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
and .

Specifically, if ''n'' is a positive integer and we add 1/''p'' to the Bernoulli number ''B''_2''n'' for every prime ''p'' such that ''p'' − 1 divides 2''n'', we obtain an integer, i.e.,
$B_ + \sum_ \frac1p \in \Z .$

This fact immediately allows us to characterize the denominators of the non-zero Bernoulli numbers ''B''_2''n'' as the product of all primes ''p'' such that ''p'' − 1 divides 2''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, ... .

The sequence of integers $B_ + \sum_ \frac1p$ 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:
:$B_=\sum_^\sum_^$
and as a corollary:
:$B_=\sum_^(-1)^jS(2n,j)$
where $S(n,j)$ are the Stirling numbers of the second kind.

Furthermore the following lemmas are needed:

Let p be a prime number then,

1. If p-1 divides 2n then,
:$\sum_^\equiv\pmod p$
2. If p-1 does not divide 2n then,
:$\sum_^\equiv0\pmod p$
Proof of (1) and (2): One has from Fermat's little theorem,
:$m^\equiv 1\pmod p$
for $m=1,2,...,p-1$.

If p-1 divides 2n then one has,
:$m^\equiv 1\pmod p$
for $m=1,2,...,p-1$.

Thereafter one has,
:$\sum_^\equiv \sum_^\pmod p$
from which (1) follows immediately.

If p-1 does not divide 2n then after Fermat's theorem one has,
:$m^\equiv m^\pmod p$
If one lets \wp= frac ( Greatest integer function) then after iteration one has,
:$m^\equiv m^\pmod p$
for $m=1,2,...,p-1$ and 0<2n-\wp(p-1).

Thereafter one has,
:$\sum_^\equiv\sum_^\pmod p$
Lemma (2) now follows from the above and the fact that ''S''(''n'',''j'')=0 for ''j''>''n''.

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

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

Proof of the theorem: 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,
:$\sum_^=3 \cdot 2^-3^-3\equiv0 \pmod 4$
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:
:$B_=I_n-\sum_$
where $I_n$ is an integer, which is the Von-Staudt Clausen theorem.T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976.

See also

* Kummer's congruence

References

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External links

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{{DEFAULTSORT:Von Staudt Clausen Theorem
Theorems in number theory