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,

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

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, ... Template: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, ... Template: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

References

• Template:Citation
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External links

• Template:MathWorld