%I #44 Jul 08 2024 13:56:48
%S 1,33554433,847288609444,1125899940397057,298023223876953126,
%T 28430288877251865252,1341068619663964900808,37778932988857102106625,
%U 717897987692699877379693,10000000298023223910507558,108347059433883722041830252,953962194872104906760006308
%N a(n) = sigma_25(n), the sum of the 25th powers of the divisors of n.
%C For k > 0, Sum_{n>=1} sigma_(4*k+1)(n) / exp(2*Pi*n) = Bernoulli(4*k+2)/(8*k+4). For k = 0, Sum_{n>=1} sigma(n)/exp(2*Pi*n) = 1/24 - 1/(8*Pi) = Bernoulli(2)/4 - 1/(8*Pi). - _Vaclav Kotesovec_, May 07 2023
%H Seiichi Manyama, <a href="/A281959/b281959.txt">Table of n, a(n) for n = 1..10000</a>
%H Peter Luschny, <a href="https://math.stackexchange.com/questions/4694898/">Is this a new representation of (some) Bernoulli numbers?</a>, MSE.
%H <a href="/index/Si#SIGMAN">Index entries for sequences related to sigma(n)</a>.
%F G.f.: Sum_{k>=1} k^25*x^k/(1-x^k).
%F a(n) == A037947(n) mod 657931.
%F a(n) = Sum_{k=1, A000005(n)} A275055(k)^25. - _Felix Fröhlich_, Feb 03 2017
%F Sum_{n>=1} a(n)/exp(2*Pi*n) = 657931/24 = Bernoulli(26)/52. - _Vaclav Kotesovec_, May 07 2023
%F From _Amiram Eldar_, Oct 29 2023: (Start)
%F Multiplicative with a(p^e) = (p^(25*e+25)-1)/(p^25-1).
%F Dirichlet g.f.: zeta(s)*zeta(s-25).
%F Sum_{k=1..n} a(k) = zeta(26) * n^26 / 26 + O(n^27). (End)
%e For n = 6: The divisors of 6 are 1, 2, 3, 6, so a(6) = sigma_25(6) = 1^25 + 2^25 + 3^25 + 6^25 = 28430288877251865252. - _Felix Fröhlich_, Feb 03 2017
%t DivisorSigma[25,Range[20]] (* _Harvey P. Dale_, Jul 08 2024 *)
%o (PARI) a(n) = sigma(n, 25) \\ _Felix Fröhlich_, Feb 03 2017
%o (Python)
%o from sympy import divisor_sigma
%o def A281959(n): return divisor_sigma(n,25) # _Chai Wah Wu_, May 07 2023
%Y Cf. A000005, A037947, A275055, A362870.
%K nonn,mult,easy
%O 1,2
%A _Seiichi Manyama_, Feb 03 2017