Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $e$, $\pi$, $\sqrt 2$, $3/2$.

share|improve this question
    
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
3  
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
2  
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

1 Answer 1

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 $n$.

share|improve this answer
    
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 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 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 at 21:24

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.