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Dec 14, 2015 at 12:41 comment added Alexey Ustinov @martin Thank you for calculations, probably $O(\sqrt p\log p)$ is a right order for error term.
Dec 14, 2015 at 12:34 comment added martin @AlexeyUstinov certainly $O(\sqrt{p}\log p)$ looks feasible: plot 1, plot 2 where first plot is $\left(\left(p_n/2\right)^3-\sum\limits_{k \in A_{p_n}}{k}\right)/\left(p_n^{5/2}\log p_n\right),$ second is $ \text{expr.} ~/\left(p_n ^{5/2}\log^{2} p_n\right).$
Dec 12, 2015 at 4:49 comment added Alexey Ustinov But your error term can be valid. In the question mathoverflow.net/questions/175822/incomplete-kloosterman-sum the error term was $O(\sqrt{p}\log p)$ indeed. In the case $f(x,y)=xy$ situation can be similar.
Dec 10, 2015 at 22:09 history edited Jan-Christoph Schlage-Puchta CC BY-SA 3.0
Corrected the error term
Dec 10, 2015 at 22:08 comment added Jan-Christoph Schlage-Puchta You are right, I was thinking in one dimension. I corrected the error term to $\mathcal{O}(\sqrt{p}\log^2 p)$.
Dec 10, 2015 at 2:20 comment added Alexey Ustinov How can you get $O(\sqrt{p}\log p)$? I can prove asymptotic formula only with $O(\sqrt{p}\log^2 p)$.
Dec 7, 2015 at 12:13 vote accept martin
Dec 7, 2015 at 12:06 history answered Jan-Christoph Schlage-Puchta CC BY-SA 3.0