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Two questions came to mind when I was reading the proof for Bertrand's Postulate (there's always a prime between $n$ and $2n$):

(1) Can we change the proof somehow to show that: $\forall x > x_{0}$, there exists a prime $p$ $\in [x, ax]$, for some $a \in (1, 2)$?

(2) Suppose the (1) is true, what is the smallest value of $x_{0}$?

I'm not sure how to prove either of them, any input would be greatly appreciated! And correct me if any of the above statement is wrong. Thank you!

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  • $\begingroup$ I gave an answer here somewhere that talked about some of the improvements on Bertrand's postulate. Let's see if I can find it... $\endgroup$
    – davidlowryduda ♦
    Sep 21, 2012 at 19:15
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    $\begingroup$ I must have remembered something longer. Here was that other answer. $\endgroup$
    – davidlowryduda ♦
    Sep 21, 2012 at 19:17
  • $\begingroup$ I'm not entirely certain how to parse statement (1). Are you asking whether $$(\exists x_0\in\mathbb{N})(\forall x\in\mathbb{N}, x>x_0)(\exists\mbox{prime }p)(\exists a\in(1,2))(p\in[x,ax])$$ (a rather trivial result) or whether $$(\forall a\in(1,2))(\exists x_0\in\mathbb{N})(\forall x\in\mathbb{N}, x>x_0)(\exists\mbox{prime }p)(p\in[x,ax])$$ or perhaps something else (perhaps changing the $\forall a$ in the last statement to $\exists a$)? $\endgroup$
    – jwodder
    Sep 21, 2012 at 19:33
  • $\begingroup$ @jwodder: thanks for the correction. I believe I meant "for some a between 1 and 2". $\endgroup$
    – Lindsay Duran
    Sep 21, 2012 at 19:40
  • $\begingroup$ @LindsayDuran: You still need to specify where in the sequence of quantifiers that $\exists a\in(1,2)$ appears. $\endgroup$
    – jwodder
    Sep 21, 2012 at 19:41

2 Answers 2

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There are better results than Bertrand's Postulate. Pierre Dusart has proven better results, some of which can be found on wikipedia.

Less specifically, for any $k > 1$, $k \in \mathbb{R}$, one can show that $\lim_{n \to \infty} \dfrac{\pi(kn) - \pi(n)}{n/\ln n} = k - 1$ using the prime number theorem, which means that for any $k>1$, there is some $x_0$ such that for $x > x_0$, there is a prime between $x$ and $kx$.

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  • $\begingroup$ Are you assuming that $k$ is an integer? $\endgroup$
    – M Turgeon
    Sep 21, 2012 at 19:26
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    $\begingroup$ @MTurgeon: No, $k$ can be any real number greater than $1$. $\endgroup$
    – davidlowryduda ♦
    Sep 21, 2012 at 19:29
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I think you would enjoy the page PRIME GAPS.

My own version of the conjecture of Shanks, actually both a little stronger and a little weaker, is $$ p_{n+1} < p_n + 3 \,\left( \log p_n \right)^2, $$ for all primes $p_n \geq 2.$ This is true as high as has been checked.

Shanks conjectured that $$ \limsup \frac{p_{n+1} - p_n}{\left( \log p_n \right)^2} = 1, $$ while Granville later corrected the number $1$ on the right hand side to $2 e^{- \gamma} \approx 1.1229,$ see CRAMER GRANVILLE. There is no hope of proving this, but I enjoy knowing what seems likely as well as what little can be proved.

Here is a table from the third edition (2004) of Unsolved Problems in Number Theory by Richard K. Guy, in which $p = p_n$ is a prime but $n$ is not calculated, then $g = p_{n+1} - p_n,$ and $p = p(g),$ so $p_{n+1} = p + g.$

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