A very short recursive formula for the $n$-th prime is (by a combination of Euclids proof and the sieve of Erathostenes):

$$p_{n+1} = \min_{x>1,\gcd(x,p_1\cdots p_n)=1} x$$

Using an approximation for the $\min$ function, we find:

$$p_{n+1} = \lim_{\rho \rightarrow \infty} -\frac{1}{\rho} \log(\sum_{x>1,\gcd(x,p_1\cdots p_n)=1} \exp(-\rho x))$$

My idea is to sum over all numbers $x>1$ but to weight them:

$$p_{n+1} =^? \lim_{\rho \rightarrow \infty} -\frac{1}{\rho} \log \left( \sum_{x\ge 2}^\infty \exp(-\rho x)\cdot \exp(-\gcd(p_1 \cdots p_n,x)^{\rho})\right )$$

Can this last equality be proven? (I have checked it numerically with SAGEMATH).

Is there any benefit in using this formula to deduce any property of the primes?

Thanks for your help.


1 Answer 1


Since $exp$ is positive for all real values we have that $ \sum \limits_{x \geq 2}^{\infty} exp(-\rho x) exp(-gcd(p_1 \cdots p_n ,x)^{\rho}) \geq exp(-p_{n+1} \rho) exp(-1)$ and $ \lim \limits_{\rho \to \infty} \frac{-1}{\rho} \log(exp(-p_{n+1} \rho) exp(-1)) = \lim \limits_{\rho \to \infty} \frac{-1}{\rho} ( -\rho p_{n+1}-1) = p_{n+1}$

Also $ \sum \limits_{x \geq 2}^{\infty} exp(-\rho x) exp(-gcd(p_1 \cdots p_n ,x)^{\rho}) \leq \sum \limits_{x=2}^{p_{n+1}-1} exp(-\rho x)exp(-2^{\rho}) + \sum \limits_{x = p_{n+1}}^{\infty} exp(-\rho x)exp(-1) \leq \sum \limits_{x=2}^{p_{n+1}-1} exp(-2^{\rho}) + \sum \limits_{x = p_{n+1}}^{\infty} exp(-\rho x)exp(-1) \leq 2p_{n} exp(-2^{\rho}) + exp(-\rho p_{n+1}-1) (1+\frac{1}{e^{\rho}}+ \frac{1}{e^{2\rho}}+...) \leq \frac{2p_n}{e^{2^{\rho}}} + exp(-\rho p_{n+1}-1) (1+\frac{2}{e^\rho}) $ for $e^\rho >> 2p_n$ we have that $\frac{2p_n}{e^{2^{\rho}}} + exp(-\rho p_{n+1}-1) (1+\frac{2}{e^\rho}) \leq exp(\rho-2^{\rho}) + exp(-\rho p_{n+1}-1) (1+\frac{2}{e^\rho}) $ and so $ \lim \limits_{\rho \to \infty} \frac{-1}{\rho} \log( exp(\rho-2^{\rho}) + exp(-\rho p_{n+1}-1) (1+\frac{2}{e^\rho})) = p_{n+1}$.

And by the squeeze theorem we have that $ \lim_{\rho \rightarrow \infty} -\frac{1}{\rho} \log \left( \sum_{x\ge 2}^\infty \exp(-\rho x)\cdot \exp(-\gcd(p_1 \cdots p_n,x)^{\rho})\right ) = p_{n+1}$

In the end your equality is correct

  • $\begingroup$ thanks for your answer (+1) $\endgroup$ Jan 4, 2021 at 5:09

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