## euler product and Goldbach conjecture..

In an anlaogy with the Euler product of the Riemann function we make:

$$\prod_{p}(1+e^{-sp})=f(s)$$ of course we have that:

$$f(p1+p2+p3)=f(p1)f(p2)f(p3)$$ f(x)=exp(-ax) if Goldbach Conjecture is true then p1+p2= even and p5+p6+p8=Odd for integer n>5? then this product should be equal to:

$$f(s)=\sum_{n=0}^{\infty}a(n)e^{-sn}$$ where the a(n) is the function that tells in how many ways and odd or even number can be descomposed as a sum of 2 or 3 primes, we now define:

$$A(x)=\sum_{n=0}^{x}a(n)$$ if a(n)=1 for every n then using Perron formula we get that A(x)=[x] (floor function) so we would have that:

$$f(s)=(1-e^{-s})$$ then if correct take numerical values to proof that the product and f(s) are equal.
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 sorry there,s a mistake if we set: $$\prod_{p} (1-e^{-sp}=f(s)= \sum_{n=0}^{\infty} a(n)e^{-sn}= \int_{0}^{\infty}A(x)e^{-sx}$$ And using the same trick Riemann did we have that: $$f(s)= \sum_{n=1}^{\infty} \frac{g(x^{1/n})}{n}$$ where $$g(x)=\pi (lnx)$$ so from this we could conclude that: $$f(x)= \frac{1}{2 \pi i}\int_{C} ds \frac{ ln \zeta(s) }{s}ln^{s} (t)$$ Where C is the real line Re(s)>c for a real c constant