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Euler numbers

From Wikipedia, the free encyclopedia
(Redirected fromEuler number)

Integers occurring in the coefficients of the Taylor series of 1/cosh t
Not to be confused withEulerian number orEuler's number.
For other uses, seeList of things named after Leonhard Euler § Numbers.

Inmathematics, theEuler numbers are asequenceEn ofintegers (sequenceA122045 in theOEIS) defined by theTaylor series expansion

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

wherecosh(t){\displaystyle \cosh(t)} is thehyperbolic cosine function. The Euler numbers are related to a special value of theEuler polynomials, namely:

En=2nEn(12).{\displaystyle E_{n}=2^{n}E_{n}({\tfrac {1}{2}}).}

The Euler numbers appear in theTaylor series expansions of thesecant andhyperbolic secant functions. The latter is the function in the definition. They also occur incombinatorics, specifically when counting the number ofalternating permutations of a set with an even number of elements.

Examples

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The odd-indexed Euler numbers are allzero. The even-indexed ones (sequenceA028296 in theOEIS) 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 (sequenceA000364 in theOEIS). This article adheres to the convention adopted above.

Explicit formulas

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In terms of Stirling numbers of the second kind

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The following two formulas express the Euler numbers in terms ofStirling numbers of the second kind:[1][2]

En=22n1=1n(1)S(n,)+1(3(14).¯(34).¯),{\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)^{\overline {\ell {\phantom {.}}}}-\left({\frac {3}{4}}\right)^{\overline {\ell {\phantom {.}}}}\right),}
E2n=42n=12n(1)S(2n,)+1(34).¯,{\displaystyle 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 {.}}}},}

whereS(n,){\displaystyle S(n,\ell )} denotes theStirling numbers of the second kind, andx.¯=(x)(x+1)(x+1){\displaystyle x^{\overline {\ell {\phantom {.}}}}=(x)(x+1)\cdots (x+\ell -1)} denotes therising factorial.

As a recursion

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The Euler numbers can be defined as an recursion:

E2n=k=1n(2n2k)E2(nk),{\displaystyle E_{2n}=-\sum _{k=1}^{n}{\binom {2n}{2k}}E_{2(n-k)},}

or alternatively:

1=k=1n(2n2k)E2k,{\displaystyle 1=-\sum _{k=1}^{n}{\binom {2n}{2k}}E_{2k},}

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

cos(x)sec(x)=1.{\displaystyle cos(x)sec(x)=1.}

As a double sum

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The following two formulas express the Euler numbers as double sums[3]

E2n=(2n+1)=12n(1)12(+1)(2n)q=0(q)(2q)2n,{\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},}
E2n=k=12n(1)k12k=02k(1)(2k)(k)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

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An explicit formula for Euler numbers is:[4]

E2n=ik=12n+1=0k(k)(1)(k2)2n+12kikk,{\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}},}

wherei denotes theimaginary unit withi2 = −1.

As a sum over partitions

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The Euler numberE2n can be expressed as a sum over the evenpartitions of2n,[5]

E2n=(2n)!0k1,,knn(Kk1,,kn)δn,mkm(12!)k1(14!)k2(1(2n)!)kn,{\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 of2n − 1,[6]

E2n=(1)n1(2n1)!0k1,,kn2n1(Kk1,,kn)δ2n1,(2m1)km(11!)k1(13!)k2((1)n(2n1)!)kn,{\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 casesK =k1 + ··· +kn and

(Kk1,,kn)K!k1!kn!{\displaystyle {\binom {K}{k_{1},\ldots ,k_{n}}}\equiv {\frac {K!}{k_{1}!\cdots k_{n}!}}}

is amultinomial coefficient. TheKronecker deltas in the above formulas restrict the sums over theks to2k1 + 4k2 + ··· + 2nkn = 2n and tok1 + 3k2 + ··· + (2n − 1)kn = 2n − 1, respectively.

As an example,

E10=10!(110!+22!8!+24!6!32!26!32!4!2+42!34!12!5)=9!(19!+31!27!+61!3!5!+13!351!45!101!33!2+71!63!11!9)=50521.{\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

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E2n is given by thedeterminant

E2n=(1)n(2n)! |12!1   14!12!1        1(2n2)!1(2n4)! 12!11(2n)!1(2n2)!14!12!|.{\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

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E2n is also given by the following integrals:

(1)nE2n=0t2ncoshπt2dt=(2π)2n+10x2ncoshxdx=(2π)2n01log2n(tanπt4)dt=(2π)2n+10π/2log2n(tanx2)dx=22n+3π2n+20π/2xlog2n(tanx)dx=(2π)2n+20πx2log2n(tanx2)dx.{\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

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W. Zhang[7] obtained the following combinational identities concerning the Euler numbers. For any primep{\displaystyle p}, we have

(1)p12Ep1{0modpif p1mod4;2modpif p3mod4.{\displaystyle (-1)^{\frac {p-1}{2}}E_{p-1}\equiv \textstyle {\begin{cases}{\phantom {-}}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 primep1(mod4){\displaystyle p\equiv 1{\pmod {4}}} and integerα1{\displaystyle \alpha \geq 1}, we have

Eϕ(pα)/20(modpα),{\displaystyle E_{\phi (p^{\alpha })/2}\not \equiv 0{\pmod {p^{\alpha }}},}

whereϕ(n){\displaystyle \phi (n)} is theEuler's totient function.

Lower bound

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

|E2n|>8nπ(4nπe)2n.{\displaystyle |E_{2n}|>8{\sqrt {\frac {n}{\pi }}}\left({\frac {4n}{\pi e}}\right)^{2n}.}

Euler zigzag numbers

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TheTaylor series ofsecx+tanx=tan(π4+x2){\displaystyle \sec x+\tan x=\tan \left({\frac {\pi }{4}}+{\frac {x}{2}}\right)} is

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

whereAn is theEuler zigzag numbers, beginning with

1, 1, 1, 2, 5, 16, 61, 272, 1385, 7936, 50521, 353792, 2702765, 22368256, 199360981, 1903757312, 19391512145, 209865342976, 2404879675441, 29088885112832, ... (sequenceA000111 in theOEIS)

For all evenn,

An=(1)n2En,{\displaystyle A_{n}=(-1)^{\frac {n}{2}}E_{n},}

whereEn is the Euler number, and for all oddn,

An=(1)n122n+1(2n+11)Bn+1n+1,{\displaystyle A_{n}=(-1)^{\frac {n-1}{2}}{\frac {2^{n+1}\left(2^{n+1}-1\right)B_{n+1}}{n+1}},}

whereBn is theBernoulli number.

For everyn,

An1(n1)!sin(nπ2)+m=0n1Amm!(nm1)!sin(mπ2)=1(n1)!.{\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

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References

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  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).Archived(PDF) from the original on 2014-04-09.
  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.doi:10.1080/00150517.1998.12428950.Archived(PDF) from the original on 2019-11-23.
  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.

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