A344302 Number of cyclic subgroups of the group (C_n)^6, where C_n is the cyclic group of order n.

1, 64, 365, 2080, 3907, 23360, 19609, 66592, 88817, 250048, 177157, 759200, 402235, 1254976, 1426055, 2130976, 1508599, 5684288, 2613661, 8126560, 7157285, 11338048, 6728905, 24306080, 12210157, 25743040, 21582653, 40786720, 21243691, 91267520, 29583457

Offset: 1

Author

Seiichi Manyama, May 14 2021

Keywords

• nonn
• mult

Comments

Inverse Moebius transform of A160895.

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, A013664, A060648, A064969, A160895, A280184, A344219, A344303, A344304, A344305, A344306.

Programs

Mathematica

f[p_, e_] := 1 + ((p^6 - 1)/(p - 1))*((p^(5*e) - 1)/(p^5 - 1)); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 30] (* Amiram Eldar, Nov 15 2022 *)

PARI

a160895(n) = sumdiv(n, d, moebius(n/d)*d^6)/eulerphi(n);
a(n) = sumdiv(n, d, a160895(d));

Formula

a(n) = Sum_{x_1|n, x_2|n, ..., x_6|n} phi(x_1)*phi(x_2)* ... *phi(x_6)/phi(lcm(x_1, x_2, ..., x_6)).

If p is prime, a(p) = 1 + (p^6 - 1)/(p - 1).

From Amiram Eldar, Nov 15 2022: (Start)

Multiplicative with a(p^e) = 1 + ((p^6 - 1)/(p - 1))*((p^(5*e) - 1)/(p^5 - 1)).

Sum_{k=1..n} a(k) ~ c * n^6, where c = (zeta(6)/6) * Product_{p prime} ((1-1/p^5)/(p^2*(1-1/p))) = 0.32592074105... . (End)