Why do $12$ and $120$ occur very often in the denominators of $\zeta(-n)$ for odd $n$?

Question

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

$\zeta(-2n) = 0$

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

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

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

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

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

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

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

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?

Answer 1

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 $120$.

Answer 2

What you observed resonates with the Von Staudt–Clausen theorem.

Indeed, by this theorem, the denominator of $(2k)\zeta(1-2k)=-B_{2k}$ equals $\prod_{p-1\mid 2k} p$. In particular, the numerator is odd, hence the denominator of $\zeta(1-2k)$ is divisible by $2\cdot 2\cdot 3=12$ for $k$ odd, and by $2\cdot 2\cdot 2\cdot 3\cdot 5=120$ for $k$ even.

Added. I misunderstood (misread) the question, and I thought the OP was only interested in the above divisibility relation. I certainly don't know how to prove that the denominator of $\zeta(1-2k)$ is infinitely often $12$ for $k$ odd and infinitely often $120$ for $k$ even. This seems an interesting but difficult question.

Answer 3

This is not an answer but rather a comment to illustrate the observation.

Consider the canonical primefactor-decomposition of the denominators of that values. You'll easily observe the relevant patterns in them. For better reading I've done them in two columns, for zeta(1-1),zeta(1-2) ; zeta(1-3),zeta(1-4) ;...;zeta(1-k),zeta(1-(k+1)); for index k=1 to some m.

at indexes      at indexes                  
k=1,3,5,7...    k=2,4,6,8,...             
=============   ===============================================================
  2            2^2  .3   <<----------------------------------------------------
  1            2^3  .3   .5
  1            2^2  .3^2      .7
  1            2^4  .3   .5
  1            2^2  .3                .11
  1            2^3  .3^2 .5   .7           .13
  1            2^2  .3   <<----------------------------------------------------
  1            2^5  .3   .5                      .17
  1            2^2  .3^3      .7                     .19
  1            2^3  .3   .5^2         .11
  1            2^2  .3                                        .23
  1            2^4  .3^2 .5   .7           .13
  1            2^2  .3   <<----------------------------------------------------
  1            2^3  .3   .5                                               .29
  1            2^2  .3^2      .7      .11                                     .31
  1            2^6  .3   .5                      .17

The rows which contain the horizontal lines denote the denominators which are exactly $12$ (and do not only contain $12$ as a factor). This patterns contain an obvious relation of the totient-value for the involved primefactors in relation to the index and are definite-ly described by the Clausen-von Staudt-theorem as mentioned in the other answer.

A tiny remark: Note, that the entry at index k=0 (zeta(1)) would contain all primefactors to infinite power.