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arXiv · 2601.03414

On the sizes of the maximal prime powers divisors of factorials

Dan Levy
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Abstract

Let p be any prime, and $p^(\nu_p(n!))$ 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 $q^(\nu_q(n!)) < p^(\nu_p(n!))$ 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 $q^(\nu_q(n!)) < p^(\nu_p(n!))$ for all $n \ge n_0$ is given by $n_0 = (p^2+p)/2$.

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Dan Levy. 2026-01-06. On the sizes of the maximal prime powers divisors of factorials. https://doi.org/10.46298/cm.17290

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