Editing Euler numbers (section)

==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]]:<ref>{{cite journal | first1=Sumit Kumar | last1= Jha | title=A new explicit formula for Bernoulli numbers involving the Euler number | journal=Moscow Journal of Combinatorics and Number Theory | volume=8 | issue=4 | pages=385–387 | year=2019 | url= https://projecteuclid.org/euclid.moscow/1572314455| doi= 10.2140/moscow.2019.8.389 | s2cid= 209973489 }}</ref><ref>{{cite web |url=https://osf.io/smw7h/ |title=A new explicit formula for the Euler numbers in terms of the Stirling numbers of the second kind |last=Jha |first=Sumit Kumar |date= 15 November 2019}}</ref>

:<math> 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), </math>
:<math> 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{.}}},</math>

where <math> S(n,\ell) </math> denotes the [[Stirling numbers of the second kind]], and <math> x^{\overline{\ell\phantom{.}}}=(x)(x+1)\cdots (x+\ell-1) </math> denotes the [[Falling and rising factorials|rising factorial]].

=== As a recursion ===
The Euler numbers can be defined as an recursion: 

<math>E_{2n}=-\sum_{{k=1}}^{n}\binom{2n}{2k}E_{2(n-k)},</math>

or alternatively:

<math>1=-\sum_{{k=1}}^{n}\binom{2n}{2k}E_{2k},</math>

Both of these recursions can be found by using the fact that.

<math>cos(x)sec(x)=1.</math>

===As a double sum===
The following two formulas express the Euler numbers as double sums<ref>{{cite journal | first1=Chun-Fu | last1= Wei | first2=Feng | last2=Qi | title=Several closed expressions for the Euler numbers | journal=Journal of Inequalities and Applications | volume=219 | issue=2015| year=2015 | doi= 10.1186/s13660-015-0738-9 | doi-access=free }}
</ref>
:<math>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}, </math>
:<math>E_{2n}=\sum_{k=1}^{2n}(-1)^{k} \frac{1}{2^{k}}\sum_{\ell=0}^{2k}(-1)^{\ell } \binom{2k}{\ell}(k-\ell)^{2n}. </math>

===As an iterated sum===
An explicit formula for Euler numbers is:<ref>{{cite web |url=https://oeis.org/A000111/a000111.pdf |archive-url=https://web.archive.org/web/20140409060145/http://oeis.org/A000111/a000111.pdf |archive-date=2014-04-09 |url-status=live |title=An Explicit Formula for the Euler zigzag numbers (Up/down numbers) from power series |last=Tang |first=Ross |date= 2012-05-11}}
</ref>

:<math>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},</math>

where {{mvar|i}} denotes the [[imaginary unit]] with {{math|''i''<sup>2</sup> {{=}} −1}}.

===As a sum over partitions===
The Euler number {{math|''E''<sub>2''n''</sub>}} can be expressed as a sum over the even [[Integer partition|partitions]] of {{math|2''n''}},<ref>{{cite journal | first1=David C. | last1= Vella | title=Explicit Formulas for Bernoulli and Euler Numbers | journal=Integers | volume=8 | issue=1 | pages=A1 | year=2008 | url= http://www.integers-ejcnt.org/vol8.html}}</ref>

:<math>  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} ,</math>

as well as a sum over the odd partitions of {{math|2''n'' − 1}},<ref>{{cite arXiv | eprint=1103.1585 | first1= J. | last1=Malenfant | title=Finite, Closed-form Expressions for the Partition Function and for Euler, Bernoulli, and Stirling Numbers| class= math.NT | year= 2011 }}</ref>

:<math> 	 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}   , </math>

where in both cases {{math|''K'' {{=}} ''k''<sub>1</sub> + ··· + ''k<sub>n</sub>''}} and
:<math> \binom K {k_1, \ldots , k_n}
          \equiv \frac{ K!}{k_1! \cdots k_n!}</math>
is a [[multinomial coefficient]]. The [[Kronecker delta]]s in the above formulas restrict the sums over the {{mvar|k}}s to {{math|2''k''<sub>1</sub> + 4''k''<sub>2</sub> + ··· + 2''nk<sub>n</sub>'' {{=}} 2''n''}} and to {{math|''k''<sub>1</sub> + 3''k''<sub>2</sub> + ··· + (2''n'' − 1)''k<sub>n</sub>'' {{=}} 2''n'' − 1}}, respectively. 

As an example,
:<math>
\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}
</math>

===As a determinant===
{{math|''E''<sub>2''n''</sub>}} is given by the [[determinant]]

:<math>
\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}
 </math>

===As an integral===
{{math|''E''<sub>2''n''</sub>}} is also given by the following integrals:
:<math>
\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}
 </math>