A358714 a(n) = phi(n)^3.

1, 1, 8, 8, 64, 8, 216, 64, 216, 64, 1000, 64, 1728, 216, 512, 512, 4096, 216, 5832, 512, 1728, 1000, 10648, 512, 8000, 1728, 5832, 1728, 21952, 512, 27000, 4096, 8000, 4096, 13824, 1728, 46656, 5832, 13824, 4096, 64000, 1728, 74088, 8000, 13824, 10648, 97336, 4096, 74088

Offset: 1

Author

Param Mayurkumar Parekh and Paavan Mayurkumar Parekh, Nov 28 2022

Keywords

• nonn
• easy
• mult

Comments

Number of solutions to gcd(x*y*z, n) = 1 such that 0 <= x,y,z <= n-1.

x*y*z == t (mod n) where t is a unit (invertible element) in Z_n. Since t is a unit, all x,y,z must be units. Here there are A000010(n) possibilities for each x,y,z so there are a total of A000010(n)^3 ways to get t as a unit.

Examples

			a(9) = A000010(9)^3 = 216.
		

Crossrefs

Cf. A000010, A127473, A361148, A335818.

Programs

Magma

[(EulerPhi(n))^3: n in [1..180]];

Mathematica

a[n_] := EulerPhi[n]^3; Array[a, 100] (* Amiram Eldar, Jan 06 2023 *)

PARI

a(n) = eulerphi(n)^3;

Formula

a(n) = A000010(n)^3.

From Amiram Eldar, Jan 06 2023: (Start)

Multiplicative with a(p^e) = (p-1)^3*p^(3*e-3).

Sum_{k=1..n} a(k) ~ c * n^4, where c = (1/4) * Product_{p prime}(1 - 3/p^2 + 3/p^3 - 1/p^4) = 0.08429696844... .

Sum_{k>=1} 1/a(k) = Product_{p prime} (1 + p^3/((p-1)^3*(p^3-1))) = 2.47619474816... (A335818). (End)