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Question: Is the conjecture as follows true or false?

For any integer $n>1$, there always exists at least one prime number $p$ with

$$n < p< n+\left(\ln\Big(\frac{n}{\ln n}\Big)+1\right)^2$$

The conjecture was checked true with $n$ up to $10^8$ and some The 80 known maximal prime gaps

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    $\begingroup$ You may want to read this Wikipedia article: en.wikipedia.org/wiki/Prime_gap It discusses both the known upper and lower bounds and the conjectures which LAGRIDA talks about in their answer. $\endgroup$
    – Wojowu
    Sep 24 '19 at 13:26
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False.

Let $n=1693182318746371$. The next prime after $n$ is $1693182318747503$.

$(\ln(\frac{n}{\ln n})+1)^2 \le1057$, but the prime gap is $1132$.

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    $\begingroup$ How did you construct this huge example? Is there some list of the largest prime gaps? $\endgroup$ Sep 24 '19 at 18:51
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    $\begingroup$ Don't know if this is how it was found, but the number appears in the tables of this page: en.m.wikipedia.org/wiki/Prime_gap $\endgroup$ Sep 27 '19 at 11:51
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Your conjecture is not compatible with some actual heuristic views:

Cramer Conjecture: $$\limsup_{n\to+\infty}\dfrac{p_{n+1}-p_n}{\log(p_n)^2}=1$$

Then if this conjecture holds, we have infinitly many intervals of size $(1+o(1))\log(n)^2$ does not contain any prime numbers.

Granvile conjecture: $$\limsup_{n\to+\infty}\dfrac{p_{n+1}-p_n}{\log(p_n)^2}\gtrsim2e^{-\gamma}\approx1.12$$ ($f(x) \gtrsim g(x) \iff f(x) \geq (1+o(1))g(x)$)

Then if Granvile's conjecture holds, we have infinitly many intervals of size $(2e^{-\gamma}+o(1))\log(n)^2$ does not contain any prime numbers.

You can see that $2e^{-\gamma} > 1$, then Granvile's conjecture holds implies that your conjecture is false.

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    $\begingroup$ I think Granville conjectured that this limsup is $\ge 2e^{-\gamma}$, not necessarily $=2e^{-\gamma}$. $\endgroup$ Sep 26 '19 at 21:24

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