For any integer $n>1$, there always exists at least one prime number $p$ with $n < p< n+\left(\ln\Big(\frac{n}{\ln n}\Big)+1\right)^2$

Question

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

Answer 1

False.

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

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

Answer 2

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.