We have the relation of the Bernoulli numbers
$$B_{2n} = (-1)^{n+1}\frac {2(2n)!} {(2\pi)^{2n}} \left(1+\frac{1}{2^{2n}}+\frac{1}{3^{2n}}+\cdots\;\right).$$
For $n>1$, the right hand sum
$$\sum_{m=2}^\infty\frac{1}{m^{2n}}=\zeta(2n)-1\approx \varepsilon$$
is de facto zero and so I think of the Bernoulli numbers as translating the double factorial to the exponential function
$$(2n)!\approx \tfrac{1}{2}\left(4\pi^2\right)^n\cdot \left|B_{2n}\right|$$
In a way, they quantify the multiplicative difference between these natural operations.
Are there relevant combinatorially transparent numbers $D_n$ such that
$$n!\approx r^n\cdot \left|D_n\right|$$
for some $r$?
Thoughts: I depending on the signs, I figure the series $\sum_{n=0}^\infty\frac{1}{n!}D_n z^n$ is non-convergent.