On the sizes of the maximal prime powers divisors of factorials
Fuente:
arXiv
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| Natura: | Preprint |
| Pubblicazione: |
2026
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| _version_ | 1866913061119459328 |
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| author | Levy, Dan |
| author_facet | Levy, Dan |
| contents | Let p be any prime, and $p^(ν_p(n!))$ the maximal power of $p$ dividing $n!$. It is proved that there exists a positive integer $n_0$, which depends only on $p$, such that $q^(ν_q(n!)) < p^(ν_p(n!))$ for all $n \ge n_0$ and all primes $q > p$. For twin primes $p$ and $q = p + 2$ it is proved that the minimal $n_0$ satisfying $q^(ν_q(n!)) < p^(ν_p(n!))$ for all $n \ge n_0$ is given by $n_0 = (p^2+p)/2$. |
| format | Preprint |
| id |
arxiv_https___arxiv_org_abs_2601_03414 |
| institution | arXiv |
| publishDate | 2026 |
| record_format | arxiv |
| spellingShingle | On the sizes of the maximal prime powers divisors of factorials Levy, Dan Number Theory 05A20, 11A51, 11B65 Let p be any prime, and $p^(ν_p(n!))$ the maximal power of $p$ dividing $n!$. It is proved that there exists a positive integer $n_0$, which depends only on $p$, such that $q^(ν_q(n!)) < p^(ν_p(n!))$ for all $n \ge n_0$ and all primes $q > p$. For twin primes $p$ and $q = p + 2$ it is proved that the minimal $n_0$ satisfying $q^(ν_q(n!)) < p^(ν_p(n!))$ for all $n \ge n_0$ is given by $n_0 = (p^2+p)/2$. |
| title | On the sizes of the maximal prime powers divisors of factorials |
| topic | Number Theory 05A20, 11A51, 11B65 |
| url | https://arxiv.org/abs/2601.03414 |