%I #53 Jul 30 2018 20:29:10
%S 0,0,3,24,180,1440,12600,120960,1270080,14515200,179625600,2634508800,
%T 37362124800,566658892800,9807557760000,167382319104000,
%U 3023343138816000,57621363351552000,1155628453883904000,25545471085854720000,587545834974658560000,13488008733331292160000
%N Sum of the sizes of the normalizers of all prime order cyclic subgroups of the alternating group A_n.
%C The first 11 terms of this sequence are the same as A317527. - _Andrew Howroyd_, Jul 30 2018
%H Andrew Howroyd, <a href="/A181967/b181967.txt">Table of n, a(n) for n = 1..200</a>
%F a(n) = n! * (A013939(n) - floor((n + 2)/4)) / 2. - _Andrew Howroyd_, Jul 30 2018
%o (GAP) List([1..7], n->Sum(Filtered( ConjugacyClassesSubgroups( AlternatingGroup(n)), x->IsPrime( Size( Representative(x))) ), x->Size(x)*Size( Normalizer( AlternatingGroup(n), Representative(x))) )); # _Andrew Howroyd_, Jul 30 2018
%o (GAP)
%o a:=function(n) local total, perm, g, p, k;
%o total:= 0; g:= AlternatingGroup(n);
%o for p in Filtered([2..n], IsPrime) do for k in [1..QuoInt(n,p)] do
%o if p>2 or IsEvenInt(k) then
%o perm:=PermList(List([0..p*k-1], i->i - (i mod p) + ((i + 1) mod p) + 1));
%o total:=total + Size(Normalizer(g, perm)) * Factorial(n) / (p^k * (p-1) * Factorial(k) * Factorial(n-k*p));
%o fi;
%o od; od;
%o return total;
%o end; # _Andrew Howroyd_, Jul 30 2018
%o (PARI) a(n)={n!*sum(p=2, n, if(isprime(p), if(p==2, n\4, n\p)))/2} \\ _Andrew Howroyd_, Jul 30 2018
%Y Cf. A181951 for the number of such subgroups.
%Y Cf. A181966 is the symmetric group case.
%Y Cf. A013939, A317527.
%K nonn
%O 1,3
%A _Olivier GĂ©rard_, Apr 04 2012
%E Some incorrect conjectures removed by _Andrew Howroyd_, Jul 30 2018
%E Terms a(9) and beyond from _Andrew Howroyd_, Jul 30 2018