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#62 by Peter Luschny at Sun Jan 29 06:03:17 EST 2023
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#61 by Joerg Arndt at Sun Jan 29 02:30:07 EST 2023
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#60 by Michel Marcus at Sat Jan 28 18:03:31 EST 2023
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#59 by Michel Marcus at Sat Jan 28 18:03:23 EST 2023
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| LINKS
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N. Bergeron, C. Reutenauer, M. Rosas, and M. Zabrocki, <a href="httphttps://arxiv.org/abs/math.CO/0502082">Invariants and Coinvariants of the Symmetric Group in Noncommuting Variables</a> , <>, arXiv:math/0502082 [math.CO], 2005; <a href="http://www.ams.org/mathscinet-getitem?mr=2398749">MR2398749</a>, Cand. J. Math 60 (2008) 266-296.
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| STATUS
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proposed
editing
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#58 by Jon E. Schoenfield at Sat Jan 28 17:52:08 EST 2023
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#57 by Jon E. Schoenfield at Sat Jan 28 17:51:53 EST 2023
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| COMMENTS
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Also the number of non-splitable set partitions (see Bergeron et al. reference) of length <= <= 4.
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| FORMULA
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O.g.f.: (1-5q+5q - 5*q + 5*q^2)/(1-6q+9q - 6*q + 9*q^2-3q - 3*q^3) = 1 - 1/(sumSum_{k=0}^..4 } q^k/(prodProduct_{i=1}^..k (} (1-i*q))).
a(n) = 6a6*a(n-1) - 9a9*a(n-2) + 3a3*a(n-3). - David Nacin, Feb 11 2012
a(n) = (1/3)*(x^(n-2) + y^(n-2) + z^(n-2)) for x=( = (2*cos (Pi/18)^))^2, y=( = (2*cos (5*Pi/18)^))^2, and z=( = (2*cos (7*Pi/18)^))^2. - Greg Dresden, Jan 28 2023
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proposed
editing
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#56 by Greg Dresden at Sat Jan 28 16:38:43 EST 2023
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#55 by Greg Dresden at Sat Jan 28 16:38:38 EST 2023
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| FORMULA
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a(n) = (1/3)*(x^(n-2) + y^(n-2) + z^(n-2)) for x=(2*cos Pi/18)^2, y=(2*cos 5*Pi/18)^2, and z=(2*cos 7*Pi/18)^2. - Greg Dresden, Jan 28 2023
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| STATUS
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approved
editing
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#54 by Susanna Cuyler at Mon May 17 08:48:46 EDT 2021
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#53 by Joerg Arndt at Mon May 17 07:41:49 EDT 2021
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