A358039 a(n) is the Euler totient function phi applied to the n-th cubefree number.

1, 1, 2, 2, 4, 2, 6, 6, 4, 10, 4, 12, 6, 8, 16, 6, 18, 8, 12, 10, 22, 20, 12, 12, 28, 8, 30, 20, 16, 24, 12, 36, 18, 24, 40, 12, 42, 20, 24, 22, 46, 42, 20, 32, 24, 52, 40, 36, 28, 58, 16, 60, 30, 36, 48, 20, 66, 32, 44, 24, 70, 72, 36, 40, 36, 60, 24, 78, 40

Offset: 1

Author

Amiram Eldar, Oct 29 2022

Keywords

• nonn

Comments

The analogous sequence with squarefree numbers is A049200.

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000
• Zhu Weiyi, On the cube free number sequences, Smarandache Notions J., Vol. 14 (2004), pp. 199-202.

Crossrefs

Cf. A000010, A002117, A004709, A049200, A358040.

Programs

Mathematica

EulerPhi[Select[Range[100], Max[FactorInteger[#][[;; , 2]]] < 3 &]]

Python

from sympy import mobius, integer_nthroot, totient
def A358039(n):
    def f(x): return n+x-sum(mobius(k)*(x//k**3) for k in range(1, integer_nthroot(x,3)[0]+1))
    m, k = n, f(n)
    while m != k:
        m, k = k, f(k)
    return totient(m) # Chai Wah Wu, Aug 06 2024

Formula

a(n) = A000010(A004709(n)).

Sum_{k=1..n} a(k) = (c/(2*zeta(3)))*n^2 + O(n^(3/2+eps)), where c = Product_{p prime} (1 - (p+1)/(p^3+p^2+1)) = 0.62583324412633345811... (Weiyi, 2004).

From Amiram Eldar, Oct 09 2023: (Start)

Sum_{n>=1} 1/(A004709(n)*a(n)) = Product_{p prime} (1 + (p^2+1)/((p-1)*p^3)) = 2.14437852780769816048... .

Sum_{n>=1} 1/a(n)^2 = Product_{p prime} (1 + (p^2+1)/((p-1)^2*p^2)) = 3.26032746607943673536... . (End)