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A258331 Sum of the cubes of the divisors of n^3. 1

%I #31 Oct 25 2023 12:17:17

%S 1,585,20440,299593,1968876,11957400,40471600,153391689,402321277,

%T 1151792460,2359720584,6123680920,10609328380,23675886000,40243825440,

%U 78536544841,118612018980,235357947045,322734750520,589861467468,827239504000,1380436541640

%N Sum of the cubes of the divisors of n^3.

%H Charles R Greathouse IV, <a href="/A258331/b258331.txt">Table of n, a(n) for n = 1..10000</a>

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

%F From _Amiram Eldar_, Nov 05 2022: (Start)

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

%F 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)

%e 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.

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

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

%o (Magma) [DivisorSigma(3, n^3): n in [1..50]]; // _Vincenzo Librandi_, May 27 2015

%o (PARI) a(n)=sigma(n^3,3) \\ _Charles R Greathouse IV_, May 27 2015

%o (Sage) [sigma(n^3, 3) for n in (1..50)] # _Bruno Berselli_, May 27 2015

%o (Python)

%o from math import prod

%o from sympy import factorint

%o 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

%Y Cf. A000578, A001158, A013668, A065827.

%K nonn,easy,mult

%O 1,2

%A _Wesley Ivan Hurt_, May 26 2015

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Last modified April 25 04:42 EDT 2024. Contains 371964 sequences. (Running on oeis4.)