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A013966
a(n) = sigma_18(n), the sum of the 18th powers of the divisors of n.
6
1, 262145, 387420490, 68719738881, 3814697265626, 101560344351050, 1628413597910450, 18014467229220865, 150094635684419611, 1000003814697527770, 5559917313492231482, 26623434909949071690, 112455406951957393130, 426880482624234915250, 1477891883850485076740
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
FORMULA
G.f.: Sum_{k>=1} k^18*x^k/(1-x^k). - Benoit Cloitre, Apr 21 2003
From Amiram Eldar, Oct 29 2023: (Start)
Multiplicative with a(p^e) = (p^(18*e+18)-1)/(p^18-1).
Dirichlet g.f.: zeta(s)*zeta(s-18).
Sum_{k=1..n} a(k) = zeta(19) * n^19 / 19. + O(n^20). (End)
MATHEMATICA
Table[DivisorSigma[18, n], {n, 50}] (* Vladimir Joseph Stephan Orlovsky, Mar 11 2009 *)
PROG
(Sage) [sigma(n, 18)for n in range(1, 13)] # Zerinvary Lajos, Jun 04 2009
(PARI) a(n)=sigma(n, 18) \\ Charles R Greathouse IV, Apr 28 2011
(Magma) [DivisorSigma(18, n): n in [1..50]]; // G. C. Greubel, Nov 03 2018
KEYWORD
nonn,mult,easy
STATUS
approved