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,

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

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 $B_{2 n} + \sum_{(p - 1) | 2 n} \frac{1}{p}$ 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_{2 n} = \sum_{j = 0}^{2 n} \frac{1}{j + 1} \sum_{m = 0}^{j} (- 1)^{m} (\binom{j}{m}) m^{2 n}

and as a corollary:

B_{2 n} = \sum_{j = 0}^{2 n} \frac{j!}{j + 1} (- 1)^{j} S (2 n, 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} (\binom{p - 1}{m}) m^{2 n} \equiv - 1 (\mod p) .

2. If Template:Math does not divide Template:Math, then

\sum_{m = 0}^{p - 1} (- 1)^{m} (\binom{p - 1}{m}) m^{2 n} \equiv 0 (\mod p) .

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

m^{p - 1} \equiv 1 (\mod p)

for Template:Math.

If Template:Math divides Template:Math, then one has

m^{2 n} \equiv 1 (\mod p)

for Template:Math. Thereafter, one has

\sum_{m = 1}^{p - 1} (- 1)^{m} (\binom{p - 1}{m}) m^{2 n} \equiv \sum_{m = 1}^{p - 1} (- 1)^{m} (\binom{p - 1}{m}) (\mod p),

from which (1) follows immediately.

If Template:Math does not divide Template:Math, then after Fermat's theorem one has

m^{2 n} \equiv m^{2 n - (p - 1)} (\mod p) .

If one lets Template:Math, then after iteration one has

m^{2 n} \equiv m^{2 n - ℘ (p - 1)} (\mod p)

for Template:Math and Template:Math.

Thereafter, one has

\sum_{m = 0}^{p - 1} (- 1)^{m} (\binom{p - 1}{m}) m^{2 n} \equiv \sum_{m = 0}^{p - 1} (- 1)^{m} (\binom{p - 1}{m}) m^{2 n - ℘ (p - 1)} (\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} (\binom{3}{m}) m^{2 n} = 3 \cdot 2^{2 n} - 3^{2 n} - 3 \equiv 0 (\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_{2 n} = I_{n} - \sum_{(p - 1) | 2 n} \frac{1}{p},

where Template:Math is an integer, as desired.^[1]^[2]

References

↑ H. Rademacher, Analytic Number Theory, Springer-Verlag, New York, 1973.
↑ T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976.

External links

Template:MathWorld

[1] H. Rademacher, Analytic Number Theory, Springer-Verlag, New York, 1973.

[2] T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976.

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Von Staudt–Clausen theorem

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Von Staudt–Clausen theorem

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