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this thread is a follow up from About Goldbach's conjecture

I actually aim at proving that the smallest typical potential primality radius of an integer $n$, written as $r_0(n)$, verifies $r_0(n)=O(\log^{2} n)$. This implies Cramer's conjecture (since if $k=\frac{p_{n+1}+p_{n}}{2}$ then one has $r_0(k)=O(\log^{2}k)$, i.e. $g_{n}:=p_{n+1}-p_{n}=2r_{0}(k)=O(\log^{2}k)=O(\log^{2}p_{n})$.

The relation we aim at proving is a consequence of both $\dfrac{2r_{0}(n)}{\epsilon_n}=O(1)$, and $\epsilon'_n=\epsilon_n(1+o(1))$ (where $\epsilon'_n:=\dfrac{P_{ord_{c}}(n)}{N_{1}(n)}$ and $\epsilon_n=\dfrac{P_{ord_{c}}(n)-2r_{0}(n)}{N_{1}(n)-1}$).

It has been established that $\dfrac{n}{\epsilon'_n}\geq \dfrac{c.n}{\log^{2} n}$ for some positive constant $c$ and $n$ large enough. From that we get $\epsilon'_n=O(\log^{2} n)$.

First, let's give a sketch of proof that $\epsilon'_n\sim\epsilon_n$:

$\dfrac{\epsilon'_n-\epsilon_n}{\epsilon_n}=\dfrac{(N_{1}(n)-1)P_{ord_{c}}(n)}{N_{1}(n)(P_{ord_{c}}(n)-2r_{0}(n))}-1$ $=\dfrac{(N_{1}(n)-1)P_{ord_{c}}(n)}{N_{1}(n)(P_{ord_{c}}(n)-2r_{0}(n))}-\dfrac{N_{1}(n)(P_{ord_{c}}(n)-2r_{0}(n))}{N_{1}(n)(P_{ord_{c}}(n)-2r_{0}(n))}$ $=\dfrac{2N_{1}(n)r_{0}(n)-P_{ord_{c}}(n)}{N_{1}(n)(P_{ord_{c}}(n)-2r_{0}(n))}$ $=-\dfrac{2N_{1}(n)r_{0}(n)-P_{ord_{c}}(n)}{2N_{1}(n)r_{0}(n)-N_{1}(n)P_{ord_{c}}(n)}$

The last quantity is, modulo the minus sign, $f(n):=\dfrac{2r_{0}(n)-O(\log^{2}(n))}{2r_{0}(n)-P_{ord_{c}}(n)}$. But whenever $n>13$, $N_1(n)>1$ so that $2r_{0}(n)-P_{ord_{c}}(n)$ is non zero. Moreover, as $P_{ord_{c}}(n)$ grows faster than $n$, one should have $\lim f(n)=0$, and thus $\epsilon'_n=\epsilon_n(1+o(1))$.

After some computations, one gets that $\lim f(n)=0$ implies $f(n)=O(\frac{1}{N_{1}(n)})$ (because if $\epsilon_n\sim\epsilon'_n$, then $r_{0}(n)=o(P_{ord_{c}}(n))$, and using this in the expression of $f(n)$ allows to conclude).

One can easily show that $\dfrac{2r_{0}(n)}{\epsilon_n}=N_1(n)\dfrac{\epsilon'_n-\epsilon_n}{\epsilon_n}+1$. As $f(n)=O(\frac{1}{N_{1}(n)})$, it comes $\dfrac{2r_{0}(n)}{\epsilon_n}=O(1)+1=O(1)$.

Finally: $r_{0}(n)=O(\log^{2} n)$.

My question is: can one prove rigorously that $\lim f(n)=0$?

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This question is very hard to read. But what is worse, you ask about some f(n) and the definition of it includes some Big-Oh. This is an odd thing to do, IMHO. –  quid Mar 3 '12 at 15:56
The definition of f(n) actually is $\dfrac{2r_{0}(n)-\frac{P_{ord_{C}}(n)}{N_{1}(n)}}{2r_{0}(n)-P_{ord_{C}}(n)}$. The $O(\log^{2}n)$ is here just to remind the order of magnitude of $\frac{P_{ord_{C}}(n)}{N_{1}(n)}$. –  Sylvain JULIEN Mar 3 '12 at 17:09
Yes, I sort of figured this. But if you write f(n):= ... in particular using colon before the =, I would definitely afterwards put the actual definition. Also, do I understand correctly that you are asking MO to prove something that will imply Cramer's conjecture? –  quid Mar 3 '12 at 18:12
I don't ask MO to prove something that will imply Cramer's conjecture, I ask MO to help me ascertain the correctness of my proof. –  Sylvain JULIEN Mar 3 '12 at 18:40
What would you do if MO would not exist? More concretly, I do not know. To be honest, I think very few people are keen to debug claimed proofs of wellknown problems. In all likelihood they are simply wrong and quite likely uninteresting. (This is a general statement, not a specifc criticism.) If you really think you have proved something interesting I think you should write it up very carefulluy, not in half-page internet posts. Perhpas you then notice a problem yourself. Or, it'd be at least simpler to check than when defs are distributed over MO questions. Sorry for the negative comment. –  quid Mar 4 '12 at 0:11
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