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January 1st, 2017, 06:09 PM  #11  
Senior Member Joined: May 2016 From: USA Posts: 591 Thanks: 251  Quote:
Perhaps I am incorrect that all you intended by your first five lines is "All primes > 3 can be expressed in the form $6n \pm 1$" and "Consider the sequence of all primes > 3 in ascending order." If I am incorrect, what did you mean? If I am correct, however, then there is no question that 2117 and 2119 are not in that sequence because they are not prime. So skipjack's example is not on point. But who is at fault for making obscure what could be clear? You are. And as skipjack says, your saying that something is clear does not make it so. It is the writer's obligation to present things as clearly as possible. You have failed to do so as demonstrated by the responses you have received.  
January 1st, 2017, 10:33 PM  #12  
Senior Member Joined: Sep 2016 From: USA Posts: 114 Thanks: 44 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
On the other hand, the number of multiples of 5 less than $N$ is approximately $\frac{N}{5}$ which you may notice as a ratio of $N$ is bounded away from 1. In fact, the density of multiples of 5 to nonmultiples of 5 is 0 which you might recognize contradicts your assertion that it is "clear" that the multiples of 5 is greater than nonmultiples of 5. I hope it is clear why this statement is false given the flaw in the previous statement.  
January 2nd, 2017, 05:35 AM  #13  
Global Moderator Joined: Dec 2006 Posts: 16,919 Thanks: 1253  Quote:
Twin primes greater than (3, 5) must have the form (6n  1, 6n + 1). From all such number pairs, MrAwojobi considers sieving out those which aren't prime pairs by removing (composite) multiples of successive odd primes. As there are infinitely many primes, there are infinitely many pairs of the form (6n  1, 6n + 1) for which at least one of these numbers is prime. It seems highly plausible that the other number will sometimes be prime as well, even if one uses only very large values of n, but this is currently unproved. The sieving method is a very simple way of finding primes, and can in principle be used to find all primes. Using this method to find twin primes will in principle find all such pairs, but examining this process doesn't seem to provide any proof that there are infinitely many of them. The attempted proof by MrAwojobi has flaws, as has been pointed out.  
January 2nd, 2017, 06:56 AM  #14 
Senior Member Joined: Aug 2010 Posts: 158 Thanks: 4 
I have noticed an error concerning a claim I made in the proof. I will reword that part of the proof and post an update.

January 4th, 2017, 03:31 AM  #15 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,225 Thanks: 186 
I think 2117 and 2119 are not in the final sequence because they have already been eliminated by 29 and 13 sieving. The way I understand his post is ... he list's the sequence of positive integers of the form $6n \pm 1$ ... then he applies the sieve (like Eratosthenes) on this list ... then he examines the numbers that remain and draws his conclusions on those remaining numbers. For example 35 gets eliminated by 5 so is NOT there to be eliminated by 7 and since 35 is gone by the time we get around to eliminating multiples of 7 , it does not add to the count of numbers eliminated by 7. So his claim ... "it should be clear that the number of multiples of 5 eliminated in the number sequence > the number of nonmultiples of 5 eliminated for any N." seems true to me However , I don't think the final claim of his post follows logically IMHO 
January 4th, 2017, 07:01 AM  #16  
Senior Member Joined: May 2016 From: USA Posts: 591 Thanks: 251  Quote:
Let's try to figure out what is being asserted. First note that N and n are different variables: n is any natural number whereas N is a natural number such that $N = 6k \pm 1,\ k \in \mathbb N^+.$ Anyway, the assertion is (assuming that I am reading this prolix stuff correctly) that in the finite sequence of numbers of the form $6n\pm 1$ from 5 to N, the number that are evenly divisible by 5 exceeds the number that are not evenly divisible by 5 but are evenly divisible by EVERY other odd prime < N. I suspect it is fairly easy to prove formally that the number of such numbers divisible by 5 exceeds the number of such numbers evenly divisible by 7 but not 5 (though the OP has not bothered to do so). It does not, however, obviously follow from that that the number of numbers in the form of $6n \pm 1$ evenly divisible by 5 exceeds the number of such numbers evenly divisible by 7 but by no lower prime plus the number of such numbers evenly divisible by 11 but by no lower prime plus the number of such numbers evenly divisible by 13 but by no lower prime, etc. However intuitive it may or may not be (and it is not intuitive to me), an appeal to intuition is not a proof. Last edited by JeffM1; January 4th, 2017 at 07:03 AM.  

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