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A281959 a(n) = sigma_25(n), the sum of the 25th powers of the divisors of n. 5
1, 33554433, 847288609444, 1125899940397057, 298023223876953126, 28430288877251865252, 1341068619663964900808, 37778932988857102106625, 717897987692699877379693, 10000000298023223910507558, 108347059433883722041830252, 953962194872104906760006308 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
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
LINKS
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)
EXAMPLE
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
PROG
(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
CROSSREFS
Sequence in context: A059277 A010813 A323661 * A194553 A246108 A184984
KEYWORD
nonn,mult,easy
AUTHOR
Seiichi Manyama, Feb 03 2017
STATUS
approved

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Last modified April 27 23:22 EDT 2024. Contains 372020 sequences. (Running on oeis4.)