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Central binomial coefficient

Summary

In mathematics the nth central binomial coefficient is the particular binomial coefficient

Pascal's triangle, rows 0 through 7. The numbers in the central column are the central binomial coefficients.
${\displaystyle {2n \choose n}={\frac {(2n)!}{(n!)^{2}}}=\prod \limits _{k=1}^{n}{\frac {n+k}{k}}{\text{ for all }}n\geq 0.}$

They are called central since they show up exactly in the middle of the even-numbered rows in Pascal's triangle. The first few central binomial coefficients starting at n = 0 are:

1, 2, 6, 20, 70, 252, 924, 3432, 12870, 48620, ...; (sequence A000984 in the OEIS)

Properties

The central binomial coefficients represent the number of combinations of a set where there are an equal number of two types of objects.

For example, ${\displaystyle n=2}$  represents AABB, ABAB, ABBA, BAAB, BABA, BBAA.

They also represent the number of combinations of A and B where there are never more B 's than A 's.

For example, ${\displaystyle n=2}$  represents AAAA, AAAB, AABA, AABB, ABAA, ABAB.

The number of factors of 2 in ${\displaystyle {\binom {2n}{n}}}$  is equal to the number of ones in the binary representation of n,[1] so 1 is the only odd central binomial coefficient.

Generating function

The ordinary generating function for the central binomial coefficients is

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

This can be proved using the binomial series and the relation
${\displaystyle {\binom {2n}{n}}=(-1)^{n}4^{n}{\binom {-1/2}{n}},}$

where ${\displaystyle \textstyle {\binom {-1/2}{n}}}$  is a generalized binomial coefficient.[2]

The central binomial coefficients have 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.[3]

The generating function of the squares of the central binomial coefficients can be written in terms of the complete elliptic integral of the first kind:[citation needed]

${\displaystyle \sum _{n=0}^{\infty }{\binom {2n}{n}}^{2}x^{n}={\frac {4}{\pi (1+{\sqrt {1-16x}})}}K\left({\frac {1-{\sqrt {1-16x}}}{1+{\sqrt {1-16x}}}}\right).}$

Asymptotic growth

The Wallis product can be written using limits:

${\displaystyle {\frac {\pi }{2}}=\lim _{n\to \infty }\prod _{k=1}^{n}{\frac {2k\cdot 2k}{(2k-1)(2k+1)}}=\lim _{n\to \infty }{\frac {4^{n}n!^{2}}{(2n-1)!!(2n+1)!!}}=\lim _{n\to \infty }4^{n}n!^{2}{\frac {2^{2n}n!^{2}}{(2n)!^{2}(2n+1)}}}$

because ${\displaystyle (2n)!=2^{n}n!(2n-1)!!}$ .

Taking the square root of both sides gives the asymptote for the central binomial coefficient:

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

The latter can also be 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.

Approximations

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

${\displaystyle {\frac {4^{n}}{2n+1}}\leq {2n \choose n}\leq 4^{n}{\text{ for all }}n\geq 0}$

Some better bounds are

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

Related sequences

The closely related Catalan numbers Cn 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.

Squaring the generating function gives

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

Comparing the coefficients of ${\displaystyle x^{n}}$  gives

${\displaystyle \sum _{k=0}^{n}{\binom {2k}{k}}{\binom {2n-2k}{n-k}}=4^{n}}$

For example, ${\displaystyle 64=1(20)+2(6)+6(2)+20(1)}$ . (sequence A000302 in the OEIS)

Other information

Half the central binomial coefficient ${\displaystyle \textstyle {\frac {1}{2}}{2n \choose n}={2n-1 \choose n-1}}$  (for ${\displaystyle n>0}$ ) (sequence A001700 in the OEIS) is seen in Wolstenholme's theorem.

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

${\displaystyle \textstyle {\binom {2n}{n}}}$  is the sum of the squares of the n-th row of Pascal's Triangle:[3]

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

For example, ${\displaystyle {\tbinom {6}{3}}=20=1^{2}+3^{2}+3^{2}+1^{2}}$ .

Erdős uses central binomial coefficients extensively in his proof of Bertrand's postulate.

Another noteworthy fact is that the power of 2 dividing ${\displaystyle (n+1)\dots (2n)}$  is exactly n.