Euler number

Template:Use American English Template:Short description Template:Confused Template:Other uses In mathematics, the Euler numbers are a sequence E_n of integers Template:OEIS defined by the Taylor series expansion $$\frac{1}{\cosh t}=\frac{2}{e^t+e^{-t}}=\sum _{n=0}^{\mathrm{\infty }}\frac{E_n}{n!}\cdot t^n,$$ where $\cosh (t)$ is the hyperbolic cosine function. The Euler numbers are related to a special value of the Euler polynomials, namely $$E_n=2^n{E}_n(\frac{1}{2}).$$

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 Template:OEIS have alternating signs. Some values are:

E0	=	1
E2	=	−1
E4	=	5
E6	=	−61
E8	=	Template:Val
E10	=	Template:Val
E12	=	Template:Val
E14	=	Template:Val
E16	=	Template:Val
E18	=	Template:Val

Some authors re-index the sequence in order to omit the odd-numbered Euler numbers with value zero, or change all signs to positive Template: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:

$E_n=2^{2n-1}\sum _{\ell =1}^n\frac{(-1{)}^{\ell }S(n,\ell )}{\ell +1}(3{\left(\frac{1}{4}\right)}^{\overline{\ell \phantom{.}}}-{\left(\frac{3}{4}\right)}^{\overline{\ell \phantom{.}}}),$

$E_{2n}=-4^{2n}\sum _{\ell =1}^{2n}(-1{)}^{\ell }\cdot \frac{S(2n,\ell )}{\ell +1}\cdot {\left(\frac{3}{4}\right)}^{\overline{\ell \phantom{.}}},$

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 by the recursion

$E_{2n}=-\sum _{k=1}^n(\genfrac{}{}{0ex}{}{2n}{2k})E_{2(n-k)},$

or equivalently

$1=-\sum _{k=1}^n(\genfrac{}{}{0ex}{}{2n}{2k})E_{2k},$

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

$E_{2n}=(2n+1)\sum _{\ell =0}^{2n}(-1{)}^{\ell }\frac{1}{2^{\ell }(\ell +1)}(\genfrac{}{}{0ex}{}{2n}{\ell })\sum _{q=0}^{\ell }(\genfrac{}{}{0ex}{}{\ell }{q})(2q-\ell {)}^{2n},$

$E_{2n}=\sum _{k=0}^{2n}(-1{)}^k\frac{1}{2^k}\sum _{\ell =0}^{2k}(-1{)}^{\ell }(\genfrac{}{}{0ex}{}{2k}{\ell })(k-\ell {)}^{2n}.$

As an iterated sum

An explicit formula for Euler numbers is

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

where Template:Mvar denotes the imaginary unit with Template:Math.

As a sum over partitions

The Euler number Template:Math can be expressed as a sum over the even partitions of Template:Math,

$E_{2n}=(2n)!\sum _{0\le k_1,\dots ,k_n\le n}(\genfrac{}{}{0ex}{}{K}{k_1,\dots ,k_n}){\delta }_{n,\sum m{k}_m}{(-\frac{1}{2!})}^{k_1}{(-\frac{1}{4!})}^{k_2}\cdots {(-\frac{1}{(2n)!})}^{k_n},$

as well as a sum over the odd partitions of Template:Math,

$E_{2n}=(-1{)}^{n-1}(2n-1)!\sum _{0\le k_1,\dots ,k_n\le 2n-1}(\genfrac{}{}{0ex}{}{K}{k_1,\dots ,k_n}){\delta }_{2n-1,\sum (2m-1)k_m}{(-\frac{1}{1!})}^{k_1}{\left(\frac{1}{3!}\right)}^{k_2}\cdots {\left(\frac{(-1{)}^n}{(2n-1)!}\right)}^{k_n},$

where in both cases Template:Math and

$(\genfrac{}{}{0ex}{}{K}{k_1,\dots ,k_n})\equiv \frac{K!}{k_1!\cdots k_n!}$

is a multinomial coefficient. The Kronecker deltas in the above formulas restrict the sums over the Template:Mvars to Template:Math and to Template:Math, respectively.

