A140585 Total number of all hierarchical orderings for all set partitions of n.

1, 4, 20, 129, 1012, 9341, 99213, 1191392, 15958404, 235939211, 3817327362, 67103292438, 1273789853650, 25973844914959, 566329335460917, 13150556885604115, 324045146807055210, 8446201774570017379, 232198473069120178475, 6715304449424099384968

Offset: 1

Author

Thomas Wieder, May 17 2008

Keywords

• nonn

Examples

			We are considering all set partitions of the n-set {1,2,3,...,n}.
For each such set partition we examine all possible hierarchical arrangements of the subsets. A hierarchy is a distribution of elements (sets in the present case) onto levels.
A distribution onto levels is "hierarchical" if a level L+1 contains at most as many elements as level L. Thus for n=4 the arrangement {1,2}:{3}{4} is not allowed.
Let the colon ":" separate two consecutive levels L and L+1.
n=2 --> 1+3=4
{1,2} {1}{2}
{1}:{2}
{2}:{1}
-----------------------
n=3 --> 1+9+10=20
1*1 3*3=9 1*10
{1,2,3} {1,2}{3} {1}{2}{3}
{1,3}{2}
{2,3}{1} {1}{2}:{3}
{3}{1}:{2}
{1,2}:{3} {2}{3}:{1}
{1,3}:{2}
{2,3}:{1} {1}:{2}:{3}
{3}:{1}:{2}
{3}:{1,2} {2}:{3}:{1}
{2}:{1,3} {1}:{3}:{2}
{1}:{2,3} {2}:{1}:{3}
{3}:{2}:{1}
-----------------------
n=4 --> 1+12+9+60+47=129
1*1 4*3=12 3*3=9 6*10=60 1*47
{1,2,3,4} {1,2,3}{4} {1,2}{3,4} {1,2}{3}{4} {1}{2}{3}{4}
{1,2,4}{3} {1,3}{2,4} {1,2}{3}:{4}
{1,3,4}{2} {1,4}{2,3} {1,2}{4}:{3} {1}{2}:{3}:{4}
{2,3,4}{1} {1}{2}:{3,4} {1}{3}:{2}:{4}
{1,2}:{3,4} {1,2}:{3}:{4} {1}{4}:{2}:{3}
{1,2,3}:{4} {1,3}:{2,4} {1,2}:{4}:{3} {1}{2}:{4}:{3}
{1,2,4}:{3} {1,4}:{2,3} {1}:{2}:{3,4} {1}{3}:{4}:{2}
{1,3,4}:{2} {3,4}:{1,2} {2}:{1}:{3,4} {1}{4}:{3}:{2}
{2,3,4}:{1} {2,4}:{1,3} {1}:{3,4}:{2}
{2,3}:{1,4} {2}:{3,4}:{1} {2}{3}:{1}:{4}
{4}:{1,2,3} {2}{4}:{1}:{3}
{3}:{1,2,4} likewise for: {2}{3}:{4}:{1}
{2}:{1,3,4} {3,4}{1}{2} {2}{4}:{3}:{1}
{1}:{2,3,4} {2,4}{1}{3}
{2,3}{1}{4} {3}{4}:{1}:{2}
{1,4}{2}{3} {3}{4}:{2}:{1}
{1,3}{2}{4}
{1}{2}:{3}{4}
{1}{3}:{2}{4}
{1}{4}:{2}{3}
{2}{3}:{1}{4}
{2}{4}:{1}{3}
{3}{4}:{1}{2}
{2}{3}{4}:{1}
{1}{3}{4}:{2}
{1}{2}{4}:{3}
{1}{2}{3}:{4}
{1}:{2}:{3}:{4}
+23 permutations
		

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 1..420 (first 160 terms from Alois P. Heinz)
• Thomas Wieder, The number of certain rankings and hierarchies formed from labeled or unlabeled elements and sets, Applied Mathematical Sciences, vol. 3, 2009, no. 55, 2707 - 2724. [_Thomas Wieder_, Nov 14 2009]

Crossrefs

Cf. A005651, A075729, A034691, A083355, A109186, A000670.

Programs

Maple

A140585 := proc(n::integer) local k, Result; Result:=0: for k from 1 to n do Result:=Result+stirling2(n,k)*A005651(k); end do; lprint(Result); end proc;
E.g.f.: series(1/mul(1-(exp(x)-1)^k/k!,k=1..10),x=0,10). # Thomas Wieder, Sep 04 2008
# second Maple program:
with(numtheory): b:= proc(k) option remember; add(d/d!^(k/d), d=divisors(k)) end: c:= proc(n) option remember; `if`(n=0, 1, add((n-1)!/ (n-k)!* b(k)* c(n-k), k=1..n)) end: a:= n-> add(Stirling2(n, k) *c(k), k=1..n): seq(a(n), n=1..30); # Alois P. Heinz, Oct 10 2008

Mathematica

Table[n!*SeriesCoefficient[1/Product[(1-(E^x-1)^k/k!),{k,1,n}],{x,0,n}],{n,1,20}] (* Vaclav Kotesovec, Sep 03 2014 *)

Formula

Stirling transform of A005651 = Sum of multinomial coefficients: a(n) = Sum_{i=1..n} S2(n,k) A005651(k).

E.g.f.: 1/Product_{k>=1} (1 - (exp(x)-1)^k/k!). - Thomas Wieder, Sep 04 2008

a(n) ~ c * n! / (log(2))^n, where c = 1/(2*log(2)) * Product_{k>=2} 1/(1-1/k!) = A247551 / (2*log(2)) = 1.82463230250447246267598544320244231645906135137... . - Vaclav Kotesovec, Sep 04 2014, updated Jan 21 2017

Extensions

More terms from Alois P. Heinz, Oct 10 2008