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?
My try if $\alpha=\sqrt{17}$ then $\lfloor n\alpha \rfloor=4n$

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    $\begingroup$ That try will fail: $\lfloor 9\sqrt{17} \rfloor=37$ $\endgroup$
    – tomi
    Jun 4 '16 at 21:29
  • 1
    $\begingroup$ $\lfloor n\alpha \rfloor$ is definitely not expressible in the form $cn$ when $\alpha$ is irrational. $\endgroup$ Jun 4 '16 at 21:40
  • $\begingroup$ you can express finite sub-sequences of $x_n = \lfloor n \alpha \rfloor$ as $y_n = c n+d$, I think all the point is to prove that some of those have $gcd(c,d ) = 1$, and that they are large enough for applying the Dirichlet theorem in arithmetic progressions. $\endgroup$
    – reuns
    Jun 4 '16 at 22:11

This is called a Beatty sequence. Here is an arxiv paper on the least prime in a Beatty sequence, by Steuding and Technau. There will indeed always be a prime in the sequence (which answers the original question), and the cited paper gives an upper bound for the least such prime for $\alpha>1$. The bound for the OP's sequence (provided $\alpha>1$) is $$p\le L^{35-16\epsilon}\alpha^{2(1-\epsilon)}p^{1+\epsilon}_{m+l}$$

where $L=\log(2\alpha)$, $p_n$ 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 $p_m\le L^{16}\alpha^2<p_{m+1}$. $\epsilon$ can be chosen arbitrarily small, but $l$ depends on $\epsilon$.

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    $\begingroup$ one of the main trick is at the beginning of page 3 : "This means that the proportion of fractional parts $\{\alpha p\} = \alpha p − \lfloor \alpha p\rfloor$ which fall in an interval $[a, b) \subset [0, 1)$ is equal to $b − a$ (the length of the interval) " and from there can we apply the Dirichlet theorem on arithmetic progressions ? there is also, just before " the sequence of numbers $\alpha p$, where $p$ runs through the prime numbers in ascending order, is uniformly distributed modulo one" $\endgroup$
    – reuns
    Jun 4 '16 at 22:24
  • $\begingroup$ I found this for a more succinct explanation xvienemsalamanca.anemat.com/wp-content/uploads/2015/10/… $\endgroup$
    – reuns
    Jun 4 '16 at 22:33

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