(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…

**Answer**

Half a suggestion for simplification: It seems a little awkward to work with the *parity* of the *floor* of $n!/11e$ when the more fundamental question is why the *fractional part* of $n!/

22e$ is always less than 1/2 or

$$\left\{\frac{n!}{22e}\right\}<\frac12.$$

Your $A_n/2$ is always an integer, so it has zero fractional part.

The fractional part of $B_n/2=P(n)/22$ is of course 22-periodic. Evaluate the polynomial $P(n)$ modulo 22 for a complete residue system modulo 22, e.g. for the integers 0 through 21.

(N.B. $P(n)$ is not strictly a polynomial — it involves a factor of $(-1)^n$ — but that does not affect its periodicity for any even period.)

$C_n/2$ is small in magnitude and its sign alternates with a period 2 which divides 22.

Although this is a small change, I feel that reworking the proof in this way may be more illuminating and remove some of the “magic” and obfuscation of the underlying ideas.

**Attribution***Source : Link , Question Author : Micah , Answer Author : Otherwise*