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The number of unitary divisors that are cubes of primes applied to the cubefull numbers.
4

%I #9 Sep 22 2024 04:27:02

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

%H Amiram Eldar, <a href="/A376364/b376364.txt">Table of n, a(n) for n = 1..10000</a>

%H Sourabhashis Das, Wentang Kuo, and Yu-Ru Liu, <a href="https://arxiv.org/abs/2409.11275">On the number of prime factors with a given multiplicity over h-free and h-full numbers</a>, arXiv:2409.11275 [math.NT], 2024. See Theorem 1.3.

%H <a href="/index/Pow#powerful">Index entries for sequences related to powerful numbers</a>.

%F a(n) = A295883(A036966(n)).

%F 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., 2024).

%t f[k_] := Module[{e = If[k == 1, {}, FactorInteger[k][[;; , 2]]]}, If[AllTrue[e, # > 2 &], Count[e, 3], Nothing]]; Array[f, 60000]

%o (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 Cf. A036966, A077761, A295883, A362974, A376362, A376363.

%K nonn,easy

%O 1,10

%A _Amiram Eldar_, Sep 21 2024