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$\zeta(s)$ has a direct link to the prime numbers (via the infinite Euler product and the non-trivial zeros in the explicit $\pi(x)$ formula). Wilson's theorem offers a (proven but very inefficient) formula to generate prime numbers.

After reading this question:

Why does the Gamma-function complete the Riemann Zeta function?

I wondered whether there could be a connection between the two and I tried the following:

1) Express Wilson's theorem in terms of the $\Gamma(s)$ (instead of the faculty) and the $cos$ (instead of the mod)

$P(s) = \cos \left( \frac{\pi}{2} \{\frac {\Gamma \left( s \right) +1 }{s}} \right)$

This function has unique integer zero's only when $s$ is a prime number.

2) Express $\Gamma(s)$ in terms of $\zeta(s)$ by e.g. using a Mellin transform:

$\zeta(s) \Gamma(s) = \int_0^\infty \frac{x^{s-1}}{e^x-1} dx$

$P(s) = \cos \left( \frac{\pi}{2} (\{\frac {\int_0^\infty \frac{x^{s-1}}{e^x-1} dx}{s \zeta(s)} + \frac{1}{s} )} \right)$

3) Link the $\zeta(s)$ back to the (encoded) prime numbers:

$\zeta(s) = \pi^{\frac{s}{2}} \dfrac{\prod_\rho \left(1- \frac{s}{\rho} \right)}{2(s-1)\Gamma(1+\frac{s}{2})}$

to make the link to the non trivial zero's ($\rho$) or simply use the infinite Euler product of prime numbers.

This is just playing with formulae, I know, but wondered if anybody knows whether a stronger link could (or does) exist?

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Nope. I don't think there's any hope of making a connection like this. The manner in which $\zeta(s)$ links with primes is very different from the mechanism driving Wilson's theorem. – David Hansen Aug 6 '11 at 20:57
I guess replacing the faculty by a Gamma function is novel, just don't tell my Dean. – Charlie Frohman Aug 7 '11 at 0:16
@Agno: gamma(1+x) = factorial(x) =/= "faculty" ;-) – Gottfried Helms Aug 7 '11 at 10:00
I know. I know. This was a stupid language mistake. In Dutch the word faculty actually has this double connotation. I obviously did mean factorial and the last thing I want to do was to upset some Deans :-) – Agno Aug 7 '11 at 13:02

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