I know from papers like Dusarts's that $$n\left(\ln n +\ln \ln n -1+ \frac{\ln \ln n -2}{\ln n}-\frac{\ln^2 \ln n -6 \ln \ln n +12}{2 \ln^2 n}\right) \leq p_n \leq n\left(\ln n +\ln \ln n -1+ \frac{\ln \ln n -2}{\ln n}-\frac{\ln^2 \ln n -6 \ln \ln n +10}{2 \ln^2 n}\right) $$
But could any one give and explicit number $n_0$ for which for all $n \geq n_0$ the above inequalities hold true ?
It does not have to be the smallest one, also if there is way, could you please show me how to get this number
I read a paper that estimate that the $n_0 \approx 3.9*10^{31}$ under the R.H. if we don't assume R.H. could we still have $n_0 \ll 10^{100}$
Thanks in advance
In this paper here, Christian Axler shows that $$p_n\ge n\left(\ln n +\ln \ln n -1+ \dfrac{\ln \ln n -2}{\ln n}-\dfrac{\ln^2 \ln n -6 \ln \ln n +11.508}{2 \ln^2 n}\right)$$ for all $n\ge 2$
And, $$p_n \leq n\left(\ln n +\ln \ln n -1+ \frac{\ln \ln n -2}{\ln n}-\frac{\ln^2 \ln n -6 \ln \ln n +10.667}{2 \ln^2 n}\right)$$ for all $n\ge 46\ 254\ 381$
I have to admit that I can't read and check all of it, because although the proof doesn't seem to require any complex analysis, it involves pretty heavy calculation.