If I understand correctly, won't one of the two numbers be even? And hence not prime?

Actually, since G is odd, G+1 is even :) However, G+2 will not be divisible by any of the smaller primes, so it is sufficient to shown that there are infinitely many primes without needing twins (it is a variation on Euclid's argument)

Hi Peter Breuer Remember that both G-1 and G+1 will be even, as G is odd, so actually neither is prime :) The value of B has been brought down to 600 during a polymath project - see the paper 'Small gaps between primes' by James Maynard :)

Peter Breuer - I defined p1 as 3 for the purpose of my observation. G is every number in the odd number line that is divisible by 3 by my definition of G.

@James Tuite - you did not read correctly. Following the sieve of Erastothenes, after division by 2 we are left with the ODD numberline. Now in the odd numberline 3 5 7 9 11 13 15 17 19 21 ... every 3rd number is divisible by 3, so the two numbers in between are odd, but not divisible by 3. I don't know how to add a separate reply to tagg you

James Tuite - just found a way to tag you so allow me to repeat. You misunderstood my definitions:

By the sieve of Eratosthenes, deleting every number that is divisible by 2, we are left with the ODD NUMBERLINE 3 5 7 9 11 13 15 ... of which every 3rd number is divisible by 3. Per my definition, this is G, and T1 and T2 the odd numbers in between each G. I call them twin numbers, because they could be twin prime or not.

I notice that by the sieve of Eratosthenes, a prime number p_n deletes numbers in our odd numberline p_n places apart.

G is a multiple of 3. Now, if we form the product of all prime numbers, starting with 3, up to p_n, we are landing on a number G=3*5*...*p_n which is divisible by 3 and by all of the prime numbers in our product.

But hence, none of the T1 T2 directly below and above this number G can be cancelled by any of the prime numbers up to p_n but only by larger ones.

But hence, if p_n were the largest prime number, this would make T1 and T2 directly above and below our number G=3*5*...*p_n twin prime. Which is a contradiction for which there cannot be a largest prime number.

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As you can imagine this emerged from contemplations about infinite twin primes.

Considering the numbers between p_n and my T1 T2 pairs, if twin primes are finite, from one largest twin prime onwards, this gap must always contain prime numbers so as to cancel at least one member of each of my T1 T2 pairs. If this is not ALWAYS possible, then our largest twin prime is no largest twin prime, which is a contradiction.

I think we all feel that by the peculiar distribution of primes through the sieve of Eratosthenes this has a nonzero probability ... but showing this we need a better picture. I have noted down several more different observations, and notes as to what must be the case for the above but that's all.

Yes, I think I have misunderstood. One thing that confuses me is that you have every third term in your sequence denoted by G, but you define G to be the product of the first n primes apart from 2? So you are looking at the next two odd numbers after that?

One other thing that I am not sure about is whether your focus here is proving that there are infinitely many primes (and your argument does this, although it is a bit harder than Euclid's argument, and you don't really need twins to do this, just one will do), or whether you are looking for a source of twin primes. If the latter, I think this will be very hit and miss. We could compare this with the product of the first n primes, including 2, plus or minus one. Numbers of the first time that happen to be prime are called primorial primes, but they are relatively rare, and it seems to be very unusual that both of these numbers are prime.

James Tuite I labelled every number in the odd numberline that is divisible by 3 with G to paint my mental picture. Then I define a product G=3*5*...*p_n. That could be any G.

Being familiar with existing proofs, my aim was to present a different proof for the infinitude of primes. Of course one does not need twin primes to prove the infinitude of primes as of that one does not need MY PROOF for it, but my proof is different (primes to be finite contradicts itself - that's funny) and my observation curious.

And thanks for the chat and input about primordial primes - so it has not been proven that they are infinite either ! That again is a different beast since the product includes the number 2 as you say ... but it gives me more to think about.

James and Peter - thank you again for your input. I formulated it way shorter and clearer.

Hi Hanna! Thanks, that is a lot more compact :) I hadn't appreciated that a Euclid-like proof would actually produce lots of new 'problematic' numbers - nice :)

James Tuite - thank you for your time. I just read another version of Euclid's proof, and my proof is in fact more similar than I thought.

Because one could say that his product 2*3*5*7*...*p_n + 1 has to be prime because it is not divisible by any prime if p_n is the largest and all are listed - which is a contradiction.

So in his case, a last prime demands one specific other prime (unless it wouldn't be the last prime). In my case a last prime demands an entire set of other primes taking the sieve of eratosthenes outwards from G ... so really an infinite amount of other primes. Will try to add this - that a last prime demands an infinite amount of other primes.

I thought Euclid had argued that either his product was prime, or if it wasn't prime, it would contain at least one prime factor "not on the list". Which I thought was considered a direct proof.