In mathematics the nth central binomial coefficient is the particular binomial coefficient
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:
The central binomial coefficient ( 2 n n ) {\displaystyle {\binom {2n}{n}}} is the number of arrangements where there are an equal number of two types of objects. For example, when n = 2 {\displaystyle n=2} , the binomial coefficient ( 2 ⋅ 2 2 ) {\displaystyle {\binom {2\cdot 2}{2}}} is equal to 6, and there are six arrangements of two copies of A and two copies of B: AABB, ABAB, ABBA, BAAB, BABA, BBAA.
The same central binomial coefficient ( 2 n n ) {\displaystyle {\binom {2n}{n}}} is also the number of words of length 2n made up of A and B within which, as one reads from left to right, there are never more B's than A's at any point. For example, when n = 2 {\displaystyle n=2} , there are six words of length 4 in which each prefix has at least as many copies of A as of B: AAAA, AAAB, AABA, AABB, ABAA, ABAB.
The number of factors of 2 in ( 2 n n ) {\displaystyle {\binom {2n}{n}}} is equal to the number of 1s in the binary representation of n.[1] As a consequence, 1 is the only odd central binomial coefficient.
The ordinary generating function for the central binomial coefficients is 1 1 − 4 x = ∑ n = 0 ∞ ( 2 n n ) x n = 1 + 2 x + 6 x 2 + 20 x 3 + 70 x 4 + 252 x 5 + ⋯ . {\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 ( 2 n n ) = ( − 1 ) n 4 n ( − 1 / 2 n ) , {\displaystyle {\binom {2n}{n}}=(-1)^{n}4^{n}{\binom {-1/2}{n}},} where ( − 1 / 2 n ) {\displaystyle \textstyle {\binom {-1/2}{n}}} is a generalized binomial coefficient.[2]
The central binomial coefficients have exponential generating function ∑ n = 0 ∞ ( 2 n n ) x n n ! = e 2 x I 0 ( 2 x ) , {\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:[4]
The asymptotic behavior can be described quite accurately:[5] ( 2 n n ) = 4 n π n ( 1 − 1 8 n + 1 128 n 2 + 5 1024 n 3 + O ( n − 4 ) ) . {\displaystyle {2n \choose n}={\frac {4^{n}}{\sqrt {\pi n}}}\left(1-{\frac {1}{8n}}+{\frac {1}{128n^{2}}}+{\frac {5}{1024n^{3}}}+O(n^{-4})\right).}
The closely related Catalan numbers Cn are given by:
A slight generalization of central binomial coefficients is to take them as Γ ( 2 n + 1 ) Γ ( n + 1 ) 2 = 1 n B ( n + 1 , n ) {\displaystyle {\frac {\Gamma (2n+1)}{\Gamma (n+1)^{2}}}={\frac {1}{n\mathrm {B} (n+1,n)}}} , with appropriate real numbers n, where Γ ( x ) {\displaystyle \Gamma (x)} is the gamma function and B ( x , y ) {\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 1 1 − 4 x = ( ∑ n = 0 ∞ ( 2 n n ) x n ) ( ∑ n = 0 ∞ ( 2 n n ) x n ) . {\displaystyle {\frac {1}{1-4x}}=\left(\sum _{n=0}^{\infty }{\binom {2n}{n}}x^{n}\right)\left(\sum _{n=0}^{\infty }{\binom {2n}{n}}x^{n}\right).} Comparing the coefficients of x n {\displaystyle x^{n}} gives ∑ k = 0 n ( 2 k k ) ( 2 n − 2 k n − k ) = 4 n . {\displaystyle \sum _{k=0}^{n}{\binom {2k}{k}}{\binom {2n-2k}{n-k}}=4^{n}.} For example, 64 = 1 ( 20 ) + 2 ( 6 ) + 6 ( 2 ) + 20 ( 1 ) {\displaystyle 64=1(20)+2(6)+6(2)+20(1)} (sequence A000302 in the OEIS).
The number lattice paths of length 2n that start and end at the origin is ∑ k = 0 n ( 2 k k ) ( 2 n − 2 k n − k ) ( 2 n 2 k ) {\displaystyle \sum _{k=0}^{n}{\binom {2k}{k}}{\binom {2n-2k}{n-k}}{\binom {2n}{2k}}} = ∑ k = 0 n ( 2 n ) ! k ! k ! ( n − k ) ! ( n − k ) ! {\displaystyle =\sum _{k=0}^{n}{\frac {(2n)!}{k!k!(n-k)!(n-k)!}}} = ∑ k = 0 n n ! n ! k ! k ! ( n − k ) ! ( n − k ) ! ( 2 n ) ! n ! n ! {\displaystyle =\sum _{k=0}^{n}{\frac {n!n!}{k!k!(n-k)!(n-k)!}}{\frac {(2n)!}{n!n!}}} = ( 2 n n ) ∑ k = 0 n ( n k ) 2 {\displaystyle ={\binom {2n}{n}}\sum _{k=0}^{n}{\binom {n}{k}}^{2}} = ( 2 n n ) 2 {\displaystyle ={\binom {2n}{n}}^{2}} (sequence A002894 in the OEIS).
Half the central binomial coefficient 1 2 ( 2 n n ) = ( 2 n − 1 n − 1 ) {\displaystyle \textstyle {\frac {1}{2}}{2n \choose n}={2n-1 \choose n-1}} (for n > 0 {\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.
( 2 n n ) {\displaystyle \textstyle {\binom {2n}{n}}} is the sum of the squares of the n-th row of Pascal's Triangle:[3]
For example, ( 6 3 ) = 20 = 1 2 + 3 2 + 3 2 + 1 2 {\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 ( n + 1 ) … ( 2 n ) {\displaystyle (n+1)\dots (2n)} is exactly n.