MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

## Primes in quasi-arithmetic progressions?

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!

-
Please reformulate the question in such a way that you don't ask for infinitely many primes p ≤ X. – François G. Dorais May 21 2010 at 18:36
@FGD: Whoops! Done. – David Hansen May 22 2010 at 18:08

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 $1-\frac{1}{\alpha}<\frac{p}{\alpha}-\lfloor \frac{p}{\alpha}\rfloor<1$. So you just need the fractional part of $p/\alpha$ to be on a fixed interval of length $\alpha$ mod1.

On a related note this paper discusses the general sequence $q\lfloor \alpha n+\beta\rfloor +a$.

-
 "...is an immediate consequence." No, it's not, because n doesn't have to be prime for $\left \lfloor \alpha n \right \rfloor$ to be prime. – David Hansen May 21 2010 at 17:45 Sorry for stating it like that, I added a clarification. – Gjergji Zaimi May 21 2010 at 18:11 Oh, now I see! Thanks! – David Hansen May 21 2010 at 19:11

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 $0 < {\epsilon}_p < 1/\alpha$, we get $p = {\alpha}n_p - {\alpha}{\epsilon}_p$ and thus ${\lfloor}{\alpha}n_p{\rfloor}$ prime for infinitely many distinct positive integers $n_p$.

-
 Yeah, I wasn't sure what the "local obstructions" for $\left\lfloor \alpha n \right \rfloor$ being prime would be exactly! – David Hansen May 21 2010 at 16:24 I didn't quite remember the details correctly, but I have fixed them now. I would have to think more about the asymptotic estimate, so I have withdrawn my remarks on that. – engelbrekt May 21 2010 at 16:34
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}