$\zeta(-n) = - \dfrac{B\_{n+1}}{n+1}$


$\zeta(-2n) = 0$


$\zeta(-1) = - \dfrac{1}{12}$


$\zeta(-3) = \dfrac{1}{120}$


$\zeta(-5) = - \dfrac{1}{252}$


$\zeta(-7) = \dfrac{1}{240}$


$\zeta(-9) = - \dfrac{691}{132}$


$\zeta(-11) = \dfrac{1}{32760}$


$\zeta(-13) = - \dfrac{3617}{12}$


What I am interested in are the sequence of denominators of these fractions.


$12, 120, 252, 240, 132, 32760, 12, 8160, 14364, 6600, 276, 65520, 12, 3480, 85932, 16320, 12, 69090840, 12, 541200, 75852, 2760, 564, 2227680, 132, 6360, 43092, 6960, 708, 3407203800, 12, 32640, 388332, 120, 9372, 10087262640$


<http://oeis.org/A006953>


They are alternately divisible by 12 and 120.


In the first $1000$ terms, $120$ occurs $53$ times, and $12$ occurs even more often. So it seems they both occur infinitely many times.


My question is: Why? What is the meaning of all this? Does that make $12$ and $120$ some kinds of special numbers?



Why do $12$ and $120$ occur very often in the denominators of $\zeta(-n)$ for odd $n$?
Von-Staudt's Second Theorem gives the exact prime decomposition of the denominator of $B\_{2k}/(2k)$; it is $$\prod\_{p:(p-1)\mid2k}p^{1+\nu\_p(2k)}$$ The product is over primes $p$ such that $p-1$ divides $2k$, and $\nu\_p(n)$ is the largest exponent $e$ such that $p^e$ divides $n$. 


If $k$ is prime, and if $2k+1$ is composite, then the only divisors of $2k$ are $1,2,k,2k$, and the only divisors of the form $p-1$ with $p$ a prime are $1$ and $2$, the primes being $2$ and $3$; moreover, $\nu\_2(2k)=1$ and $\nu\_3(2k)=0$. Thus, in all such cases, the denominator of $B\_{2k}/(2k)$ is 12. 


If $k\equiv1\bmod3$ (and $k>1$), then $2k+1$ is composite. As there are infinitely many primes $k\equiv1\bmod3$, there are infinitely many $k$ such that the denominator of $B\_{2k}/(2k)$ is $12$. 


A similar argument for $k=2q$ with $q$ prime and $q\equiv1\bmod{15}$ shows that there are infinitely many $k$ such that the denominator of $B\_{2k}/(2k)$ is $120$. 


