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September 16th, 2015, 05:11 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, 05:46 AM  #2 
Math Team Joined: Apr 2010 Posts: 2,778 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.

September 16th, 2015, 05:47 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, 05:53 AM  #4  
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41  Quote:
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, 05:58 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. 
September 16th, 2015, 06:14 AM  #6  
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41  Quote:
Even if something is highly likely does not mean that it always work except if you find a mathematical proof.  
September 16th, 2015, 09:49 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, 02:20 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 $\frac{P+1}{2}+\frac{P3}{2}=P1<P$, in general $\frac{P+1}{2}+\frac{Pk}{2}<P$ for $3\leq k\leq P2$ and $k$ odd. That is, we have $\frac{P1}{2}$ choices $\{\frac{P+1}{2}, \frac{P+1}{2}+1, \frac{P+1}{2}+2, \ldots, \frac{P+1}{2}+\frac{P3}{2}\}$ so that: (1) all such numbers (call $N$) are less than $P$ as required; (2) obviously from (1) we get that at least one such $2N$ will be tested, that is, we need to prove that at least one of them of the form $2N=2\left(\frac{P+1}{2}+\frac{Pk}{2}\right)=2Pk+1$ is such that for fixed $k$ we have $2Pk+1p_i$ composite for each of the $i>2$ 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 $N<P$ so that $2Np_i$ is composite for all the $i$ primes. Now, suppose that for each $k$ there is at least one prime number, contrary to the conjecture. Then, we form $\frac{P1}{2}$ unique prime numbers, so below $P$ this would entail a density of prime numbers at least $$ \frac{P}{\frac{P1}{2}+i} $$ Let $i=3$, that is, the smallest possible value so that the denominator is $\frac{P5}{2}$. As $P$ grows, the value $\frac{P5}{2}$ must be asymptotically such that $\frac{P5}{2}\simeq\ln(P)$ or $\frac{P5}{2}<\ln(P)$ accordingly to the Theorem of the distribution of primes. But that is easy to see is not the case, that is, we have much more primes than expected if the conjecture is false for not so large $P$. It is just a naive argument indicating that the conjecture must be true, excluding, perhaps, some particular small cases. For large $i$ it is intuitively clear that $\frac{P1}{2}+i\simeq\frac{P}{2}$ and the same conclusion follows. 
September 17th, 2015, 03:20 AM  #9  
Math Team Joined: Apr 2010 Posts: 2,778 Thanks: 361  Quote:
 
September 17th, 2015, 05:51 PM  #10  
Senior Member Joined: Dec 2007 Posts: 687 Thanks: 47  Quote:
Anyway, this won't change the conclusion I wrote above after small modifications.  

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