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a(n) = sigma_10(n), the sum of the 10th powers of the divisors of n.
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%I #41 Oct 29 2023 02:36:32

%S 1,1025,59050,1049601,9765626,60526250,282475250,1074791425,

%T 3486843451,10009766650,25937424602,61978939050,137858491850,

%U 289537131250,576660215300,1100586419201,2015993900450,3574014537275,6131066257802,10250010815226,16680163512500,26585860217050

%N a(n) = sigma_10(n), the sum of the 10th powers of the divisors of n.

%C 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).

%C 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

%H Vincenzo Librandi, <a href="/A013958/b013958.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Si#SIGMAN">Index entries for sequences related to sigma(n)</a>.

%F G.f.: Sum_{k>=1} k^10*x^k/(1-x^k). - _Benoit Cloitre_, Apr 21 2003

%F L.g.f.: -log(Product_{k>=1} (1 - x^k)^(k^9)) = Sum_{n>=1} a(n)*x^n/n. - _Ilya Gutkovskiy_, May 06 2017

%F From _Amiram Eldar_, Oct 29 2023: (Start)

%F Multiplicative with a(p^e) = (p^(10*e+10)-1)/(p^10-1).

%F Dirichlet g.f.: zeta(s)*zeta(s-10).

%F Sum_{k=1..n} a(k) = zeta(11) * n^11 / 11 + O(n^12). (End)

%t lst={};Do[AppendTo[lst,DivisorSigma[10,n]],{n,5!}];lst (* _Vladimir Joseph Stephan Orlovsky_, Mar 11 2009 *)

%t DivisorSigma[10,Range[20]] (* _Harvey P. Dale_, Jan 04 2012 *)

%o (Sage) [sigma(n,10)for n in range(1,18)] # _Zerinvary Lajos_, Jun 04 2009

%o (PARI) a(n)=sigma(n,10) \\ _Charles R Greathouse IV_, Apr 28, 2011

%o (Magma) [DivisorSigma(10,n): n in [1..20]]; // _Bruno Berselli_, Apr 10 2013

%Y Cf. A000203, A001157-A001160, A013669, A013954-A013972, A017665-A017712.

%K nonn,mult,easy

%O 1,2

%A _N. J. A. Sloane_