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!

5$\begingroup$ Please reformulate the question in such a way that you don't ask for infinitely many primes p ≤ X. $\endgroup$ – François G. Dorais♦ May 21 '10 at 18:36

$\begingroup$ @FGD: Whoops! Done. $\endgroup$ – David Hansen May 22 '10 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 $q\lfloor \alpha n+\beta\rfloor +a$.

$\begingroup$ "...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. $\endgroup$ – David Hansen May 21 '10 at 17:45

$\begingroup$ Sorry for stating it like that, I added a clarification. $\endgroup$ – Gjergji Zaimi May 21 '10 at 18:11
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 $n_p$.

$\begingroup$ Yeah, I wasn't sure what the "local obstructions" for $\left\lfloor \alpha n \right \rfloor$ being prime would be exactly! $\endgroup$ – David Hansen May 21 '10 at 16:24

$\begingroup$ 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 I would have to think more about the asymptotic estimate, so I have withdrawn my remarks on that. $\endgroup$ – engelbrekt May 21 '10 at 16:34