1,2

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

Inverse Mobius transform of A001014. - R. J. Mathar, Oct 13 2011

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Index entries for sequences related to sigma(n).

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

From Amiram 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^6). (End)

A013954 := proc(n)

numtheory[sigma][6](n) ;

end proc: # R. J. Mathar, Oct 13 2011

lst={}; Do[AppendTo[lst, DivisorSigma[6, n]], {n, 5!}]; lst (* Vladimir Joseph Stephan Orlovsky, Mar 11 2009 *)

DivisorSigma[6, Range[30]] (* Harvey P. Dale, May 11 2025 *)

(SageMath) [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

Cf. A000203, A001014, A001157-A001160, A013665, A013955-A013972, A017665-A017712.

Sequence in context: A034680 A351301 A017675 * A294301 A343508 A116277

Adjacent sequences: A013951 A013952 A013953 * A013955 A013956 A013957

nonn,mult,easy

approved