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

Publicités