Roland Waters

2022-06-20

Does there exist a positive irrational number $\alpha $, such that for any positive integer $n$ the number $\lfloor n\alpha \rfloor $ is not a prime?

Abigail Palmer

Beginner2022-06-21Added 30 answers

This is called a Beatty sequence. There will indeed always be a prime in the sequence. The bound for the OP's sequence (provided $\alpha >1$) is

$p\le {L}^{35-16\u03f5}{\alpha}^{2(1-\u03f5)}{p}_{m+l}^{1+\u03f5}$

where $L=\mathrm{log}(2\alpha )$, ${p}_{n}$ denotes the numerator of the ${n}^{\text{th}}$ convergent to the regular continued fraction expansion of $\alpha $, and $m$ is the unique integer such that ${p}_{m}\le {L}^{16}{\alpha}^{2}<{p}_{m+1}$. $\u03f5$ can be chosen arbitrarily small, but l depends on $\u03f5$.

$p\le {L}^{35-16\u03f5}{\alpha}^{2(1-\u03f5)}{p}_{m+l}^{1+\u03f5}$

where $L=\mathrm{log}(2\alpha )$, ${p}_{n}$ denotes the numerator of the ${n}^{\text{th}}$ convergent to the regular continued fraction expansion of $\alpha $, and $m$ is the unique integer such that ${p}_{m}\le {L}^{16}{\alpha}^{2}<{p}_{m+1}$. $\u03f5$ can be chosen arbitrarily small, but l depends on $\u03f5$.