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I am wondering if anyone is aware of any results related to the following? I need a few good updates/ideas/hints ...

For large enough $x$'s $$x-\sqrt{x}< p_{\pi(x)}\leq x \tag{1}$$

A few explanatory notes:

  • $\pi(x)$ is the prime counting function
  • $p_n$ - is the $n$'s prime with $p_1=2$
  • one example, $p_{\pi(\sqrt{12})}=p_{\pi(3)}=p_2=3$
  • if $x=p_n$ then $\pi(p_n)=n$ and $p_{\pi(p_n)}=p_n$.

Examples and progress ...

Observation 1. $(1)$ is true for the prime numbers

$$p_n - \sqrt{p_n}< p_{\pi(p_n)}=p_n \leq p_n$$ It also works for $x=p_n+1$ because $p_{\pi(x)}=p_{\pi(p_n)}=p_n$. Given $x-p_n=1<\sqrt{x}$ we conclude $$x-\sqrt{x}<p_n=p_{\pi(x)}<p_n+1=x$$

Observation 2. RHS of $(1)$ is true, i.e. $$p_{\pi(x)}\leq x$$

From the definition of $\pi(x)$, if $\pi(x)=n$ then $p_1<p_2<...<p_n\leq x$, thus $p_{\pi(x)}\leq x$.

Observation 3. Heuristics for LHS of $(1)$ $$\lim\limits_{x\to\infty} \frac{p_{\pi(x)}}{x-\sqrt{x}}=1$$

From $$\frac{p_{\pi(x)}}{x-\sqrt{x}}=\frac{p_{\pi(x)}}{x}\cdot \frac{x}{x-\sqrt{x}}$$ $\frac{x}{x-\sqrt{x}}\to 1$ and $\frac{p_{\pi(x)}}{x}\to 1$ (an application of PNT). As a result, for some very small $\varepsilon>0$ and suficiently large $x$'s $$p_{\pi(x)}>(1-\varepsilon)\cdot (x-\sqrt{x})>x\cdot(1-\varepsilon)-\sqrt{x} \tag{2}$$ Also a few examples $$2-\sqrt{2}<p_{\pi(2)}=p_1=2\leq 2$$ $$3-\sqrt{3}<p_{\pi(3)}=p_2=3\leq 3$$ $$4-\sqrt{4}<p_{\pi(4)}=p_2=3< 4$$ $$5-\sqrt{5}<p_{\pi(5)}=p_3=5\leq 5$$ $$6-\sqrt{6}<p_{\pi(6)}=p_3=5< 6$$ $$...$$ $$14-\sqrt{14}<p_{\pi(14)}=p_6=13< 14$$

To some extent, this is a re-formulation of the Oppermann's conjecture, with $(2)$ as a faint light at the end of the tunnel. $\varepsilon$ is spoiling everything.

Another application is, from $(1)$: $$x-p_{\pi(x)}<\sqrt{x}$$ for $x=p_n - 1$ we have $\pi(p_n - 1)=n-1$, thus $$p_n - 1 - p_{n-1}<\sqrt{p_n - 1} \Rightarrow \\ p_n - p_{n-1}<\frac{\sqrt{p_n - 1}}{\sqrt{p_{n-1}}}\cdot \sqrt{p_{n-1}}+1<\sqrt{2}\cdot\sqrt{p_{n-1}}+1$$ and this is (a little bit better than) Andrica's conjecture.

rtybase
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  • nobody knows. At all. https://en.wikipedia.org/wiki/Prime_gap#Upper_bounds – Will Jagy Feb 14 '21 at 23:03
  • @WillJagy I suspect so ... but $(2)$ is so close. Even if we can prove it for special cases, like square numbers ($x=n^2$), that will be a good progress. – rtybase Feb 14 '21 at 23:06
  • Another close encounter, coming from this is $$x-\pi(x)<p_{\pi(x)}\leq x$$ for large enough $x$'s. I can even narrow it to$$x-\frac{x}{\log{x}}<p_{\pi(x)}\leq x$$ – rtybase Feb 14 '21 at 23:16
  • If $x=n^2$ then you’d have that there is always a prime $p$ such that $(n-1/2)^2<p<n^2.$ I’m not sure if it is known to be true when $1/2$ is replaced by $1.$ – Thomas Andrews Feb 14 '21 at 23:31
  • Yeah, if this is true, then Legendre’s Conjecture is true, which Wikipedia says was unsolved as of last year. https://en.wikipedia.org/wiki/Legendre%27s_conjecture – Thomas Andrews Feb 14 '21 at 23:35

1 Answers1

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If this is true, then Legendre’s Conjecture would be true for large enough $n.$ Legendre’s Conjecture is currently unsolved, even for large enough $n.$

This is because when $x=(n+1)^2,$ your conjecture says there is a prime between $(n+1)^2-(n+1)=n^2+n>n^2$ and $(n+1)^2.$

Thomas Andrews
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