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arXiv 2403.08306math.NT
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关于$n^2$与${(n+1)}^2$之间素数个数的研究

On the number of prime numbers between $n^2$ and ${(n+1)}^2$

Jimin Li, Haonan Li

AI总结:

本文通过比较互质数密度并建立不同区间内互质数比例的关系,证明了在任意$n^2$与${(n+1)}^2$之间至少存在2个素数。

摘要:

设$p_{r+1}-1>n \geq p_r-1$,基于序列$\{1,2,3\cdots\\ M_r(M_r=p_1p_2\cdots p_r)\}$,我们比较了互质数的密度,并建立了从1到连续平方数范围内互质数比例之间的相关性。然后,我们推导了$n^2 \sim {(n+1)}^2$区间内互质数个数与$1 \sim n^2$区间内互质数比例之间的关系,证明了在任意$n^2$与${(n+1)}^2$之间至少存在一个素数。通过将研究扩展到$1 \sim M_r^2$的范围,我们建立了从${M_r}^2$到连续平方数范围内向后互质数比例之间的关系;此外,我们还建立了小区间内互质数比例与整个区间内互质数比例之间的关系。最后,结论表明,$n^2$与${(n+1)}^2$之间的互质数个数大于$n\prod_{i=1}^{r}{(1-\frac{1}{p_i}})$,从而证明了在$n^2$与${(n+1)}^2$之间至少存在2个素数。

英文摘要:

Let $p_{r+1}-1>n \geq p_r-1$, based on a sequence $\{1,2,3\cdots\ M_r(M_r=p_1p_2\cdots p_r)\}$, we compare the density of coprime numbers and establish a correlation between the proportions of coprime numbers in the ranges from 1 to consecutive square numbers. Then, we derive the relationship between the number of coprimes in the interval of $n^2 \sim {(n+1)}^2$ and the proportion of coprimes in the interval of $1 \sim n^2$, proving that there is at least one prime number between any $n^2$ and ${(n+1)}^2$. By extending our research to the range of $1 \sim M_r^2$, we establish the relationship between the proportions of backwards coprime numbers in the ranges from ${M_r}^2$ to consecutive square numbers; furthermore, we establish a relationship between the proportions of coprimes in small interval and the whole interval. Then, in conclusion, the number of coprimes between $n^2$ and ${(n+1)}^2$ is greater than $n\prod_{i=1}^{r}{(1-\frac{1}{p_i}})$, thus proving that there are at least 2 prime numbers between $n^2$ and ${(n+1)}^2$.

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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