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The number of prime factors of the cubefree numbers, counted with multiplicity.
1

%I #16 Sep 21 2024 14:47:41

%S 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,

%T 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,

%U 2,1,4,2,2,2,1,4,2,3,2,2,2,1,3,3,4,1,3

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

%H Amiram Eldar, <a href="/A368779/b368779.txt">Table of n, a(n) for n = 1..10000</a>

%H Rafael Jakimczuk and Matilde Lalín, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL25/Lalin/lalin2.html">The Number of Prime Factors on Average in Certain Integer Sequences</a>, Journal of Integer Sequences, Vol. 25 (2022), Article 22.2.3.

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

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

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

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

%o (Python)

%o from sympy import mobius, integer_nthroot, primeomega

%o def A368779(n):

%o def f(x): return n+x-sum(mobius(k)*(x//k**3) for k in range(1, integer_nthroot(x,3)[0]+1))

%o m, k = n, f(n)

%o while m != k:

%o m, k = k, f(k)

%o return primeomega(m) # _Chai Wah Wu_, Aug 06 2024

%Y Cf. A001222, A004709, A072047.

%K nonn,easy

%O 1,4

%A _Amiram Eldar_, Jan 05 2024