login
A124578
Define p(alpha,2) to be the number of H-conjugacy classes where H is an infant subgroup ( similar to Young subgroups of S_n) of type alpha of the hyperoctahedral group B_n. Then a(n) = sum p(alpha,2) where |alpha| = n and alpha has at most n parts.
1
2, 16, 150, 1784, 25460
OFFSET
1,1
COMMENTS
p((n,0),2) = A000712. B_n can also be thought of as the signed permutation group. B_3 acts on the alphabet {1,2,3,bar{1}, bar{2}, bar{3}}. An infant subgroup of type (2,1) will be the subgroup which stabilizes the sets {1,bar{1}, 2, bar{2}} and {3,bar{3}}.
REFERENCES
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.
LINKS
FORMULA
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,2) we do the following. sum_alpha p(alpha,2)x^alpha = prod_(b in Phi) prod_(k = 1)^infinity 1/(1- x^(C(b) times k ))^2 = prod_(b in Phi) prod_(k = 1)^infinity (1+ x^(k times C(b)) + x^(2k times C(b)) + x^(3k times C(b)) + ....)^2
EXAMPLE
E.g p((2,1),2) = # H-conjugacy classes of B_3 where H = Inft((2,1)) isom B_2 times B_1 . Then a(3) = p((3),2) + p((2,1),2) + p((2,0,1),2) + p((1,2),2) + p((1,1,1),2)+ p((1,0,2),2)+ p((0,3),2) + p((0,2,1),2) + p((0,1,2),2) + p((0,0,3),2) =10 + 16 + 16 + 16 + 24 + 16 + 10 + 16 + 16 +10 = 150
CROSSREFS
Sequence in context: A337793 A103885 A262266 * A332566 A085510 A290785
KEYWORD
more,nonn
AUTHOR
Richard Bayley (r.t.bayley(AT)qmul.ac.uk), Nov 12 2006
STATUS
approved