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 $n$ 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 $2 n$ , then we obtain an integer; that is,

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

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, ... (sequence A002445 in the OEIS).

The sequence of integers $B_{2n}+\sum _{(p-1)|2n}{\frac {1}{p}}$ is

1, 1, 1, 1, 1, 1, 2, -6, 56, -528, 6193, -86579, 1425518, -27298230, ... (sequence A000146 in the OEIS).

Proof

A proof of the 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}}

and as a corollary:

B_{2n}=\sum _{j=0}^{2n}{\frac {j!}{j+1}}(-1)^{j}S(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 $2 n$ , then

\sum _{m=0}^{p-1}{(-1)^{m}{p-1 \choose m}m^{2n}}\equiv {-1}{\pmod {p}}.

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

\sum _{m=0}^{p-1}{(-1)^{m}{p-1 \choose m}m^{2n}}\equiv 0{\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 $2 n$ , 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}{\binom {p-1}{m}}m^{2n}\equiv \sum _{m=1}^{p-1}(-1)^{m}{\binom {p-1}{m}}{\pmod {p}},

from which (1) follows immediately.

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

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

If one lets $℘ = ⌊ 2 n / (p - 1) ⌋$ , then after iteration one has

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

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

Thereafter, one has

\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}}.

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 the second kind are integers.

Now we are ready to prove the 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}{\binom {3}{m}}m^{2n}=3\cdot 2^{2n}-3^{2n}-3\equiv 0{\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_{2n}=I_{n}-\sum _{(p-1)|2n}{\frac {1}{p}},

where $I n$ 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.

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 die Reine und Angewandte Mathematik, 21: 372–374, ISSN 0075-4102, ERAM 021.0672cj

External links

Weisstein, Eric W. "von Staudt-Clausen Theorem". 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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Proof

See also

References

External links