Abstract
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 \(x\rightarrow \infty \).
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Acknowledgements
This work is supported in part by the National Natural Science Foundation of China (Grant Nos. 11771252 and 11771176).
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Ma, J., Sun, H. On a sum involving the divisor function. Period Math Hung 83, 185–191 (2021). https://doi.org/10.1007/s10998-020-00378-3
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DOI: https://doi.org/10.1007/s10998-020-00378-3