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A051636 Number of "labeled" cyclic subgroups of alternating group A_n.

%I A051636 #29 Jun 04 2021 09:41:18
%S A051636 1,1,2,8,32,167,947,6974,53426,454682,4303532,50366912,553031624,
%T A051636 6760260236,90333982832,1369522152392,20986020606632,350528387240264,
%U A051636 5751957395258096,111685506968916032,2139383543480892032,41770889787378732752,869742098042083451264
%N A051636 Number of "labeled" cyclic subgroups of alternating group A_n.
%H A051636 Alois P. Heinz, <a href="/A051636/b051636.txt">Table of n, a(n) for n = 1..140</a>
%H A051636 L. Naughton and G. Pfeiffer, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL16/Naughton/naughton2.html">Integer Sequences Realized by the Subgroup Pattern of the Symmetric Group</a>, J. Int. Seq. 16 (2013) #13.5.8.
%F A051636 a(n) = 1/2*Sum_{pi} (1+(-1)^(k_2+k_4+...)) * n!/(k_1!*1^k_1*k_2!*2^k_2*...*k_n!*n^k_n*phi(lcm{i:k_i != 0})), where pi runs through all partitions k_1+2*k_2+...+n*k_n=n and phi is Euler's function.
%p A051636 b:= proc(n, i, m, t) option remember; `if`(n=0, (1+(-1)^t)/numtheory
%p A051636       [phi](m), add(1/j!/i^j*b(n-i*j, i-1, ilcm(m, `if`(j=0, 1, i)),
%p A051636        irem(t+j*irem(i+1, 2), 2)), j=`if`(i=1, n, 0..n/i)))
%p A051636     end:
%p A051636 a:= n-> n!*b(n$2, 1, 0)/2:
%p A051636 seq(a(n), n=1..25);  # _Alois P. Heinz_, Jul 03 2018
%t A051636 f[list_] :=Total[list]!/(Apply[Times, list]*Apply[Times, Map[Length, Split[list]]!])/EulerPhi[Apply[LCM, list]]; Table[Total[Map[f,
%t A051636    Select[IntegerPartitions[n],EvenQ[Length[Select[#, EvenQ[#] &]]] &]]], {n, 1, 21}] (* _Geoffrey Critzer_, Oct 03 2015 *)
%t A051636 b[n_, i_, m_, t_] := b[n, i, m, t] = If[n == 0, (1 + (-1)^t)/
%t A051636      EulerPhi[m], If[i == 0, 0, Sum[1/j!/i^j*b[n - i*j, i - 1, LCM[m,
%t A051636      If[j == 0, 1, i]], Mod[t+j*Mod[i+1, 2], 2]], {j, Range[0, n/i]}]]];
%t A051636 a[n_] := n! b[n, n, 1, 0]/2;
%t A051636 Array[a, 25] (* _Jean-François Alcover_, Jun 04 2021, after _Alois P. Heinz_ *)
%o A051636 (PARI) \\ permcount is number of permutations of given type.
%o A051636 permcount(v) = {my(m=1,s=0,k=0,t); for(i=1,#v,t=v[i]; k=if(i>1&&t==v[i-1],k+1,1); m*=t*k;s+=t); s!/m}
%o A051636 a(n)={my(s=0); forpart(p=n, if(sum(i=1, #p, p[i]-1)%2==0, s+=permcount(p) / eulerphi(lcm(Vec(p))))); s} \\ _Andrew Howroyd_, Jul 03 2018
%Y A051636 Row sums of A303728.
%Y A051636 Cf. A000010, A051625.
%K A051636 easy,nonn
%O A051636 1,3
%A A051636 _Vladeta Jovovic_