(A follow-up of sorts to this question.)

The quantity $\left\lfloor \frac{n!}{11e}\right\rfloor$ is always even, which can be proved as follows.

Using the sum for $\frac{1}{e}$, we split the fraction up into three parts:

- $A_n=\sum_{k=0}^{n-11} (-1)^k\frac{n!}{11k!}$ is a multiple of the even integer $10! \binom{n}{11}$, and so can be ignored.
- $B_n=\sum_{k=n-10}^n (-1)^k\frac{n!}{11k!}=\frac{1}{11}\sum_{k=0}^{10} (-1)^{n-k}(n)_k$. This is a finite sum of falling factorials, all of which are polynomial in $n$ with integer coefficients. So $B_n$ is of the form $\frac{P(n)}{11}$ where $P(n)$ is a polynomial in $n$ with integer coefficients.
- $C_n = \sum_{k=n+1}^{\infty} (-1)^k\frac{n!}{11k!}$ is an alternating series whose terms decrease monotonically in absolute value, and so $|C_n|<\frac{n!}{11(n+1)!}<\frac{1}{11}$.

Putting all this together, we can see that:

- Since $B_n$ is always an integer multiple of $\frac{1}{11}$ and $|C_n|<\frac{1}{11}$, $C_n$ can only affect the value of $\left\lfloor \frac{n!}{11e}\right\rfloor$ when $B_n$ is an integer. In this case it will change the parity when $C_n$ is negative (i.e., $n$ is even) and leave it alone when $C_n$ is positive.
- Since $P(n)=11B_n$ is a polynomial with integer coefficients, $B_n$'s integer status is $11$-periodic, which means that whether $C_n$ affects the parity of $\left\lfloor \frac{n!}{11e}\right\rfloor$ is $22$-periodic.
- Similarly, the parity of $\lfloor B_n \rfloor$ is also $22$-periodic.

So the parity of $\left\lfloor \frac{n!}{11e}\right\rfloor$ is $22$-periodic. Moreover, we can compute its first $22$ values to be: $$ 0, 0, 0, 0, 0, 4, 24, 168, 1348, 12136, 121360, 1334960, 16019530, 208253902, 2915554640, 43733319612, 699733113794, 11895462934514, 214118332821268, 4068248323604100, 81364966472082010, 1708664295913722230 $$ (a sequence which does not appear to be in OEIS).

All of these are even, and so $\left\lfloor \frac{n!}{11e}\right\rfloor$ must be even for all $n$.

This is not a very satisfying proof, though; in the end, it looks like we need a random $2^{22}$-fold coincidence to go our way in order to get the result we want. (In fact, that's the only place we used the specific value of $11$ in our proof at all; the rest of the proof shows that the parity of $\left\lfloor \frac{n!}{ke}\right\rfloor$ is $2k$-periodic for all positive integers $k$.) Even though $11$ was chosen arbitrarily, it looks like the heuristic probability of everything lining up falls off rapidly enough that it's surprising it all works out for any $k$ which is even that large.

Can someone convince me that this fact is less surprising than it looks? I would take either a completely different proof that established the result with less case analysis, or a reason why the parity of these numbers can be expected to be non-independent...