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%I #11 Apr 05 2012 03:01:26
%S 1,1,6,36,240,2160,20160,241920,2903040,39916800,578793600,9580032000,
%T 161902540800,3007651046400,58845346560000,1234444603392000,
%U 26854400821248000,624231436308480000,15083992450695168000,385614968295997440000
%N Sum of the sizes of normalizers of all the cyclic subgroups of Alternating Group of order n.
%C For each cyclic subgroup of the Alternate group on n symbols, add the size of its normalizer (permutations leaving the subgroup invariant by conjugation).
%C a(7) is remarquable because it is equal to the size of Alt(8).
%F a(n) = n!/2 * A046682(n).
%e Decomposing by number of cyclic subgroups * size of normalizer of subgroups
%e a(5) = 1*60 + 4*15 + 6*10 + 0*60 + 10*6 = 240.
%e a(6) = 1*360 + 8*45 + (18*20+18*20) + 8*45 + 10*36 = 2160.
%Y Cf. A053529, A181950.
%K nonn,easy
%O 1,3
%A _Olivier Gérard_, Apr 04 2012
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