A281959 a(n) = sigma_25(n), the sum of the 25th powers of the divisors of n.

1, 33554433, 847288609444, 1125899940397057, 298023223876953126, 28430288877251865252, 1341068619663964900808, 37778932988857102106625, 717897987692699877379693, 10000000298023223910507558, 108347059433883722041830252, 953962194872104906760006308

Offset: 1

Author

Seiichi Manyama, Feb 03 2017

Keywords

• nonn
• mult
• easy

Comments

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

Since n^25 == n^5 (mod 25), it follows that sigma_25(n) == sigma_5(n) (mod 25). In fact, sigma_25(n) == sigma_(5) (mod 13200), where 13200 = (2^4)*3*(5^2)*11 = A006863(10). - Peter Bala, Jan 12 2025

Examples

			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
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..10000
• Peter Luschny, Is this a new representation of (some) Bernoulli numbers?, Mathematics Stack Exchange.
• Index entries for sequences related to sigma(n).

Crossrefs

Cf. A000005, A037947, A275055, A362870.

Programs

Mathematica

DivisorSigma[25,Range[20]] (* Harvey P. Dale, Jul 08 2024 *)

PARI

a(n) = sigma(n, 25) \\ Felix Fröhlich, Feb 03 2017

Python

from sympy import divisor_sigma
def A281959(n): return divisor_sigma(n,25) # Chai Wah Wu, May 07 2023

Formula

G.f.: Sum_{k>=1} k^25*x^k/(1-x^k).

a(n) == A037947(n) mod 657931.

a(n) = Sum_{k=1..A000005(n)} A275055(k)^25. - Felix Fröhlich, Feb 03 2017

Sum_{n>=1} a(n)/exp(2*Pi*n) = 657931/24 = Bernoulli(26)/52. - Vaclav Kotesovec, May 07 2023

From Amiram Eldar, Oct 29 2023: (Start)

Multiplicative with a(p^e) = (p^(25*e+25)-1)/(p^25-1).

Dirichlet g.f.: zeta(s)*zeta(s-25).

Sum_{k=1..n} a(k) = zeta(26) * n^26 / 26 + O(n^27). (End)