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A355265
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Bicubeful numbers.
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2
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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
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OFFSET
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1,1
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COMMENTS
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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
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LINKS
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FORMULA
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A number k is bicubeful iff it is divisible by the 6th power of an integer > 1.
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EXAMPLE
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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.
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MAPLE
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with(NumberTheory):
isBicubeful := n -> irem(Radical(n), LargestNthPower(n, 3)) <> 0:
select(isBicubeful, [`$`(1..2752)]);
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MATHEMATICA
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bicubQ[n_] := AnyTrue[FactorInteger[n][[;; , 2]], # > 5 &]; Select[Range[3000], bicubQ] (* Amiram Eldar, Jul 13 2022 *)
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PROG
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(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)))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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