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 $p_{N+1}$?