# sharp upper bound for alpha_{n}

As mentioned in the previous article, it appears that the upper bound $\alpha_{n}=O_{\varepsilon}(n^{1/2+\varepsilon})$ would follow from the following reasonable assumption:

$N_{2}(n)$ is « rather close » to $N_{1}(n)\dfrac{n-\sqrt{2n-3}}{P_{ord_{C}}(n)-\sqrt{2n-3}}$

This follows from the very definition of what a typical primality radius is. Indeed, writing

$N_{2}(n)=N_{1}(n)\dfrac{n-\sqrt{2n-3}}{P_{ord_{C}(n)}-\sqrt{2n-3}}+O(f(n))$

one gets:

$\dfrac{n.N_{1}(n)}{P_{ord_{C}(n)}}(1+\dfrac{\alpha_{n}}{n})=N_{1}(n)\dfrac{n-\sqrt{2n-3}}{P_{ord_{C}(n)}-\sqrt{2n-3}}+O(f(n))$

Hence

$1+\dfrac{\alpha_{n}}{n}=\dfrac{P_{ord_{C}(n)}}{n}\left(\dfrac{n-\sqrt{2n-3}}{P_{ord_{C}(n)}-\sqrt{2n-3}}\right)+O(f(n)\dfrac{P_{ord_{C}(n)}}{n.N_{1}(n)})$

So $\dfrac{\alpha_{n}}{n}=\dfrac{P_{ord_{C}(n)}}{n}\dfrac{n-\sqrt{2n-3}}{P_{ord_{C}(n)}-\sqrt{2n-3}}-\dfrac{n(P_{ord_{C}(n)}-\sqrt{2n-3})}{n(P_{ord_{C}(n)}-\sqrt{2n-3})}+O(f(n)\dfrac{\log^{2} n}{n})$

Thus $\alpha_{n}=\dfrac{(n-P_{ord_{C}(n)})\sqrt{2n-3}}{P_{ord_{C}(n)}-\sqrt{2n-3}}+O(f(n)\log^{2} n)$

Finally $\vert\alpha_{n}\vert=(\sqrt{2n})^{1+\varepsilon}+O(f(n)\log^{2}n)$

Assuming $f(n)=\sqrt{n}$ (which remains to be proven), one gets $\vert\alpha_{n}\vert=O_{\varepsilon}(n^{1/2+\varepsilon})$. Maybe some kind of central limit theorem could shed a new light on the true nature of $f(n)$.