Von Staudt–Clausen theorem

Template:Short description In number theory, the von Staudt–Clausen theorem is a result determining the fractional part of Bernoulli numbers, found independently by Template:Harvs and Template:Harvs.

Specifically, if Template:Mvar is a positive integer and we add Template:Math to the Bernoulli number Template:Math for every prime Template:Mvar such that Template:Math divides Template:Math, then we obtain an integer; that is,

$B_{2n}+\sum _{(p-1)|2n}\frac{1}{p}\in \mathbb{\mathbb{Z}}.$

This fact immediately allows us to characterize the denominators of the non-zero Bernoulli numbers Template:Math as the product of all primes Template:Mvar such that Template:Math divides Template:Math; consequently, the denominators are square-free and divisible by 6.

These denominators are

6, 30, 42, 30, 66, 2730, 6, 510, 798, 330, 138, 2730, 6, 870, 14322, 510, 6, 1919190, 6, 13530, ... (sequence A002445 in the OEIS).

The sequence of integers $B_{2n}+\sum _{(p-1)|2n}\frac{1}{p}$ is

1, 1, 1, 1, 1, 1, 2, -6, 56, -528, 6193, -86579, 1425518, -27298230, ... (sequence A000146 in the OEIS).

Proof

A proof of the Von Staudt–Clausen theorem follows from an explicit formula for Bernoulli numbers which is:

$B_{2n}=\sum _{j=0}^{2n}\frac{1}{j+1}\sum _{m=0}^j(-1{)}^m(\genfrac{}{}{0ex}{}{j}{m})m^{2n}$

and as a corollary:

$B_{2n}=\sum _{j=0}^{2n}\frac{j!}{j+1}(-1{)}^{j}S(2n,j)$

where Template:Math are the Stirling numbers of the second kind.

Furthermore the following lemmas are needed:

Let Template:Mvar be a prime number; then

1. If Template:Math divides Template:Math, then

$\sum _{m=0}^{p-1}(-1{)}^m(\genfrac{}{}{0ex}{}{p-1}{m})m^{2n}\equiv -1(\mathrm{mod}p).$

2. If Template:Math does not divide Template:Math, then

$\sum _{m=0}^{p-1}(-1{)}^m(\genfrac{}{}{0ex}{}{p-1}{m})m^{2n}\equiv 0(\mathrm{mod}p).$

Proof of (1) and (2): One has from Fermat's little theorem,

$m^{p-1}\equiv 1(\mathrm{mod}p)$

for Template:Math.

If Template:Math divides Template:Math, then one has

$m^{2n}\equiv 1(\mathrm{mod}p)$

for Template:Math. Thereafter, one has

$\sum _{m=1}^{p-1}(-1{)}^m(\genfrac{}{}{0ex}{}{p-1}{m})m^{2n}\equiv \sum _{m=1}^{p-1}(-1{)}^m(\genfrac{}{}{0ex}{}{p-1}{m})(\mathrm{mod}p),$

from which (1) follows immediately.

If Template:Math does not divide Template:Math, then after Fermat's theorem one has

$m^{2n}\equiv m^{2n-(p-1)}(\mathrm{mod}p).$

If one lets Template:Math, then after iteration one has

$m^{2n}\equiv m^{2n-\mathrm{\wp }(p-1)}(\mathrm{mod}p)$

for Template:Math and Template:Math.

Thereafter, one has

$\sum _{m=0}^{p-1}(-1{)}^m(\genfrac{}{}{0ex}{}{p-1}{m})m^{2n}\equiv \sum _{m=0}^{p-1}(-1{)}^m(\genfrac{}{}{0ex}{}{p-1}{m})m^{2n-\mathrm{\wp }(p-1)}(\mathrm{mod}p).$

Lemma (2) now follows from the above and the fact that Template:Math for Template:Math.

(3). It is easy to deduce that for Template:Math and Template:Math, Template:Mvar divides Template:Math.

(4). Stirling numbers of the second kind are integers.

Now we are ready to prove the theorem.

If Template:Math is composite and Template:Math, then from (3), Template:Math divides Template:Math.

For Template:Math,

$\sum _{m=0}^3(-1{)}^m(\genfrac{}{}{0ex}{}{3}{m})m^{2n}=3\cdot 2^{2n}-3^{2n}-3\equiv 0(\mathrm{mod}4).$

If Template:Math is prime, then we use (1) and (2), and if Template:Math is composite, then we use (3) and (4) to deduce

$B_{2n}=I_n-\sum _{(p-1)|2n}\frac{1}{p},$

where Template:Math is an integer, as desired.

See also

• Kummer's congruence