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

1,4

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.

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]

Mathematica

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

Prog

(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

Crossrefs

Cf. A001222, A004709, A072047.

Sequence in context: A118459 A274354 A083870 * A052299 A071854 A183025

Adjacent sequences: A368776 A368777 A368778 * A368780 A368781 A368782

Keyword

nonn,easy

Author

Amiram Eldar, Jan 05 2024

Status

approved