It is well known that:


$$\zeta(s):=\prod\_{n=1}^{\infty} \frac{1}{1-p\_n^{-s}} \qquad \Re(s) \gt 1$$


with $p\_n =$ the $n$-th prime. It also known that:


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


where $B\_{2n}$ is the $2n$-th Bernoulli number.


Now define the function:


$$f(k,N,x):= \zeta(2k) - \left(\prod\_{n=1}^{N} \frac{1}{1-p\_n^{-2k}}\right)\cdot \left(\frac{1}{1-x^{-2k}}\right) \qquad \Re(s) \gt 1$$


where $k, N \in \mathbb{N}$ and $x$ is the unknown next prime ($p\_{N+1}$) to be computed.


I found numerically that solving $x$ in $f(k,N,x)=0$ for some $N$, yields an increasingly accurate approximation of $p\_{N+1}$ when $k$ increases. For example take the first 6 primes and try to derive the 7th prime (17):



> 
> $f(6, 6, x) = 0 \rightarrow x = 16.64054...$
> 
> 
> $f(12, 6, x) = 0 \rightarrow x = 16.95214...$
> 
> 
> $f(24, 6, x) = 0 \rightarrow x =16.99830...$
> 
> 
> 


The key question is how high $k$ needs to go to ensure that $x=p\_{N+1}$ after rounding. In the following Maple code, I simply used $k=2N$ and it already correctly generates all 'next' primes up to $N=60$:



```
Digits:=600
for N from 1 to 60 do ithprime(N), ithprime(N+1), round(fsolve(f(2*N, N, x), x = 0 .. 300)) end do

```

I immediately acknowledge that this is a highly inefficient and impractical algorithm to generate primes. However, is there more to say about the minimally required value of $k$ (maybe as a function of $N$) to ensure that rounding $x$ will correctly yield $p\_{N+1}$?



Could computing the next prime in a finite Euler product be made rigorous?
$2k=1+p\_N$ works for $N>1$, but $2k\le 0.56 \, p\_N$ will fail if
$p\_{N+2}=p\_{N+1}+2$.


With $q=p\_{N+1}$, we have
$$
\frac{1}{1-q^{-2k}} < \frac{1}{1-x^{-2k}} = \frac{1}{1-q^{-2k}} \prod\_{p>q} \frac{1}{1-p^{-2k}} .
$$
It follows that
$$
q^{-2k} < x^{-2k} < q^{-2k} + \sum\_{j\ge 2} (q+j)^{-2k} < q^{-2k} +\frac{1}{(q+1)^{2k-1}(2k-1)}.
$$
Taking logarithms, and using $\log(1+y)\le y$, we get
$$
-2k \log q < -2k \log x < -2k\log q + \frac{q+1}{\exp\{(1-o(1)) 2k/q\} (2k-1)}.
$$
Dividing by $-2k$ and exponentiating, we have
$$
q > x > q - \frac{q(q+1)}{\exp\{(1-o(1))2k/q\} 2k (2k-1)}.
$$
We want the last expression to be less than $1/2$. Since $q/p\_N \to 1 $ as $N\to \infty$, we need $k\ge (1+o(1)) c \, p\_N$,
where $c=0.45...$ is the solution to $e^{2c}4 c^2=2$.
So $2k=1+p\_N$ works for large $N$. We can check with a computer that it also works for small $N$.
Similar calculations show that when $p\_{N+2}=p\_{N+1}+2$, it is necessary that $k>0.28 \, p\_N$ when $N$ is large.


**Edit:** Using the inequality $\log(1+y)\le y$ was somewhat wasteful and not necessary. Also, as the OP points out in the comments, we can use the ceiling function instead of rounding, since $x<q$. With those two modifications, we find that $k\ge \frac{1}{3}p\_N$ works for $N\ge 1$, but $k\le 0.19 \, p\_N$ fails at twin primes $p\_{N+2}=p\_{N+1}+2$.


