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 Specifically, if "n" is a positive integer and we add 1/"p" to the Bernoulli number "B" for every prime "p" such that "p" − 1 divides 2"n", we obtain an integer, i.e., formula_1 This fact immediately allows us to characterize the denominators of the non-zero Bernoulli numbers "B" 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 The sequence of integers formula_2 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 formula_5 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 formula_9. If p-1 divides 2n then one has, for formula_9. 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 formula_14 (Greatest integer function) then after iteration one has, for formula_9 and formula_17 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 formula_20 is an integer, which is the Von-Staudt Clausen theorem.