A355265 Bicubeful numbers.

64, 128, 192, 256, 320, 384, 448, 512, 576, 640, 704, 729, 768, 832, 896, 960, 1024, 1088, 1152, 1216, 1280, 1344, 1408, 1458, 1472, 1536, 1600, 1664, 1728, 1792, 1856, 1920, 1984, 2048, 2112, 2176, 2187, 2240, 2304, 2368, 2432, 2496, 2560, 2624, 2688, 2752

Offset: 1

Author

Peter Luschny, Jul 12 2022

Keywords

• nonn

Comments

Let lp(n, e) denote the largest positive integer b such that b^e divides n. For example for e = 1, 2, 3, 4 the sequences (lp(n, e), n >= 1) are A000027, A000188, A053150, and A053164. Let rad(n) = A007947(n) be the squarefree kernel of n. k is in this sequence if lp(n, 3) does not divide rad(n). The case e = 1 gives A013929, and the case e = 2 is A046101.

The asymptotic density of this sequence is 1 - 1/zeta(6) = 1 - 945/Pi^6 = 0.017047... . - Amiram Eldar, Jul 13 2022

Examples

			n = 512 = 2^9, rad(n) = 2, lp(n, 3) = 8 since n/8^3 = 1. But 8 does not divide 2.
n = 704 = 2^6*11, rad(n) = 22, lp(n, 3) = 4 since n/4^3 = 11. But 4 does not divide 22.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A007947, A000188, A053150, A053164, A013929, A046101 (biquadrateful).

Cf. A013664, A343359.

Programs

Maple

with(NumberTheory):
isBicubeful := n -> irem(Radical(n), LargestNthPower(n, 3)) <> 0:
select(isBicubeful, [`$`(1..2752)]);

Mathematica

bicubQ[n_] := AnyTrue[FactorInteger[n][[;; , 2]], # > 5 &]; Select[Range[3000], bicubQ] (* Amiram Eldar, Jul 13 2022 *)

Python

from itertools import count, islice
from sympy import factorint
def A355265_gen(startvalue=1): # generator of terms >= startvalue
    return filter(lambda n:any(map(lambda m:m>5,factorint(n).values())),count(max(startvalue,1)))
A355265_list = list(islice(A355265_gen(),30)) # Chai Wah Wu, Jul 12 2022

Formula

A number k is bicubeful iff it is divisible by the 6th power of an integer > 1.