OFFSET
1,2
COMMENTS
If the canonical factorization of n into prime powers is the product of p^e(p) then sigma_k(n) = Product_p ((p^((e(p)+1)*k))-1)/(p^k-1).
Sum_{d|n} 1/d^k is equal to sigma_k(n)/n^k. So sequencesA017665-A017712also give the numerators and denominators of sigma_k(n)/n^k for k = 1..24. The power sums sigma_k(n) are in sequencesA000203(k=1),A001157-A001160(k=2,3,4,5),A013954-A013972for k = 6,7,...,24. - Ahmed Fares (ahmedfares(AT)my-deja ), Apr 05 2001
LINKS
Vincenzo Librandi,Table of n, a(n) for n = 1..1000
FORMULA
G.f.: Sum_{k>=1} k^13*x^k/(1-x^k). -Benoit Cloitre,Apr 21 2003
Dirichlet g.f.: zeta(s-13)*zeta(s). -Ilya Gutkovskiy,Sep 10 2016
Empirical: Sum_{n>=1} a(n)/exp(2*Pi*n) = 1/24. -Simon Plouffe,Mar 01 2021
FromAmiram Eldar,Oct 29 2023: (Start)
Multiplicative with a(p^e) = (p^(13*e+13)-1)/(p^13-1).
Sum_{k=1..n} a(k) = zeta(14) * n^14 / 14 + O(n^15). (End)
MAPLE
MATHEMATICA
DivisorSigma[13, Range[30]] (*Vincenzo Librandi,Sep 10 2016 *)
PROG
(Sage) [sigma(n, 13)for n in range(1, 16)] #Zerinvary Lajos,Jun 04 2009
(Magma) [DivisorSigma(13, n): n in [1..20]]; //Vincenzo Librandi,Sep 10 2016
(PARI) my(N=99, q='q+O('q^N)); Vec(sum(n=1, N, n^13*q^n/(1-q^n))) \\Altug Alkan,Sep 10 2016
(PARI) a(n) = sigma(n, 13); \\Michel Marcus,Sep 10 2016
CROSSREFS
KEYWORD
nonn,mult,easy
AUTHOR
STATUS
approved