A102232 Number of preferential arrangements of n labeled elements when at least k=three ranks are required.

0, 0, 0, 6, 60, 510, 4620, 47166, 545580, 7086750, 102246540, 1622630526, 28091563500, 526858340190, 10641342954060, 230283190945086, 5315654681915820, 130370767029004830, 3385534663256583180, 92801587319327886846, 2677687796244383154540, 81124824998504071784670

Offset: 0

Author

Thomas Wieder, Jan 01 2005

Keywords

• nonn

Comments

The labeled case for k=2 is given by A052875. The unlabeled case for k=3 is given by A000295 = Eulerian numbers 2^n - n - 1. The unlabeled case for k=2 is given by A000225 = 2^n - 1.

Examples

			Let 1,2,3 denote three labeled elements. Let | denote a separation between two ranks. E.g. if element 1 is on rank (also called level) one, element 3 is on rank two and element 2 is on rank three, then we have the ranking 1|3|2.
For n=3 we have obviously a(3)=6 possible rankings: 2|3|1, 3|2|1, 1|2|3, 2|1|3, 3|1|2, 1|3|2.
For n=4 we have a(4) = 60 possible rankings, e.g. (elements 1 and 3 are on the same rank in the first two examples) 31|2|4, 2|4|31, 4|1|3|2.
		

Crossrefs

Cf. A000670, A025875, A000295.

Programs

Maple

series((1-exp(z))^3/(exp(z)-2),z=0,30);
spec := [S,
{
   B = Set(Z, 1 <= card),
   C = Sequence(B, 2 <= card),
   S = Prod(B, C)
}, labeled]:
struct := n -> combstruct[count](spec, size = n);
seq(struct(n), n = 0..21); # Peter Luschny, Jul 22 2014

Mathematica

m = 22; CoefficientList[(1-E^(z))^3/(E^z-2) + O[z]^m, z] Range[0, m-1]! (* Jean-François Alcover, Jun 11 2019 *)

Formula

E.g.f.: (1-exp(z))^3/(exp(z)-2).

Extensions

More terms from Peter Luschny, Jul 22 2014