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

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, 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, 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

Offset: 1

Author

Amiram Eldar, Sep 21 2024

Keywords

• nonn
• easy

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000
• Sourabhashis Das, Wentang Kuo, and Yu-Ru Liu, On the number of prime factors with a given multiplicity over h-free and h-full numbers, Journal of Number Theory, Vol. 267 (2025), pp. 176-201; arXiv preprint, arXiv:2409.11275 [math.NT], 2024. See Theorem 1.3.
• Index entries for sequences related to powerful numbers.

Crossrefs

Cf. A036966, A077761, A295883, A362974, A376362, A376363.

Programs

Mathematica

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

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), ", ")));}

Formula

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

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).