On the sizes of the maximal prime powers divisors of factorials

Summary · qwen2.5:32b

The article proves that for any prime $p$, there exists a threshold integer $n_0$ such that the maximal power of any larger prime $q > p$ dividing $n!$ is less than that of $p$ for all $n \ge n_0$, establishing a dominance order among prime powers in factorials. For twin primes $p$ and $q = p + 2$, the minimal such threshold $n_0$ is specifically $(p^2+p)/2$.

Excerpt

arXiv:2601.03414v5 Announce Type: replace 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 $n_0 = (p^2+p)/2$.