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The formula I am having some problems is this one $$ f\left( x \right) =\sin \left(\frac { 2(x-3)!-x+1 }{ 2x } \pi \right) =\sin \left(\frac { 2\,\Gamma (x-2)-x+1 }{ 2x } \pi \right) $$

Although this might seem too general it is a variation of Wilson's theorem which (in my opinion) has slightly better properties than the general Wilson's theorem formula.

If you look at the graph you'll see that the values for $f(x)$ where $x$ is prime is 0 while for $x$ nonprime tends to be converging toward $-1$ and in almost a perfect fashion and rapidly. My idea is this: turn this into infinite Taylor series and add $x$ to it, that way we could get a good approximation of prime counting function to infinity.

In other words:

$$ \lim _{ x\rightarrow \infty }{ x+\sum _{ n\epsilon N }^{ }{ f(n) } } =\lim _{ x\rightarrow \infty }{ \Pi (x) } $$

The problem I am having is actually converting it into Taylor series. I really didn't have any experience with this kind of equations so I was wondering is there any trick to do this, or should I just let this one pass

Edited: Changed the formula, forgot to multiply by $\pi$

vonbrand
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user1398593
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1 Answers1

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Well it seems I will answer my own question. There is a similar formula that clearly says the nature of prime numbers and it goes like this:

$$\pi (m)=\sum _{ j=2 }^{ m }{ \frac { \sin ^{ 2 }{ (\pi \frac { { ((j-1)!) }^{ 2 } }{ j } ) } }{ \sin ^{ 2 }{ (\frac { \pi }{ j } ) } } } $$

This function really is a prime counting function but you can see that it is a little tricky to develop it into a working formula since it should be integrated over ${ 1 }_{ N }$ .

However since I'm already answering I'm going to show what I planned on doing with the formula:

$$\pi (x)\quad \approx \quad \sum _{ n=1 }^{ n=x }{ [\quad \cos { (\frac { 2*(n-3)!+1 }{ 2n } \pi ) } -\cos { (\frac { 2*(n-3)!+1 }{ n } \pi ) } \quad ] } $$

user1398593
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