Based on my previous answer and your help


[Is there a procedure for extracting first integer $q\_0$ from $\sum\limits\_{k=0}^{\infty}\frac{1}{q\_k^z}$, all $0<q\_0<q\_1<...$ integers, $z$ complex?](https://mathoverflow.net/q/300543/113386)


This formula derived in the question


$$ \ln(p\_n)=-\lim\limits\_{\operatorname{Re}(s) \to +\infty} \frac{\ln\left (\zeta(s)\prod\limits\_{k=1}^{n-1}(1-p\_k^{-s})-1 \right )}{s} $$


can actually turn into


$$ p\_n=\lim\limits\_{m \to +\infty} \left ( \frac{|B\_{2m}|(2\pi)^{2m}}{2(2m)!}\prod\limits\_{k=1}^{n-1}(1-p\_k^{-2m})-1 \right )^{-\frac1{2m}} $$


using


$$\zeta(2m) = \frac{(-1)^{m+1}B\_{2m}(2\pi)^{2m}}{2(2m)!}$$


Do you see any problem in using real limit instead of complex one? (I don't.)


This recurrent formula is not known to me and if correct, it looks like something worth some attention.


The formula is more to the point in logarithm form of course:


$$ \ln(p\_n)=-\lim\limits\_{m \to +\infty} \frac1{2m}\ln \left ( \frac{|B\_{2m}|(2\pi)^{2m}}{2(2m)!}\prod\limits\_{k=1}^{n-1}(1-p\_k^{-2m})-1 \right ) $$


with a form that more clearly displays the precision it needs


$$ \ln(p\_n)=-2\pi-\lim\limits\_{m \to +\infty} \frac1{2m} \ln \left ( \frac{|B\_{2m}|}{2(2m)!}\prod\limits\_{k=1}^{n-1}(1-p\_k^{-2m})-\frac{1}{(2\pi)^{2m}} \right ) $$


which gives if we conveniently replace the internal expressions with $a\_{2m}$ and $b\_{2m}$:


$$ \ln(p\_n)=-2\pi-\lim\limits\_{m \to +\infty} \frac1{2m} \ln \left ( a\_{2m}\prod\limits\_{k=1}^{n-1}(1-p\_k^{-2m})-b\_{2m} \right ) $$



Recursive formula for n-th prime derived from a previous question
Based on


1. [Keller - A recursion equation for prime
numbers](http://arxiv.org/abs/0711.3940)
2. [Kawalec - The recurrence
formulas for primes and non-trivial zeros of the Riemann zeta
function](http://arxiv.org/abs/2009.02640)


it is obvious that they meant to use or even used real values. The novelty here is that the expression has a form that you can split in two parts, one depending on $m \gg 1$ and another on $n$ where it all becomes very illustrative as some a sort of an internal sieving machine.


I would still opt of logarithm because root looks quite clumsy to deal with:


$$ \ln(p\_n)=-2\pi-\lim\limits\_{m \to +\infty} \frac1{2m} \ln \left ( a\_{2m}\prod\limits\_{k=1}^{n-1}(1-p\_k^{-2m})-b\_{2m} \right ) $$


Additionally, the other summable version one of the authors is mentioning you get directly:


$$\ln(p\_n)=-\lim\limits\_{m \to +\infty} \frac1{2m}\ln \left ( \prod\limits\_{k=1}^{n-1}(1-p\_k^{-2m})-\frac{2(2m)!}{|B\_{2m}|(2\pi)^{2m}} \right )$$


which is cute.


