A368779 The number of prime factors of the cubefree numbers, counted with multiplicity.

0, 1, 1, 2, 1, 2, 1, 2, 2, 1, 3, 1, 2, 2, 1, 3, 1, 3, 2, 2, 1, 2, 2, 3, 1, 3, 1, 2, 2, 2, 4, 1, 2, 2, 1, 3, 1, 3, 3, 2, 1, 2, 3, 2, 3, 1, 2, 2, 2, 1, 4, 1, 2, 3, 2, 3, 1, 3, 2, 3, 1, 1, 2, 3, 3, 2, 3, 1, 2, 1, 4, 2, 2, 2, 1, 4, 2, 3, 2, 2, 2, 1, 3, 3, 4, 1, 3

Offset: 1

Author

Amiram Eldar, Jan 05 2024

Keywords

• nonn
• easy

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000
• Rafael Jakimczuk and Matilde Lalín, The Number of Prime Factors on Average in Certain Integer Sequences, Journal of Integer Sequences, Vol. 25 (2022), Article 22.2.3.

Crossrefs

Cf. A001222, A004709, A072047.

Programs

Mathematica

f[n_] := Module[{e = FactorInteger[n][[;; , 2]]}, If[AllTrue[e, # < 3 &], Total[e], Nothing]]; f[1] = 0; Array[f, 100]

PARI

lista(max) = {my(e); for(k = 1, max, e = factor(k)[,2]; if(k == 1 || vecmax(e) < 3, print1(vecsum(e), ", ")));}

Python

from sympy import mobius, integer_nthroot, primeomega
def A368779(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 primeomega(m) # Chai Wah Wu, Aug 06 2024

Formula

a(n) = A001222(A004709(n)).

Sum_{A004709(k) <= x} a(k) = (1/zeta(3)) * x * log(log(x)) + O(x) (Jakimczuk and Lalín, 2022). [corrected Sep 21 2024]