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 2011-11-27, 21:39 #1 davar55     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.)
 2011-11-27, 22:10 #2 Brian-E     "Brian" Jul 2007 The Netherlands 1100110001012 Posts My hand-waving 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 long-in-the-future (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 Brian-E on 2011-11-27 at 22:11 Reason: changed erroneous "false" to "true"
 2011-11-28, 05:03 #3 Primeinator     "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.
2011-11-28, 06:10   #4
CRGreathouse

Aug 2006

32·5·7·19 Posts

Quote:
 Originally Posted by Primeinator If there are a finite number of twin primes...then I don't know. My mathematical knowledge is limited.
If there are only finitely many twin primes, then "as p gets large" in the problem specification doesn't make sense.

2011-11-28, 15:06   #5
Primeinator

"Kyle"
Feb 2005
Somewhere near M52..

3×5×61 Posts

Quote:
 Originally Posted by CRGreathouse If there are only finitely many twin primes, then "as p gets large" in the problem specification doesn't make sense.
This is true. My wording was not very good. However, I must be honest and say that my mathematical muscles are not strong enough to attempt a good explanation at this problem.

2011-11-28, 18:32   #6
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26·131 Posts

Quote:
 Originally Posted by davar55 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.)
I'm guessing you have a code like this:

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)

2011-11-28, 18:49   #7
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts

Quote:
 Originally Posted by science_man_88 I'm guessing you have a code like this: 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) already.
oh and for an easier read append:

Code:
b=vector(#b/3,n,[b[(n-1)*3+1],b[(n-1)*3+2],b[(n-1)*3+3]])

2011-11-28, 23:52   #8
CRGreathouse

Aug 2006

32×5×7×19 Posts

Quote:
 Originally Posted by Primeinator This is true. My wording was not very good. However, I must be honest and say that my mathematical muscles are not strong enough to attempt a good explanation at this problem.
It's not your wording but that of the OP. I'm just saying that you can drop the part of your post about "if the twin prime conjecture is false" because then there is no question in the first place.

2011-11-29, 03:22   #9
Christenson

Dec 2010
Monticello

34038 Posts

Quote:
 Originally Posted by Primeinator This is true. My wording was not very good. However, I must be honest and say that my mathematical muscles are not strong enough to attempt a good explanation at this problem.
Assume there is a last twin prime, q-final. It is going to be a very large number, with more than a few tens of digits, so we can use the average properties of the number of pairs of primes that sum to the number with (waves hands) little error, and there will be a (very) slight bias towards the larger prime in the pair having more choices of prime addend pairs.

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....

2011-11-29, 05:52   #10
axn

Jun 2003

11×449 Posts

Quote:
 Originally Posted by davar55 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?
My prediction : The central form "2p+2" will have twice as many pairs as the other two forms, asymptotically. EDIT:- IOW, the ratio tends to 1:2:1

Last fiddled with by axn on 2011-11-29 at 06:13

2011-11-30, 20:55   #11
Brian-E

"Brian"
Jul 2007
The Netherlands

7×467 Posts

Quote:
 Originally Posted by axn My prediction : The central form "2p+2" will have twice as many pairs as the other two forms, asymptotically. EDIT:- IOW, the ratio tends to 1:2:1
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 counter-intuitive) 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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