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 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 These denominators are The sequence of integers is

Proof

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.