A258331 Sum of the cubes of the divisors of n^3.

1, 585, 20440, 299593, 1968876, 11957400, 40471600, 153391689, 402321277, 1151792460, 2359720584, 6123680920, 10609328380, 23675886000, 40243825440, 78536544841, 118612018980, 235357947045, 322734750520, 589861467468, 827239504000, 1380436541640

Offset: 1

Author

Wesley Ivan Hurt, May 26 2015

Keywords

• nonn
• easy
• mult

Examples

			For n=2, the divisors of 2^3 = 8 are 1, 2, 4 and 8. The sum of the cubes of these divisors is 1^3+2^3+4^3+8^3 = 585, therefore a(2) = 585.
		

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000578, A001158, A013668, A065827.

Programs

Magma

[DivisorSigma(3, n^3): n in [1..50]]; // Vincenzo Librandi, May 27 2015

Maple

with(numtheory): A258331:=n->sigma[3](n^3): seq(A258331(n), n=1..50);

Mathematica

Table[DivisorSigma[3, n^3], {n, 50}]

PARI

a(n)=sigma(n^3,3) \\ Charles R Greathouse IV, May 27 2015

Python

from math import prod
from sympy import factorint
def A258331(n): return prod((p**((3*e+1)*3)-1)//(p**3-1) for p,e in factorint(n).items()) # Chai Wah Wu, Oct 25 2023

Sage

[sigma(n^3, 3) for n in (1..50)] # Bruno Berselli, May 27 2015

Formula

a(n) = sigma_3(n^3) = A001158(A000578(n)).

From Amiram Eldar, Nov 05 2022: (Start)

Multiplicative with a(p^e) = (p^(9*e + 3) - 1)/(p^3 - 1).

Sum_{k=1..n} a(k) ~ c * n^10, where c = (zeta(10)/10) * Product_{p prime} (1 + 1/p^4 + 1/p^7) = 0.1087440273... . (End)