Timeline for Generalising Dirichlet's theorem in arithmetic progressions-prime combinatorics
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 6, 2011 at 16:22 | vote | accept | CommunityBot | ||
| Dec 6, 2011 at 16:22 | history | bounty ended | Asterios Gkantzounis | ||
| Dec 5, 2011 at 20:28 | comment | added | Asterios Gkantzounis | I give an answer with the hint for the proof | |
| Dec 5, 2011 at 5:07 | comment | added | Aaron Meyerowitz | It is up to you. If, as you say, It is not hard to see that if there is not such a choice for any M Dirichlet's theorem in arithmetic progressions is an easy consequence. Then a hint in a comment should be enough to let the readers here see how easy it is. Maybe it is pretty easy and I just don't see it. | |
| Dec 4, 2011 at 19:23 | comment | added | Asterios Gkantzounis | Do you think that i should write an answer and write down them? | |
| Dec 4, 2011 at 19:22 | comment | added | Asterios Gkantzounis | I know that (C) implies (D).I have a very simple proof for that and for all the other :" to the previous question, if we take 2 arithmetic progressions for each pi>M, instead of 1 with the same condition then we have something stronger than the twin prime conjecture and polignac's conjecture in general, If we take 3 arithmetic progressions we have something stronger than the problem that infinitely many times exists every logical form with 3 primes ( for example p, p+2,p+6), etc. To these problems it is enough to take infinitely many M, not all..." | |
| Dec 4, 2011 at 16:59 | comment | added | Aaron Meyerowitz | Let (C) denote your CONJECTURE and (D) Dirichlets theorem. I may well be missing something, but I said : "I don't see why (something slightly stronger than) (C) implies (D), are you saying that a counter example to (D) provides a counter example to (ssst) (C)?" Your reply is " of course we could have a counter-example but I think that (C) implies (D). In your question you said MOTIVATION it is not hard to see that (C) implies (D). SO do you think or know that (C) implies (D). Are you saying that we could maybe or we would obviously have a counterexample? Can you say why? | |
| Dec 4, 2011 at 15:30 | comment | added | Asterios Gkantzounis | I think that if my conjecture is true then it means that something smaller than h(n) ( in the meaning that you dont use every prime ) is not bigger than p^2/M for every n . | |
| Dec 4, 2011 at 15:13 | comment | added | Asterios Gkantzounis | Yes of course in this case we could have a counter example but if my first conjecture is true the Dirichlets theorem is a direct consequence. | |
| Dec 4, 2011 at 7:43 | comment | added | Aaron Meyerowitz | I have to admit that I don't see how a your first conjecture establishes Dirichlet's theorem in arithmetic progressions. Are you saying that if there was some prime p and an r with gcd(p,r)=1 yet no primes of the form pn+r (or only finitely many) then one could produce a counter example to your conjecture? | |
| Dec 4, 2011 at 7:38 | comment | added | Aaron Meyerowitz | In your question you asked is there some M such that... then you should be able to use any larger M and maybe use a larger prime M which simply excludes one large prime. Anyway, it is a weaker requirement (since I left out relatively prime) so true it would be a stronger result but the answer is still probably no AND an even weaker requirement is to just have $h(n)$ bigger than p^2/M infinitely often. | |
| Dec 3, 2011 at 21:05 | comment | added | Asterios Gkantzounis | thank you for your answer, the motivation to require p_i relatively prime to M has to do with the relation to diriclhlet's theorem and the related questions. I am sorry but i cant ubderstand why this version of my question is weaker, in my opinion it is stronger because it is more easy (maybe) to prove that the A_i's cant cover all the naturals if every k_i is greater than p_i^2/M than for all except finitely... | |
| Dec 3, 2011 at 6:09 | history | answered | Aaron Meyerowitz | CC BY-SA 3.0 |