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I observed this phenomenon and checked this up to some(very small) $n$, but it seems so astoundingly trivial and consistent. I'm up for any proof/counterexample and opinion, thanks. Sorry, everyone that I made several mistakes in quoting the question, it was my bad! Now I made the final revision and corrected it. Now it's all fine. No more change I promise!

user18724
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  • I edited the question @Regret – user18724 Mar 05 '15 at 09:04
  • Side note: "astoundingly trivial" is an oxymoron. Either it's astounding or it is trivial (or neither, but it can't be both). – barak manos Mar 05 '15 at 09:11
  • Sorry everyone that I made several mistakes in quoting the question, it was my bad! Now I made the final revision and corrected it. Now it's all fine. No more change I promise! @Regret – user18724 Mar 05 '15 at 09:21
  • Sorry everyone that I made several mistakes in quoting the question, it was my bad! Now I made the final revision and corrected it. Now it's all fine. No more change I promise! @Winther – user18724 Mar 05 '15 at 09:21
  • Sorry everyone that I made several mistakes in quoting the question, it was my bad! Now I made the final revision and corrected it. Now it's all fine. No more change I promise! @barakmanos – user18724 Mar 05 '15 at 09:22
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    What is ${2_n}?$ – gammatester Mar 05 '15 at 09:22
  • It's $2n$ not $2_n$ sorry for the typo! @gammatester – user18724 Mar 05 '15 at 09:26
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    First convince yourself that $\pi(p_{m}) = m$ then it follows that the the number you are after is $\pi(p_{2n}) - \pi(p_{n}) - 1 = 2n-n-1 = n-1$ – Winther Mar 05 '15 at 09:26
  • Ok, I got the picture now, it's very simple and useful, thanks. @Winther – user18724 Mar 05 '15 at 09:34
  • What do you denote $p_n$? – Bernard Mar 05 '15 at 09:54
  • $p_n$ is a standard notation for $n$th prime, where $p$ is a prime. Is there, however, anything else that $p_n$ could mean, in mathematics? I don't think so. @Bernard – user18724 Mar 05 '15 at 11:21
  • Then the question seems to be a tautology: between the $n$th prime and the $2n$-th prime, there are exactly $n-1$ primes, and this has nothing to see with primes (you can replace primes with candies). Or do I miss something? – Bernard Mar 05 '15 at 11:41
  • Thanks for the comment. If I would have realized the solution/fact that you and other responded with, I wouldn't even ask the question in the first place. @Bernard – user18724 Mar 05 '15 at 13:25

2 Answers2

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With your revisions, and assuming that by $p_n$ you mean the $n$th prime number, you're asking if it's always true that between the $n$th and $2n$th items in a sequence, there are $n-1$ items.

There's nothing specific to prime numbers here; this is simply counting. To get from the $n$th item to the $2n$th item, we would have to count off $n$ items, so if we want the count of items strictly between the two, we must subtract one; hence, $n-1$ items.

AakashM
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The claim is far from true, because the index $2p_n$ is much bigger than $n-1$ in general. We can easily give some examples, i.e., $n=5$ and $x=78$. We have $p_n=p_5=11$ and $p_{2p_n}=p_{2\cdot 11}=p_{22}=79$. So $p_n<x<p_{2p_n}$ is satisfied, but $\pi(x)=\pi(78)=21\neq n-1=4$.

Edit: The answer and the comments refer to the original version of the question that $\pi(x)=n-1$ for all $x$ with $p_n<x<p_{2p_n}$. But also, between $p_5$ and $p_{2p_5}$ are more than $n-1=4$ primes.

Dietrich Burde
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  • Sorry, everyone that I made several mistakes in quoting the question, it was my bad! Now I made the final revision and corrected it. Now it's all fine. No more change I promise! @DietrichBurde – user18724 Mar 05 '15 at 09:23