congruence of Clausen and von Staudt

Let $B_k$ denote the $k$th Bernoulli number:

$$B_0=1,B_1=-\frac{1}{2},B_2=\frac{1}{6},B_3=0,B_4=-\frac{1}{30},\mathrm{\dots }$$

In fact, $B_k=0$ for all odd $k\ge 3$, so we will only consider $B_k$ for even $k$. The following is a well-known congruence, due to Thomas Clausen and Karl von Staudt.

Theorem (Congruence of Clausen and von Staudt).

For an even integer $k\mathrm{\ge }\mathrm{2}$,

$$B_k\equiv -\sum _{p\mathit{\text{ prime}},(p-1)|k}\frac{1}{p}mod\mathbb{Z}$$

where the sum is over all primes $p$ such that $\mathrm{(}p\mathrm{-}\mathrm{1}\mathrm{)}$ divides $k$. In other words, there exists an integer $n_k$ such that

$$B_k=n_k-\sum _{p\mathit{\text{ prime}},(p-1)|k}\frac{1}{p}.$$

For example:

$$B_2=\frac{1}{6}=1-\frac{1}{2}-\frac{1}{3},B_4=-\frac{1}{30}=1-\frac{1}{2}-\frac{1}{3}-\frac{1}{5}.$$

Sometimes the theorem is stated in this alternative form:

Corollary.

For an even integer $k\mathrm{\ge }\mathrm{2}$ and any prime $p$ the product $p\mathit{}{B}_k$ is $p$-integral, that is, $p\mathit{}{B}_k$ is a rational number $t\mathrm{/}s$ (in lowest terms) such that $p$ does not divide $s$. Moreover:

$$p{B}_k\equiv \{\begin{array}{cc}-1modp,\mathit{\text{ if }}(p-1)\mathit{\text{ divides }}k\mathit{\text{;}}\hfill & \\ 0modp,\mathit{\text{ if }}(p-1)\mathit{\text{ does not divide }}k.\hfill & \end{array}$$