Timeline for Prime races in two competing arithmetic progressions - error bound
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
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| Jan 22, 2023 at 2:51 | comment | added | Vincent Granville | Thank you Ofir. If I you use $O((n \log n)^{r+\epsilon})$ for the bound, it seems you can prove this weak version of GRH, if my reasoning is correct: no root for $L(s,\chi_4)$ if $\Re(s)>r$. I want to avoid GRH. I assume getting the strength $r=\frac{1}{2}$ without GRH is hopeless, so looking for something weaker. Maybe $r=3/4$, not using GRH. | |
| Jan 22, 2023 at 1:39 | comment | added | Ofir Gorodetsky | PNT in arithmetic progressions shows $p_{i,n} \sim 2n\log n$ and with a bit of work, $(*)$ (or rather GRH in the form $\pi(x;4,i) = \mathrm{Li}(x)/2 + O(\sqrt{x} \log x)$) implies that $p_{1,n}-p_{3,n} = O(\sqrt{n}(\log n)^{5/2})$, if I didn't mess the power of log. Using conjectural bounds for $\pi(x;4,i)-\mathrm{Li}(x)/2$ one can do a bit better. The conjecture from the previous comment is related to probabilistic models for $(\sum_{\rho} x^{\rho}/\rho)/\sqrt{x}$, whose maximum one wants to understand. See the literature referred to in p. 484 (Monach and Montgomery). | |
| Jan 22, 2023 at 1:35 | comment | added | Ofir Gorodetsky | The usually studied quantities are $\pi(x;q,a):=\#\{p \le x: p \equiv a \bmod q\}$ so that $p_{i,n}$ ($i\in \{1,3\}$) is the smallest number with $\pi(p_{i,n};4,i)=n$. Under GRH, $(*)\, \pi(x;4,1)-\pi(x;4,3)= O(\sqrt{x}\log x)$ (see Corollary 13.8 in Montgomery and Vaughan's book). It is known that this difference is $\Omega(\sqrt{x}(\log x)^{-1}\log \log \log x)$ (p. 481 of said book) and it is conjectured that the difference is $O(\sqrt{x}(\log x)^{-1}(\log \log \log x)^2)$ and that this is best possible (see p. 484 of said book). All of this follows from the explicit formula. | |
| Jan 22, 2023 at 0:30 | history | asked | Vincent Granville | CC BY-SA 4.0 |