A064969 Number of cyclic subgroups of the group C_n X C_n X C_n (where C_n is the cyclic group of order n).

1, 8, 14, 36, 32, 112, 58, 148, 131, 256, 134, 504, 184, 464, 448, 596, 308, 1048, 382, 1152, 812, 1072, 554, 2072, 807, 1472, 1184, 2088, 872, 3584, 994, 2388, 1876, 2464, 1856, 4716, 1408, 3056, 2576, 4736, 1724, 6496, 1894, 4824, 4192, 4432, 2258, 8344

Offset: 1

Author

Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Oct 30 2001

Keywords

• nonn
• mult
• easy

Comments

Inverse Moebius transform of A160889. - Vladeta Jovovic, Nov 21 2009

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• László Tóth, Menon's identity and arithmetical sums representing functions of several variables, Rend. Sem. Mat. Univ. Politec. Torino, 69 (2011), 97-110 and arXiv, arXiv:1103.5861 [math.NT], 2011.
• László Tóth, On the number of cyclic subgroups of a finite abelian group, arXiv:1203.6201 [math.GR], 2012.

Crossrefs

Cf. A000010, A002117, A064803, A060648, A060724, A064767, A160889, A280184, A330595, A344219, A344302, A344303.

Programs

Maple

with(numtheory):
# define Jordan totient function J(r,n)
J(r,n) := add(d^r*mobius(n/d), d in divisors(n)):
seq(add(J(3,d)/phi(d), d in divisors(n)), n = 1..50); # Peter Bala, Jan 23 2024

Mathematica

a[n_] := Sum[EulerPhi[i] EulerPhi[j] (EulerPhi[k] / EulerPhi[LCM[i, j, k]]), {i, Divisors[n]}, {j, Divisors[n]}, {k, Divisors[n]}];
Array[a, 48] (* Jean-François Alcover, Dec 13 2018, after Vladeta Jovovic *)
f[p_, e_] := 1 + (p^2 + p + 1)*((p^(2*e) - 1)/(p^2 - 1)); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 50] (* Amiram Eldar, Nov 15 2022 *)

PARI

a(n) = sumdiv(n, i, sumdiv(n, j, sumdiv(n, k, eulerphi(i)*eulerphi(j)*eulerphi(k)/eulerphi(lcm(lcm(i, j), k))))); \\ Michel Marcus, Dec 14 2018

PARI

a160889(n) = sumdiv(n, d, moebius(n/d)*d^3)/eulerphi(n);
a(n) = sumdiv(n, d, a160889(d)); \\ Seiichi Manyama, May 12 2021

Formula

a(n) = Sum_{i|n, j|n, k|n} phi(i)*phi(j)*phi(k)/phi(lcm(i, j, k)), where phi is Euler totient function (cf. A000010).

From Amiram Eldar, Nov 15 2022: (Start)

Multiplicative with a(p^e) = 1 + (p^2 + p + 1)*((p^(2*e) - 1)/(p^2 - 1)).

Sum_{k=1..n} a(k) ~ c * n^3, where c = (zeta(3)/3) * Product_{p prime} (1 + 1/p^2 + 1/p^3) = A002117 * A330595 / 3 = 0.700772... . (End)

a(n) = Sum_{d divides n} J_3(d)/phi(d) = Sum_{1 <= i, j, k <= n} 1/phi(n/gcd(i,j,k,n)), where the Jordan totient function J_3(n) = A059376(n). - Peter Bala, Jan 23 2024

Extensions

Formula and more terms from Vladeta Jovovic, Oct 30 2001