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
Keywords
Links
- Andrew Howroyd, Table of n, a(n) for n = 1..200
Crossrefs
Programs
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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
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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
Formula
a(n) = n! * (A013939(n) - floor((n + 2)/4)) / 2. - Andrew Howroyd, Jul 30 2018
Extensions
Some incorrect conjectures removed by Andrew Howroyd, Jul 30 2018
Terms a(9) and beyond from Andrew Howroyd, Jul 30 2018
Comments