TITLE: Is $\lfloor n!/e\rfloor$ always even for $n\in\mathbb N$?
QUESTION [63 upvotes]: I checked several thousand natural numbers and observed that $\lfloor n!/e\rfloor$ seems to always be an even number. Is it indeed true for all $n\in\mathbb N$? How can we prove it?
Are there any positive irrational numbers $a\ne e$ such that $\lfloor n!/a\rfloor$ is even for all $n\in\mathbb N$?

REPLY [12 votes]: Following @Vladimir's comment, I can show that $a=3e$ has this property. I don't find the proof very enlightening, though...
We have
$$
\frac{n!}{3e} = \sum_{k=0}^n \frac{1}{3}\frac{n!}{k!}(-1)^k + E
$$
where $E$ is an error term that is less than $1$ in absolute value and also small by comparison with the other terms — so it won't affect the parity of the floor except in the case where it's negative and the initial sum is an integer.
In that initial sum, all but the last three individual terms will be even integers, as they will be multiples of $\frac{n(n-1)(n-2)}{3}=2\binom{n}{3}$. So we can neglect them.
The last three terms will take the form
$$
(-1)^n \frac{1-n+n(n-1)}{3}=(-1)^n \frac{(n-1)^2}{3}
$$
What this all boils down to is that we need $\left\lfloor (-1)^n \frac{(n-1)^2}{3}\right \rfloor$ to be even except when $\frac{(n-1)^2}{3}$ is actually an integer and also $n$ is even (which is the case where the error term is negative): that is, we want to check that $\left\lfloor (-1)^n \frac{(n-1)^2}{3}\right \rfloor$ is odd when $n \equiv 4 \pmod{6}$ and even otherwise. Which... it is, but it's not clear what the unifying principle is here, if any. And doing this for $11e$ seems possible, but horrifying. (One easier approach to proving this for $a=11e$ would be to just notice that, by this kind of argument, we need only consider the congruence class of $n$ modulo $22$, and then explicitly checking that $\lfloor 0!/(11e)\rfloor, \lfloor 1!/(11e)\rfloor,\dots,\lfloor 21!/(11e)\rfloor$ are all even. But that's if anything even less enlightening.)