On the sizes of the maximal prime powers divisors of factorials
Dan Levy
TL;DR
The paper investigates the relative sizes of the maximal prime power divisors of factorials by studying the exponents $\nu_p(n!)$ in the factorization of $n!$. It shows there exists a threshold $n_0(p)$, depending only on the fixed prime $p$, such that for every $n \ge n_0(p)$ and every prime $q>p$, $q^{\nu_q(n!)} < p^{\nu_p(n!)}$, using Legendre's formula together with the alternative form $\nu_p(n!) = \frac{n - s_p(n)}{p-1}$ and a monotone function $h_p(x)$ derived from base-$p$ digit sums. For the special case of twin primes ($q=p+2$), the paper determines the exact minimal threshold $n_0(p) = \frac{p^2 + p}{2}$ by analyzing the ratio $r(n,p) = \frac{n - s_p(n)}{n - s_{p+2}(n)}$ and locating its global minimum, ensuring the required dominance holds for all $n \ge n_0(p)$. These results tie into questions in finite group theory (e.g., the orders of Sylow subgroups of symmetric groups) and illustrate how digit-sum techniques complemented by elementary inequalities can yield precise asymptotic dominance conclusions. The work blends analytic bounds, combinatorial digit-sum arguments, and computational checks to establish both general and twin-prime-specific thresholds.
Abstract
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 $n_0 = (p^2+p)/2$.