A336278 a(n) = Sum_{k=1..n} mu(k)*k^4.

1, -15, -96, -96, -721, 575, -1826, -1826, -1826, 8174, -6467, -6467, -35028, 3388, 54013, 54013, -29508, -29508, -159829, -159829, 34652, 268908, -10933, -10933, -10933, 446043, 446043, 446043, -261238, -1071238, -1994759, -1994759, -808838, 527498, 2028123

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, A336279.

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

Programs

Mathematica

Array[Sum[MoebiusMu[k]*k^4, {k, #}] &, 35] (* Michael De Vlieger, Jul 15 2020 *)
Accumulate[Table[MoebiusMu[x]x^4,{x,40}]] (* Harvey P. Dale, Jan 14 2021 *)

PARI

a(n) = sum(k=1, n, moebius(k)*k^4); \\ Michel Marcus, Jul 15 2020

Python

from functools import lru_cache
@lru_cache(maxsize=None)
def A336278(n):
    if n <= 1:
        return 1
    c, j = 1, 2
    k1 = n//j
    while k1 > 1:
        j2 = n//k1 + 1
        c -= (j2*(j2**2*(j2*(6*j2 - 15) + 10) - 1)-j*(j**2*(j*(6*j - 15) + 10) - 1))//30*A336278(k1)
        j, k1 = j2, n//j2
    return c-(n*(n**2*(n*(6*n + 15) + 10) - 1)-j*(j**2*(j*(6*j - 15) + 10) - 1))//30 # Chai Wah Wu, Apr 04 2023

Formula

Partial sums of A334660.

From Seiichi Manyama, Apr 03 2023: (Start)

G.f. A(x) satisfies x = Sum_{k>=1} k^4 * (1 - x^k) * A(x^k).

Sum_{k=1..n} k^4 * a(floor(n/k)) = 1. (End)