# An recursive formula for the n-th prime?

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?

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