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A368779
The number of prime factors of the cubefree numbers, counted with multiplicity.
1
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
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
KEYWORD
nonn,easy
AUTHOR
Amiram Eldar, Jan 05 2024
STATUS
approved