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

${\displaystyle {\frac {1}{\cosh t}}={\frac {2}{e^{t}+e^{-t}}}=\sum _{n=0}^{\infty }{\frac {E_{n}}{n!}}\cdot t^{n}}$,

where ${\displaystyle \cosh(t)}$ is the

${\displaystyle E_{n}=2^{n}E_{n}({\tfrac {1}{2}}).}$

The Euler numbers appear in the

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

Following two formulas express the Euler numbers in terms of Stirling numbers of the second kind^{[1] [2]}

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

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

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

As a double sum

Following two formulas express the Euler numbers as double sums^[3]

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

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

As an iterated sum

An explicit formula for Euler numbers is:^[4]

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

where i denotes the imaginary unit with i² = −1.

As a sum over partitions

The Euler number E_2n can be expressed as a sum over the even partitions of 2n,^[5]

${\displaystyle E_{2n}=(2n)!\sum _{0\leq k_{1},\ldots ,k_{n}\leq n}{\binom {K}{k_{1},\ldots ,k_{n}}}\delta _{n,\sum mk_{m}}\left(-{\frac {1}{2!}}\right)^{k_{1}}\left(-{\frac {1}{4!}}\right)^{k_{2}}\cdots \left(-{\frac {1}{(2n)!}}\right)^{k_{n}},}$

as well as a sum over the odd partitions of 2n − 1,^[6]

${\displaystyle E_{2n}=(-1)^{n-1}(2n-1)!\sum _{0\leq k_{1},\ldots ,k_{n}\leq 2n-1}{\binom {K}{k_{1},\ldots ,k_{n}}}\delta _{2n-1,\sum (2m-1)k_{m}}\left(-{\frac {1}{1!}}\right)^{k_{1}}\left({\frac {1}{3!}}\right)^{k_{2}}\cdots \left({\frac {(-1)^{n}}{(2n-1)!}}\right)^{k_{n}},}$

where in both cases K = k₁ + ··· + k_n and

${\displaystyle {\binom {K}{k_{1},\ldots ,k_{n}}}\equiv {\frac {K!}{k_{1}!\cdots k_{n}!}}}$

is a

E_2n is given by the determinant

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

As an integral

E_2n is also given by the following integrals:

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

Congruences

W. Zhang^[7] obtained the following combinational identities concerning the Euler numbers, for any prime ${\displaystyle p}$, we have

${\displaystyle (-1)^{\frac {p-1}{2}}E_{p-1}\equiv \textstyle {\begin{cases}0\mod p&{\text{if }}p\equiv 1{\bmod {4}};\\-2\mod p&{\text{if }}p\equiv 3{\bmod {4}}.\end{cases}}}$

W. Zhang and Z. Xu^[8] proved that, for any prime ${\displaystyle p\equiv 1{\pmod {4}}}$ and integer ${\displaystyle \alpha \geq 1}$, we have

${\displaystyle E_{\phi (p^{\alpha })/2}\not \equiv 0{\pmod {p^{\alpha }}}}$

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

Asymptotic approximation

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

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

Euler zigzag numbers

The Taylor series of ${\displaystyle \sec x+\tan x=\tan \left({\frac {\pi }{4}}+{\frac {x}{2}}\right)}$ is

${\displaystyle \sum _{n=0}^{\infty }{\frac {A_{n}}{n!}}x^{n},}$

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,

${\displaystyle A_{n}=(-1)^{\frac {n}{2}}E_{n},}$

where E_n is the Euler number; and for all odd n,

${\displaystyle A_{n}=(-1)^{\frac {n-1}{2}}{\frac {2^{n+1}\left(2^{n+1}-1\right)B_{n+1}}{n+1}},}$

where B_n is the Bernoulli number.

For every n,

${\displaystyle {\frac {A_{n-1}}{(n-1)!}}\sin {\left({\frac {n\pi }{2}}\right)}+\sum _{m=0}^{n-1}{\frac {A_{m}}{m!(n-m-1)!}}\sin {\left({\frac {m\pi }{2}}\right)}={\frac {1}{(n-1)!}}.}$^{[citation needed]}

See also

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

References