A336279 a(n) = Sum_{k=1..n} mu(k)*k^5.

1, -31, -274, -274, -3399, 4377, -12430, -12430, -12430, 87570, -73481, -73481, -444774, 93050, 852425, 852425, -567432, -567432, -3043531, -3043531, 1040570, 6194202, -242141, -242141, -242141, 11639235, 11639235, 11639235, -8871914, -33171914, -61801065

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, A336277, A336278.

Cf. A008683, A055615, A070891, A344429, A344430.

Programs

Mathematica

Array[Sum[MoebiusMu[k]*k^5, {k, #}] &, 32] (* Michael De Vlieger, Jul 15 2020 *)

PARI

a(n) = sum(k=1, n, moebius(k)*k^5); \\ Michel Marcus, Jul 15 2020

Python

from functools import lru_cache
@lru_cache(maxsize=None)
def A336279(n):
    if n <= 1:
        return 1
    c, j = 1, 2
    k1 = n//j
    while k1 > 1:
        j2 = n//k1 + 1
        c -= (j2**2*(j2**2*(j2*(2*j2 - 6) + 5) - 1)-j**2*(j**2*(j*(2*j - 6) + 5) - 1))//12*A336279(k1)
        j, k1 = j2, n//j2
    return c-(n**2*(n**2*(n*(2*n + 6) + 5) - 1)-j**2*(j**2*(j*(2*j - 6) + 5) - 1))//12 # Chai Wah Wu, Apr 04 2023

Formula

From Seiichi Manyama, Apr 03 2023: (Start)

G.f. A(x) satisfies x = Sum_{k>=1} k^5 * (1 - x^k) * A(x^k).

Sum_{k=1..n} k^5 * a(floor(n/k)) = 1. (End)