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A192983 a(n) is the number of pairs (g, h) of elements of the symmetric group S_n such that g and h have conjugates that commute. 1

%I #41 Apr 21 2015 12:42:43

%S 1,4,24,264,5640,151200,5722920,282868992,18371308032,1504791561600,

%T 148978034686800,18007146260231040,2528615024682544512,

%U 426310052282058252672,81830910530970671616000,18305445786667543107072000,4570435510076312321728158720

%N a(n) is the number of pairs (g, h) of elements of the symmetric group S_n such that g and h have conjugates that commute.

%C a(n) / n!^2 is the probability that two permutation in S_n, chosen independently and uniformly at random, have conjugates that commute.

%C Apparently n | a(n), and, for n>1, n*(n-1) | a(n). - Alexander R. Povolotsky, Sep 30 2011

%H Simon R. Blackburn, John R. Britnell, and Mark Wildon, <a href="http://arxiv.org/abs/1108.1784">The probability that a pair of elements of a finite group are conjugate</a>, arXiv:1108.1784, 2011

%H J. R. Britnell and M. Wildon, <a href="http://dx.doi.org/10.1515/JGT.2009.013">Commuting elements in conjugacy classes: an application of Hall's Marriage Theorem to group theory</a>, J. Group Theory, 12 (2009), 795-802.

%H Mark Wildon, <a href="http://www.ma.rhul.ac.uk/~uvah099/other.html#ConjugacyProbability">Haskell source code</a> for computing values of the sequence.

%e For n = 3 the probability that two elements of S_3 have conjugates that commute is a(3)/3!^2 = 2/3. Proof: only the transpositions and three cycles fail to have conjugates that commute; the probability of choosing one permutation from each of these classes is 2*1/2*1/3 = 1/3.

%o (Haskell) See links for code.

%Y Cf. A087132 (the sum of squares of the sizes of the conjugacy classes of S_n).

%K nonn

%O 1,2

%A _Mark Wildon_, Aug 03 2011

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