A181967 Sum of the sizes of the normalizers of all prime order cyclic subgroups of the alternating group A_n.

0, 0, 3, 24, 180, 1440, 12600, 120960, 1270080, 14515200, 179625600, 2634508800, 37362124800, 566658892800, 9807557760000, 167382319104000, 3023343138816000, 57621363351552000, 1155628453883904000, 25545471085854720000, 587545834974658560000, 13488008733331292160000

Offset

1,3

Comments

The first 11 terms of this sequence are the same as A317527. - Andrew Howroyd, Jul 30 2018

Links

Andrew Howroyd, Table of n, a(n) for n = 1..200

Formula

a(n) = n! * (A013939(n) - floor((n + 2)/4)) / 2. - Andrew Howroyd, Jul 30 2018

Prog

(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
(GAP)
a:=function(n) local total, perm, g, p, k;
  total:= 0; g:= AlternatingGroup(n);
  for p in Filtered([2..n], IsPrime) do for k in [1..QuoInt(n, p)] do
     if p>2 or IsEvenInt(k) then
       perm:=PermList(List([0..p*k-1], i->i - (i mod p) + ((i + 1) mod p) + 1));
       total:=total + Size(Normalizer(g, perm)) * Factorial(n) / (p^k * (p-1) * Factorial(k) * Factorial(n-k*p));
     fi;
  od; od;
  return total;
end; # Andrew Howroyd, Jul 30 2018
(PARI) a(n)={n!*sum(p=2, n, if(isprime(p), if(p==2, n\4, n\p)))/2} \\ Andrew Howroyd, Jul 30 2018

Crossrefs

Cf. A181951 for the number of such subgroups.

Cf. A181966 is the symmetric group case.

Cf. A013939, A317527.

Sequence in context: A073985 A197209 A317527 * A144087 A110347 A213100

Adjacent sequences: A181964 A181965 A181966 * A181968 A181969 A181970

Keyword

nonn

Author

Olivier Gérard, Apr 04 2012

Extensions

Some incorrect conjectures removed by Andrew Howroyd, Jul 30 2018

Terms a(9) and beyond from Andrew Howroyd, Jul 30 2018

Status

approved