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What is a good lower bound on $\pi(2n)-\pi(n)$? Bertrand's postulate gives $1$. It is expected to be as I understand of form $\frac{c\cdot n}{\log n}$ from Prime Number Theorem.

  1. Does the ratio always hold for all large enough $n$ with some $c$ always between $0$ and $c_0$ for some absolute constant $c_0$?

  2. How often does it fail as far as we know?

Turbo
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1 Answers1

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This is partly answered here:

Primes between $n$ and $2n$

It follows from the prime-number theorem that $$ \lim_{n \to \infty} \frac{\pi(2n) - \pi(n)}{n/\log n} = 2 - 1 = 1,$$ so the number of primes between $n$ and $2n$, which is $\pi(2n) - \pi(n)$, is actually asymptotic to $\frac{n}{\log n}$ which gets arbitrarily large.

Dietrich Burde
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  • However the best gap seems to be of form $n^{0.525}$ and with that I get only $O(n^{0.475})$. What do I miss? – Turbo Dec 30 '18 at 16:18
  • @1.. The prime number theorem guarantees that there aren't too many large gaps close together (and also, not too many small gaps). – Daniel Fischer Sep 20 '20 at 15:09