Let $\chi$ be be 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I am aware of Heath-Brown's unpublished note, which, assuming the Burgess bound is optimal, proves that:$$\chi(p)=-1 \hspace{5 mm} \text{ for } \hspace{5 mm} q^{1/4\sqrt{e}}< p< q^{1/4},$$ $$\chi(p)=1 \hspace{6 mm} \text{ for } \hspace{5 mm} q^{1/4}< p< q^{1/2\sqrt{e}}.$$

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

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I am aware of Heath-Brown's unpublished note, which, assuming the Burgess bound is optimal, proves that: \begin{align*} & \chi(p)=-1 & \text{for} && q^{1/4\sqrt{e}}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

Let $\chi$ be be quadrtic charachetr 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I am aware of Heath-Brown's unpublished note, which, assuming the Burgess bound is optimal, proves that:$$\chi(p)=-1 \hspace{5 mm} \text{ for } \hspace{5 mm} q^{1/4\sqrt{e}}< p< q^{1/4},$$ $$\chi(p)=1 \hspace{6 mm} \text{ for } \hspace{5 mm} q^{1/4}< p< q^{1/2\sqrt{e}}.$$

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

Let $\chi$ be be 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I am aware of Heath-Brown's unpublished note, which, assuming the Burgess bound is optimal, proves that:$$\chi(p)=-1 \hspace{5 mm} \text{ for } \hspace{5 mm} q^{1/4\sqrt{e}}< p< q^{1/4},$$ $$\chi(p)=1 \hspace{6 mm} \text{ for } \hspace{5 mm} q^{1/4}< p< q^{1/2\sqrt{e}}.$$

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