- #1
eljose
- 492
- 0
In an anlaogy with the Euler product of the Riemann function we make:
[tex] \prod_{p}(1+e^{-sp})=f(s) [/tex] of course we have that:
[tex] f(p1+p2+p3)=f(p1)f(p2)f(p3) [/tex] 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:
[tex] f(s)=\sum_{n=0}^{\infty}a(n)e^{-sn} [/tex] 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:
[tex] A(x)=\sum_{n=0}^{x}a(n) [/tex] if a(n)=1 for every n then using Perron formula we get that A(x)=[x] (floor function) so we would have that:
[tex] f(s)=(1-e^{-s}) [/tex] then if correct take numerical values to proof that the product and f(s) are equal.
[tex] \prod_{p}(1+e^{-sp})=f(s) [/tex] of course we have that:
[tex] f(p1+p2+p3)=f(p1)f(p2)f(p3) [/tex] 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:
[tex] f(s)=\sum_{n=0}^{\infty}a(n)e^{-sn} [/tex] 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:
[tex] A(x)=\sum_{n=0}^{x}a(n) [/tex] if a(n)=1 for every n then using Perron formula we get that A(x)=[x] (floor function) so we would have that:
[tex] f(s)=(1-e^{-s}) [/tex] then if correct take numerical values to proof that the product and f(s) are equal.