Editing Euler numbers (section)

===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>