In number theory, the von Staudt–Clausen theorem is a result determining the fractional part of Bernoulli numbers, found independently by
and .
Specifically, if is a positive integer and we add to the Bernoulli number for every prime such that divides , then we obtain an integer; that is,
<math> B_{2n} + \sum_{(p-1)|2n} \frac1p \in \Z . </math>
This fact immediately allows us to characterize the denominators of the non-zero Bernoulli numbers as the product of all primes such that divides ; 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, ... .
The sequence of integers <math>B_{2n} + \sum_{(p-1)|2n} \frac1p</math> is
: 1, 1, 1, 1, 1, 1, 2, -6, 56, -528, 6193, -86579, 1425518, -27298230, ... .
Proof
A proof of the Von Staudt–Clausen theorem follows from an explicit formula for Bernoulli numbers which is:
:<math> B_{2n}=\sum_{j=0}^{2n}{\frac{1}{j+1\sum_{m=0}^{j}{(-1)^{m}{j\choose m}m^{2n </math>
and as a corollary:
:<math> B_{2n}=\sum_{j=0}^{2n}{\frac{j!}{j+1(-1)^jS(2n,j) </math>
where are the Stirling numbers of the second kind.
Furthermore the following lemmas are needed:
Let be a prime number; then
1. If divides , then
: <math> \sum_{m=0}^{p-1}{(-1)^m{p-1\choose m} m^{2n\equiv{-1}\pmod p. </math>
2. If does not divide , then
: <math> \sum_{m=0}^{p-1}{(-1)^m{p-1\choose m} m^{2n\equiv0\pmod p. </math>
Proof of (1) and (2): One has from Fermat's little theorem,
: <math> m^{p-1} \equiv 1 \pmod{p} </math>
for .
If divides , then one has
: <math> m^{2n} \equiv 1 \pmod{p} </math>
for . Thereafter, one has
: <math> \sum_{m=1}^{p-1} (-1)^m \binom{p-1}{m} m^{2n} \equiv \sum_{m=1}^{p-1} (-1)^m \binom{p-1}{m} \pmod{p},</math>
from which (1) follows immediately.
If does not divide , then after Fermat's theorem one has
: <math> m^{2n} \equiv m^{2n-(p-1)} \pmod{p}. </math>
If one lets , then after iteration one has
: <math> m^{2n} \equiv m^{2n-\wp(p-1)} \pmod{p} </math>
for and .
Thereafter, one has
: <math> \sum_{m=0}^{p-1} (-1)^m \binom{p-1}{m} m^{2n} \equiv \sum_{m=0}^{p-1} (-1)^m \binom{p-1}{m} m^{2n-\wp(p-1)} \pmod{p}. </math>
Lemma (2) now follows from the above and the fact that for .
(3). It is easy to deduce that for and , divides .
(4). Stirling numbers of the second kind are integers.
Now we are ready to prove the theorem.
If is composite and , then from (3), divides .
For ,
: <math> \sum_{m=0}^{3} (-1)^m \binom{3}{m} m^{2n} = 3 \cdot 2^{2n} - 3^{2n} - 3 \equiv 0 \pmod{4}. </math>
If is prime, then we use (1) and (2), and if is composite, then we use (3) and (4) to deduce
:<math> B_{2n} = I_n - \sum_{(p-1)|2n} \frac{1}{p}, </math>
where is an integer, as desired.
See also
- Kummer's congruence