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%I #24 Dec 13 2015 07:18:02
%S 1,1,2,3,3,4,4
%N The smallest positive integer k such that the symmetric group S_n is a product of k cyclic groups.
%C Since S_{n+1} is a product of a subgroup isomorphic to S_n and the cyclic group <(1,2,3,...,n+1)> we have a(n+1) <= a(n) + 1. On the other hand it is not clear that a(n) <= a(n+1) for all n. A lower bound is given by A226143(n) = ceiling(log(m)(n!)), m = A000793(n), a sequence that is not nondecreasing.
%C This sequence was suggested by a posting of _L. Edson Jeffery_ on the seqfans mailing list on May 24, 2013.
%C Cardinality of the smallest subset(s) X of S_n such that every permutation in S_n can be expressed as a product of some elements in X. - _Joerg Arndt_, Dec 13 2015
%H Miklós Abért, <a href="http://dx.doi.org/10.1112/S0024609302001042">Symmetric groups as products of Abelian subgroups</a>, Bull. Lond. Math. Soc., Volume 34, Issue 04, July 2002, pp. 451-456.
%e a(7) = 4 since a factorization of S_7 is given by C_1*C_2*C_3*C_4 where
%e C_1 = <(1,2,3,4)(5,6,7)>,
%e C_2 = <(1,4,6)(2,3,5,7)>,
%e C_3 = <(1,2,5,7)(3,4,6)>,
%e C_4 = <(1,3,5,6,7)(2,4)>,
%e and a brute force computation shows that S_7 is not a product of 3 or fewer cyclic subgroups.
%Y Cf. A051625, A225788.
%K nonn,hard,more,nice
%O 1,3
%A _W. Edwin Clark_, May 27 2013
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