A344306 Number of cyclic subgroups of the group (C_n)^10, where C_n is the cyclic group of order n.

1, 1024, 29525, 524800, 2441407, 30233600, 47079209, 268698112, 581150417, 2500000768, 2593742461, 15494720000, 11488207655, 48209110016, 72082541675, 137573433856, 125999618779, 595098027008, 340614792101, 1281250393600

Offset: 1

Author

Seiichi Manyama, May 14 2021

Keywords

• nonn
• mult

Comments

Inverse Moebius transform of A160957.

Links

• Amiram Eldar, 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, A013668, A160957, A060648, A064969, A280184, A344219, A344302, A344303, A344304, A344305.

Programs

Mathematica

f[p_, e_] := 1 + ((p^10 - 1)/(p - 1))*((p^(9*e) - 1)/(p^9 - 1)); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 30] (* Amiram Eldar, Nov 15 2022 *)

PARI

a160957(n) = sumdiv(n, d, moebius(n/d)*d^10)/eulerphi(n);
a(n) = sumdiv(n, d, a160957(d));

Formula

a(n) = Sum_{x_1|n, x_2|n, ..., x_10|n} phi(x_1)*phi(x_2)* ... *phi(x_10)/phi(lcm(x_1, x_2, ..., x_10)).

If p is prime, a(p) = 1 + (p^10 - 1)/(p - 1).

From Amiram Eldar, Nov 15 2022: (Start)

Multiplicative with a(p^e) = 1 + ((p^10 - 1)/(p - 1))*((p^(9*e) - 1)/(p^9 - 1)).

Sum_{k=1..n} a(k) ~ c * n^10, where c = (zeta(10)/10) * Product_{p prime} ((1-1/p^9)/(p^2*(1-1/p))) = 0.1944248708... . (End)