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A258331
Sum of the cubes of the divisors of n^3.
1
1, 585, 20440, 299593, 1968876, 11957400, 40471600, 153391689, 402321277, 1151792460, 2359720584, 6123680920, 10609328380, 23675886000, 40243825440, 78536544841, 118612018980, 235357947045, 322734750520, 589861467468, 827239504000, 1380436541640
OFFSET
1,2
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
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)
EXAMPLE
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.
MAPLE
with(numtheory): A258331:=n->sigma[3](n^3): seq(A258331(n), n=1..50);
MATHEMATICA
Table[DivisorSigma[3, n^3], {n, 50}]
PROG
(Magma) [DivisorSigma(3, n^3): n in [1..50]]; // Vincenzo Librandi, May 27 2015
(PARI) a(n)=sigma(n^3, 3) \\ Charles R Greathouse IV, May 27 2015
(Sage) [sigma(n^3, 3) for n in (1..50)] # Bruno Berselli, 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
CROSSREFS
KEYWORD
nonn,easy,mult
AUTHOR
Wesley Ivan Hurt, May 26 2015
STATUS
approved