Euler numbers

In mathematics, the Euler numbers are a sequence En of integers defined by the Taylor series expansion


 * $$\frac{1}{\cosh t} = \frac{2}{e^{t} + e^ {-t} } = \sum_{n=0}^\infty \frac{E_n}{n!} \cdot t^n$$,

where $$\cosh (t)$$ is the hyperbolic cosine function. The Euler numbers are related to a special value of the Euler polynomials, namely:
 * $$E_n=2^nE_n(\tfrac 12).$$

The Euler numbers appear in the Taylor series expansions of the secant and hyperbolic secant functions. The latter is the function in the definition. They also occur in combinatorics, specifically when counting the number of alternating permutations of a set with an even number of elements.

Examples
The odd-indexed Euler numbers are all zero. The even-indexed ones have alternating signs. Some values are:

Some authors re-index the sequence in order to omit the odd-numbered Euler numbers with value zero, or change all signs to positive. This article adheres to the convention adopted above.
 * E0 ||=||align=right| 1
 * E2 ||=||align=right| −1
 * E4 ||=||align=right| 5
 * E6 ||=||align=right| −61
 * E8 ||=||align=right| $1,385$
 * E10 ||=||align=right| $−50521$
 * E12 ||=||align=right| $2,702,765$
 * E14 ||=||align=right| $−199,360,981$
 * E16 ||=||align=right| $19,391,512,145$
 * E18 ||=||align=right| $−2,404,879,675,441$
 * }
 * E12 ||=||align=right| $i$
 * E14 ||=||align=right| $k$
 * E16 ||=||align=right| $A_{n}$
 * E18 ||=||align=right| $n$
 * }
 * E16 ||=||align=right| $E_{n}$
 * E18 ||=||align=right| $n$
 * }
 * }

In terms of Stirling numbers of the second kind
Following two formulas express the Euler numbers in terms of Stirling numbers of the second kind


 * $$ E_{n}=2^{2n-1}\sum_{\ell=1}^{n}\frac{(-1)^{\ell}S(n,\ell)}{\ell+1}\left(3\left(\frac{1}{4}\right)^{(\ell)}-\left(\frac{3}{4}\right)^{(\ell)}\right), $$
 * $$ E_{2n}=-4^{2n}\sum_{\ell=1}^{2n}(-1)^{\ell}\cdot \frac{S(2n,\ell)}{\ell+1}\cdot \left(\frac{3}{4}\right)^{(\ell)},$$

where $$ S(n,\ell) $$ denotes the Stirling numbers of the second kind, and $$ x^{(\ell)}=(x)(x+1)\cdots (x+\ell-1) $$ denotes the rising factorial.

As a double sum
Following two formulas express the Euler numbers as double sums
 * $$E_{2n}=(2 n+1)\sum_{\ell=1}^{2n} (-1)^{\ell}\frac{1}{2^{\ell}(\ell +1)}\binom{2 n}{\ell}\sum _{q=0}^{\ell}\binom{\ell}{q}(2q-\ell)^{2n}, $$
 * $$E_{2n}=\sum_{k=1}^{2n}(-1)^{k} \frac{1}{2^{k}}\sum_{\ell=0}^{2k}(-1)^{\ell } \binom{2k}{\ell}(k-\ell)^{2n}. $$

As an iterated sum
An explicit formula for Euler numbers is:


 * $$E_{2n}=i\sum _{k=1}^{2n+1} \sum _{\ell=0}^k \binom{k}{\ell}\frac{(-1)^\ell(k-2\ell)^{2n+1}}{2^k i^k k},$$

where $B_{n}$ denotes the imaginary unit with $i^{2} = −1$.

As a sum over partitions
The Euler number $E_{2n}$ can be expressed as a sum over the even partitions of $2n$,


 * $$ E_{2n} = (2n)! \sum_{0 \leq k_1, \ldots, k_n \leq n} \binom K {k_1, \ldots, k_n}

\delta_{n,\sum mk_m} \left( -\frac{1}{2!} \right)^{k_1} \left( -\frac{1}{4!} \right)^{k_2} \cdots \left( -\frac{1}{(2n)!} \right)^{k_n} ,$$

as well as a sum over the odd partitions of $2n − 1$,


 * $$ 	 E_{2n} = (-1)^{n-1} (2n-1)! \sum_{0 \leq k_1, \ldots, k_n \leq 2n-1}

\binom K {k_1, \ldots, k_n} \delta_{2n-1,\sum (2m-1)k_m }  \left( -\frac{1}{1!} \right)^{k_1}  \left( \frac{1}{3!} \right)^{k_2} \cdots \left( \frac{(-1)^n}{(2n-1)!} \right)^{k_n} , $$

where in both cases $K = k_{1} + ··· + k_{n}$ and
 * $$ \binom K {k_1, \ldots, k_n}

\equiv \frac{ K!}{k_1! \cdots k_n!}$$ is a multinomial coefficient. The Kronecker deltas in the above formulas restrict the sums over the ⇭⇭⇭s to $2k_{1} + 4k_{2} + ··· + 2nk_{n} = 2n$ and to $k_{1} + 3k_{2} + ··· + (2n − 1)k_{n} = 2n − 1$, respectively.

