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A180499
n^3 + n-th cubefree number.
1
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
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
KEYWORD
nonn,easy
AUTHOR
Jonathan Vos Post, Jan 20 2011
STATUS
approved