Sum of Euler Numbers by Binomial Coefficient

Theorem

Let $n \in \Z_{>0}$ be a (strictly) positive integer.

Then:

 $\displaystyle E_{2 n}$ $=$ $\displaystyle \sum_{k \mathop = 0}^{n - 1} \dbinom {2 n} {2 k} E_{2 n - 2 k}$ $\displaystyle$ $=$ $\displaystyle \binom {2 n} 2 E_{2 n - 2} + \binom {2 n} 4 E_{2 n - 4} + \binom {2 n} 6 E_{2 n - 6} + \cdots + 1$

where $E_n$ denotes the $n$th Euler number.