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).

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