Von Staudt–Clausen theorem

In number theory, the von Staudt–Clausen theorem is a result determining the fractional part of Bernoulli numbers, found independently by Karl von Staudt (1840) and Thomas Clausen (1840).

Specifically, if we add 1/p to B_n for every prime p such that p − 1 divides n, we obtain an integer.

This fact immediately allows us to characterize the denominators of the non-zero Bernoulli numbers B_n as the product of all primes p such that p − 1 divides n; 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 OEIS)

Formulation

The von Staudt–Clausen theorem has two parts. The first one describes how the denominators of the Bernoulli numbers can be computed. Paraphrasing the words of Clausen it can be stated as:

“The denominator of the 2nth Bernoulli number can be found as follows: Add to all divisors of 2n, 1, 2, a, a', ..., 2n the unity, which gives the sequence 2, 3, a + 1, a' + 1, ..., 2n + 1. Select from this sequence only the prime numbers 2, 3, p, p', etc. and build their product.”

The second part of the von Staudt–Clausen theorem is a representation of the Bernoulli numbers. This representation is given for the first few nonzero Bernoulli numbers in the next table.

Von Staudt–Clausen representation of Bn
B0	=	1
B1	=	− 1/2
B2	=	1 − 1/2 − 1/3
B4	=	1 − 1/2 − 1/3 − 1/5
B6	=	1 − 1/2 − 1/3 − 1/7
B8	=	1 − 1/2 − 1/3 − 1/5
B10	=	1 − 1/2 − 1/3 − 1/11

Proof

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

$B_{2n}=\sum_{j=0}^{2n}{\frac{1}{j+1}}\sum_{m=0}^{j}{(-1)^{m}{j\choose m}m^{2n}} \!$

where $S(n,j) \!$ are the Stirling numbers of the second kind.

Furthermore following lemmas are needed:

Let p be a prime number then,

1.If p-1 divides 2n then,

$\sum_{m=0}^{p-1}{(-1)^m{p-1\choose m} m^{2n}}\equiv{-1}\pmod p \!$

2. If if p-1 does not divide 2n then,

$\sum_{m=0}^{p-1}{(-1)^m{p-1\choose m} m^{2n}}\equiv0\pmod p \!$

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

$m^{p-1}\equiv 1 \pmod p \!$

for $m=1,2,...,p-1 \!$.

If p-1 divides 2n then one has,

$m^{2n}\equiv 1 \pmod p \!$

for $m=1,2,...,p-1 \!$.

Thereafter one has,

$\sum_{m=1}^{p-1}{(-1)^m{p-1\choose m} m^{2n}}\equiv \sum_{m=1}^{p-1}{(-1)^m{p-1\choose m}}\pmod p \!$

from which (1) follows immediately.

If p-1 does not divide 2n then after Fermat's theorem one has,

$m^{2n}\equiv m^{2n-(p-1)}\pmod p \!$

If one lets $\wp=[\frac{2n}{p-1}] \!$ (Greatest integer function) then after iteration one has,

$m^{2n}\equiv m^{2n-\wp(p-1)}\pmod p \!$

for $m=1,2,...,p-1 \!$ and $0<2n-\wp(p-1)<p-1 \!$.

Thereafter one has,

$\sum_{m=0}^{p-1}{(-1)^m{p-1\choose m} m^{2n}}\equiv\sum_{m=0}^{p-1}{(-1)^m{p-1\choose m} m^{2n-\wp(p-1)}}\pmod p \!$

(2) follows from above and the fact that for j>n S(n,j)=0.

(3).It is easy to deduce the fact that for a>2 and b>2, ab divides (ab-1)!.

Proof: Now we are ready to prove Von-Staudt Clausen theorem,
If j+1 is composite and j>3 then from (3), j+1 divides j!.
For j=3,

$\sum_{m=0}^{3}{(-1)^m{3\choose m}m^{2n}}=3.2^{2n}-3^{2n}-3\equiv0\pmod 4 \!$

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

If j+1 is prime then we use (1) and (2) and if j+1 is composite then we use (3) and (4) to deduce:

$B_{2n}=I_n-\sum_{p-1|2n}{\frac{1}{p}} \!$

where $I_n \!$ is an integer.
which is Von-Staudt Clausen theorem.

See also

• Kummer's congruence

References

• Clausen, Thomas (1840), "Theorem", Astronomische Nachrichten 17 (22): 351–352, doi:10.1002/asna.18400172204 
• Rado, R. (1934), "A New Proof of a Theorem of V. Staudt", J. London Math. Soc. 9 (2): 85–88, doi:10.1112/jlms/s1-9.2.85 
• von Staudt, Ch. (1840), "Beweis eines Lehrsatzes, die Bernoullischen Zahlen betreffend", Journal für Reine und Angewandte Mathematik 21: 372–374, ISSN 0075-4102, JFM 021.0672cj, http://resolver.sub.uni-goettingen.de/purl?GDZPPN002142562 

External links

• Weisstein, Eric W., "von Staudt-Clausen Theorem" from MathWorld.