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.

  • $\begingroup$ Well, Stirling's formula says that we could take $r = 1/e$ and $D_n = n^n \sqrt{2\pi n}$. $\endgroup$ – Antonio Vargas Feb 21 '14 at 14:54
  • $\begingroup$ Another example for the case in your post is $$(2n)! \approx \frac{1}{2} \left(\frac{\pi}{2}\right)^{2n+1} |E_{2n}|,$$ where $E_{n}$ are the Euler numbers. $\endgroup$ – Antonio Vargas Feb 21 '14 at 15:01

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