Primes in $\lfloor a^{n} \rfloor$

Motivated by the question Is there any result, that says that $\lfloor e^{n} \rfloor$ is never a prime for $n>2$?, take a real number $a>1$ and consider the sequence $\lfloor a^{n} \rfloor$.

• Does it always contains a prime number?
• Does it always contain an infinite number of primes?

As mentioned in the comments, to avoid trivial cases, $a$ cannot be an integer.

As this is probably too hard to have been answered in general, I'd be interested in partial results for interesting numbers such as $e$, $\pi$, $\sqrt 2$, $3/2$.

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If $a$ is an integer, the sequence contains at most one prime. </belabor-the-obvious> –  Daniel Fischer Sep 19 '13 at 12:45
Without restrictions for $a$ it is wrong. Take $a=4$ or $a=p^2$. –  gammatester Sep 19 '13 at 12:46
Related to en.wikipedia.org/wiki/Mills%27_constant but different. –  lhf Sep 19 '13 at 12:47
There are some very hard questions here. It is known that $[(4/3)^n]$ is composite infinitely often, likewise $[(3/2)^n]$, but these are nontrivial results of Forman and Shapiro, and I think $[(5/2)]^n$ hasn't been settled (although Dubickas has done work on this kind of problem, and I don't recall exactly what progress he made). Proving there are primes is harder than proving there are composites. This looks like good reading: vddb.laba.lt/fedora/get/… –  Gerry Myerson Sep 19 '13 at 13:04
I don't feel my comment really answers the question, and I don't have time to track down references right now, but perhaps I will think about it some more and post an answer in a day or two. Or you could track the papers down, and post an answer yourself. –  Gerry Myerson Sep 19 '13 at 13:10