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A376364 The number of unitary divisors that are cubes of primes applied to the cubefull numbers.

%I A376364 #12 Mar 25 2025 20:08:52
%S A376364 0,1,0,1,0,0,0,1,0,2,0,0,1,1,0,0,1,0,1,2,0,0,1,1,1,1,0,0,1,0,0,2,0,2,
%T A376364 1,0,1,0,1,1,0,1,1,0,1,1,0,1,0,2,0,1,0,2,1,0,1,1,0,0,0,1,0,0,1,1,2,1,
%U A376364 0,0,0,1,1,0,1,1,3,1,1,0,1,1,0,1,0,0,1,2,0,2,0,0,1,2,1,0,0,0,1,1,2,2,1,0,2
%N A376364 The number of unitary divisors that are cubes of primes applied to the cubefull numbers.
%H A376364 Amiram Eldar, <a href="/A376364/b376364.txt">Table of n, a(n) for n = 1..10000</a>
%H A376364 Sourabhashis Das, Wentang Kuo, and Yu-Ru Liu, <a href="https://doi.org/10.1016/j.jnt.2024.08.007">On the number of prime factors with a given multiplicity over h-free and h-full numbers</a>, Journal of Number Theory, Vol. 267 (2025), pp. 176-201; <a href="https://arxiv.org/abs/2409.11275">arXiv preprint</a>, arXiv:2409.11275 [math.NT], 2024. See Theorem 1.3.
%H A376364 <a href="/index/Pow#powerful">Index entries for sequences related to powerful numbers</a>.
%F A376364 a(n) = A295883(A036966(n)).
%F A376364 Sum_{A036966(k) <= x} a(k) = c * sqrt(x) * (log(log(x)) + B - log(3) - L(3, 6)) + O(x^(1/3)/log(x)), where c = A362974, B is Mertens's constant (A077761), L(h, r) = Sum_{p prime} 1/(p^(r/h - 1) * (p - p^(1 - 1/h) + 1)), and L(3, 6) = 0.67060646664392140547... (Das et al., 2025).
%t A376364 f[k_] := Module[{e = If[k == 1, {}, FactorInteger[k][[;; , 2]]]}, If[AllTrue[e, # > 2 &], Count[e, 3], Nothing]]; Array[f, 60000]
%o A376364 (PARI) lista(kmax) = {my(e, is); for(k = 1, kmax, e = factor(k)[, 2]; is = 1; for(i = 1, #e, if(e[i] < 3, is = 0; break)); if(is, print1(#select(x -> x == 3, e), ", ")));}
%Y A376364 Cf. A036966, A077761, A295883, A362974, A376362, A376363.
%K A376364 nonn,easy
%O A376364 1,10
%A A376364 _Amiram Eldar_, Sep 21 2024