TITLE: Recursion relation for Euler numbers
QUESTION [6 upvotes]: I am trying to solve the following:

The Euler numbers $E_n$ are defined by the power series expansion
$$\frac{1}{\cos z}=\sum_{n=0}^\infty \frac{E_n}{n!}z^n\text{ for }|z|<\pi/2$$
(a)  Show that $E_n=0$ when $n$  is an odd integer.
(b) Establish a recursion relation for the sequence.

Through research it seems that part a seems well established but I couldn't find a proof.
Part (b) I found a formula, but couldn't find how to establish it. Any help appreciated.

REPLY [3 votes]: The exponential generating function of the Euler numbers is $\mathrm{sech}(x)$. Since $\mathrm{sech}(x)$ is even, the odd Euler numbers are $0$. In this answer, it is shown that
$$
\begin{align}
1
&=\cosh(x)\,\mathrm{sech}(x)\\[9pt]
&=\sum_{k=0}^\infty\frac1{(2k)!}x^{2k}\sum_{n=0}^\infty\frac{\mathrm{E}_{2n}}{(2n)!}x^{2n}\\
&=\sum_{n=0}^\infty\left(\sum_{k=0}^n\frac1{(2n-2k)!}\frac{\mathrm{E}_{2k}}{(2k)!}\right)x^{2n}\\
&=\sum_{n=0}^\infty\left(\sum_{k=0}^n\binom{2n}{2k}\mathrm{E}_{2k}\right)\frac{x^{2n}}{(2n)!}\tag{1}
\end{align}
$$
Equating coefficients of $x^{2n}$, we get $\color{#C00000}{\mathrm{E}_0=1}$ and
$$
\color{#C00000}{\mathrm{E}_{2n}=-\sum_{k=0}^{n-1}\binom{2n}{2k}\mathrm{E}_{2k}}\tag{2}
$$

Since $\sec(x)=\mathrm{sech}(ix)$,
$$
\begin{align}
\sec(x)
&=\sum_{n=0}^\infty(-1)^n\frac{\mathrm{E}_{2n}}{(2n)!}x^{2n}\\
&=\sum_{n=0}^\infty\frac{\tilde{\mathrm{E}}_{2n}}{(2n)!}x^{2n}\tag{3}
\end{align}
$$
where $\tilde{\mathrm{E}}_{2n}=(-1)^n\mathrm{E}_{2n}$. Therefore, if $\sec(x)$ is the exponential generating function for $\tilde{\mathrm{E}}_n$, $(2)$ would become $\color{#00A000}{\tilde{\mathrm{E}}_0=1}$ and
$$
\color{#00A000}{\tilde{\mathrm{E}}_{2n}=\sum_{k=0}^{n-1}(-1)^{n-k+1}\binom{2n}{2k}\tilde{\mathrm{E}}_{2k}}\tag{4}
$$