A363497 a(n) = Sum_{k=0..n} floor(sqrt(k))^3.

0, 1, 2, 3, 11, 19, 27, 35, 43, 70, 97, 124, 151, 178, 205, 232, 296, 360, 424, 488, 552, 616, 680, 744, 808, 933, 1058, 1183, 1308, 1433, 1558, 1683, 1808, 1933, 2058, 2183, 2399, 2615, 2831, 3047, 3263, 3479, 3695, 3911, 4127, 4343, 4559, 4775, 4991, 5334

Offset: 0

Author

Hans J. H. Tuenter, Jun 05 2023

Keywords

• nonn
• easy

Comments

Partial sums of the third powers of the terms of A000196.

Links

• Karl-Heinz Hofmann, Table of n, a(n) for n = 0..10000

Crossrefs

Sums of powers of A000196: A022554 (1st), A174060 (2nd), this sequence (3rd), A363498 (4th), A363499 (5th).

Programs

Mathematica

Table[(n + 1) #^3 - (1/60) # (# + 1) (3 # - 1) (12 #^2 + 7 # - 4) &[Floor@ Sqrt[n]], {n, 0, 50}] (* Michael De Vlieger, Jun 10 2023 *)

PARI

a(n) = sum(k=0, n, sqrtint(k)^3); \\ Michel Marcus, Jun 06 2023

Python

from math import isqrt
A363497 = [0]
for n in range(1,50): A363497.append(A363497[-1] + isqrt(n)**3)
print(A363497) # Karl-Heinz Hofmann, Jun 14 2023

Python

from math import isqrt
def A363497(n):return (m:=isqrt(n))**3*(n+1)-(m*(m+1)*(3*m-1)*(12*m**2+7*m-4))//60
# Karl-Heinz Hofmann, Jun 14 2023

Formula

a(n) = (n+1)*m^3 - (1/60)*m*(m+1)*(3*m-1)*(12*m^2+7*m-4), where m = floor(sqrt(n)).