A175926 Sum of divisors of cubes.
1, 15, 40, 127, 156, 600, 400, 1023, 1093, 2340, 1464, 5080, 2380, 6000, 6240, 8191, 5220, 16395, 7240, 19812, 16000, 21960, 12720, 40920, 19531, 35700, 29524, 50800, 25260, 93600, 30784, 65535, 58560, 78300, 62400, 138811, 52060, 108600, 95200
Offset: 1
Links
- Seiichi Manyama, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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Magma
[ SumOfDivisors(n^3) : n in [1..100]]; // Vincenzo Librandi, Apr 14 2011
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Mathematica
DivisorSigma[1,#]&/@((Range[40])^3) (* Harvey P. Dale, Aug 30 2015 *) f[p_, e_] := (p^(3*e + 1) - 1)/(p - 1); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 50] (* Amiram Eldar, Sep 10 2020 *)
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PARI
a(n) = sigma(n^3); \\ Amiram Eldar, Nov 05 2022
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Python
from math import prod from sympy import factorint def A175926(n): return prod((p**(3*e+1)-1)//(p-1) for p,e in factorint(n).items()) # Chai Wah Wu, Oct 25 2023
Formula
a(n) = A000203(n^3). - R. J. Mathar, Mar 31 2011
Multiplicative with a(p^e) = (p^(3e+1)-1)/(p-1). - R. J. Mathar, Mar 31 2011
Sum_{k>=1} 1/a(k) = 1.11535899887110289127674868460900333554265894187008102863022551119560512446... - Vaclav Kotesovec, Sep 20 2020
Sum_{k=1..n} a(k) ~ c * n^4, where c = (zeta(4)/4) * Product_{p prime} (1 + 1/p^2 + 1/p^3) = 0.4732277044... . - Amiram Eldar, Nov 05 2022
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