Formula for primes

Willan's Formula is a formula that can find the nth prime number.

$1+\sum _{i=1}^{2^n}\lfloor (\frac{n}{\sum _{j=1}^i\lfloor (\cos \pi \frac{(j-1)!+j}{j}{)}^2\rfloor }{)}^{\frac{1}{n}}\rfloor$

Proof

Let's first start with the $\frac{(j-1)!+1}{j}$.

Wilson's theorem says if $(j-1)!+1$ is divisible by $j$, than $j$ is either a prime number or $1$, meaning when $j$ is prime, $\frac{(j-1)!+1}{j}$ is an integer.

It would be much easier if the formula gives a number instead of checking if the number is an integer, and we can do this with the $\lfloor (\cos \pi \frac{(j-1)!+j}{j}{)}^2\rfloor$ part.

The reason the formula has $\pi$ multiplied by the $\frac{(j-1)!+1}{j}$ part is because when $\frac{(j-1)!+1}{j}$ is an integer, $\cos (\frac{(j-1)!+1}{j}\pi )$ will give $1$ or $-1$.

When squaring the result then $\cos (\frac{(j-1)!+1}{j}\pi )$ will equal $1$ when $\frac{(j-1)!+1}{j}$ is an integer.

By flooring this, the only results are $1$ when $\frac{(j-1)!+1}{j}$ is an integer and $0$ when it isn't, leaving$\lfloor (\cos \pi \frac{(j-1)!+j}{j}{)}^2\rfloor$.

The $\sum _{j=1}^i\lfloor (\cos \pi \frac{(j-1)!+j}{j}{)}^2\rfloor$ will add $1$s for the primes $1$ - $i$ and and will sum up to the $(\mathrm{\#}\text{primes}\le i)+1$.

The $\lfloor (\frac{n}{\sum _{j=1}^i\lfloor (\cos \pi \frac{(j-1)!+j}{j}{)}^2\rfloor }{)}^{\frac{1}{n}}\rfloor$ in short will give $1$ if $n>(\mathrm{\#}\text{primes}\le i)$ and $0$ when $n\ge (\mathrm{\#}\text{primes}\le i)$.

Take the $p(x)$ of both sides where $p(x)$ is the nth prime number:

$1$ when $nth$ $\text{prime}>i$ $\Rightarrow i<nth$ $\text{prime}$

$0$ when $nth$ $\text{prime}\le i$ $\Rightarrow i\ge nth$ $\text{prime}$

$\sum _{i=1}^{2^n}\lfloor (\frac{n}{\sum _{j=1}^i\lfloor (\cos \pi \frac{(j-1)!+j}{j}{)}^2\rfloor }{)}^{\frac{1}{n}}\rfloor$ gives the number $-1$, and the $-1$ is because when $i<nth$ $\text{prime}$ reaches $i=nth$ $\text{prime}$, the function doesn't add 1. The formula adds up to $2^n$ is because Bertrand's postulate says $2^n$ is bigger than the nth prime number.

And finally, $1$ is added because of the $-1$.