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Let $\chi$ be a quadratic character mod $q$. I am interested in finding the best result for how large $N$ should be such that it is guaranteed that

$$\sum_{p=1}^{N} \chi(p) \log p= o(N).$$

I am aware of Heath-Brown's unpublished note, which, assuming the Burgess bound is optimal, proves that: \begin{align*} & \chi(p)=-1 & \text{for} && q^{1/4\sqrt{e}}< p< q^{1/4}, \\ & \chi(p)=1 & \text{for} && q^{1/4}< p< q^{1/2\sqrt{e}}. \end{align*}

But it is not clear to me how large the character sum over primes should be to guarantee cancellation. We may assume there are no Siegel zeros.

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  • $\begingroup$ I think the precise form of the "no Siegel zeroes" assumption could matter a lot here. $\endgroup$
    – Will Sawin
    Commented Sep 30 at 14:56
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    $\begingroup$ If 'no Siegel zero' means 'no exceptional zero', you have cancellation when $q=N^{o(1)}$. Indeed, this is addressed by Theorem 11.16 of Montgomery-Vaughan and Exercise 11.3.1.1 in page 382. $\endgroup$
    – Ofir Gorodetsky
    Commented Sep 30 at 15:01
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    $\begingroup$ Both Theorem 11.16 and Exercise 11.3.1.1, in the case of no exceptional zero, boil down to equation (11.26). On GRH you can of course take $N$ to be much smaller. $\endgroup$
    – Ofir Gorodetsky
    Commented Sep 30 at 15:31
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    $\begingroup$ Is $\sum_{p = 1}^N$ meant to be a sum only over primes? $\endgroup$
    – LSpice
    Commented Sep 30 at 20:12
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    $\begingroup$ Re, then does $\sum_{p = 1}^N$ mean the sum over the primes in $[1, N]$, or over the first $N$ primes? $\endgroup$
    – LSpice
    Commented Oct 1 at 3:39

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