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
Keywords
Links
- Seiichi Manyama, Table of n, a(n) for n = 1..10000
Programs
-
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 *)
-
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
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).