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 B2n for every prime p such that p − 1 divides 2n, then we obtain an integer; that is,
This fact immediately allows us to characterize the denominators of the non-zero Bernoulli numbers B2n as the product of all primes p such that p − 1 divides 2n; consequently, the denominators are square-free and divisible by 6.
These denominators are
The sequence of integers is
A proof of the Von Staudt–Clausen theorem follows from an explicit formula for Bernoulli numbers which is:
and as a corollary:
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
2. If p – 1 does not divide 2n, then
Proof of (1) and (2): One has from Fermat's little theorem,
for m = 1, 2, ..., p – 1.
If p – 1 divides 2n, then one has
for m = 1, 2, ..., p – 1. Thereafter, one has
from which (1) follows immediately.
If p – 1 does not divide 2n, then after Fermat's theorem one has
If one lets ℘ = ⌊ 2n / (p – 1) ⌋, then after iteration one has
for m = 1, 2, ..., p – 1 and 0 < 2n – ℘(p – 1) < p – 1.
Thereafter, one has
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,
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
where In is an integer, as desired. [1] [2]
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 B2n for every prime p such that p − 1 divides 2n, then we obtain an integer; that is,
This fact immediately allows us to characterize the denominators of the non-zero Bernoulli numbers B2n as the product of all primes p such that p − 1 divides 2n; consequently, the denominators are square-free and divisible by 6.
These denominators are
The sequence of integers is
A proof of the Von Staudt–Clausen theorem follows from an explicit formula for Bernoulli numbers which is:
and as a corollary:
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
2. If p – 1 does not divide 2n, then
Proof of (1) and (2): One has from Fermat's little theorem,
for m = 1, 2, ..., p – 1.
If p – 1 divides 2n, then one has
for m = 1, 2, ..., p – 1. Thereafter, one has
from which (1) follows immediately.
If p – 1 does not divide 2n, then after Fermat's theorem one has
If one lets ℘ = ⌊ 2n / (p – 1) ⌋, then after iteration one has
for m = 1, 2, ..., p – 1 and 0 < 2n – ℘(p – 1) < p – 1.
Thereafter, one has
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,
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
where In is an integer, as desired. [1] [2]