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 September 16th, 2015, 04:11 AM #1 Banned Camp   Joined: Dec 2013 Posts: 1,117 Thanks: 41 Conjecture about consecutive primes Hi eveybody, For any finite sequence of consecutive primes P(n) (n>=2) of length l > 2 it always exists at least one even positive integer 2m such as : 2m = P(n)+C(j) with C(j) is composite for any P(n) and m < p(1)*p(2)*....*p(n) Example : m=49 2m=98 setP(n)=(3,5,7,11,13} m<3*5*7*11*13 3+95=98 5+93=98 7+91=98 11+87=98 13+85=98 All the C(j)={95,93,91,87,85} are composite. Any counterexample? Any proof? Thanks for your comments. September 16th, 2015, 04:46 AM #2 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 For a proof you could try to show a prime gap of p(n) - p(1) exists below p(1) * ... * p(n) for all n > 1. Thanks from mobel September 16th, 2015, 04:47 AM #3 Global Moderator   Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Let me see if I understand the problem. You have a sequence of 3 or more consecutive (odd?) primes and you want to know if for any such there is a positive integer m less than the product of the primes such that any of the primes plus 2m is composite. Right? September 16th, 2015, 04:53 AM   #4
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Quote:
 Originally Posted by CRGreathouse Let me see if I understand the problem. You have a sequence of 3 or more consecutive (odd?) primes and you want to know if for any such there is a positive integer m less than the product of the primes such that any of the primes plus 2m is composite. Right?
Right!
3 or more consecutive odd primes.

For sure odds P(2)=3 is the starting point
2 is prime for sure but it will be superfluous to indicate it because 2+something = 2m then something will always be composite. September 16th, 2015, 04:58 AM #5 Global Moderator   Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms If I understand correctly it seems highly likely. You're looking for a small number of composites with a large range to search. With 3 primes starting at p the 'chance' of a prime given a nearby odd is ~2/log p so the chance for three composites is around (1 - 2/log p)^3, which is more than 1/2 once you hit a few tens of thousands. For p above 10^6 you have > 10^18 chances, each more than 50% likely to work. As you go higher or increase the number of primes it only gets easier. Thanks from mobel September 16th, 2015, 05:14 AM   #6
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 Originally Posted by CRGreathouse If I understand correctly it seems highly likely. You're looking for a small number of composites with a large range to search. With 3 primes starting at p the 'chance' of a prime given a nearby odd is ~2/log p so the chance for three composites is around (1 - 2/log p)^3, which is more than 1/2 once you hit a few tens of thousands. For p above 10^6 you have > 10^18 chances, each more than 50% likely to work. As you go higher or increase the number of primes it only gets easier.
Yes but this is not a proof.
Even if something is highly likely does not mean that it always work except if you find a mathematical proof. September 16th, 2015, 08:49 AM #7 Global Moderator   Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Sure -- just checking that we're all on the same page here. I checked that there are no counterexamples of triples less than 10^9, so there's no need to worry about small cases, just the general/large cases. September 17th, 2015, 01:20 AM #8 Senior Member   Joined: Dec 2007 Posts: 687 Thanks: 47 Let $p_1,\ldots,p_n$ be a list of consecutive odd primes, where $n>2$ and $\displaystyle P=\prod_{i=1}^np_i$. Clearly $P+1$ is even. Now, notice that $\frac{P+1}{2}+\frac{P-3}{2}=P-12$ prime numbers in sequence; (3) this satisfies the conditions of the problem, that for any product $P$ of at least three odd primes we find \$N
September 17th, 2015, 02:20 AM   #9
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Quote:
 Originally Posted by CRGreathouse ...any of the primes plus 2m is composite.
Probably doesn't change much about a proof, but that's not how I read the problem. I read it as '...any of the primes plus a composite is 2m.' September 17th, 2015, 04:51 PM   #10
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 Originally Posted by Hoempa Probably doesn't change much about a proof, but that's not how I read the problem. I read it as '...any of the primes plus a composite is 2m.'
Yes Hoempa, I read Mobel in the same way, CRG probably made a typo.

Anyway, this won't change the conclusion I wrote above after small modifications. Tags conjecture, consecutive, primes Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post uvkajed Number Theory 2 August 5th, 2015 10:51 AM miket Number Theory 5 May 15th, 2013 05:35 PM greg1313 Number Theory 3 September 25th, 2012 06:36 PM ibougueye Number Theory 1 August 13th, 2012 08:24 PM Bogauss Number Theory 32 March 1st, 2012 06:30 AM

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