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A180499

n^3 + n-th cubefree number.

2, 10, 30, 68, 130, 222, 350, 521, 739, 1011, 1343, 1741, 2211, 2759, 3392, 4114, 4932, 5852, 6880, 8022, 9284, 10673, 12193, 13852, 15654, 17606, 19714, 21985, 24423, 27035, 29827, 32805, 35975, 39343

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OFFSET

1,1

COMMENTS

First differs from n^3 + n (A034262) at n=8 because 8 is the first positive integer which is not cubefree.

What cubes appear in this sequence?

No cubes appear in this sequence: the n-th cubefree number is asymptotically zeta(3)*n, putting members of this sequence strictly between n^3 and (n+1)^3. (Lacking effective error bounds this actually only shows that there are finitely many.) - Charles R Greathouse IV, Jan 21 2011

LINKS

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

FORMULA

a(n) = n^3 + A004709(n) = A000578(n) + A004709(n).

EXAMPLE

a(8) = 8^3 + 8th number that is not divisible by any cube > 1 = 8^3 + 9 = 521.

MATHEMATICA

#[[1]]+#[[2]]^3&/@Module[{cf=Select[Range[50], Max[FactorInteger[#][[All, 2]]] < 3&]}, Thread[{cf, Range[Length[cf]]}]] (* Harvey P. Dale, Jun 28 2020 *)

PROG

(Python)

from sympy import mobius, integer_nthroot

def A180499(n):

def f(x): return n+x-sum(mobius(k)*(x//k**3) for k in range(1, integer_nthroot(x, 3)[0]+1))

m, k = n, f(n)

while m != k:

m, k = k, f(k)

return n**3+m # Chai Wah Wu, Aug 12 2024

CROSSREFS

Cf. A000578, A004709, A034262, A161203.

Sequence in context: A162524 A065137 A034262 * A167214 A034827 A328532