OFFSET
1,2
COMMENTS
Conjecture: a(n) changes sign infinitely often.
LINKS
Seiichi Manyama, Table of n, a(n) for n = 1..10000
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
MATHEMATICA
Array[Sum[MoebiusMu[k]*k^2, {k, #}] &, 47] (* Michael De Vlieger, Jul 15 2020 *)
PROG
(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
CROSSREFS
KEYWORD
easy,sign
AUTHOR
Donald S. McDonald, Jul 15 2020
STATUS
approved