Hello,

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 $\lim f(n)=0$?

definitionof it includes some Big-Oh. This is an odd thing to do, IMHO. – quid Mar 3 '12 at 15:56