%I #31 Apr 04 2023 22:06:32

%S 1,-15,-96,-96,-721,575,-1826,-1826,-1826,8174,-6467,-6467,-35028,

%T 3388,54013,54013,-29508,-29508,-159829,-159829,34652,268908,-10933,

%U -10933,-10933,446043,446043,446043,-261238,-1071238,-1994759,-1994759,-808838,527498,2028123

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

%C Conjecture: a(n) changes sign infinitely often.

%H Seiichi Manyama, <a href="/A336278/b336278.txt">Table of n, a(n) for n = 1..10000</a>

%F Partial sums of A334660.

%F From _Seiichi Manyama_, Apr 03 2023: (Start)

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

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

%t Array[Sum[MoebiusMu[k]*k^4, {k, #}] &, 35] (* _Michael De Vlieger_, Jul 15 2020 *)

%t Accumulate[Table[MoebiusMu[x]x^4,{x,40}]] (* _Harvey P. Dale_, Jan 14 2021 *)

%o (PARI) a(n) = sum(k=1, n, moebius(k)*k^4); \\ _Michel Marcus_, Jul 15 2020

%o (Python)

%o from functools import lru_cache

%o @lru_cache(maxsize=None)

%o def A336278(n):

%o if n <= 1:

%o return 1

%o c, j = 1, 2

%o k1 = n//j

%o while k1 > 1:

%o j2 = n//k1 + 1

%o c -= (j2*(j2**2*(j2*(6*j2 - 15) + 10) - 1)-j*(j**2*(j*(6*j - 15) + 10) - 1))//30*A336278(k1)

%o j, k1 = j2, n//j2

%o 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

%Y Cf. A002321, A068340, A336276, A336277, A336279.

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

%K easy,sign

%O 1,2

%A _Donald S. McDonald_, Jul 15 2020