As an example,

$\begin{array}{cc}{E}_{10}& =10!(-\frac{1}{10!}+\frac{2}{2!8!}+\frac{2}{4!6!}-\frac{3}{2{!}^{2}6!}-\frac{3}{2!4{!}^2}+\frac{4}{2{!}^{3}4!}-\frac{1}{2{!}^5})\\ & =9!(-\frac{1}{9!}+\frac{3}{1{!}^{2}7!}+\frac{6}{1!3!5!}+\frac{1}{3{!}^3}-\frac{5}{1{!}^{4}5!}-\frac{10}{1{!}^{3}3{!}^2}+\frac{7}{1{!}^{6}3!}-\frac{1}{1{!}^9})\\ & =-50521.\end{array}$

As a determinant

Template:Math is given by the determinant

$\begin{array}{cc}{E}_{2n}& =(-1{)}^n(2n)!\left|\begin{array}{ccccc}\frac{1}{2!}& 1& & & \\ \frac{1}{4!}& \frac{1}{2!}& 1& & \\ ⋮& & \ddots & \ddots & \\ \frac{1}{(2n-2)!}& \frac{1}{(2n-4)!}& & \frac{1}{2!}& 1\\ \frac{1}{(2n)!}& \frac{1}{(2n-2)!}& \cdots & \frac{1}{4!}& \frac{1}{2!}\end{array}\right|.\end{array}$

As an integral

Template:Math is also given by the following integrals:

$\begin{array}{cc}(-1{)}^n{E}_{2n}& ={\int }_0^{\mathrm{\infty }}\frac{t^{2n}}{\cosh \frac{\pi t}{2}}dt={\left(\frac{2}{\pi }\right)}^{2n+1}{\int }_0^{\mathrm{\infty }}\frac{x^{2n}}{\cosh x}dx\\ & ={\left(\frac{2}{\pi }\right)}^{2n}{\int }_0^1{\log }^{2n}(\tan \frac{\pi t}{4})dt={\left(\frac{2}{\pi }\right)}^{2n+1}{\int }_0^{\pi /2}{\log }^{2n}(\tan \frac{x}{2})dx\\ & =\frac{2^{2n+3}}{{\pi }^{2n+2}}{\int }_0^{\pi /2}x{\log }^{2n}(\tan x)dx={\left(\frac{2}{\pi }\right)}^{2n+2}{\int }_0^{\pi }\frac{x}{2}{\log }^{2n}(\tan \frac{x}{2})dx.\end{array}$

Congruences

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

$(-1{)}^{\frac{p-1}{2}}{E}_{p-1}\equiv \{\begin{array}{cc}\phantom{-}0modp& \text{if }p\equiv 1mod4;\\ -2modp& \text{if }p\equiv 3mod4.\end{array}$

W. Zhang and Z. Xu proved that, for any prime $p\equiv 1(\mathrm{mod}4)$ and integer $\alpha \ge 1$, we have

$E_{\varphi (p^{\alpha })/2}\equiv ̸0(\mathrm{mod}{p}^{\alpha }),$

where $\varphi (n)$ 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 $\sec x+\tan x=\tan (\frac{\pi }{4}+\frac{x}{2})$ is

$\sum _{n=0}^{\mathrm{\infty }}\frac{A_n}{n!}{x}^n,$

where Template:Mvar 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, ... Template:OEIS

For all even Template:Mvar,

$A_n=(-1{)}^{\frac{n}{2}}{E}_n,$

where Template:Mvar is the Euler number, and for all odd Template:Mvar,

$A_n=(-1{)}^{\frac{n-1}{2}}\frac{2^{n+1}(2^{n+1}-1)B_{n+1}}{n+1},$

where Template:Mvar is the Bernoulli number.

For every n,

$\frac{A_{n-1}}{(n-1)!}\sin \left(\frac{n\pi }{2}\right)+\sum _{m=0}^{n-1}\frac{A_m}{m!(n-m-1)!}\sin \left(\frac{m\pi }{2}\right)=\frac{1}{(n-1)!}.$Template:Cn

See also

• Bell number
• Bernoulli number
• Dirichlet beta function
• Euler–Mascheroni constant

External links

• Template:Springer
• Template:MathWorld