Recurrence Relation for Euler Numbers

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Theorem

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

Then:

\(\ds E_{2 n}\) \(=\) \(\ds -\sum_{k \mathop = 0}^{n - 1} \dbinom {2 n} {2 k} E_{2 k}\)
\(\ds \) \(=\) \(\ds -\paren {\binom {2 n} 0 E_0 + \binom {2 n} 2 E_2 + \binom {2 n} 4 E_4 + \cdots + \binom {2 n} {2 n - 2} E_{2 n - 2} }\)

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


Proof

\(\ds \forall n \in \Z_{>0}: \, \) \(\ds \sum_{k \mathop = 0}^n \binom {2 n} {2 k} E_{2 k}\) \(=\) \(\ds 0\) Sum of Euler Numbers by Binomial Coefficients Vanishes
\(\ds \leadsto \ \ \) \(\ds \sum_{k \mathop = 0}^{n - 1} \dbinom {2 n} {2 k} E_{2 k} + \dbinom {2 n} {2 n} E_{2 n}\) \(=\) \(\ds 0\) separating out case where $k = n$
\(\ds \leadsto \ \ \) \(\ds E_{2 n}\) \(=\) \(\ds -\sum_{k \mathop = 0}^{n - 1} \dbinom {2 n} {2 k} E_{2 k}\) Binomial Coefficient with Self: $\dbinom {2 n} {2 n} = 1$

$\blacksquare$


Also presented as

Recurrence Relation for Euler Numbers can also be presented using the alternative form of the Euler numbers:

\(\ds {E_n}^*\) \(=\) \(\ds -\sum_{k \mathop = 0}^{n - 1} \paren {-1}^{n - 1} \dbinom {2 n} {2 k} {E_k}^*\)
\(\ds \) \(=\) \(\ds \binom {2 n} 2 {E_1}^* - \binom {2 n} 4 {E_2}^* + \cdots + \paren {-1}^{n - 2} \binom {2 n} {2 n - 2} {E_{n - 1} }^*\)


Examples

$\begin{array}{r|cccccccccc} E_k & \dbinom n 0 & & \dbinom n 2 & & \dbinom n 4 & & \dbinom n 6 & & \dbinom n 8 & & \dbinom n {10} \\ \hline E_0 = +1 & 1 E_0 & & & & & & & & & & & = 1 \\ E_2 = -1 & 1 E_0 & + & 1 E_2 & & & & & & & & & = 0 \\ E_4 = +5 & 1 E_0 & + & 6 E_2 & + & 1 E_4 & & & & & & & = 0 \\ E_6 = -61 & 1 E_0 & + & 15 E_2 & + & 15 E_4 & + & 1 E_6 & & & & & = 0 \\ E_8 = +1385 & 1 E_0 & + & 28 E_2 & + & 70 E_4 & + & 28 E_6 & + & 1 E_8 & & & = 0 \\ E_{10} = -50521 & 1 E_0 & + & 45 E_2 & + & 210 E_4 & + & 210 E_6 & + & 45 E_8 & + & 1 E_{10} & = 0 \\ \end{array}$


Also see


Sources

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