Are the Bernoulli denominators always divisible by these corresponding primes? I was wondering whether it has been proven/disproven yet or at least conjectured that the bernoulli denominator of $B_{2n}$ is divisible by $2n+1$ if and only if $2n+1$ is prime?
If not, must the denominator always share a common factor with the value $2n+1$?
Or the question could be written as is denom($B_{2n}) = \prod\limits_{p:(p-1)|(2n)} p    = (2n+1)m$ where m is an integer always true if and only if $2n+1$ is prime?
 A: No.  For example, the denominator of $B_{560}$ is $15037922004270$, which is divisible by $561 = 3 \times 11 \times 17$.
Hmm: is there a relation to Carmichael numbers?
EDIT: By Korselt's criterion, composite positive integer $x$ is a Carmichael number iff $x$ is square-free and $p-1 \mid x-1$ for all primes $p$ dividing $x$. Moreover every Carmichael number is odd.
By von Staudt-Clausen, the denominator of $B_{2n}$ is the product of all the primes $p$ such that $p-1 \mid 2n$.
So if $x=2n+1$ is a Carmichael number, the primes $p$ dividing $x$ are all factors of the denominator of $B_{2n}$, and the product of those primes, which is
$x$ itself since $x$ is square-free, divides that denominator.
On the other hand, every odd composite $x=2n+1$ such that $x$ divides the denominator of $B_{2n}$ must be square-free (because the denominator of $B_{2n}$ is square-free), and the primes $p$ dividing $x$ must divide the denominator of $B_{2n}$ so $p-1 \mid x-1$, i.e. $x$ must be Carmichael.
So the result is: $2n+1$ divides the denominator of $B_{2n}$ iff $2n+1$ is either prime or Carmichael. 
