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 $N=60$:

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 $p_{N+1}$?

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    $\begingroup$ this is cool. its like an analytic approach to finding primes $\endgroup$ Jul 10, 2020 at 17:20
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    $\begingroup$ 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 $p_{N+2}=p_{N+1}+2$. $\endgroup$ Jul 10, 2020 at 23:44

1 Answer 1


$2k=1+p_N$ 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 $p_{N+2}=p_{N+1}+2$.

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    $\begingroup$ 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 $2$? $\endgroup$
    – Agno
    Jul 11, 2020 at 14:59
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    $\begingroup$ 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 $2k\le 0.38 \, p_N$. $\endgroup$ Jul 11, 2020 at 16:00

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