Upper bound for r0(n) (2)

Writing:

\mathcal{N}_{n}(x)=k_{n}x(1+\frac{\alpha_{n}(x)}{x})

and using the same kind of statistical heuristics as in « Primality radius »,

one gets:

\alpha_{n}(x)=O(\sqrt{x}\log^{2} n)

hence:

r_{0}(n)=\dfrac{P_{ord_{C(n)}}}{2N_{1}(n)}+O(\sqrt{r_{0}(n)}\log^{2} n)

that is:

r_{0}(n)=O(\log^{2} n)+O(\sqrt{r_{0}(n)}\log^{2} n)

Writing:

r_{0}(n)=O(\log^{m} n)

one gets:

r_{0}(n)=O(\log^{2} n)+O(\log^{2+m/2} n)

so that:

m=4

and r_{0}(n)=O(\log^{4} n).

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