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 A102232 Number of preferential arrangements of n labeled elements when at least k=three ranks are required. 1
 0, 0, 0, 6, 60, 510, 4620, 47166, 545580, 7086750, 102246540, 1622630526, 28091563500, 526858340190, 10641342954060, 230283190945086, 5315654681915820, 130370767029004830, 3385534663256583180, 92801587319327886846, 2677687796244383154540, 81124824998504071784670 (list; graph; refs; listen; history; text; internal format)
 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 FORMULA G.f.: -(exp(z)^3-3*exp(z)^2+3*exp(z)-1)/(-2+exp(z)). 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(-(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); seq(struct(n), n = 0..21); # Peter Luschny, Jul 22 2014 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

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Last modified May 22 00:25 EDT 2019. Contains 323472 sequences. (Running on oeis4.)