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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Von Staudt–Clausen theorem ： ウィキペディア英語版 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 and . 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'', we obtain an integer, i.e., $B\_\; +\; \backslash sum\_\; \backslash frac1p\; \backslash in\; \backslash Z\; .$ 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, ... . == Proof == A proof of the Von Staudt–Clausen theorem follows from an explicit formula for Bernoulli numbers which is: :$B\_=\backslash sum\_^\}\backslash sum\_^m^\}\; \backslash !$ and as a corollary: :$B\_=\backslash sum\_^\}(-1)^jS(2n,j)\; \backslash !$ where $S(n,j)\; \backslash !$ 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, :$\backslash sum\_^\}\backslash equiv\backslash pmod\; p\; \backslash !$ 2. If p-1 does not divide 2n then, :$\backslash sum\_^\}\backslash equiv0\backslash pmod\; p\; \backslash !$ Proof of (1) and (2): One has from Fermat's little theorem, :$m^\backslash equiv\; 1\backslash pmod\; p\; \backslash !$ for $m=1,2,...,p-1\; \backslash !$. If p-1 divides 2n then one has, :$m^\backslash equiv\; 1\backslash pmod\; p\; \backslash !$ for $m=1,2,...,p-1\; \backslash !$. Thereafter one has, :$\backslash sum\_^\}\backslash equiv\; \backslash sum\_^\backslash equiv\; m^\backslash pmod\; p\; \backslash !$ If one lets $\backslash wp=()\; \backslash !$ (Greatest integer function) then after iteration one has, :$m^\backslash equiv\; m^\backslash pmod\; p\; \backslash !$ for $m=1,2,...,p-1\; \backslash !$ and Thereafter one has, :$\backslash sum\_^\}\backslash equiv\backslash sum\_^\}\backslash pmod\; p\backslash !$ 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, :$\backslash sum\_^\}=3\; \backslash cdot\; 2^-3^-3\backslash equiv0\; \backslash pmod\; 4\; \backslash !$ 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: :$B\_=I\_n-\backslash sum\_\}\; \backslash !$ where $I\_n\; \backslash !$ is an integer, which is the Von-Staudt Clausen theorem.〔H. Rademacher, Analytic Number Theory, Springer-Verlag, New York, 1973.〕〔T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 197.〕 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Von Staudt–Clausen theorem」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.