A362148 Numbers that are neither cubefree nor cubefull.

24, 40, 48, 54, 56, 72, 80, 88, 96, 104, 108, 112, 120, 135, 136, 144, 152, 160, 162, 168, 176, 184, 189, 192, 200, 208, 224, 232, 240, 248, 250, 264, 270, 272, 280, 288, 296, 297, 304, 312, 320, 324, 328, 336, 344, 351, 352, 360, 368, 375, 376, 378, 384, 392, 400

Offset: 1

Author

Bernard Schott, Apr 09 2023

Keywords

• nonn
• easy

Comments

In fact, every cubefull number > 1 is noncubefree, but the converse is false.

This sequence = A046099 \ A036966 and lists these counterexamples.

Numbers k such that for some primes p and q, k is divisible by p^3*q but not by q^3. - Robert Israel, Apr 28 2023

The asymptotic density of this sequence is 1 - 1/zeta(3) = 0.168092... - Charles R Greathouse IV, Apr 28 2023

From Amiram Eldar, Sep 17 2023: (Start)

Numbers k such that A360539(k) > 1 and A360540(k) > 1.

Equivalently, numbers that have in their prime factorization at least one exponent that is smaller than 3 and at least one exponent that is larger than 2. (End)

Examples

			24 = 2^3 * 3 is noncubefree as it is divisible by the cube 2^3, but it is not cubefull because 3 divides 24 but 3^3 does not divide 24, hence 24 is a term.
648 = 2^4 * 3^3 is noncubefree as it is divisible by the cube 3^3, but it is also cubefull because primes 2 and 3 divide 648, and 2^3 and 3^3 divide also 648, so 648 is not a term.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..10000

Crossrefs

Intersection of A046099 (not cubefree) and A362147 (not cubefull)

Cf. A004709 (cubefree), A036966 (cubefull), A360539, A360540.

Programs

Maple

filter:= proc(n) local F;
F:= ifactors(n)[2][..,2];
  min(F) < 3 and max(F) >= 3
end proc:
select(filter, [$1..400]); # Robert Israel, Apr 28 2023

Mathematica

Select[Range[500], Min[(e = FactorInteger[#][[;; , 2]])] < 3 && Max[e] > 2 &] (* Amiram Eldar, Apr 09 2023 *)

PARI

isok(k) = (k>1) && (vecmax(factor(k)[, 2])>2) && (vecmin(factor(k)[, 2])<=2); \\ Michel Marcus, Apr 19 2023

Formula

Equals A362147 \ A004709.

Sum_{n>=1} 1/a(n) = 1 + zeta(s) - zeta(s)/zeta(3*s) - Product_{p prime}(1 + 1/(p^(2*s)*(p^s-1))), s > 1. - Amiram Eldar, Sep 17 2023