OFFSET
1,2
COMMENTS
In general, 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).
This formula can best be understood as a statement about the divided Bernoulli numbers b(n) = B(n) / n. Then you can say: If v is twice an odd number greater than 1 (i.e., v = 4*n + 2, a term of A016825 that is greater than 2), then b(v) = 2 * Sum_{j>=1} sigma_{v - 1}(j) / exp(2*Pi*j) = A358625(v) / A075180(v - 1). - Peter Luschny, May 08 2023
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.
FORMULA
G.f.: Sum_{k>=1} k^29 * x^k / (1-x^k).
Dirichlet g.f.: zeta(s-29)*zeta(s).
Sum_{k=1..n} a(k) ~ zeta(30) * n^30 / 30.
Sum_{n>=1} a(n)/exp(2*Pi*n) = 1723168255201/171864 = Bernoulli(30)/60.
Multiplicative with a(p^e) = (p^(29*e+29)-1)/(p^29-1). - Amiram Eldar, Oct 29 2023
MAPLE
with(NumberTheory): seq(SumOfDivisors(k, 29), k = 1..20);
MATHEMATICA
DivisorSigma[29, Range[20]]
PROG
(PARI) for(n=1, 20, print1(direuler( p=2, n, 1 / (1 - X) /(1 - p^29*X))[n], ", "))
(Python)
from sympy import divisor_sigma
def A362870(n): return divisor_sigma(n, 29) # Chai Wah Wu, May 07 2023
CROSSREFS
KEYWORD
nonn,easy,mult
AUTHOR
Vaclav Kotesovec, May 07 2023
STATUS
approved