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,
<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 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 <math>B_{2n} + \sum_{(p-1)|2n} \frac1p</math> is
- 1, 1, 1, 1, 1, 1, 2, -6, 56, -528, 6193, -86579, 1425518, -27298230, ... (sequence A000146 in the OEIS).
ProofEdit
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 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
- <math> \sum_{m=0}^{p-1}{(-1)^m{p-1\choose m} m^{2n}}\equiv{-1}\pmod p. </math>
2. If Template:Math does not divide Template:Math, 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 Template:Math.
If Template:Math divides Template:Math, then one has
- <math> m^{2n} \equiv 1 \pmod{p} </math>
for Template:Math. 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 Template:Math does not divide Template:Math, then after Fermat's theorem one has
- <math> m^{2n} \equiv m^{2n-(p-1)} \pmod{p}. </math>
If one lets Template:Math, then after iteration one has
- <math> m^{2n} \equiv m^{2n-\wp(p-1)} \pmod{p} </math>
for Template:Math and Template:Math.
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 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,
- <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 Template:Math is prime, then we use (1) and (2), and if Template:Math is composite, then we use (3) and (4) to deduce
- <math> B_{2n} = I_n - \sum_{(p-1)|2n} \frac{1}{p}, </math>
where Template:Math is an integer, as desired.<ref>H. Rademacher, Analytic Number Theory, Springer-Verlag, New York, 1973.</ref><ref>T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976.</ref>
See alsoEdit
ReferencesEdit
<references />
External linksEdit
- {{#invoke:Template wrapper|{{#if:|list|wrap}}|_template=cite web
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