$\nu_3(n!)$ and $\nu_4(n!)$ ($\nu$ being the valuation function) seems to have the same value most of the time (on "small numbers"). Here is a table with a few values:
$\begin{array} \textbf{n} &1&2&3&4&5&6&7&8&9&10&100&1000&10000&.... \\ \hline \nu_3(n!) &0&0&1&1&1&2&2&2&4&4&48&498&4996&... \\ \nu_4(n!) &0&0&0&1&1&2&2&3&3&4&48&497&4997&... \\ \end{array}$
Is it still the case when $n \rightarrow \infty $ ? Is there a bound on the difference ? How often are they equal ?
From a density perspective, they seem to have almost the same coverage:
$\nu_3(n!)=\sum\limits_{k\geq 1} \lfloor \frac{n}{3^k} \rfloor$
$\nu_4(n!)=\lfloor \frac{n+2}{8} \rfloor+\sum\limits_{k\geq 1} \lfloor \frac{n}{4^k} \rfloor$
$\sum\limits_{k\geq 1}\frac{1}{3^k}=\frac{1}{2}$
$\frac{1}{8}+\sum\limits_{k\geq 1}\frac{1}{4^k}=\frac{1}{8}+\frac{1}{3}=\frac{11}{24}\simeq\frac{1}{2}$
It feels like I am missing an obvious link, but the more I dig the less find it obvious...
EDIT: bounds added (Thx Trebor)
$\frac{1}{3}-\frac{1}{n}\leq \frac{\nu_3(n!)}{n}\leq \frac{1}{3-1}$
$\frac{1}{4}-\frac{1}{n}+\frac{1}{8}-\frac{3}{4n}\leq \frac{\nu_4(n!)}{n}\leq \frac{1}{4-1}+\frac{1}{8}+\frac{1}{4n}$
and for $n \rightarrow \infty $
$.333...\leq \frac{\nu_3(n!)}{n}\leq .5$
$.375\leq \frac{\nu_4(n!)}{n}\leq .458333...$
EDIT2: Thanks to the comments
$\nu_4(n!)\simeq\sum\limits_{k\geq 1} \lfloor \frac{n}{2\cdot4^k}+\frac{1}{4} \rfloor+\sum\limits_{k\geq 1} \lfloor \frac{n}{4^k} \rfloor$
So the density is indeed $\frac{1}{6}+\frac{1}{3}=\frac{1}{2}$