1, b_1:a_2, a_2:b_1, b_1|a_2, b_2:b_1.
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..398 (terms 0..150 from Alois P. Heinz)
- N. J. A. Sloane and Thomas Wieder, The Number of Hierarchical Orderings, arXiv:math/0307064 [math.CO], 2003; Order 21 (2004), 83-89.
Programs
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Maple
SetSeqUnnL := [T, {T=Set(S,card>=1), S=Sequence(U,card>=1), U=Union(a,b),a=Atom, b=Atom},labeled]; seq(count(SetSeqUnnL,size=j),j=1..20); A104533 := proc(n::integer) local i,j,prttnlst,prttn,liste,ZahlVerschiedenerTeile,H,Mltplztt; Mltplztt:=vector[1000]; prttnlst:=partition(n); H := 0; for i from 1 to nops(prttnlst) do prttn := prttnlst[i]; liste := convert(prttn,multiset); ZahlVerschiedenerTeile := nops(liste); for j from 1 to ZahlVerschiedenerTeile do Mltplztt[j] := op(2,op(j,liste)); od; H := H + (n!/mul(Mltplztt[j]!,j=1..ZahlVerschiedenerTeile)) * 2^n; od; print(n,H); end proc; -
Mathematica
CoefficientList[Exp[2 x/(1 - 2 x)] + O[x]^21, x]*Range[0, 20]! (* or: *) a[0] = 1; a[n_] := 2^n*n!*Hypergeometric1F1[n + 1, 2, 1]/E; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Nov 10 2017 *)
Formula
a(n) = 2^n*A000262(n) = 2^n*n!*Sum_{k=0..n} C(n-1,k)/(k+1)!. - Paul Barry, Apr 28 2007
With p(n) = the number of integer partitions of n, d(i) = the number of different parts of the i-th partition of n, m(i, j) = multiplicity of the j-th part of the i-th partition of n, sum_{i=1}^{p(n)} = sum over i and prod_{j=1}^{d(i)} = product over j one has: a(n)=sum_{i=1}^{p(n)} n!/(prod_{j=1}^{d(i)} m(i, j)!) * 2^(n)
E.g.f.: E(0)/2, where E(k)= 1 + 1/(1 - 2*x/(2*x + (k+1)*(1-2*x)/E(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 09 2013
E.g.f.: E(0) - 1, where E(k) = 2 + 2*x/((2*k+1)*(1-2*x) - 2*x/E(k+1) ); (continued fraction ). - Sergei N. Gladkovskii, Dec 31 2013
Extensions
Edited by N. J. A. Sloane, May 06 2008, at the suggestion of Joerg Arndt
Comments