Central binomial coefficient

The central binomial coefficients satisfy the recurrence

${\displaystyle {\binom {2(n+1)}{n+1}}={\frac {4n+2}{n+1}}\cdot {\binom {2n}{n}}.}$

Since ${\displaystyle \textstyle {\binom {-1/2}{n+1}}={\frac {-1/2-n}{n+1}}\cdot {\binom {-1/2}{n}}}$ we find

${\displaystyle {\binom {2n}{n}}=(-1)^{n}4^{n}{\binom {-1/2}{n}}}$

Together with the binomial series we obtain the generating function

${\displaystyle {\frac {1}{\sqrt {1-4x}}}=\sum _{n=0}^{\infty }{\binom {2n}{n}}x^{n}=1+2x+6x^{2}+20x^{3}+70x^{4}+252x^{5}+\cdots }$

and exponential generating function

${\displaystyle \sum _{n=0}^{\infty }{\binom {2n}{n}}{\frac {x^{n}}{n!}}=e^{2x}I_{0}(2x),}$

where I₀ is a modified Bessel function of the first kind.[1]

The Wallis product can be written in asymptotic form for the central binomial coefficient:

${\displaystyle {2n \choose n}\sim {\frac {4^{n}}{\sqrt {\pi n}}}.}$

The latter can also be easily established by means of Stirling's formula. On the other hand, it can also be used as a means to determine the constant ${\displaystyle {\sqrt {2\pi }}}$ in front of the Stirling formula, by comparison.

Simple bounds that immediately follow from ${\displaystyle 4^{n}=(1+1)^{2n}=\sum _{k=0}^{2n}{\binom {2n}{k}}}$ are

${\displaystyle {\frac {4^{n}}{2n+1}}\leq {2n \choose n}\leq 4^{n}{\text{ for all }}n\geq 1}$

Some better bounds are[2]

${\displaystyle {\frac {4^{n}}{\sqrt {4n}}}\leq {2n \choose n}\leq {\frac {4^{n}}{\sqrt {3n+1}}}{\text{ for all }}n\geq 1}$

and, if more accuracy is required,

${\displaystyle {2n \choose n}={\frac {4^{n}}{\sqrt {\pi n}}}\left(1-{\frac {c_{n}}{n}}\right){\text{ where }}{\frac {1}{9}}<c_{n}<{\frac {1}{8}}}$ for all ${\displaystyle n\geq 1.}$

The only central binomial coefficient that is odd is 1. More specifically, the number of factors of 2 in ${\displaystyle {\binom {2n}{n}}}$ is equal to the number of ones in the binary representation of n.[3] Half the central binomial coefficient ${\displaystyle {\frac {1}{2}}{2n \choose n}={2n-1 \choose n-1}}$ (for ${\displaystyle n>0}$) (sequence A001700 in the OEIS) is seen in Wolstenholme's theorem.

By the Erdős squarefree conjecture, proven in 1996, no central binomial coefficient with n > 4 is squarefree.

The central binomial coefficient ${\displaystyle {2n \choose n}}$ equals the sum of the squares of the elements in row n of Pascal's triangle.[1]

The closely related Catalan numbers C_n are given by:

${\displaystyle C_{n}={\frac {1}{n+1}}{2n \choose n}={2n \choose n}-{2n \choose n+1}{\text{ for all }}n\geq 0.}$

A slight generalization of central binomial coefficients is to take them as ${\displaystyle {\frac {\Gamma (2n+1)}{\Gamma (n+1)^{2}}}={\frac {1}{n\mathrm {B} (n+1,n)}}}$, with appropriate real numbers n, where ${\displaystyle \Gamma (x)}$ is the gamma function and ${\displaystyle \mathrm {B} (x,y)}$ is the beta function.

The powers of two that divide the central binomial coefficients are given by Gould's sequence, whose nth element is the number of odd integers in row n of Pascal's triangle.

• Koshy, Thomas (2008), Catalan Numbers with Applications, Oxford University Press, ISBN 978-0-19533-454-8.

• Central binomial coefficient at PlanetMath.
• Binomial coefficient at PlanetMath.
• Pascal's triangle at PlanetMath.
• Catalan numbers at PlanetMath.

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