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 sequences A017665-A017712 also give the numerators and denominators of sigma_k(n)/n^k for k = 1..24. The power sums sigma_k(n) are in sequences A000203 (k=1), A001157-A001160 (k=2,3,4,5), A013954-A013972 for k = 6,7,...,24. - Ahmed Fares (ahmedfares(AT)my-deja.com), Apr 05 2001
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..1000
FORMULA
G.f.: Sum_{k>=1} k^17*x^k/(1-x^k). - Benoit Cloitre, Apr 21 2003
Dirichlet g.f.: zeta(s-17)*zeta(s). - Ilya Gutkovskiy, Sep 10 2016
Empirical: Sum_{n>=1} a(n)/exp(2*Pi*n) = 43867/28728, also equals Bernoulli(18)/36. - Simon Plouffe, May 06 2023
From Amiram Eldar, Oct 29 2023: (Start)
Multiplicative with a(p^e) = (p^(17*e+17)-1)/(p^17-1).
Sum_{k=1..n} a(k) = zeta(18) * n^18 / 18 + O(n^19). (End)
MATHEMATICA
DivisorSigma[17, Range[20]] (* Harvey P. Dale, May 30 2013 *)
PROG
(Sage) [sigma(n, 17)for n in range(1, 14)] # Zerinvary Lajos, Jun 04 2009
(Magma) [DivisorSigma(17, 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^17*q^n/(1-q^n))) \\ Altug Alkan, Sep 10 2016
(PARI) a(n) = sigma(n, 17); \\ Amiram Eldar, Oct 29 2023
CROSSREFS
KEYWORD
nonn,easy,mult
AUTHOR
STATUS
approved