A360658 a(1) = 1; a(n) = -Sum_{k=2..n} k^3 * a(floor(n/k)).

1, -8, -35, -27, -152, 91, -252, -252, -252, 873, -458, -674, -2871, 216, 3591, 3591, -1322, -1322, -8181, -9181, 80, 12059, -108, -108, -108, 19665, 19665, 16921, -7468, -37843, -67634, -67634, -31697, 12520, 55395, 55395, 4742, 66473, 125792, 125792, 56871, -26478

Offset: 1

Author

Seiichi Manyama, Apr 01 2023

Keywords

• sign

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..10000

Crossrefs

Partial sums of A359531.

Cf. A092149, A359478, A360390.

Cf. A336277.

Programs

Mathematica

f[p_, e_] := If[e == 1, -p^3, 0]; f[2, e_] := Switch[e, 1, -9, 2, 8, , 0]; s[1] = 1; s[n] := Times @@ f @@@ FactorInteger[n]; Accumulate[Array[s, 100]] (* Amiram Eldar, May 10 2023 *)

Python

from functools import lru_cache
@lru_cache(maxsize=None)
def A360658(n):
    if n <= 1:
        return 1
    c, j = 0, 2
    k1 = n//j
    while k1 > 1:
        j2 = n//k1 + 1
        c -= ((j2*(j2-1))**2-(j*(j-1))**2>>2)*A360658(k1)
        j, k1 = j2, n//j2
    return c-((n*(n+1))**2-((j-1)*j)**2>>2) # Chai Wah Wu, Apr 01 2023

Formula

Sum_{k=1..n} k^3 * a(floor(n/k)) = 0 for n > 1.

G.f. A(x) satisfies x * (1 - x) = Sum_{k>=1} k^3 * (1 - x^k) * A(x^k).