Integers occurring in the coefficients of the Taylor series of 1/cosh t
Inmathematics , theEuler numbers are asequence En ofintegers (sequenceA122045 in theOEIS ) defined by theTaylor series expansion
1 cosh t = 2 e t + e − t = ∑ n = 0 ∞ E n n ! ⋅ t n {\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:
E n = 2 n E n ( 1 2 ) . {\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.
The odd-indexed Euler numbers are allzero . The even-indexed ones (sequenceA028296 in theOEIS ) have alternating signs. Some values are:
E 0 = 1 E 2 = −1 E 4 = 5 E 6 = −61 E 8 = 1385 E 10 = −50521 E 12 = 2702 765 E 14 = −199360 981 E 16 = 19391 512 145 E 18 = −2404 879 675 441
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.
In terms of Stirling numbers of the second kind [ edit ] The following two formulas express the Euler numbers in terms ofStirling numbers of the second kind :[ 1] [ 2]
E n = 2 2 n − 1 ∑ ℓ = 1 n ( − 1 ) ℓ S ( n , ℓ ) ℓ + 1 ( 3 ( 1 4 ) ℓ . ¯ − ( 3 4 ) ℓ . ¯ ) , {\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),} E 2 n = − 4 2 n ∑ ℓ = 1 2 n ( − 1 ) ℓ ⋅ S ( 2 n , ℓ ) ℓ + 1 ⋅ ( 3 4 ) ℓ . ¯ , {\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 , ℓ ) {\displaystyle S(n,\ell )} denotes theStirling numbers of the second kind , andx ℓ . ¯ = ( x ) ( x + 1 ) ⋯ ( x + ℓ − 1 ) {\displaystyle x^{\overline {\ell {\phantom {.}}}}=(x)(x+1)\cdots (x+\ell -1)} denotes therising factorial .
The Euler numbers can be defined as an recursion:
E 2 n = − ∑ k = 1 n ( 2 n 2 k ) E 2 ( n − k ) , {\displaystyle E_{2n}=-\sum _{k=1}^{n}{\binom {2n}{2k}}E_{2(n-k)},}
or alternatively:
1 = − ∑ k = 1 n ( 2 n 2 k ) E 2 k , {\displaystyle 1=-\sum _{k=1}^{n}{\binom {2n}{2k}}E_{2k},}
Both of these recursions can be found by using the fact that.
c o s ( x ) s e c ( x ) = 1. {\displaystyle cos(x)sec(x)=1.}
The following two formulas express the Euler numbers as double sums[ 3]
E 2 n = ( 2 n + 1 ) ∑ ℓ = 1 2 n ( − 1 ) ℓ 1 2 ℓ ( ℓ + 1 ) ( 2 n ℓ ) ∑ q = 0 ℓ ( ℓ q ) ( 2 q − ℓ ) 2 n , {\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},} E 2 n = ∑ k = 1 2 n ( − 1 ) k 1 2 k ∑ ℓ = 0 2 k ( − 1 ) ℓ ( 2 k ℓ ) ( k − ℓ ) 2 n . {\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}.} An explicit formula for Euler numbers is:[ 4]
E 2 n = i ∑ k = 1 2 n + 1 ∑ ℓ = 0 k ( k ℓ ) ( − 1 ) ℓ ( k − 2 ℓ ) 2 n + 1 2 k i k k , {\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 withi 2 = −1 .
As a sum over partitions [ edit ] The Euler numberE 2n can be expressed as a sum over the evenpartitions of2n ,[ 5]
E 2 n = ( 2 n ) ! ∑ 0 ≤ k 1 , … , k n ≤ n ( K k 1 , … , k n ) δ n , ∑ m k m ( − 1 2 ! ) k 1 ( − 1 4 ! ) k 2 ⋯ ( − 1 ( 2 n ) ! ) k n , {\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]
E 2 n = ( − 1 ) n − 1 ( 2 n − 1 ) ! ∑ 0 ≤ k 1 , … , k n ≤ 2 n − 1 ( K k 1 , … , k n ) δ 2 n − 1 , ∑ ( 2 m − 1 ) k m ( − 1 1 ! ) k 1 ( 1 3 ! ) k 2 ⋯ ( ( − 1 ) n ( 2 n − 1 ) ! ) k n , {\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 =k 1 + ··· +kn and
( K k 1 , … , k n ) ≡ K ! k 1 ! ⋯ k n ! {\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 thek s to2k 1 + 4k 2 + ··· + 2nkn = 2n and tok 1 + 3k 2 + ··· + (2n − 1)kn = 2n − 1 , respectively.
