# 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: 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

• 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. – Wojowu Sep 24 at 13:26

False.

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

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

• How did you construct this huge example? Is there some list of the largest prime gaps? – Thomas Dybdahl Ahle Sep 24 at 18:51
• 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 – Dror Speiser Sep 27 at 11:51

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.

• I think Granville conjectured that this limsup is $\ge 2e^{-\gamma}$, not necessarily $=2e^{-\gamma}$. – Micah Milinovich Sep 26 at 21:24