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Euler numbers

Integers occurring in the coefficients of the Taylor series of 1/cosh t


Integers occurring in the coefficients of the Taylor series of 1/cosh t

In mathematics, the Euler numbers are a sequence En of integers defined by the Taylor series expansion :\frac{1}{\cosh t} = \frac{2}{e^{t} + e^ {-t} } = \sum_{n=0}^\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^nE_n(\tfrac 12).

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 have alternating signs. Some values are: :{| |E0 ||=||align=right| 1 |- |E2 ||=||align=right| −1 |- |E4 ||=||align=right| 5 |- |E6 ||=||align=right| −61 |- |E8 ||=||align=right| |- |E10 ||=||align=right| |- |E12 ||=||align=right| |- |E14 ||=||align=right| |- |E16 ||=||align=right| |- |E18 ||=||align=right| |} Some authors re-index the sequence in order to omit the odd-numbered Euler numbers with value zero, or change all signs to positive . 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}\left(3\left(\frac{1}{4}\right)^{\overline{\ell\phantom{.}}}-\left(\frac{3}{4}\right)^{\overline{\ell\phantom{.}}}\right), : 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_^{n}\binom{2n}{2k}E_{2(n-k)},

or equivalently

1=-\sum_^{n}\binom{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}=(2 n+1)\sum_{\ell=0}^{2n} (-1)^{\ell}\frac{1}{2^{\ell}(\ell +1)}\binom{2 n}{\ell}\sum {q=0}^{\ell}\binom{\ell}{q}(2q-\ell)^{2n}, :E{2n}=\sum_{k=0}^{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

: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 .

As a sum over partitions

The Euler number E2n can be expressed as a sum over the even partitions of 2n,

: 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,

: 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 and : \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 and to , respectively.

As an example, : \begin{align} 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{align}

As a determinant

E2n is given by the determinant

: \begin{align} 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{align}

As an integral

E2n is also given by the following integrals: : \begin{align} (-1)^n E_{2n} & = \int_0^\infty \frac{t^{2n}}{\cosh\frac{\pi t}2}; dt =\left(\frac2\pi\right)^{2n+1} \int_0^\infty \frac{x^{2n}}{\cosh x}; dx\[8pt] &=\left(\frac2\pi\right)^{2n} \int_0^1\log^{2n}\left(\tan \frac{\pi t}{4} \right),dt =\left(\frac2\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(\frac2\pi\right)^{2n+2} \int_0^\pi \frac{x}{2} \log^{2n} \left(\tan \frac{x}{2} \right),dx.\end{align}

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 \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 proved that, for any prime p \equiv 1 \pmod{4} and integer \alpha\geq 1 , we have : E_{\phi(p^{\alpha})/2}\not \equiv 0 \pmod{p^{\alpha}}, where \phi(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_{2 n}| 8 \sqrt { \frac{n}{\pi} } \left(\frac{4 n}{ \pi e}\right)^{2 n}.

Euler zigzag numbers

The Taylor series of \sec x + \tan x = \tan\left(\frac\pi4 + \frac x2\right) is

:\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, ...

For all even n, :A_n = (-1)^\frac{n}{2} E_n, where En is the Euler number, and for all odd n, :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, :\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)!}.

References

References

  1. (2019). "A new explicit formula for Bernoulli numbers involving the Euler number". Moscow Journal of Combinatorics and Number Theory.
  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. (2008). "Explicit Formulas for Bernoulli and Euler Numbers". Integers.
  4. (2011). "Finite, Closed-form Expressions for the Partition Function and for Euler, Bernoulli, and Stirling Numbers".
  5. (1998). "Some identities involving the Euler and the central factorial numbers". Fibonacci Quarterly.
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