Denote by $\pi(x,a,q)$ the number of primes $p\le x$ of the form $p=qk+a$ and $E(x,a,q)=\phi(q)^{-1}\mathrm{Li}(x)-\pi(x,a,q)$. What is the strongest conjectured bound on $E(x,a,q)$ in terms of $x,q$?

  • 7
    $\begingroup$ $x/\log x$ is the wrong approximation. You need to replace it with $Li(x) = \int_2^x dt/\log t$. Then the dependence on $x$ on the error term will be $x^{1/2}\log x$ by GRH. I am sure what dependence on $q$ will be, but an expert will probably answer you. $\endgroup$ – Felipe Voloch Jan 5 '11 at 22:46
  • $\begingroup$ I meant to say "I'm not sure"... $\endgroup$ – Felipe Voloch Jan 5 '11 at 22:47
  • $\begingroup$ Related: mathoverflow.net/questions/834/… $\endgroup$ – David E Speyer Jan 5 '11 at 23:13

As Felipe notes, the main term should be li(x)/\phi(q). Replacing this in your definition of $E(x,a,q)$ above we have that, the best that is know even on the GRH is that $E(x,a,q) = O(x^{1/2}\log x)$. This estimate doesn't get better when $q$ is large (compared to $x$) and a lot more is believed to be true. For instance Montgomery conjectures that $E(x,a,q) = O_{\epsilon}( x^{1/2+\epsilon}/ q^{1/2})$ (see Conjecture 13.9 in his book on Multiplicative Number Theory).

Edit: This is really is the best one can hope for, as Friedlander and Granville have shown (using Maier's Matrix Method) that the $O_{\epsilon}(x^{1/2+\epsilon})$ cannot be replaced with a term of the form $O_{A}(x^{1/2} \log^{A}(x))$.

See: Friedlander and Granville, Limitations to the equi-distribution of primes, III, Compositio Math. 81 (1992), 19-32.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.