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

0,4

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.

Links

Table of n, a(n) for n=0..21.

Formula

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

Example

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.

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 *)

Crossrefs

Cf. A000670, A025875, A000295.

Sequence in context: A061495 A220411 A248217 * A121113 A213269 A091710

Adjacent sequences: A102229 A102230 A102231 * A102233 A102234 A102235

Keyword

nonn

Author

Thomas Wieder, Jan 01 2005

Extensions

More terms from Peter Luschny, Jul 22 2014

Status

approved