Mean binomial coefficient

In mathematics , the -th binomial mean coefficient for a nonnegative integer is given by ${\displaystyle n}$ ${\displaystyle n}$

${\displaystyle {2n \choose n}={\frac {(2n)!}{(n!)^{2}}}}$

The name "mean binomial coefficient" comes from the fact that these binomial coefficients lie exactly in the middle of the line in Pascal's triangle :

${\displaystyle \mathbf {1} }$

${\displaystyle 1}$

${\displaystyle 1}$

${\displaystyle 1}$

${\displaystyle \mathbf {2} }$

${\displaystyle 1}$

${\displaystyle 1}$

${\displaystyle 3}$

${\displaystyle 3}$

${\displaystyle 1}$

${\displaystyle 1}$

${\displaystyle 4}$

${\displaystyle \mathbf {6} }$

${\displaystyle 4}$

${\displaystyle 1}$

${\displaystyle 1}$

${\displaystyle 5}$

${\displaystyle 10}$

${\displaystyle 10}$

${\displaystyle 5}$

${\displaystyle 1}$

${\displaystyle 1}$

${\displaystyle 6}$

${\displaystyle 15}$

${\displaystyle \mathbf {20} }$

${\displaystyle 15}$

${\displaystyle 6}$

${\displaystyle 1}$

${\displaystyle 1}$

${\displaystyle 7}$

${\displaystyle 21}$

${\displaystyle 35}$

${\displaystyle 35}$

${\displaystyle 21}$

${\displaystyle 7}$

${\displaystyle 1}$

${\displaystyle 1}$

${\displaystyle 8}$

${\displaystyle 28}$

${\displaystyle 56}$

${\displaystyle \mathbf {70} }$

${\displaystyle 56}$

${\displaystyle 28}$

${\displaystyle 8}$

${\displaystyle 1}$

The first mean binomial coefficients are therefore (sequence A000984 in OEIS ):

1, 2, 6, 20, 70, 252, 924, 3432, 12870, 48620, ...

Representations

It applies

${\displaystyle {2n \choose n}=2^{2n}\cdot {\frac {1\cdot 3\cdot 5\cdots (2n-1)}{2\cdot 4\cdot 6\cdots (2n)}}.}$

The fraction is related to the Valais product .

After the Vandermonde convolution applies

${\displaystyle {2n \choose n}=\sum _{k=0}^{n}{n \choose k}^{2}.}$

Estimates

With the help of the Stirling formula one obtains for the estimation: ${\displaystyle n\geq 1}$

${\displaystyle {\frac {1}{2}}{\frac {4^{n}}{\sqrt {\pi n}}}<{2n \choose n}<{\frac {4^{n}}{\sqrt {\pi n}}}}$

So the following applies (for notation see Landau symbol ):

${\displaystyle {2n \choose n}\in \Theta \left({\frac {4^{n}}{\sqrt {n}}}\right)}$

More accurate:

${\displaystyle \lim _{n\rightarrow \infty }\left({2n \choose n}\left({\frac {4^{n}}{\sqrt {\pi n}}}\right)^{-1}\right)=1}$

Generating function

The generating function is

${\displaystyle {\frac {1}{\sqrt {1-4x}}}=\mathbf {1} +\mathbf {2} x+\mathbf {6} x^{2}+\mathbf {20} x^{3}+\mathbf {70} x^{4}+\mathbf {252} x^{5}+\cdots .}$

Number theoretic properties

According to Wolstenholme's theorem, applies to prime numbers ${\displaystyle p\geq 5}$

${\displaystyle {2p \choose p}\equiv 2\mod p^{3}}$

(for the symbolism see congruence (number theory) ).

Also, there are no odd numbers except . ${\displaystyle 1}$

Furthermore, the numbers for are never square-free , see Sárkőzy's theorem . ${\displaystyle n>4}$

Integral representation

An integral representation is as follows:

${\displaystyle {\binom {2n}{n}}={\frac {2^{2n+1}}{\pi }}\int \limits _{0}^{\infty }{\frac {\mathrm {d} x}{(x^{2}+1)^{n+1}}}}$

Representation with gamma function

${\displaystyle {\binom {2n}{n}}=(-4)^{n}{\frac {\sqrt {\pi }}{\Gamma ({\frac {1}{2}}-n)\Gamma (1+n)}}}$

Rows

${\displaystyle \sum _{n=0}^{\infty }{\frac {\binom {2n}{n}}{(-4)^{n}}}={\frac {1}{\sqrt {2}}}}$

In general, the following applies (if the series diverges for regularized values ​​calculated with the gamma function):

${\displaystyle \sum _{n=0}^{\infty }{\frac {\binom {2n}{n}}{(p/q)^{n}}}={\sqrt {\frac {p}{p-4q}}}}$ With ${\displaystyle q\in \mathbb {N} ,p\in \mathbb {Z} /\{0\}}$

In addition, the following applies to partial sums (sequence A285388 in OEIS ):

${\displaystyle \lim _{n\rightarrow \infty }\sum _{k=0}^{n^{2}-1}{\frac {2k \choose k}{4^{k}n}}=\lim _{n\rightarrow \infty }{\frac {n{2n^{2} \choose n^{2}}}{2^{2n^{2}-1}}}=\lim _{m\rightarrow \infty }{\frac {{\sqrt {m}}{2m \choose m}}{2^{2m-1}}}={\frac {2}{\sqrt {\pi }}}={\frac {1}{\Gamma ({\frac {3}{2}})}}=\sum _{n=0}^{\infty }{(-1)^{n}{\frac {\Gamma ({\frac {n+1}{2}})}{\Gamma (1+{\frac {n}{2}})}}}}$

Series of reciprocals

The following applies:

${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{\binom {2n}{n}}}={\frac {1}{27}}(2\pi {\sqrt {3}}+9)=0{,}7363998587187\ldots }$

The individual decimal places form sequence A073016 in OEIS .