As an example,

\begin{align} E_{10} & = 10! \left( - \frac{1}{10!} + \frac{2}{2!\,8!} + \frac{2}{4!\,6!}	- \frac{3}{2!^2\, 6!}- \frac{3}{2!\,4!^2} +\frac{4}{2!^3\, 4!} - \frac{1}{2!^5}\right) \\[6pt] & = 9! \left( - \frac{1}{9!} + \frac{3}{1!^2\,7!} + \frac{6}{1!\,3!\,5!}	+\frac{1}{3!^3}- \frac{5}{1!^4\,5!} -\frac{10}{1!^3\,3!^2} + \frac{7}{1!^6\, 3!} - \frac{1}{1!^9}\right) \\[6pt] & = -50\,521. \end{align} $$

As a determinant
$E_{2n}$ is given by the determinant



\begin{align} E_{2n} &=(-1)^n (2n)!~ \begin{vmatrix}  \frac{1}{2!}& 1 &~& ~&~\\ \frac{1}{4!}& \frac{1}{2!} & 1 &~&~\\ \vdots & ~ &  \ddots &\ddots & ~\\ \frac{1}{(2n-2)!}& \frac{1}{(2n-4)!}& ~&\frac{1}{2!} & 1\\ \frac{1}{(2n)!}&\frac{1}{(2n-2)!}& \cdots & \frac{1}{4!} & \frac{1}{2!}\end{vmatrix}.

\end{align} $$

As an integral
$E_{2n}$ is also given by the following integrals:

\begin{align} (-1)^n E_{2n} & = \int_0^\infty \frac{t^{2n}}{\cosh\frac{\pi t}2}\; dt =\left(\frac2\pi\right)^{2n+1} \int_0^\infty \frac{x^{2n}}{\cosh x}\; dx\\[8pt] &=\left(\frac2\pi\right)^{2n} \int_0^1\log^{2n}\left(\tan \frac{\pi t}{4} \right)\,dt =\left(\frac2\pi\right)^{2n+1}\int_0^{\pi/2} \log^{2n}\left(\tan \frac{x}{2} \right)\,dx\\[8pt] &= \frac{2^{2n+3}}{\pi^{2n+2}} \int_0^{\pi/2} x \log^{2n} (\tan x)\,dx = \left(\frac2\pi\right)^{2n+2} \int_0^\pi \frac{x}{2} \log^{2n} \left(\tan \frac{x}{2} \right)\,dx.\end{align} $$

Congruences
W. Zhang obtained the following combinational identities concerning the Euler numbers, for any prime $$ p $$, we have

(-1)^{\frac{p-1}{2}} E_{p-1} \equiv \textstyle\begin{cases} 0 \mod p &\text{if }p\equiv 1\bmod 4; \\ -2 \mod p & \text{if }p\equiv 3\bmod 4. \end{cases} $$ W. Zhang and Z. Xu proved that, for any prime $$ p \equiv 1 \pmod{4} $$ and integer $$ \alpha\geq 1 $$, we have
 * $$ E_{\phi(p^{\alpha})/2}\not \equiv 0 \pmod{p^{\alpha}} $$

where $$\phi(n)$$ is the Euler's totient function.

Asymptotic approximation
The Euler numbers grow quite rapidly for large indices as they have the following lower bound


 * $$ |E_{2 n}| > 8 \sqrt { \frac{n}{\pi} } \left(\frac{4 n}{ \pi e}\right)^{2 n}. $$

Euler zigzag numbers
The Taylor series of $$\sec x + \tan x = \tan\left(\frac\pi4 + \frac x2\right)$$ is


 * $$\sum_{n=0}^{\infty} \frac{A_n}{n!}x^n,$$

where ⇭⇭⇭ is the Euler zigzag numbers, beginning with
 * 1, 1, 1, 2, 5, 16, 61, 272, 1385, 7936, 50521, 353792, 2702765, 22368256, 199360981, 1903757312, 19391512145, 209865342976, 2404879675441, 29088885112832, ...

For all even ⇭⇭⇭,
 * $$A_n = (-1)^\frac{n}{2} E_n,$$

where ⇭⇭⇭ is the Euler number; and for all odd ⇭⇭⇭,
 * $$A_n = (-1)^\frac{n-1}{2}\frac{2^{n+1}\left(2^{n+1}-1\right)B_{n+1}}{n+1},$$

where ⇭⇭⇭ is the Bernoulli number.

For every n,
 * $$\frac{A_{n-1}}{(n-1)!}\sin{\left(\frac{n\pi}{2}\right)}+\sum_{m=0}^{n-1}\frac{A_m}{m!(n-m-1)!}\sin{\left(\frac{m\pi}{2}\right)}=\frac{1}{(n-1)!}.$$