摘要
令ρ3(n)=∑n=|m|3-|l|3,(m,l)=1 1.本文研究了和式R3(x)=∑n≤xρ3(n)= A3x2/3+B3x1/2+E3(x),并且在黎曼假设下,得到E3(x)=0(x4/15+ε),从而进一步改进了前人的结果.
Abstract
Letρ3(n)=∑n=|m|3-|l|3,(m,l)=1 1.In this paper we study the sum R3(x)=∑n≤xρ3(n)=A3x2/3+B3x1/2+E3(x), and we shall prove that if the Riemman Hypothesis (RH) is true, then E3(x)=O(x4/15+ε), which improves the previous result.
关键词
指数和 /
渐近公式 /
黎曼猜想
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张德瑜;翟文广;.
两个互素的立方数之差. 数学学报, 2006, 49(5): 1181-118 https://doi.org/10.12386/A2006sxxb0146
De Yu ZHANG; Wen Guang ZHAI.
On Differences of Two Relative Prime Cubes. Acta Mathematica Sinica, Chinese Series, 2006, 49(5): 1181-118 https://doi.org/10.12386/A2006sxxb0146
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脚注
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