A336277 a(n) = Sum_{k=1..n} mu(k)*k^3.

1, -7, -34, -34, -159, 57, -286, -286, -286, 714, -617, -617, -2814, -70, 3305, 3305, -1608, -1608, -8467, -8467, 794, 11442, -725, -725, -725, 16851, 16851, 16851, -7538, -34538, -64329, -64329, -28392, 10912, 53787, 53787, 3134, 58006, 117325, 117325, 48404

Offset: 1

Author

Donald S. McDonald, Jul 15 2020

Keywords

• easy
• sign

Comments

Conjecture: a(n) changes sign infinitely often.

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A002321, A068340, A336276, A336278, A336279.

Cf. A008683, A055615, A070891.

Programs

Mathematica

Array[Sum[MoebiusMu[k]*k^3, {k, #}] &, 41] (* Michael De Vlieger, Jul 15 2020 *)
Accumulate[Table[MoebiusMu[n] n^3,{n,50}]] (* Harvey P. Dale, Aug 15 2024 *)

PARI

a(n) = sum(k=1, n, moebius(k)*k^3); \\ Michel Marcus, Jul 15 2020

Python

from functools import lru_cache
@lru_cache(maxsize=None)
def A336277(n):
    if n <= 1:
        return 1
    c, j = 1, 2
    k1 = n//j
    while k1 > 1:
        j2 = n//k1 + 1
        c -= ((j2*(j2-1))**2-(j*(j-1))**2>>2)*A336277(k1)
        j, k1 = j2, n//j2
    return c-((n*(n+1))**2-((j-1)*j)**2>>2) # Chai Wah Wu, Apr 04 2023

Formula

Partial sums of A334659.

G.f. A(x) satisfies x = Sum_{k>=1} k^3 * (1 - x^k) * A(x^k). - Seiichi Manyama, Apr 01 2023

Sum_{k=1..n} k^3 * a(floor(n/k)) = 1. - Seiichi Manyama, Apr 03 2023