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%I #15 Jun 11 2019 10:45:08
%S 0,0,0,6,60,510,4620,47166,545580,7086750,102246540,1622630526,
%T 28091563500,526858340190,10641342954060,230283190945086,
%U 5315654681915820,130370767029004830,3385534663256583180,92801587319327886846,2677687796244383154540,81124824998504071784670
%N Number of preferential arrangements of n labeled elements when at least k=three ranks are required.
%C 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 E.g.f.: -(exp(z)^3-3*exp(z)^2+3*exp(z)-1)/(-2+exp(z)).
%e 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 For n=3 we have obviously a(3)=6 possible rankings:
%e 2|3|1, 3|2|1, 1|2|3, 2|1|3, 3|1|2, 1|3|2.
%e 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)
%e 31|2|4, 2|4|31, 4|1|3|2.
%p series(-(exp(z)^3-3*exp(z)^2+3*exp(z)-1)/(-2+exp(z)),z=0,30);
%p spec := [S,
%p {
%p B = Set(Z, 1 <= card),
%p C = Sequence(B, 2 <= card),
%p S = Prod(B, C)
%p }, labeled]:
%p struct := n -> combstruct[count](spec, size = n);
%p seq(struct(n), n = 0..21); # _Peter Luschny_, Jul 22 2014
%t m = 22; CoefficientList[-(E^(3z) - 3*E^(2z) + 3*E^z - 1)/(-2 + E^z) + O[z]^m, z] Range[0, m-1]! (* _Jean-François Alcover_, Jun 11 2019 *)
%Y Cf. A000670, A025875, A000295.
%K nonn
%O 0,4
%A _Thomas Wieder_, Jan 01 2005
%E More terms from _Peter Luschny_, Jul 22 2014
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