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Search: id:A124577
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| A124577 |
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Define p(alpha) to be the number of H-conjugacy classes where H is a Young subgroup of type alpha of the symmetric group S_n. Then a(n) = sum p(alpha) where |alpha| = n and alpha has at most n parts. |
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3
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OFFSET
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1,2
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COMMENT
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p((0,n)) = A000041, p((1,n)) = A000070, p((2,n) = A093695
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REFERENCES
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Richard Bayley, Relative Character Theory and the Hyperoctahedral Group, Ph.D. thesis, Queen Mary College, University of London, to be published 2007.
Steve Donkin, Invariant functions on Matrices, Math. Proc. Camb. Phil. Soc. 113 (1993) 23-43.
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LINKS
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Richard Bayley, Homepage.
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FORMULA
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Let x = x_1x_2x_3... and x^alpha = x_1^(alpha_1)x_2^(alpha_2)x_3^(alpha_3).... Let Phi = set of all primitive necklaces. If b is a primitive necklace then C(b) = Content(b) = (beta_1, beta_2,beta_3,.....) where beta_i = the number of times i occurs in b. For example if b=[11233] then C(b) = (2,1,2). To generate the p(alpha) we do the following. sum_alpha p(alpha)x^alpha = prod_(b in Phi) prod_(k = 1)^infinity 1/(1- x^(c(b) times k )) = prod_(b in Phi) prod_(k = 1)^infinity (1+ x^(k times C(b)) + x^(2k times C(b)) + x^(3k times C(b)) + ....)
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EXAMPLE
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E.g p((2,1)) = # H-conjugacy classes of S_3 where H = Yng((2,1)) isom S_2 times S_1 . Then a(3) = p((3)) + p((2,1)) + p((2,0,1)) + p((1,2)) + p((1,1,1))+ p((1,0,2)+ p((0,2,1)) + p((0,1,2)) + p((0,0,3)) = 3+4+4+4+6+4+3+4+4+3 = 39
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PROGRAM
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(GAP)
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CROSSREFS
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Cf, A124578, A000041, A000070, A093695.
Sequence in context: A058191 A113347 A031972 this_sequence A006678 A145709 A034661
Adjacent sequences: A124574 A124575 A124576 this_sequence A124578 A124579 A124580
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KEYWORD
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more,nonn
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AUTHOR
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Richard Bayley (r.t.bayley(AT)qmul.ac.uk), Nov 05 2006
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