Sequence of numbers ((2n) choose (n))
Pascal's triangle, rows 0 through 7. The numbers in the central column are the central binomial coefficients.
In mathematics the n th central binomial coefficient is the particular binomial coefficient
(
2
n
n
)
=
(
2
n
)
!
(
n
!
)
2
for all
n
≥ ≥ -->
0.
{\displaystyle {2n \choose n}={\frac {(2n)!}{(n!)^{2}}}{\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 )
Combinatorial interpretations and other properties
The central binomial coefficients give the number of possible number of assignments of n -a-side sports teams from 2n players, taking into account the playing area side
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.
Generating function
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 I 0 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]
∑ ∑ -->
n
=
0
∞ ∞ -->
(
2
n
n
)
2
x
n
=
2
π π -->
K
(
4
x
)
.
{\displaystyle \sum _{n=0}^{\infty }{\binom {2n}{n}}^{2}x^{n}={\frac {2}{\pi }}K(4{\sqrt {x}}).}
Asymptotic growth
The asymptotic behavior can be described quite accurately:[ 5]
(
2
n
n
)
=
4
n
π 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 C n are given by:
C
n
=
1
n
+
1
(
2
n
n
)
=
(
2
n
n
)
− − -->
(
2
n
n
+
1
)
for all
n
≥ ≥ -->
0.
{\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
Γ Γ -->
(
2
n
+
1
)
Γ 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 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 n th 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]
(
2
n
n
)
=
∑ ∑ -->
k
=
0
n
(
n
k
)
2
{\displaystyle {2n \choose n}=\sum _{k=0}^{n}{\binom {n}{k}}^{2}}
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 .
See also
References
^ Sloane, N. J. A. (ed.). "Sequence A000120" . The On-Line Encyclopedia of Integer Sequences . OEIS Foundation.
^ Stanley, Richard P. (2012), Enumerative Combinatorics , vol. 1 (2 ed.), Cambridge University Press, Example 1.1.15, ISBN 978-1-107-60262-5
^ a b Sloane, N. J. A. (ed.). "Sequence A000984 (Central binomial coefficients)" . The On-Line Encyclopedia of Integer Sequences . OEIS Foundation.
^ Sloane, N. J. A. (ed.). "Sequence A002894" . The On-Line Encyclopedia of Integer Sequences . OEIS Foundation.
^ Luke, Yudell L. (1969). The Special Functions and their Approximations, Vol. 1 . New York, NY, USA: Academic Press, Inc. p. 35.
Koshy, Thomas (2008), Catalan Numbers with Applications , Oxford University Press, ISBN 978-0-19533-454-8 .
External links