cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A075729 Number of different hierarchical orderings that can be formed from n labeled elements: these are divided into groups and the elements in each group are then arranged in a "preferential arrangement" or "weak order" as in A000670.

Original entry on oeis.org

1, 1, 4, 23, 173, 1602, 17575, 222497, 3188806, 50988405, 899222457, 17329515172, 362164300173, 8155216185781, 196789115887252, 5064722539020379, 138457553073641465, 4006059432756066914, 122284085809137076203, 3926775294104305483621, 132313462760902116605534
Offset: 0

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Author

Thomas Wieder and N. J. A. Sloane, Oct 06 2002

Keywords

Comments

If all individuals form a single society ("uniparate society"), then the number of different hierarchies for that single society is equal to the ordered Bell number Bell_ordered(n) (A000670).
Represent a labeled pre-order (quasi-order, topology, A000798) as a directed graph. a(n) is the number of such digraphs in which the underlying graph of each component is complete. a(3)=23 because there are 29 such digraphs but o->o<-o and o<-o->o are not counted. Each has 3 labelings. 29 - 6 = 23. - Geoffrey Critzer, Jul 30 2014

Examples

			a(3) = 23: Let the n = 3 individuals be named 1, 2 and 3. Let a pair of parentheses () indicate a society and let square brackets [] denote a set of disparate societies. Finally, let the ranks be ordered from left to right and separated by a colon, e.g., (1,2:3) is a society with individual 3 on top and individuals 1 and 2 on the same bottom rank.
Then the hierarchical ordering for n = 3 is composed of the following sets: [(1),(2),(3)], [(1,2)(3)], [(3,2)(1)], [(3,1)(2)], [(1:2)(3)], [(3:2)(1)], [(1:3)(2)], [(2:1)(3)], [(2:3)(1)], [(3:1)(2)], [(3:2:1)], [(1:3:2)], [(2:1:3)], [(1:2:3)], [(3:1:2)], [(2:3:1)], [(1,3:2)], [(3,2:1)], [(2,1:3)], [(3:1,2)], [(1:2,3)], [(2:3,1)], [(1,2,3)].
		

Crossrefs

Cf. A000670, A075744. See A075900 for the unlabeled case.

Programs

  • Maple
    A075729 := n->n!*exp(1/4/ln(2)-3/4)/2/sqrt(Pi)/(2*ln(2))^(1/4)*exp(-n*ln(ln(2)))*exp(sqrt(2*n/ln(2)))*n^(-3/4);
    with(combstruct); SetSeqSetL := [T, {T=Set(S), S=Sequence(U,card >= 1), U=Set(Z,card >=1)},labeled]; seq(count(SetSeqSetL,size=j),j=1..12);
    # alternative Maple program:
    b:= proc(n) option remember: `if`(n<2, 1,
          (2*n-1)*b(n-1) -(n-1)*(n-2)*b(n-2))
        end:
    a:= n-> add(b(k)*Stirling2(n,k), k=0..n):
    seq(a(n), n=0..20);  # Alois P. Heinz, May 22 2018
  • Mathematica
    Range[0, 20]!CoefficientList[Series[E^(1/(2 - E^x) - 1), {x, 0, 20}], x] (* Robert G. Wilson v, Jul 13 2004 *)
    Fubini[n_, r_] := Sum[k!*Sum[(-1)^(i+k+r)(i+r)^(n-r)/(i!*(k-i-r)!), {i, 0, k-r}], {k, r, n}]; Fubini[0, 1] = 1; a[0] = 1; a[n_] := a[n] = (n-1)! Sum[a[n-k] Fubini[k, 1]/((n-k)! (k-1)!), {k, 1, n}]; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Mar 31 2016 *)
    Table[Sum[BellY[n, k, PolyLog[-Range[n], 1/2]/2], {k, 0, n}], {n, 0, 20}] (* Vladimir Reshetnikov, Nov 09 2016 *)
    With[{nn=20},CoefficientList[Series[Exp[1/(2-Exp[x])-1],{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, Aug 26 2022 *)
  • Maxima
    a(n):= sum(sum(stirling2(n,k)*k!*binomial(k-1,m-1), k,m,n)/m!, m,1,n) /* Vladimir Kruchinin, Aug 10 2010 */

Formula

E.g.f.: exp(f(x)-1) where f(x) = 1/(2-exp(x)) = e.g.f. for A000670.
STIRLINGi transform of A000262.
a(n) = (n-1)! * Sum_k=1^n a(n-k)*b(k)/((n-k)!*(k-1)!); a(n) = a(n) + C(n-1, k-1)*a(n-k)*b(k) (where b(n) = A000670(n)). - Thomas Wieder, Dec 31 2002
a(n) = (Sum_{j=1..n} m(j))*(n!*Product_{j=1..n} B(j)^m(j))/(Product_{j=1..n} (m(j))!*(j!)^m(j)), where the sum is over all (m(1),m(2),...,m(n)) such that Sum_{j=1..n} (j*m(j)) = n. - Thomas Wieder, May 18 2003
a(n) is asymptotic to exp(1/(4*log(2))-3/4) /(2*sqrt(Pi*sqrt(2*log(2)))) *n!*exp(-log(log(2))*n)*exp(sqrt(2*n /log(2))) /n^(3/4). Calculated using the Maple package "algolib", using the command "equivalent(exp(1/(2-exp(x))-1), x, n);". - Thomas Wieder, Nov 12 2002
a(n) = Sum_{k=0..n} A079641(n,k)*A000110(k). - Vladeta Jovovic, Sep 25 2006
a(n) = sum(sum(stirling2(n,k)*k!*C(k-1,m-1), k=m..n)/m!, m=1..n). - Vladimir Kruchinin, Aug 10 2010