Is $\lfloor n!/e\rfloor$ always even for $n\in\mathbb N$? 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 $n\in\mathbb N$?
 A: Note that:
$$e^{-1}=\sum_{k=0}^\infty \frac{(-1)^k}{k!}$$
Then:
$$\frac{n!}e=n!e^{-1} = \left(\sum_{k=0}^{n} (-1)^k\frac{n!}{k!}\right) + \sum_{k=n+1}^{\infty} (-1)^{k}\frac{n!}{k!}$$
Show that if $a_n=\sum_{k=n+1}^{\infty} (-1)^{k}\frac{n!}{k!}$ then $0<|a_{n}|<1$ and $a_n>0$ if and only if $n$ is odd.
So the when $n$ is odd, the value is:
$$\left\lfloor\frac{n!}{e}\right\rfloor=\sum_{k=0}^{n} (-1)^k\frac{n!}{k!}\tag{1}$$
When $n$ is even it is one less:
$$\left\lfloor\frac{n!}{e}\right\rfloor=-1+\sum_{k=0}^{n} (-1)^k\frac{n!}{k!}\tag{2}$$
Now, almost all of these terms are even. The last term $n!/n!=1$ is odd. When $n$ is odd, the second-to-last term $n!/(n-1)!$ is odd, also. But all other terms are even.
So for $n$ odd, there are two odd terms in the sum, $k=n,n-1$.
For $n$ even, there are two odd terms in the sum, $-1$ and $k=n.$

The trick, then, is to show that the $a_n$ has these properties:
$$\begin{align}
&0<|a_n|<1\\
&a_n>0\iff  n\text{ is odd}
\end{align}$$
To show these, we note that  $\frac{n!}{k!}$ is strictly decreasing for $k>n$ and $(-1)^k\frac{n!}{k!}$ is alternating. In general, any alternating sum of a decreasing series converges to a value strictly between $0$ and the first term of the sequence, which in this case is $\frac{(-1)^{n+1}}{n+1}.$
A: The number of derangements of $[n]=\{1,\ldots,n\}$ is 
$$d_n=n!\sum_{k=0}^n\frac{(-1)^k}{k!}\;,$$
so
$$\frac{n!}e-d_n=n!\sum_{k>n}\frac{(-1)^k}{k!}\;,$$
which is less than $\frac1{n+1}$ in absolute value. Thus for $n\ge 1$, $d_n$ is the integer nearest $\frac{n!}e$, and
$$d_n=\begin{cases}
\left\lfloor\frac{n!}e\right\rfloor,&\text{if }n\text{ is odd}\\\\
\left\lceil\frac{n!}e\right\rceil,&\text{if }n\text{ is even}\;.
\end{cases}$$
The recurrence $d_n=nd_{n-1}+(-1)^n$ is also well-known. We have $d_0=1$, so an easy induction shows that $d_n$ is odd when $n$ is even, and even when $n$ is odd. Thus, for odd $n$ we have $\left\lfloor\frac{n!}e\right\rfloor=d_n$ is even, and for even $n$ we have $\left\lfloor\frac{n!}e\right\rfloor=\left\lceil\frac{n!}e\right\rceil-1$ is again even.
A: 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.)
