Sum of Euler Numbers by Binomial Coefficient

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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.


Proof


Sources