# Euler numbers

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In mathematics, the Euler numbers are a sequence En 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 hyperbolic cosine function. The Euler numbers are related to a special value of the Euler polynomials, namely:

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

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:

 E0 = 1 E2 = −1 E4 = 5 E6 = −61 E8 = 1385 E10 = −50521 E12 = 2702765 E14 = −199360981 E16 = 19391512145 E18 = −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

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 E2n 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 + ··· + kn and

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

is a multinomial coefficient. The Kronecker deltas in the above formulas restrict the sums over the ks to 2k1 + 4k2 + ··· + 2nkn = 2n and to k1 + 3k2 + ··· + (2n − 1)kn = 2n − 1, respectively.

As an example,

{\displaystyle {\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

E2n 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

E2n 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 An 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 En 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 Bn 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]

## References

1. ^ Jha, Sumit Kumar (2019). "A new explicit formula for Bernoulli numbers involving the Euler number". Moscow Journal of Combinatorics and Number Theory. 8 (4): 385–387. doi:10.2140/moscow.2019.8.389. S2CID 209973489.
2. ^ Jha, Sumit Kumar (15 November 2019). "A new explicit formula for the Euler numbers in terms of the Stirling numbers of the second kind".
3. ^ Wei, Chun-Fu; Qi, Feng (2015). "Several closed expressions for the Euler numbers". Journal of Inequalities and Applications. 219 (2015). doi:10.1186/s13660-015-0738-9.
4. ^ Tang, Ross (2012-05-11). "An Explicit Formula for the Euler zigzag numbers (Up/down numbers) from power series" (PDF).
5. ^ Vella, David C. (2008). "Explicit Formulas for Bernoulli and Euler Numbers". Integers. 8 (1): A1.
6. ^ Malenfant, J. (2011). "Finite, Closed-form Expressions for the Partition Function and for Euler, Bernoulli, and Stirling Numbers". arXiv:1103.1585 [math.NT].
7. ^ Zhang, W.P. (1998). "Some identities involving the Euler and the central factorial numbers" (PDF). Fibonacci Quarterly. 36 (4): 154–157.
8. ^ Zhang, W.P.; Xu, Z.F. (2007). "On a conjecture of the Euler numbers". Journal of Number Theory. 127 (2): 283–291. doi:10.1016/j.jnt.2007.04.004.