arXiv 1406.48012014-06-19math.NT

与一个与素数分布有关的猜想等价的两个陈述

Two statements that are equivalent to a conjecture related to the distribution of prime numbers

Germán Paz

备注：: 16 pages, 3 figures (version 2 includes them also as ancillary files, no changes to version 1 have been made), Mathematica code; keywords: Andrica's conjecture, Brocard's conjecture, Legendre's conjecture, Oppermann's conjecture, prime numbers, triangular numbers. arXiv admin note: text overlap with arXiv:1310.1323

摘要：

让$n\in\mathbb{Z}^+$。在[8]中，我们提出了一个问题，即任何大于$n^2$且小于$(n+1)^2$的$n$连续整数序列是否至少包含一个素数，并且我们证明了对于每个$n\leq 1,193,806,023$实际上都是这种情况。此外，我们证明了一个肯定的答案，前一个问题的所有$n$将意味着勒让德的，Brocard的，Andrica的，和Oppermann的布局，以及假设，对于每一个$n$总是有一个素数的间隔$[n,n+2\lfloor\sqrt{n}\rfloor-1]$。令$π[n+g(n),n+f(n)+g(n)]$表示区间$[n+g(n),n+f(n)+g(n)]$中素数的数量。本文证明了[8]中的猜想等价于$$π[n+g(n),n+f(n)+g(n)]\ge 1\text{, }\forall n\in\mathbb{Z}^+\text{,}$$的陈述，其中$$f(n)=\left(\frac{n-\lfloor\sqrt{n}\rfloor^2-\lfloor\sqrt{n}\rfloor-β}{|n-\lfloor\sqrt{n}\rfloor^2-\lfloor\sqrt{n}\rfloor-β|}\right)(1-\lfloor\sqrt{n}\rfloor)\text{, }g(n)=\left\lfloor1-\sqrt{n}+\lfloor\sqrt{n}\rfloor\right\rfloor\text{,}$$和$β$是任意实数，使得$1<β<2$ .我们还证明，猜想中的问题是等价的声明，$$π[S_n,S_n+\lfloor\sqrt{S_n}\rfloor-1]\ge 1\text{, }\forall n\in\mathbb{Z}^+\text{,}$$其中$$S_n=n+\frac{1}{2}\left\lfloor\frac{\sqrt{8n+1}-1}{2}\right\rfloor^2-\frac{1}{2}\left\lfloor\frac{\sqrt{8n+1}-1}{2}\right\rfloor+1\text{.}$$我们使用这最后的结果，以创建图的$h(n)=π[S_n,S_n+\lfloor\sqrt{S_n}\rfloor-1]$的许多值的$n$。

英文摘要：

Let $n\in\mathbb{Z}^+$. In [8] we ask the question whether any sequence of $n$ consecutive integers greater than $n^2$ and smaller than $(n+1)^2$ contains at least one prime number, and we show that this is actually the case for every $n\leq 1,193,806,023$. In addition, we prove that a positive answer to the previous question for all $n$ would imply Legendre's, Brocard's, Andrica's, and Oppermann's conjectures, as well as the assumption that for every $n$ there is always a prime number in the interval $[n,n+2\lfloor\sqrt{n}\rfloor-1]$. Let $π[n+g(n),n+f(n)+g(n)]$ denote the amount of prime numbers in the interval $[n+g(n),n+f(n)+g(n)]$. Here we show that the conjecture described in [8] is equivalent to the statement that $$π[n+g(n),n+f(n)+g(n)]\ge 1\text{, }\forall n\in\mathbb{Z}^+\text{,}$$ where $$f(n)=\left(\frac{n-\lfloor\sqrt{n}\rfloor^2-\lfloor\sqrt{n}\rfloor-β}{|n-\lfloor\sqrt{n}\rfloor^2-\lfloor\sqrt{n}\rfloor-β|}\right)(1-\lfloor\sqrt{n}\rfloor)\text{, }g(n)=\left\lfloor1-\sqrt{n}+\lfloor\sqrt{n}\rfloor\right\rfloor\text{,}$$ and $β$ is any real number such that $1<β<2$. We also prove that the conjecture in question is equivalent to the statement that $$π[S_n,S_n+\lfloor\sqrt{S_n}\rfloor-1]\ge 1\text{, }\forall n\in\mathbb{Z}^+\text{,}$$ where $$S_n=n+\frac{1}{2}\left\lfloor\frac{\sqrt{8n+1}-1}{2}\right\rfloor^2-\frac{1}{2}\left\lfloor\frac{\sqrt{8n+1}-1}{2}\right\rfloor+1\text{.}$$ We use this last result in order to create plots of $h(n)=π[S_n,S_n+\lfloor\sqrt{S_n}\rfloor-1]$ for many values of $n$.

机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。

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