A344219 Number of cyclic subgroups of the group (C_n)^5, where C_n is the cyclic group of order n.

1, 32, 122, 528, 782, 3904, 2802, 8464, 9923, 25024, 16106, 64416, 30942, 89664, 95404, 135440, 88742, 317536, 137562, 412896, 341844, 515392, 292562, 1032608, 488907, 990144, 803804, 1479456, 732542, 3052928, 954306, 2167056, 1964932, 2839744, 2191164, 5239344, 1926222

Offset: 1

Author

Seiichi Manyama, May 12 2021

Keywords

• nonn
• mult

Comments

Inverse Moebius transform of A160893.

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..10000
• László Tóth, On the number of cyclic subgroups of a finite abelian group, arXiv: 1203.6201 [math.GR], 2012.

Crossrefs

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

Programs

Mathematica

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 *)

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])))))));

PARI

a160893(n) = sumdiv(n, d, moebius(n/d)*d^5)/eulerphi(n);
a(n) = sumdiv(n, d, a160893(d));

Formula

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)).

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

From Amiram Eldar, Nov 15 2022: (Start)

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

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)