A336276 a(n) = Sum_{k=1..n} mu(k)*k^2.

1, -3, -12, -12, -37, -1, -50, -50, -50, 50, -71, -71, -240, -44, 181, 181, -108, -108, -469, -469, -28, 456, -73, -73, -73, 603, 603, 603, -238, -1138, -2099, -2099, -1010, 146, 1371, 1371, 2, 1446, 2967, 2967, 1286, -478, -2327, -2327, -2327, -211, -2420

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, A336277, A336278, A336279.

Cf. A008683, A055615, A070891, A360390, A361983.

Programs

Mathematica

Array[Sum[MoebiusMu[k]*k^2, {k, #}] &, 47] (* Michael De Vlieger, Jul 15 2020 *)

PARI

a(n) = sum(k=1, n, moebius(k)*k^2); \\ Michel Marcus, Jul 15 2020

Python

from functools import lru_cache
@lru_cache(maxsize=None)
def A336276(n):
    if n <= 1:
        return 1
    c, j = 1, 2
    k1 = n//j
    while k1 > 1:
        j2 = n//k1 + 1
        c -= (j2*(j2-1)*((j2<<1)-1)-j*(j-1)*((j<<1)-1))//6*A336276(k1)
        j, k1 = j2, n//j2
    return c-(n*(n+1)*((n<<1)+1)-j*(j-1)*((j<<1)-1))//6 # Chai Wah Wu, Apr 04 2023

Formula

Partial sums of A334657.

G.f. A(x) satisfies x = Sum_{k>=1} k^2 * (1 - x^k) * A(x^k). - Seiichi Manyama, Apr 01 2023

Sum_{k=1..n} k^2 * a(floor(n/k)) = 1. - Seiichi Manyama, Apr 03 2023