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Let $2 = p_1 < p_2 < \dots$ be the increasing sequence of all prime numbers. For large $n$, which one is larger $2^{p_n}$ or $p_{2^n}$?

I need some help on this please.

I have tried using the Prime number theorem since $$\pi(x) \approx \frac{x}{\log x}$$

I used this expression for $x=2^{p_n}$ and $x=p_{2^n}$ but I couldn't make any progress.

$p_{2^n}$ is the $2^n$-th prime number.

Bill Dubuque
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Josh97
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    Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Shaun Feb 05 '21 at 22:45
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    In addition to what @Shaun said, please use informative titles. And what is $p_{2^n}$ supposed to be? Please check that you wrote out everything correctly. – The Pointer Feb 05 '21 at 23:02
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    The $2^n$th prime, surely. Y'all harshing on this thread so much haha To the OP, this will probably help you: https://math.stackexchange.com/questions/2773138/asymptotic-expression-for-the-nth-prime-number – John Samples Feb 05 '21 at 23:03
  • For $n = 5, p_n = 11, 2^{(11)} = 2048,$ and $p_{2^5} = p_{(32)}$, which is nowhere near $2048.$ @JohnSamples I suspect, but can not easily prove, that you have it backwards. – user2661923 Feb 05 '21 at 23:11
  • @user2661923 Oh hahaha I didn't even realize the "dual meaning" of that part of my comment. It was directed to "The Pointer" when he asks "what is $p_{2^n}$. Very misleading, oops. I still think this is the cleanest proof. – John Samples Feb 06 '21 at 01:25
  • After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?. – Shaun Feb 06 '21 at 01:31

1 Answers1

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From the prime number theorem, for sufficiently large $n$, the $n$th prime is less than $n^2$.

Furthermore, all primes except $2$ or $3$ are $1 \text{ or } 5 \pmod 6$ (which means intuitively, that at most $1/3$ of numbers are prime). So $p_n > 3n -c$ for some constant $c$.

$$p_{2^n} < (2^n)^2 = 2^{2n}, \text{ while } 2^{p_n} > 2^{3n-c} = 2^{3n} / 2^c.$$

As $n \rightarrow \infty$, $2^{3n} / 2^c$ will be bigger than $2^{2n}$. Therefore, for sufficiently large $n$, $2^{p_n}$ is bigger

David Lui
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