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 October 20th, 2015, 11:46 AM #1 Banned Camp   Joined: Dec 2013 Posts: 1,117 Thanks: 41 Twin primes conjecture Hi, r=int(sqrt(n)) Conjecture : For each n>2 there is always at least one pair of twin-prime numbers between n!+n (not included) and n!+(r+1)^n (included)? Any counterexample? Is there a way to prove it if the conjecture is true?
 October 20th, 2015, 11:58 AM #2 Member   Joined: Jun 2015 From: Ohio Posts: 99 Thanks: 19 It isn't even known if there are infinitely many twin primes, so you would probably need to prove that before you were able to tackle this. That has been unproved for quite a long time, so I assume the problem is very challenging. Thanks from CRGreathouse and mobel
 October 20th, 2015, 01:14 PM #3 Banned Camp   Joined: Dec 2013 Posts: 1,117 Thanks: 41 I assume that there are infinite twin primes. Trying to find a thin range where there at least one prime we can consider that as n goes to infinite (r+1)^n is far less < n!+n. So m=(n!+(r+1)^n)/(n!+n) will be very < 2. m will surely converges to some value between 1 and 2. My goal as I said before if to find a thin range where there is at least one prime. I do not think that my conjecture will work for a twin prime but for a prime I will not discard it. Thank you for your comment.
October 20th, 2015, 09:57 PM   #4
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 Originally Posted by mobel Hi, r=int(sqrt(n)) Conjecture : For each n>2 there is always at least one pair of twin-prime numbers between n!+n (not included) and n!+(r+1)^n (included)?
This is highly likely to be true, but beyond current mathematical technology to prove.

October 20th, 2015, 10:00 PM   #5
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 Originally Posted by mobel I assume that there are infinite twin primes.
You can do that, and it's not unreasonable, but it doesn't help proving the conjecture at hand because (a priori) we don't know anything about how they might be distributed.

Quote:
 Originally Posted by mobel Trying to find a thin range where there at least one prime we can consider that as n goes to infinite (r+1)^n is far less < n!+n. So m=(n!+(r+1)^n)/(n!+n) will be very < 2. m will surely converges to some value between 1 and 2.
Yes, it converges to 1.

 October 21st, 2015, 12:13 PM #6 Banned Camp   Joined: Dec 2013 Posts: 1,117 Thanks: 41 Hard to prove but not impossible.
October 21st, 2015, 01:13 PM   #7
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 Originally Posted by mobel Hard to prove but not impossible.
Neither part of that statement has been proven -- but I tend to agree.

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