Do Bernoulli Numbers Always Have Unique Prime Factors in Their Denominators?

  • Thread starter Thread starter Damidami
  • Start date Start date
  • Tags Tags
    Bernoulli
Damidami
Messages
93
Reaction score
0
Is this a known bernoulli number conjeture/theorem?:
The denominators of B_n (when expressed as an irreducible fraction), doesn't contain as a factor powers of prime numbers (ex. isn't divided by 5^2)

Example:

B_2 = 1/6
6 = 2*3

B_4 = -1/30
30 = 2*3*5

B_{24} = -236364091/2730
2730 = 2*3*5*7*13

I know Ramanujan proved that the denominator contain 2 and 3 as a factor one and only once, but I hadn't heard that any prime on the factorization of the denominator happens only once.
 
Physics news on Phys.org
Nevermind.

http://en.wikipedia.org/wiki/Von_Staudt%E2%80%93Clausen_theorem"

Aditional information is welcome.
 
Last edited by a moderator:
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...
Back
Top