A360390 a(1) = 1; a(n) = -Sum_{k=2..n} k^2 * a(floor(n/k)).

1, -4, -13, -9, -34, 11, -38, -38, -38, 87, -34, -70, -239, 6, 231, 231, -58, -58, -419, -519, -78, 527, -2, -2, -2, 843, 843, 647, -194, -1319, -2280, -2280, -1191, 254, 1479, 1479, 110, 1915, 3436, 3436, 1755, -450, -2299, -2783, -2783, -138, -2347, -2347, -2347, -2347, 254, -422

Offset

1,2

Links

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

Formula

Sum_{k=1..n} k^2 * a(floor(n/k)) = 0 for n > 1.

G.f. A(x) satisfies x * (1 - x) = Sum_{k>=1} k^2 * (1 - x^k) * A(x^k).

Mathematica

f[p_, e_] := If[e == 1, -p^2, 0]; f[2, e_] := Switch[e, 1, -5, 2, 4, _, 0]; s[1] = 1; s[n_] := Times @@ f @@@ FactorInteger[n]; Accumulate[Array[s, 100]] (* Amiram Eldar, May 10 2023 *)

Prog

(Python)
from functools import lru_cache
@lru_cache(maxsize=None)
def A360390(n):
    if n <= 1:
        return 1
    c, j = 0, 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*A360390(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 01 2023

Crossrefs

Partial sums of A359485.

Cf. A092149, A359478, A360658.

Cf. A336276.

Sequence in context: A264341 A356799 A144290 * A101181 A160249 A173800

Adjacent sequences: A360387 A360388 A360389 * A360391 A360392 A360393

Keyword

sign

Author

Seiichi Manyama, Apr 01 2023

Status

approved