与 必被 实筛.现观察闭区间B: 内的奇数.
(1) . 故知:闭区间B内的奇数均小于 .
(2)∵ ∴
而 .
故知闭区间B内必含 的奇倍数 .
又因 被 实筛,故知, 必被 实筛而被 虚筛.
(3)与 相邻的 的奇倍数为: .且∵ ∴ .
故知:
.
.
故知,闭区间B内有且仅有一个 的奇倍数 ,且为 虚筛.
且因 ,
故 .
故知,在闭区间 内至少有 个同时又属于闭区间B内的受 虚筛的连续奇数.故由推论四知,在这 个大于 ,小于 的连续奇数中,至少有二个 ,至少有一个 即 .
故知: 当 时, .
由1、2、知: .
由一、二、知:当 时, .
故由Ⅰ.Ⅱ.知:当 时, .(证毕)
参考文献:
①张 忠 在 与 间至少有二个素数.连云港师范高等专科学校学报.2001.4.
②闵嗣鹤、严士健 初等数论[M]北京.人民教育出版社出版 1982.
③张 忠 论同余式方程组正整数解的构造式[ J ],南通职业大学学报,2000.14(4).
④张 忠 谈谈素数分布的两个规律[ J ] 《中华学术论坛》2003. 9总第六期
Elementary discussion on the two laws of the distribution of prime numbers
ZHANG Zhong
(Nantong South-City Zhong-Yian Bookshop 226001 NT Jiangsu,China)
Abstract: In this paper, we use the theory of congruence, group the integer segment ( ) which consist of ordered continuous integers into residue segment classes of module , and also use the n-dimensional sieve method, analyse the segment  which in the minimal nonnegative complete system of residue segment classes of module real-sieved or false-sieved by . And then,we obtained the conclusions: there are at least two numbers prime to among continuous integers, and from this,obtain the two laws of the distribution of prime numbers: 1. there are at least two prime numbers between and ; 2.the difference of a prime number and its successor .
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