Jun 262012
Sylvester’s theorem the binomial coefficient
\begin{equation}{n\choose k}=\frac{n(n-1)(n-2)\cdots(n-k+1)}{k!}\end{equation}
has a prime divisor \(p\) greater than \(k\). In other words, if \(n\geq 2k\), then the product of \(k\) consecutive integers greater than \(k\) is divisible by a prime greater than \(k\). Note that for \(n=2k\) we obtain precisely Bertrand’s postulate.
Theorem
- if \(n>6\) is an integer, then there always exists at least two prime numbers \(p\), one is the form \(4k+1\) and the other \(4k+3\), such that \(n<p<2n\).
- if \(k\) is a positive integer, then there always exists a positive integer \(N\) such that for every integer \(n>N\), there exist at least \(k\) prime numbers \(p\) such that \(n<p<2n\).
- 对任意正整数 \(n>6\) , 至少存在一个 \(4k+1\) 型和一个 \(4k+3\) 型素数 \(p\) 使得 \(n<p<2n\).
- 对任意正整数 \(k\), 存在正整数 \(N\), 使得只要正整数 \(n>N\), 就存在 \(k\) 个素数 \(p\) 使得 \(n<p<2n\).
注意, 定理 \(2\) 已经隐含在 Ramanujan 给出的 Bertrand’s postulate 的证明中 Ramanujan’s proof of Bertrand’s postulate.