September 11th, 2017, 07:01 AM  #1 
Newbie Joined: Jan 2010 Posts: 9 Thanks: 0  Goldbach
The formula that generates whole prime numbers : y=(2^(x1)1)/x ( formula1) If y is an integer , then x must be absolutely a prime number . the set of x for any value of integer y ; x = { 3,5,7,11,13,....} and it generates all the prime numbers . The question is that for the set of prime numbers ( x1 , x2) does the formula generates all the even numbers or not ? y1 = (2^(x11) 1 ) /x1 + (2^(x21) 1 ) /x2 for ( x1 , x2) = ( 3,3) then y1 = 2 ; for ( x1 , x2) = ( 3,5) then y2 = 4 ; for ( x1 , x2) = ( 5,5) then y3 = 6 ; for ( x1 , x2) = (5,7) then y4 = 8 ; .................... The result for whole prime sets of ( x1 , x2) then you can generate all the even number's set . P.S.: For the proof of formula1 and to learn more about it please contact me . For example formula1 must be always divided by 3 . METE UZUN TEL: +905315540733 email: meteuzun@hotmail.com 
September 11th, 2017, 09:47 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 18,433 Thanks: 1462 
Though $(2^{3411}1)/341$ is an integer, 341 = 11 × 31 is composite.


Tags 
goldbach 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Goldbach Theorem  surya  Number Theory  26  January 13th, 2013 02:30 AM 
Goldbach Theorem  surya  Number Theory  28  January 1st, 2013 12:15 PM 
A formula for Goldbach?  goodjobbro  Number Theory  12  August 25th, 2012 10:00 AM 
Goldbach's conjecture  ibougueye  Number Theory  1  August 13th, 2012 09:28 PM 
Goldbach conjecture  brunojo  Number Theory  109  June 17th, 2009 08:08 PM 