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  • $\begingroup$ 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... $\endgroup$ Commented Dec 3, 2011 at 21:05
  • $\begingroup$ 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. $\endgroup$ Commented Dec 4, 2011 at 7:38
  • $\begingroup$ 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? $\endgroup$ Commented Dec 4, 2011 at 7:43
  • $\begingroup$ 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. $\endgroup$ Commented Dec 4, 2011 at 15:13
  • $\begingroup$ 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 . $\endgroup$ Commented Dec 4, 2011 at 15:30