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
Inverse Mobius transform ofA001014.-R. J. Mathar,Oct 13 2011
LINKS
Vincenzo Librandi,Table of n, a(n) for n = 1..1000
FORMULA
G.f.: Sum_{k>=1} k^6*x^k/(1-x^k). -Benoit Cloitre,Apr 21 2003
L.g.f.: -log(Product_{k>=1} (1 - x^k)^(k^5)) = Sum_{n>=1} a(n)*x^n/n. -Ilya Gutkovskiy,May 06 2017
FromAmiram Eldar,Oct 29 2023: (Start)
Multiplicative with a(p^e) = (p^(6*e+6)-1)/(p^6-1).
Dirichlet g.f.: zeta(s)*zeta(s-6).
Sum_{k=1..n} a(k) = zeta(7) * n^7 / 7 + O(n^8). (End)
MAPLE
MATHEMATICA
lst={}; Do[AppendTo[lst, DivisorSigma[6, n]], {n, 5!}]; lst (*Vladimir Joseph Stephan Orlovsky,Mar 11 2009 *)
PROG
(Sage) [sigma(n, 6)for n in range(1, 24)] #Zerinvary Lajos,Jun 04 2009
(PARI) a(n)=sigma(n, 6) \\Charles R Greathouse IV,Apr 28, 2011
(Magma) [DivisorSigma(6, n): n in [1..30]]; //Bruno Berselli,Apr 10 2013
CROSSREFS
KEYWORD
nonn,mult,easy
AUTHOR
STATUS
approved