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On a sum involving the divisor function
Periodica Mathematica Hungarica ( IF 0.8 ) Pub Date : 2021-02-11 , DOI: 10.1007/s10998-020-00378-3 Jing Ma , Huayan Sun
中文翻译:
涉及除数函数的总和
更新日期:2021-02-11
Periodica Mathematica Hungarica ( IF 0.8 ) Pub Date : 2021-02-11 , DOI: 10.1007/s10998-020-00378-3 Jing Ma , Huayan Sun
Let d(n) be the divisor function and denote by [t] the integral part of the real number t. In this short note, we prove that \(\sum _{n\le x} \ d\Big (\Big [\frac{x}{n}\Big ]\Big ) = x\sum _{m\ge 1}\frac{\ d(m)}{m(m+1)} + O_{\varepsilon }\big (x^{11/23+\varepsilon }\big )\) for \(x\rightarrow \infty \).
中文翻译:
涉及除数函数的总和
令d(n)为除数函数,并用[ t ]表示实数t的整数部分。在此简短说明中,我们证明\(\ sum _ {n \ le x} \ d \ Big(\ Big [\ frac {x} {n} \ Big] \ Big)= x \ sum _ {m \ ge 1} \ frac {\ d(m)} {m(m + 1)} + O _ {\ varepsilon} \ big(x ^ {11/23 + \ varepsilon} \ big)\)表示\(x \ rightarrow \冒犯\)。