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

This sequence not necessarily contains a prime number.

We will prove that there is a number $a\in [4,5]$ such that $\lfloor a^n \rfloor$ is even for all $n$.

The idea is that for $a\in[4,5]$, the set of possible values of $a^2$ is the interval $[4^2,5^2]$, which contains a even number. So, in the interval $[4,5]$ there is a subrange of values for $a$ such that $\lfloor a^2 \rfloor$ is even. Now, in that subrange, there will exist another subrange such that $\lfloor a^3 \rfloor$ is even for $a$ in this subrange, and so on. If we intersect all these subranges and use the nested intervals theorem, we will find a real $a$ such that all $\lfloor a\rfloor, \lfloor a^2\rfloor,\lfloor a^3\rfloor...$ are even.

Of course this is just the idea, see below the formal proof (I hope it is not too confusing):

Starting with the interval $I_1=[4.1,4.9]$ (we choose this initial interval to make sure $a$ is not an integer), we will construct inductively a infinite sequence of closed intervals $I_1,I_2,...$ satisfying the following conditions:

1) $I_{n} \subset I_{n-1}$

2) If $I_n=[a_n,b_n]$, then $b_n^{n+1}-a_n^{n+1}\ge3$

3) The interval $[a_n^n,b_n^n]$ is contained in a interval of the form $[2k,2k+0.9]$ for some integer $k$.

The basis of induction is easily verified. Now, given $I_1,I_2,...I_n$, let's construct $I_{n+1}$:

Since $b_n^{n+1}-a_n^{n+1}\ge3$, the interval $[a_n^{n+1},b_n^{n+1}]$ must contain a interval of the form $[2k,2k+1]$ for some integer $k$. Then define $I_{n+1}$ to be $[a_{n+1},b_{n+1}]$ where $a_{n+1}=\sqrt[n+1]{2k}$ and $b_{n+1}=\sqrt[n+1]{2k+0.9}$.

3) Clearly holds for this interval

1) Holds because $a_{n+1}=\sqrt[n+1]{2k}\ge \sqrt[n+1]{a_n^{n+1}}=a_n$, and

$b_{n+1}=\sqrt[n+1]{2k+0.9}<\sqrt[n+1]{2k+1}\le \sqrt[n+1]{b_n^{n+1}}=b_n$

Finally, 2) holds because

$b_{n+1}^{n+2}-a_{n+1}^{n+2}= (2k+1)b_{n+1}-2ka_{n+1}=2k(b_{n+1}-a_{n+1})+b_{n+1}>b_{n+1}>4$ (because

$I_{n+1}\subset[4,5]$)

So we are done with our induction. Now, by the nested intervals theorem, there must exist a real number $a$ such that $a\in I_n$ for all $n$. But if $a\in I_n$, $then a^n\in [a_n^n,b_n^n]$. Since the last is contained in a interval of the form $[2k,2k+1]$, it follows that for this $a$, $\lfloor a^n \rfloor$ is even for all $n$.

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Thanks for your answer. This process seems to converge to $a=4$, which is known not to work (see the comments to the question). –  lhf Oct 22 '14 at 10:29
I hadn't consider this possibility, but it can be solved. You just have to change the initial interval to $[4.1,4.9]$ instead of $[4,4.9]$. I will edit the answer. –  Rodrigo Oct 22 '14 at 16:21
By running this algorithm in my computer I found for instance $a=4.282492249616384717906420795860777879382613851721238001477026538069515061851‌​9034186279663751127347493$, for which $\lfloor a^n \rfloor$ is even for $1\le n \le 151$. With more decimal digits we get a bigger range for which it is even, and with infinitely many digits we get even values for all $n$. –  Rodrigo Oct 22 '14 at 21:24