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

1,2

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.

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)

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 *)

Prog

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

Crossrefs

Cf. A000010, A013664, A060648, A064969, A160895, A280184, A344219, A344303, A344304, A344305, A344306.

Sequence in context: A302414 A303181 A017486 * A105918 A224042 A188840

Adjacent sequences: A344299 A344300 A344301 * A344303 A344304 A344305

Keyword

nonn,mult

Author

Seiichi Manyama, May 14 2021

Status

approved