Proof for an explicit formula for the even Euler numbers

Question

The EULER numbers $E_n$, $n \in \mathbb{N}$, are defined via the TAYLOR expansion of the hyperbolic secant:

\begin{equation} \text{sech}(x) = \sum_{n=0}^{\infty} \frac{E_n}{n!} x^n = \sum_{n=0}^{\infty} \frac{E_{2n}}{(2n)!} x^{2n} \end{equation}

(the odd ones $E_{2n+1}$ being zero due to the hyberbolic secant's being an even function). As close cousins of the celebrated BERNOULLI numbers they, too, show up everywhere in Analysis, Combinatorics and Number Theory, and as for those there are many explicit finite expressions for them. One of these is

\begin{equation} E_{2n} = \sum_{k=1}^n \frac{(-1)^k}{2^{k-1}} \sum_{m=0}^{k-1} (-1)^m \binom{2k}{m} (k-m)^{2n}. \end{equation}

I have been unable to locate a proof. Does somebody have a reference for such a one?

Answer 1

@Sam Hopkins: The formula there reads

\begin{equation*} E_{2n}=\sum_{k=1}^{2n }(-1)^k\frac{1}{2^k}\sum_{\ell=0}^{2k}(-1)^{\ell}\binom{2k}{\ell}(k-\ell)^{2n} \end{equation*}

which is slightly different from the formula I gave and which is from

NIST Handbook of Mathematical Functions (Olver, F.W.J. et al, Eds.) (CUP 2010), 

where it is formula 24.6.4 on p. 591, given, however, without clear reference. These formulas are easily seen to be the same, since the $k \ge n+1$ produce zeros, and the $m \le k-1$ and the $m \ge k+1$ match.

Wei et al. mention Guo et al. [5], as a source of the formula, which in this way is seen to have appeared earlier in the NIST Handbook and might be as well already quite old. At any rate, if one strips away the cloud of Bell polynomials evoked in the Wei and Guo papers one is left, for a proof, with the power series calculations

\begin{equation*} \text{sech}(x)=\frac{1}{\cosh(x)}=\frac{1}{1+2\sinh^2(x/2)}= \sum_{k=0}^{\infty}(-1)^k 2^{k}\sinh^{2k}(x/2) \end{equation*}

by expanding $\sinh(t)^m=2^{-t}(\exp(t)-\exp(-t))^m$ via the binomial and exponential series. The Bell polynomial stuff is needed for the general formalism of inverting a power serie $1-g(t)$ with $\text{ord}(g) \ge 1$, but not in this concrete situation.