A339804 a(n) = Sum_{k=1..floor(n/2)} k * (n-k) * floor((n-k)/k).
0, 1, 4, 13, 22, 50, 68, 116, 162, 236, 278, 437, 498, 634, 794, 1018, 1118, 1450, 1574, 1975, 2276, 2598, 2774, 3519, 3834, 4273, 4746, 5490, 5772, 6887, 7214, 8163, 8856, 9586, 10330, 12072, 12540, 13443, 14382, 16244, 16806, 18861, 19480, 21192, 22954, 24267
Offset: 1
Programs
-
Mathematica
Table[Sum[k (n - k)*Floor[(n - k)/k], {k, Floor[n/2]}], {n, 50}]
-
PARI
a(n) = sum(k=1, n\2, k*(n-k)*((n-k)\k)); \\ Michel Marcus, Dec 19 2020
-
Python
from math import isqrt def A339804(n): return (n*(1-n**2)+((s:=isqrt(n))**4<<1)+s**3*(3*(1-n))+s**2*(1-3*n) + sum((q:=n//k)*(-6*k**2+n*(3*((k<<1)+q+1))-q*((q<<1)+3)-1) for k in range(1,s+1)))//6 # Chai Wah Wu, Oct 27 2023
Formula
a(n) ~ n^3*(Pi^2-2-4*zeta(3))/12. - Rok Cestnik, Dec 19 2020
Comments