Some more similar series are:

${\displaystyle {\begin{aligned}\sum _{n=1}^{\infty }{\frac {1}{n{\binom {2n}{n}}}}&={\frac {1}{9}}\pi {\sqrt {3}}&=&\,0{,}60459\ldots \\\sum _{n=1}^{\infty }{\frac {1}{n^{2}{\binom {2n}{n}}}}&={\frac {1}{18}}\pi ^{2}&=&\,0{,}54831\dots \\\sum _{n=1}^{\infty }{\frac {1}{n^{3}{\binom {2n}{n}}}}&={\frac {1}{18}}\pi {\sqrt {3}}\left(\psi _{1}({\tfrac {1}{3}})-\psi _{1}({\tfrac {2}{3}})\right)-{\frac {4}{3}}\zeta (3)&{}&{}\\\sum _{n=1}^{\infty }{\frac {1}{n^{4}{\binom {2n}{n}}}}&={\frac {17}{3240}}\pi ^{4}&=&\,0{,}51109\ldots \\\sum _{n=1}^{\infty }{\frac {1}{n^{5}{\binom {2n}{n}}}}&={\frac {1}{432}}\pi {\sqrt {3}}\left(\psi _{3}({\tfrac {1}{3}})-\psi _{3}({\tfrac {2}{3}})\right)-{\frac {19}{3}}\zeta (5)+{\frac {1}{9}}\zeta (3)\pi ^{2}&{}&{}\end{aligned}}}$

see. Follow A073010 in OEIS , Follow A086463 in OEIS , -, Follow A086464 in OEIS , -. Here referred to the digamma function , the Trigammafunktion and generally the th polygamma function ; the Riemann zeta function and the circle number . ${\displaystyle \ \psi _{1}}$ ${\displaystyle \ \psi _{2}}$ ${\displaystyle \ \psi _{n}}$ ${\displaystyle n}$ ${\displaystyle \ \zeta (x)}$ ${\displaystyle \ \pi }$

In general, the following formula applies:

${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{k}{\binom {2n}{n}}}}={\frac {1}{2}}\,\cdot \,{}_{k+1}F_{k}\left(\underbrace {1,\ldots ,1} _{k+1};{\tfrac {3}{2}},\underbrace {2,\ldots ,2} _{k-1};{\tfrac {1}{4}}\right)}$

for , where denotes the generalized hypergeometric function ; see. ${\displaystyle k\geq 1}$ ${\displaystyle {}_{m}F_{n}(a_{1},\ldots ,a_{m};b_{1},\ldots ,b_{n};x)}$

The corresponding alternating series also converge to the following limit values:

${\displaystyle {\begin{aligned}\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{\binom {2n}{n}}}&={\frac {1}{25}}\left(5+4{\sqrt {5}}\cdot \operatorname {arcsch} (2)\right)&=&\,0{,}37216357638560161555577\ldots \\\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n{\binom {2n}{n}}}}&={\frac {2}{5}}{\sqrt {5}}\cdot \operatorname {arcsch} (2)&=&\;0{,}430408940964\ldots \\\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n^{2}{\binom {2n}{n}}}}&=2\left(\operatorname {arcsch} (2)\right)^{2}&=&\;0{,}463129641154\ldots \\\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n^{3}{\binom {2n}{n}}}}&={\frac {2}{5}}\zeta (3)&=&\;0{,}48082276126\ldots \end{aligned}}}$

see. Episode A086465 in OEIS , episode A086466 in OEIS , episode A086467 in OEIS , episode A086468 in OEIS .

In general, the following can be written:

${\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n^{k}{\binom {2n}{n}}}}={\frac {1}{2}}\,\cdot \,{}_{k+1}F_{k}\left(\underbrace {1,\ldots ,1} _{k+1};{\tfrac {3}{2}},\underbrace {2,\ldots ,2} _{k-1};{\tfrac {-1}{4}}\right).}$

Related terms

The Catalan numbers are closely related to the mean binomial coefficients . You are given by ${\displaystyle C_{n}}$

${\displaystyle C_{n}={\frac {1}{n+1}}{2n \choose n}}$

generalization

In Pascal's triangle, only the lines with an even-numbered index have a clear middle entry, whereas the lines with an odd-numbered index have two entries in the middle. However, since these two entries always match, they are occasionally included in the definition of the mean binomial coefficient, which is then:

${\displaystyle {m \choose {\lfloor {\frac {m}{2}}\rfloor }}}$for . ${\displaystyle m\in \mathbb {N} _{0}}$

The first definition is obtained by looking at the even numbers here . ${\displaystyle m}$

Web links

• Eric W. Weisstein : Central Binomial Coefficients . In: MathWorld (English).

Individual evidence

1. ^ VH Moll: Some Questions in the Evaluation of Definite Integrals. MAA Short Course, San Antonio, TX. Jan. 2006. Archived copy ( memento of the original from April 2, 2008 in the Internet Archive ) Info: The archive link was inserted automatically and has not yet been checked. Please check the original and archive link according to the instructions and then remove this notice. 
2. ^ S. Plouffe: The Art of Inspired Guessing. Aug 7, 1998; http://www.lacim.uqam.ca/~plouffe/inspired.html