# Euler numbers

In mathematics, the Euler numbers are a sequence E$_n$ of integers (sequence A122045 in the OEIS) 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:
    $E_{n}=2^{n}E_{n}({\tfrac {1}{2}}).$
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 (sequence A028296 in the OEIS) have alternating signs. Some values are:
    E$_0$=1 E$_2$=−1 E$_4$=5 E$_6$=−61 E$_8$=1385 E$_{10}$=−50521 E$_{12}$=2702765 E$_{14}$=−199360981 E$_{16}$=19391512145 E$_{18}$=−2404879675441
Some authors re-index the sequence in order to omit the odd-numbered Euler numbers with value zero, or change all signs to positive (sequence A000364 in the OEIS). This article adheres to the convention adopted above.

## Explicit formulas
### In terms of Stirling numbers of the second kind

The following two formulas express the Euler numbers in terms of Stirling numbers of the second kind:
    $E_{n}=2^{2n-1}\sum _{\ell =1}^{n}{\frac {(-1)^{\ell }S(n,\ell )}{\ell +1}}\left(3\left({\frac {1}{4}}\right)^{\overline {\ell {\phantom {.}}}}-\left({\frac {3}{4}}\right)^{\overline {\ell {\phantom {.}}}}\right),$
    $E_{2n}=-4^{2n}\sum _{\ell =1}^{2n}(-1)^{\ell }\cdot {\frac {S(2n,\ell )}{\ell +1}}\cdot \left({\frac {3}{4}}\right)^{\overline {\ell {\phantom {.}}}},$
where $S(n,\ell )$ denotes the Stirling numbers of the second kind, and $x^{\overline {\ell {\phantom {.}}}}=(x)(x+1)\cdots (x+\ell -1)$ denotes the rising factorial.

### As a recursion

The Euler numbers can be defined as an recursion:
$E_{2n}=-\sum _{k=1}^{n}{\binom {2n}{2k}}E_{2(n-k)},$
or alternatively:
$1=-\sum _{k=1}^{n}{\binom {2n}{2k}}E_{2k},$
Both of these recursions can be found by using the fact that.
$cos(x)sec(x)=1.$

### As a double sum

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

### 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 i 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 ks to 2k$_1$ + 4k$_2$ + ··· + 2nk$_n$ = 2n and to k$_1$ + 3k$_2$ + ··· + (2n − 1)k$_n$ = 2n − 1, respectively.
As an example,
    ${\begin{aligned}E_{10}&=10!\left(-{\frac {1}{10!}}+{\frac {2}{2!\,8!}}+{\frac {2}{4!\,6!}}-{\frac {3}{2!^{2}\,6!}}-{\frac {3}{2!\,4!^{2}}}+{\frac {4}{2!^{3}\,4!}}-{\frac {1}{2!^{5}}}\right)\\[6pt]&=9!\left(-{\frac {1}{9!}}+{\frac {3}{1!^{2}\,7!}}+{\frac {6}{1!\,3!\,5!}}+{\frac {1}{3!^{3}}}-{\frac {5}{1!^{4}\,5!}}-{\frac {10}{1!^{3}\,3!^{2}}}+{\frac {7}{1!^{6}\,3!}}-{\frac {1}{1!^{9}}}\right)\\[6pt]&=-50\,521.\end{aligned}}$

### As a determinant

E$_{2n}$ is given by the determinant
    ${\begin{aligned}E_{2n}&=(-1)^{n}(2n)!~{\begin{vmatrix}{\frac {1}{2!}}&1&~&~&~\\{\frac {1}{4!}}&{\frac {1}{2!}}&1&~&~\\\vdots &~&\ddots ~~&\ddots ~~&~\\{\frac {1}{(2n-2)!}}&{\frac {1}{(2n-4)!}}&~&{\frac {1}{2!}}&1\\{\frac {1}{(2n)!}}&{\frac {1}{(2n-2)!}}&\cdots &{\frac {1}{4!}}&{\frac {1}{2!}}\end{vmatrix}}.\end{aligned}}$

### As an integral

E$_{2n}$ is also given by the following integrals:
    ${\begin{aligned}(-1)^{n}E_{2n}&=\int _{0}^{\infty }{\frac {t^{2n}}{\cosh {\frac {\pi t}{2}}}}\;dt=\left({\frac {2}{\pi }}\right)^{2n+1}\int _{0}^{\infty }{\frac {x^{2n}}{\cosh x}}\;dx\\[8pt]&=\left({\frac {2}{\pi }}\right)^{2n}\int _{0}^{1}\log ^{2n}\left(\tan {\frac {\pi t}{4}}\right)\,dt=\left({\frac {2}{\pi }}\right)^{2n+1}\int _{0}^{\pi /2}\log ^{2n}\left(\tan {\frac {x}{2}}\right)\,dx\\[8pt]&={\frac {2^{2n+3}}{\pi ^{2n+2}}}\int _{0}^{\pi /2}x\log ^{2n}(\tan x)\,dx=\left({\frac {2}{\pi }}\right)^{2n+2}\int _{0}^{\pi }{\frac {x}{2}}\log ^{2n}\left(\tan {\frac {x}{2}}\right)\,dx.\end{aligned}}$

## Congruences

W. Zhang obtained the following combinational identities concerning the Euler numbers. For any prime $p$, we have
    $(-1)^{\frac {p-1}{2}}E_{p-1}\equiv \textstyle {\begin{cases}{\phantom {-}}0\mod p&{\text{if }}p\equiv 1{\bmod {4}};\\-2\mod p&{\text{if }}p\equiv 3{\bmod {4}}.\end{cases}}$
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.

## Lower bound

The Euler numbers grow quite rapidly for large indices, as they have the lower bound
    $|E_{2n}|>8{\sqrt {\frac {n}{\pi }}}\left({\frac {4n}{\pi e}}\right)^{2n}.$

## Euler zigzag numbers

The Taylor series of $\sec x+\tan x=\tan \left({\frac {\pi }{4}}+{\frac {x}{2}}\right)$ is
    $\sum _{n=0}^{\infty }{\frac {A_{n}}{n!}}x^{n},$
where A$_n$ 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, ... (sequence A000111 in the OEIS)
For all even n,
    $A_{n}=(-1)^{\frac {n}{2}}E_{n},$
where E$_n$ is the Euler number, and for all odd n,
    $A_{n}=(-1)^{\frac {n-1}{2}}{\frac {2^{n+1}\left(2^{n+1}-1\right)B_{n+1}}{n+1}},$
where B$_n$ is the Bernoulli number.
For every n,
    ${\frac {A_{n-1}}{(n-1)!}}\sin {\left({\frac {n\pi }{2}}\right)}+\sum _{m=0}^{n-1}{\frac {A_{m}}{m!(n-m-1)!}}\sin {\left({\frac {m\pi }{2}}\right)}={\frac {1}{(n-1)!}}.$