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this question is moreorless related to
For $N$ a positive integer, let's define the set $A(N)$ as $\{m, \ \ \forall n\geq N, \ \ g_{n}:=(p_{n+1}-p_{n})<(\log p_{n})^{m}\}$. Let's assume that for all positive integer $H$, $A(H)$ is a non empty set. Then for any given positive integer $N$, $A(N)$ admits an infimum $m_{0}(N)$.

Let's furthermore assume Andrica's conjecture. This conjecture states that $\forall n>0, \ \ g_{n}<2\sqrt{p_{n}}+1$.
Due to the fact that $m_{0}(N)$ is the infimum of $A(N)$, there exists an integer $k\geq N$ such that $g_{k}=(\log p_{k})^{m_{0}(N)}$. Now, suppose that $\forall n\geq N, \ \ (\log p_{n})^{m_{0}(N)}<2\sqrt{p_{n}}+1$. Equivalently, $\forall x\geq p_{N}, \ \ (\log x)^{m_{0}(N)}\lt 2\sqrt{x}+1$.
So $\forall x\geq p_{N}, \ \ \log ((\log x)^{m_{0}(N)})\lt \log (2\sqrt{x}+1)$.`
Equivalently $\forall x\geq p_{N}, \ \ m_{0}(N)\log \log x\lt \log 2+\log (\sqrt{x}+1/2)$ and finally $\forall x\geq p_{N}, \ \ m_{0}(N)\lt \frac{\log 2+\log (\sqrt{x}+1/2)}{\log \log x}$.

As $m_{0}(N)$ doesn't depend on $x$ this means that $\forall x\geq p_{N},\ \ \frac{\log 2+\log (\sqrt{x}+1/2)}{\log \log x}\gt m_{0}(N)$. So one has $m_{0}(N)\lt y_{0}$, where $y_{0}=1.98772705...$ is the minimum of the function $f:x\gt e\mapsto \dfrac{\log 2+\log(\sqrt{x}+1/2)}{\log\log x}$, provided $p_{N}\lt x_{0}$ (where $x_{0}$ is such that $y_{0}=f(x_{0})$). As $67\lt x_{0}\lt 68$, one gets $N\leq 19$.

Furthermore, as $g_{n}\geq 2$, one must have $(\log p_{n})^{m_{0}(N)}\geq 2$. Since $m_{0}(N)>1$, one can take $n$ such that $\log p_{n}\geq 2$, hence $n\geq 5$. So that if for all integer $N$ $A(N)$ is a non empty set, and if Andrica's conjecture holds true, then (a stronger form of) Cramer's conjecture holds true, namely $c=0$, with $c$ defined in the link given at the beginning of this post.

Edit: I think I took Gerhard Paseman and quid's comments into account. So, is the proof correct now? Thank you in advance.

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Isn't the pk more relevant than the pN, and why should this by less than x0 and what if it is not? In any case, from a meta point of view this can hardly work. Also, you might consider recalling Cramer's conjecture rather then linking to it in a mildly confusing way. Finally, this is a slightly off-topic use of MO. Voting to close, at least in the current form. – user9072 Aug 15 '11 at 18:45
I also think the proof is not valid. I don't see why "one has $m_0(N) \lt y_0$ ", unless $y_0$ were a maximum. Gerhard "Sometimes Knows Where Up Is" Paseman, 2011.08.15 – Gerhard Paseman Aug 15 '11 at 19:17
There are two other places where I think gaps reside. The most important one is that you need a nonempty set for any N in order to talk about the stuff relating to the conjecture, so this is already assuming something stronger (asymptotically) than Andrica's conjecture. Although I disagree with quid about the question being off-topic, the question in its current form is not suitable for MathOverflow. I recommend re-editing before reconsidering. Gerhard "Ask Me About System Design" Paseman, 2011.08.15 – Gerhard Paseman Aug 15 '11 at 19:36
@Gerhard Paseman: are you now convinced that one has $m_{0}(N)<y_{0}$ ? – Sylvain JULIEN Aug 15 '11 at 21:08
Nope. Still has gaps. Just because a power of a log of x is less than a multiple of sqrt(x) for all x > N does not mean you can now pull N down to 69. You still can't assert that m_0 is less than y_0 on the basis of your reasoning. On your next revision, pretend there is a prime p_n such that g_n is between 3.4 log p_n and 3.5 log p_n, while still having Andrica's conjecture hold. Gerhard "Ask Me About System Design" Paseman, 2011.08.15 – Gerhard Paseman Aug 15 '11 at 21:15

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