## Re: [PrimeNumbers] Re: TWIN PRIMES conjecture by Dimitris Valianatos GREECE

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• ... That s a good remark. It also implies that either there is an infinite number of twin primes, or the formula is wrong. ;-) Maximilian PS: ;-) means (or
Message 1 of 8 , Feb 22, 2012
On Wed, Feb 22, 2012 at 6:07 PM, Jack Brennen <jfb@...> wrote:

> **
> A = For all twin primes (p,q)=(12n+5,12n+7), product of (1+2/p)
> B = For all twin primes (p,q)=(12n+11,12n+13), product of (1+2/p)
>
> The hypothesis is that 5*B == 2*A*sqrt(Pi).
>

That's a good remark.

It also implies that either there is an infinite number of twin primes, or
the formula is wrong.

;-)

Maximilian

PS: ";-)" means "(or both)" - of course.

[Non-text portions of this message have been removed]
• ... We may also obtain Pi by taking products over primes, having regard to their residues modulo 3 or 4: prod(prime p 2, 1 - if(p%4==1,1,-1)/p) = 4/Pi
Message 2 of 8 , Feb 22, 2012
Jack Brennen <jfb@...> wrote:

> My guess is that it's a close miss, just because other known
> formulas for Pi such as the Wallis product are taken over all
> integers, not just over a subset of a subset of the integers.

We may also obtain Pi by taking products over primes,
having regard to their residues modulo 3 or 4:

prod(prime p > 2, 1 - if(p%4==1,1,-1)/p) = 4/Pi
prod(prime p > 2, 1 + if(p%4==1,1,-1)/p) = 2/Pi
prod(prime p > 3, 1 - if(p%3==1,1,-1)/p) = 2*sqrt(3)/Pi
prod(prime p > 3, 1 + if(p%3==1,1,-1)/p) = 3*sqrt(3)/(2*Pi)

What is hard to believe about Dimitris' numerology
is that he takes a residue-dependent product over
only /twin/ primes.

David
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