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, we obtain an integer, i.e., 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 6, 30, 42, 30, 66, 2730, 6, 510, 798, 330, 138, 2730, 6, 870, 14322, 510, 6, 1919190, 6, 13530, ... . The sequence of integers 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: and as a corollary: where 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 . If p-1 divides 2n then one has, for . 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 (Greatest integer function) then after iteration one has, for and . 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 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, 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 is an integer, which is the Von-Staudt Clausen theorem. See also Kummer's congruence References External links Theorems in number theory