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

Offset: 1

Author

Jonathan Vos Post, Jan 20 2011

Keywords

• nonn
• easy

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

Examples

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

Links

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

Crossrefs

Cf. A000578, A004709, A034262, A161203.

Programs

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 *)

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

Formula

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