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Yet another formula for pi(x) (prime number counting function)

start with wilson's therorem : p is prime iff p divides (p-1)! + 1

let G(x) be the gamma function

then p is prime iff sin(pi*(G(x)+1)/x) = 0

let f be the function x -> sin(pi*(G(x)+1)/x)

Since f(x) = sin(pi*G(x)/x+pi/x) and because G(x)/x is integer (and even) when x is an integer <> 4,

for any integer x, non prime and different from 4, f(x) = sin(pi/x)

Then let h(x) = f(x) / sin(pi/x) = sin(pi*(G(x)+1)/x) / sin(pi/x)

h(x) = 0 if x is prime

h(x) = 1 for any non prime integer x >4

h(4) = -1

therefore, for x >= 5,

1) Is there a way to convert this sum into an integral and have a cool expression of pi(x) ?

2) is there a way to differentiate this sum and then have an expression

of d pi(x) / dx ?

3) since sin(pi*x) = pi*x*product[n=1..inf, 1-(x/n)^2)],

and G(x)=(x-1)*G(x-1),

is there a way to express h(x) as an infinite product, and then simplify it and simplfy the above expression of pi(x) ?

start with wilson's therorem : p is prime iff p divides (p-1)! + 1

let G(x) be the gamma function

then p is prime iff sin(pi*(G(x)+1)/x) = 0

let f be the function x -> sin(pi*(G(x)+1)/x)

Since f(x) = sin(pi*G(x)/x+pi/x) and because G(x)/x is integer (and even) when x is an integer <> 4,

for any integer x, non prime and different from 4, f(x) = sin(pi/x)

Then let h(x) = f(x) / sin(pi/x) = sin(pi*(G(x)+1)/x) / sin(pi/x)

h(x) = 0 if x is prime

h(x) = 1 for any non prime integer x >4

h(4) = -1

therefore, for x >= 5,

**pi(x) = x-2+sum[k=5..x, h(k)]**__Questions :__1) Is there a way to convert this sum into an integral and have a cool expression of pi(x) ?

2) is there a way to differentiate this sum and then have an expression

of d pi(x) / dx ?

3) since sin(pi*x) = pi*x*product[n=1..inf, 1-(x/n)^2)],

and G(x)=(x-1)*G(x-1),

is there a way to express h(x) as an infinite product, and then simplify it and simplfy the above expression of pi(x) ?

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