2 added 420 characters in body

EDIT: this doesn't really work. I'm still a good human being.

Evenly enough, it seems possible to get the number of zeros in the binary expansion of $n!$

One can get a fairly accurate expression for $$\log_2 \; n! = \frac{\log \; n!}{\log 2}$$ from using extra terms in Stirling's formula. Taking the floor of that and adding 1 gives the total number of digits in base two..

Legendre's formula $$v_2(n!) = \left\lfloor \frac{n}{2} \right\rfloor + \left\lfloor \frac{n}{4} \right\rfloor + \cdots$$ has a companion,

$$v_p(n!) = \frac{n - S_p(n)}{p-1}$$ where $S_p(n)$ is the sum of the digits when $n$ is written in base $p.$ As all the digits in a base two expansion are $1,$ we find that $S_2(n)$ is simply the count of 1's in the base two expansion of $n.$

Alright, some people, who shall remain nameless, have attempted to cast aspersions on the reputation of your humble servant, pointing out that the number of ones in the binary expansion of $n!$ is not the same as the number of ones in the binary expansion of $n$ itself. I try so hard. Don't change the light bulb, I'll just sit here in the dark.

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Evenly enough, it seems possible to get the number of zeros in the binary expansion of $n!$

One can get a fairly accurate expression for $$\log_2 \; n! = \frac{\log \; n!}{\log 2}$$ from using extra terms in Stirling's formula. Taking the floor of that and adding 1 gives the total number of digits in base two..

Legendre's formula $$v_2(n!) = \left\lfloor \frac{n}{2} \right\rfloor + \left\lfloor \frac{n}{4} \right\rfloor + \cdots$$ has a companion,

$$v_p(n!) = \frac{n - S_p(n)}{p-1}$$ where $S_p(n)$ is the sum of the digits when $n$ is written in base $p.$ As all the digits in a base two expansion are $1,$ we find that $S_2(n)$ is simply the count of 1's in the base two expansion of $n.$