NOTE ON THE DISTRIBUTION OF THE INTERVALS BETWEEN PRIME NUMBERS
NOTE ON THE DISTRIBUTION OF THE INTERVALS BETWEEN PRIME NUMBERS
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关于素数之间的间隔分布的注释
DOI：
10.1093/qmath/os-17.1.46
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发表时间：
1946
影响因子：
0.7
通讯作者：
Lord Cherwell
中科院分区：
文献类型：
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作者：
Lord Cherwell
在本文中，试图列出一些关于素数对、三元组等的频率的结果，这些结果可以通过概率方法从素数分布可以被视为“随机”的假设中导出。这种方法显然排除了任何数学意义上的严格证明的尝试。然而，有趣的是，通过这种简单的方法获得的结果与我们从表格中（必然是有限的）枚举发现的事实相当接近，并且在两种情况下，至少出现了与通过更复杂的方法导出的公式相同的公式，正如我所指出的那样。*如果 P 是奇素数，则只有当 TO 是偶数时，P-\-n 才能是素数。在下文中，P 和 p 始终表示奇数素数 (p# P)，而 n 始终表示偶数。如果我们假设素数是随机分布的，则很容易证明，对于除以 n 的每个不同奇数素数 p，P-\-n 为素数的概率会增加一个因子 (p—l)/(p—2)。例如，让我们比较 P+ 2p 不能被 p 整除的概率与 P+ 2 不能被 p 整除的概率。如果我们对 P 一无所知，则概率显然相等，即每个 (p-1)/p。但是，如果我们知道 P 是素数，那么 P 和 P-\-2p 都不能被 pf 整除。另一方面，由于 p—l 中间奇数中的一个必须能被 p 整除，所以它不被 P+ 2 的机会是 (p—2)/(p—1)。由于 P-\-2p 的机会不是
IN this note an attempt is made to set out some of the results concerning the frequency of prime-pairs,-triplets, and so forth which can be derived by probability methods from the assumption that the distribution of the prime numbers may be treated as' random'. This approach clearly precludes any attempt at rigorous proof in the mathematical sense. Nevertheless, it seems interesting that results obtained by such simple means fit fairly closely the'facts found by our (necessarily limited) enumeration from tables, and that in two cases at least formulae emerge which are identical, as has been pointed out to me, with those derived by more elaborate methods.* If P is an odd prime number P-\-n can only be prime if TO is even. In what follows P and p always denote odd primes (p# P) and n always an even number.If we assume a random distribution of primes, it is easy to show that the probability of P-\-n being a prime is enhanced by a factor (p—l)/(p—2) for each different odd prirrle p which divides n. Let us compare for instance the probability of P+ 2p not being divisible by p with the probability of P+ 2 not being divisible by p. If we know nothing about P, the probabilities are clearly equal, namely, each (p—1)/p. But, if we know that P is a prime, then neither P nor P-\-2p can be divisible by pf On the other hand, since one of the p—l intermediate odd numbers must be divisible by p, the chances of its not being P+ 2 are (p—2)/(p—1). As the chance of P-\-2p not