As an example,
E 10 = 10 ! ( − 1 10 ! + 2 2 ! 8 ! + 2 4 ! 6 ! − 3 2 ! 2 6 ! − 3 2 ! 4 ! 2 + 4 2 ! 3 4 ! − 1 2 ! 5 ) = 9 ! ( − 1 9 ! + 3 1 ! 2 7 ! + 6 1 ! 3 ! 5 ! + 1 3 ! 3 − 5 1 ! 4 5 ! − 10 1 ! 3 3 ! 2 + 7 1 ! 6 3 ! − 1 1 ! 9 ) = − 50 521. {\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}}} E 2n is given by thedeterminant
E 2 n = ( − 1 ) n ( 2 n ) ! | 1 2 ! 1 1 4 ! 1 2 ! 1 ⋮ ⋱ ⋱ 1 ( 2 n − 2 ) ! 1 ( 2 n − 4 ) ! 1 2 ! 1 1 ( 2 n ) ! 1 ( 2 n − 2 ) ! ⋯ 1 4 ! 1 2 ! | . {\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}}} E 2n is also given by the following integrals:
( − 1 ) n E 2 n = ∫ 0 ∞ t 2 n cosh π t 2 d t = ( 2 π ) 2 n + 1 ∫ 0 ∞ x 2 n cosh x d x = ( 2 π ) 2 n ∫ 0 1 log 2 n ( tan π t 4 ) d t = ( 2 π ) 2 n + 1 ∫ 0 π / 2 log 2 n ( tan x 2 ) d x = 2 2 n + 3 π 2 n + 2 ∫ 0 π / 2 x log 2 n ( tan x ) d x = ( 2 π ) 2 n + 2 ∫ 0 π x 2 log 2 n ( tan x 2 ) d x . {\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}}} W. Zhang[ 7] obtained the following combinational identities concerning the Euler numbers. For any primep {\displaystyle p} , we have
( − 1 ) p − 1 2 E p − 1 ≡ { − 0 mod p if p ≡ 1 mod 4 ; − 2 mod p if p ≡ 3 mod 4 . {\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 primep ≡ 1 ( mod 4 ) {\displaystyle p\equiv 1{\pmod {4}}} and integerα ≥ 1 {\displaystyle \alpha \geq 1} , we have
E ϕ ( p α ) / 2 ≢ 0 ( mod p α ) , {\displaystyle E_{\phi (p^{\alpha })/2}\not \equiv 0{\pmod {p^{\alpha }}},} whereϕ ( n ) {\displaystyle \phi (n)} is theEuler's totient function .
The Euler numbers grow quite rapidly for large indices, as they have the lower bound
| E 2 n | > 8 n π ( 4 n π e ) 2 n . {\displaystyle |E_{2n}|>8{\sqrt {\frac {n}{\pi }}}\left({\frac {4n}{\pi e}}\right)^{2n}.} Euler zigzag numbers [ edit ] TheTaylor series ofsec x + tan x = tan ( π 4 + x 2 ) {\displaystyle \sec x+\tan x=\tan \left({\frac {\pi }{4}}+{\frac {x}{2}}\right)} is
∑ n = 0 ∞ A n n ! x n , {\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 ,
A n = ( − 1 ) n 2 E n , {\displaystyle A_{n}=(-1)^{\frac {n}{2}}E_{n},} whereEn is the Euler number, and for all oddn ,
A n = ( − 1 ) n − 1 2 2 n + 1 ( 2 n + 1 − 1 ) B n + 1 n + 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 ,
A n − 1 ( n − 1 ) ! sin ( n π 2 ) + ∑ m = 0 n − 1 A m m ! ( n − m − 1 ) ! sin ( m π 2 ) = 1 ( n − 1 ) ! . {\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 ] ^ 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 . ^ Jha, Sumit Kumar (15 November 2019)."A new explicit formula for the Euler numbers in terms of the Stirling numbers of the second kind" . ^ 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 . ^ 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. ^ Vella, David C. (2008)."Explicit Formulas for Bernoulli and Euler Numbers" .Integers .8 (1): A1. ^ Malenfant, J. (2011). "Finite, Closed-form Expressions for the Partition Function and for Euler, Bernoulli, and Stirling Numbers".arXiv :1103.1585 [math.NT ]. ^ 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. ^ 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 .
Possessing a specific set of other numbers
Expressible via specific sums