A344219 - OEIS

%I #31 Nov 15 2022 09:17:04

%S 1,32,122,528,782,3904,2802,8464,9923,25024,16106,64416,30942,89664,

%T 95404,135440,88742,317536,137562,412896,341844,515392,292562,1032608,

%U 488907,990144,803804,1479456,732542,3052928,954306,2167056,1964932,2839744,2191164,5239344,1926222

%N Number of cyclic subgroups of the group (C_n)^5, where C_n is the cyclic group of order n.

%C Inverse Moebius transform of A160893.

%H Seiichi Manyama, <a href="/A344219/b344219.txt">Table of n, a(n) for n = 1..10000</a>

%H László Tóth, <a href="http://arxiv.org/abs/1203.6201">On the number of cyclic subgroups of a finite abelian group</a>, arXiv: 1203.6201 [math.GR], 2012.

%F a(n) = Sum_{x_1|n, x_2|n, x_3|n, x_4|n, x_5|n} phi(x_1)*phi(x_2)*phi(x_3)*phi(x_4)*phi(x_5)/phi(lcm(x_1, x_2, x_3, x_4, x_5)).

%F If p is prime, a(p) = 1 + (p^5 - 1)/(p - 1).

%F From _Amiram Eldar_, Nov 15 2022: (Start)

%F Multiplicative with a(p^e) = 1 + ((p^5 - 1)/(p - 1))*((p^(4*e) - 1)/(p^4 - 1)).

%F Sum_{k=1..n} a(k) ~ c * n^5, where c = (zeta(5)/5) * Product_{p prime} (1 + 1/p^2 + 1/p^3 + 1/p^4 + 1/p^5) = 0.3939461744... . (End)

%t f[p_, e_] := 1 + ((p^5 - 1)/(p - 1))*((p^(4*e) - 1)/(p^4 - 1)); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 40] (* _Amiram Eldar_, Nov 15 2022 *)

%o (PARI) a(n) = sumdiv(n, i, sumdiv(n, j, sumdiv(n, k, sumdiv(n, l, sumdiv(n, m, eulerphi(i)*eulerphi(j)*eulerphi(k)*eulerphi(l)*eulerphi(m)/eulerphi(lcm([i, j, k, l, m])))))));

%o (PARI) a160893(n) = sumdiv(n, d, moebius(n/d)*d^5)/eulerphi(n);

%o a(n) = sumdiv(n, d, a160893(d));

%Y Cf. A000010, A013663, A060648, A064969, A160893, A280184.

%K nonn,mult

%O 1,2

%A _Seiichi Manyama_, May 12 2021