that $T_{2k} = E_{2k + 1} = 0$, for all $k \geq 0$. The remaining nontrivial values can be expressed in terms of special values of Euler polynomials. For the tangent numbers, we have $$T_{2k + 1} = (-1)^k 2^{2k + 1} E_{2k + 1}(1), \quad k \geq 0.
  \label{5.5}$$ This follows easily, by comparing the Taylor expansion of $1 + \tanh t$ $$1 + \tanh t = \frac{2 e^{2t}}{e^{2t} + 1}
    = 1 + \sum_{k = 0}^\infty (-1)^k T_{2k + 1} \frac{t^{2k + 1}}{(2k + 1)!}$$ and (\[5.4\]), with $x = 1$ and $2t$, instead of $t$. Similarly, by comparing the Taylor expansion of ${\operatorname{sech}}t$ $${\operatorname{sech}}t = \frac{2 e^t}{e^{2t} + 1}
    = \sum_{k = 0}^\infty (-1)^k E_{2k} \frac{t^{2k}}{(2k)!}$$ and (\[5.4\]), with $x = 1/2$ and $2t$, instead of $t$, we get $$E_{2k} = (-1)^k 2^{2k} E_{2k} \left( \frac{1}{2} \right), \quad k \geq 0.
  \label{5.6}$$

The following identities will also be needed in the proof of (\[5.3\]).

Let $d \geq 0$ be a non-negative integer. Then $$\begin{aligned}
  \sum_{\ell = 0}^{d + 1} (-1)^\ell \binom{d + 1}{\ell} E_n(\ell) = 0,
  \label{5.7} \\
  \sum_{\ell = 0}^{d + 1} (-1)^\ell \binom{d + 1}{\ell} E_n(\ell + 1) = 0,
  \label{5.8}\end{aligned}$$ for all $n = 