# Could computing the next prime in a finite Euler product be made rigorous?

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 $$p_{N+1}$$?

• this is cool. its like an analytic approach to finding primes Jul 10, 2020 at 17:20
• It's not too difficult to show that $k=p_N$ works, and that we need $k\ge c p_N$ for some positive constant $c$ in case $p_{N+2}=p_{N+1}+2$. Jul 10, 2020 at 23:44

$$2k=1+p_N$$ 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 $$x. With those two modifications, we find that $$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$$.
• Many thanks, Andreas! One more observation that could maybe help sharpening these bounds. Numerical evidence suggests (I have no proof) that when $k$ increases, $x$ always approaches $p_{N+1}$ from 'below'. So, if this is true, could using the ceiling instead of rounding $x$ improve the bound by a factor $2$?
• Yes, the last display of my answer shows that $x<q=p_{N+1}$. Using the ceiling instead of rounding will always work if $2k \ge 0.71 \, p_N$, but will fail at twin primes if $2k\le 0.38 \, p_N$. Jul 11, 2020 at 16:00