20111127, 21:39  #1 
May 2004
New York City
3×17×83 Posts 
Gold Twins
Just a question relating Twin Primes to Goldbach's Conjecture:
Suppose p and q = p+2 are twin primes. Then we automatically have three consecutive even integers 2p, 2p+2 = p+q, and 2p+4 = 2q which can easily be written as the sum of two (maximal) primes, i.e. which satisfy the Goldbach Conjecture. My question is just: which one "wins" or "loses"? IOW as p gets large and the lesser part of a TP pair, if we count the number of prime pairs that sum to 2p, 2p+2, 2p+4, is there any tendancy for one of these three to have more or fewer solutions than the others? (Please note: I haven't even looked at large primes for this.) 
20111127, 22:10  #2 
"Brian"
Jul 2007
The Netherlands
110011000101_{2} Posts 
My handwaving solution is a guess in the sense that it is based on no theoretical knowledge whatsoever. But since everyone can participate in this puzzle section I'll risk ridicule by writing it down.
I think the answer depends on the truth or otherwise of the conjecture that there are infinitely many twin primes. If the conjecture is false, then the answer will be dependent on longinthefuture (currently nowhere near feasable) calculations of all the twin prime pairs (p,q) and all the divisions of 2p, 2p+2, 2p+4 into sums of primes. In short: we don't know yet. If the conjecture is true then I think that none of the three "win" as p and q tend to infinity. The tiny advantage of the highest number of the trio  in general larger even numbers have more ways of being expressed as the sum of two primes  will "tend to no advantage" as p tends to infinity. Last fiddled with by BrianE on 20111127 at 22:11 Reason: changed erroneous "false" to "true" 
20111128, 05:03  #3 
"Kyle"
Feb 2005
Somewhere near M52..
3×5×61 Posts 
I agree with Brian.
If there are an infinite number of twin primes, then I don't expect that either of the three possibilities is more significant as p and q tend towards infinity. If there are a finite number of twin primes...then I don't know. My mathematical knowledge is limited. 
20111128, 06:10  #4 
Aug 2006
3^{2}·5·7·19 Posts 

20111128, 15:06  #5  
"Kyle"
Feb 2005
Somewhere near M52..
3×5×61 Posts 
Quote:


20111128, 18:32  #6  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:
Code:
b=[];c=0;forprime(x=1,1000,if(isprime(x+2),b=concat(b,[2*x,2*x+2,2*x+4])));for(a=1,#b,forprime(y=1,100,forprime(z=1,y,if(z+y==b[a],c=c+1)));b[a]=c;c=0) 

20111128, 18:49  #7  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
Code:
b=vector(#b/3,n,[b[(n1)*3+1],b[(n1)*3+2],b[(n1)*3+3]]) 

20111128, 23:52  #8  
Aug 2006
3^{2}×5×7×19 Posts 
Quote:


20111129, 03:22  #9  
Dec 2010
Monticello
3403_{8} Posts 
Quote:
My mathematical nose (admittedly not terribly discerning) says that this might be related to the points in which the distribution of primes crosses its first approximation, the first point once estimated to be under skewe's number, now known to be under 10^400 or so...having to do with the zeroes of Riemann's Zeta function. Might be worth a bit of numerical experimentation...Goldbach's has been verified to 10^17 or so.... 

20111129, 05:52  #10  
Jun 2003
11×449 Posts 
Quote:
Last fiddled with by axn on 20111129 at 06:13 

20111130, 20:55  #11 
"Brian"
Jul 2007
The Netherlands
7×467 Posts 
This answer really intrigues me as I know you always post from a knowledgeable starting point. I'd love to know what you base your (to me counterintuitive) prediction on, even if it's just a hint which might point me in the right direction. Either from you, or from the OP or anyone else who sees more in this puzzle than I do.

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