Note on a sum involving the divisor function
Fuente:
arXiv
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| Format: | Preprint |
| Publié: |
2025
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| _version_ | 1866908347224031232 |
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| author | Wu, Liuying |
| author_facet | Wu, Liuying |
| contents | Let $d(n)$ be the divisor function and denote by $[t]$ the integral part of the real number $t$. In this paper, we prove that $$\sum_{n\leq x^{1/c}}d\left(\left[\frac{x}{n^c}\right]\right)=d_cx^{1/c}+\mathcal{O}_{\varepsilon,c} \left(x^{\max\{(2c+2)/(2c^2+5c+2),5/(5c+6)\}+\varepsilon}\right),$$ where $d_c=\sum_{k\geq1}d(k)\left(\frac{1}{k^{1/c}}-\frac{1}{(k+1)^{1/c}}\right)$ is a constant. This result constitutes an improvement upon that of Feng. |
| format | Preprint |
| id |
arxiv_https___arxiv_org_abs_2505_01645 |
| institution | arXiv |
| publishDate | 2025 |
| record_format | arxiv |
| spellingShingle | Note on a sum involving the divisor function Wu, Liuying Number Theory Let $d(n)$ be the divisor function and denote by $[t]$ the integral part of the real number $t$. In this paper, we prove that $$\sum_{n\leq x^{1/c}}d\left(\left[\frac{x}{n^c}\right]\right)=d_cx^{1/c}+\mathcal{O}_{\varepsilon,c} \left(x^{\max\{(2c+2)/(2c^2+5c+2),5/(5c+6)\}+\varepsilon}\right),$$ where $d_c=\sum_{k\geq1}d(k)\left(\frac{1}{k^{1/c}}-\frac{1}{(k+1)^{1/c}}\right)$ is a constant. This result constitutes an improvement upon that of Feng. |
| title | Note on a sum involving the divisor function |
| topic | Number Theory |
| url | https://arxiv.org/abs/2505.01645 |