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A102232
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Number of preferential arrangements of n labeled elements when at least k=three ranks are required.
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2
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0, 0, 0, 6, 60, 510, 4620, 47166, 545580, 7086750, 102246540, 1622630526, 28091563500, 526858340190, 10641342954060, 230283190945086, 5315654681915820, 130370767029004830, 3385534663256583180, 92801587319327886846, 2677687796244383154540, 81124824998504071784670
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OFFSET
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0,4
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COMMENTS
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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.
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LINKS
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FORMULA
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E.g.f.: -(exp(z)^3-3*exp(z)^2+3*exp(z)-1)/(-2+exp(z)).
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EXAMPLE
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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.
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MAPLE
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series(-(exp(z)^3-3*exp(z)^2+3*exp(z)-1)/(-2+exp(z)), 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);
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MATHEMATICA
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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 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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