Is there a "good" reason why $\left\lfloor \frac{n!}{11e}\right\rfloor$ is always even?

Solution 1:

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.