most recent 30 from http://mathoverflow.net 2013-05-19T04:08:09Z http://mathoverflow.net/feeds/question/25499 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/25499/primes-in-quasi-arithmetic-progressions David Hansen 2010-05-21T15:59:53Z 2011-01-21T14:22:42Z

<p>Suppose $\alpha > 1$ is irrational. Are there infinitely many primes of the form $\left\lfloor \alpha n \right\rfloor$? Is the number of $p \leq X$ of this form $\sim \alpha^{-1} X (\log{X})^{-1}$? I know this is the kind of thing the circle method was born to do, but I cannot for the life of me find a reference for this!</p>

http://mathoverflow.net/questions/25499/primes-in-quasi-arithmetic-progressions/25500#25500 engelbrekt 2010-05-21T16:21:01Z 2010-05-21T16:37:45Z

<p>Imre Ruzsa observed that since $p/\alpha$ is equidistributed modulo $1$, we have infinitely many primes $p$ for which the fractional part of $p/\alpha$ is less than $1/\alpha$. Writing $p/\alpha = n_p - {\epsilon}_p$ with $n_p$ an integer and $0 &lt; {\epsilon}_p &lt; 1/\alpha$, we get $p = {\alpha}n_p - {\alpha}{\epsilon}_p$ and thus ${\lfloor}{\alpha}n_p{\rfloor}$ prime for infinitely many distinct positive integers $n_p$.</p>

http://mathoverflow.net/questions/25499/primes-in-quasi-arithmetic-progressions/25510#25510 Gjergji Zaimi 2010-05-21T17:42:31Z 2010-05-21T18:07:49Z

<p>I think the uniform distribution mod1 of $\{p/\alpha\}$ is due to Vinogradov, and the asmptotic for primes in a Beatty sequence $\sim \frac{\pi(x)}{\alpha}$ is an immediate consequence. Indeed for $p$ to be equal to some $\lfloor k\alpha\rfloor$ it is equivalent to $1-\frac{1}{\alpha}&lt;\frac{p}{\alpha}-\lfloor \frac{p}{\alpha}\rfloor&lt;1$. So you just need the fractional part of $p/\alpha$ to be on a fixed interval of length $\alpha$ mod1.</p> <p>On a related note <a href="http://arxiv.org/abs/0708.1015" rel="nofollow">this</a> paper discusses the general sequence $q\lfloor \alpha n+\beta\rfloor +a$.</p>