cp's OEIS Frontend

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

%I A102232 #24 Jan 12 2025 10:18:07
%S A102232 0,0,0,6,60,510,4620,47166,545580,7086750,102246540,1622630526,
%T A102232 28091563500,526858340190,10641342954060,230283190945086,
%U A102232 5315654681915820,130370767029004830,3385534663256583180,92801587319327886846,2677687796244383154540,81124824998504071784670
%N A102232 Number of preferential arrangements of n labeled elements when at least k=three ranks are required.
%C A102232 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.
%F A102232 E.g.f.: (1-exp(z))^3/(exp(z)-2).
%e A102232 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.
%e A102232 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.
%e A102232 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.
%p A102232 series((1-exp(z))^3/(exp(z)-2),z=0,30);
%p A102232 spec := [S,
%p A102232 {
%p A102232    B = Set(Z, 1 <= card),
%p A102232    C = Sequence(B, 2 <= card),
%p A102232    S = Prod(B, C)
%p A102232 }, labeled]:
%p A102232 struct := n -> combstruct[count](spec, size = n);
%p A102232 seq(struct(n), n = 0..21); # _Peter Luschny_, Jul 22 2014
%t A102232 m = 22; CoefficientList[(1-E^(z))^3/(E^z-2) + O[z]^m, z] Range[0, m-1]! (* _Jean-François Alcover_, Jun 11 2019 *)
%Y A102232 Cf. A000670, A025875, A000295.
%K A102232 nonn
%O A102232 0,4
%A A102232 _Thomas Wieder_, Jan 01 2005
%E A102232 More terms from _Peter Luschny_, Jul 22 2014