In combinatorial mathematics, the Catalan numbers are a sequence of natural numbers that occur in various counting problems, often involving recursively defined objects. They are named after the French-Belgian mathematician Eugène Charles Catalan, though they were previously discovered in the 1730s by Minggatu.
The -th Catalan number can be expressed directly in terms of the central binomial coefficients by
Cn=
1 | |
n+1 |
{2n\choosen}=
(2n)! | |
(n+1)!n! |
=
n | |
\prod\limits | |
k=2 |
n+k | |
k |
forn\ge0.
The first Catalan numbers for are
.
An alternative expression for is
Cn={2n\choosen}-{2n\choosen+1}
n\ge0,
\tbinom{2n}{n+1}=\tfrac{n}{n+1}\tbinom{2n}n
Another alternative expression is
Cn=
1 | |
2n+1 |
{2n+1\choosen},
The Catalan numbers satisfy the recurrence relations
C0=1 and Cn
n | |
=\sum | |
i=1 |
Ci-1Cn-i forn>0
C0=1 and Cn=
2(2n-1) | |
n+1 |
Cn-1 forn>0.
Asymptotically, the Catalan numbers grow asin the sense that the quotient of the -th Catalan number and the expression on the right tends towards 1 as approaches infinity.
This can be proved by using the asymptotic growth of the central binomial coefficients, by Stirling's approximation for
n!
The only Catalan numbers that are odd are those for which ; all others are even. The only prime Catalan numbers are and .[1]
The Catalan numbers have the integral representations[2] [3]
C | ||||
|
4 | |
\int | |
0 |
| ||||
x |
infty | |
\sum | |
n=0 |
Cn | |
4n |
=2
This has a simple probabilistic interpretation. Consider a random walk on the integer line, starting at 0. Let -1 be a "trap" state, such that if the walker arrives at -1, it will remain there. The walker can arrive at the trap state at times 1, 3, 5, 7..., and the number of ways the walker can arrive at the trap state at time
2k+1
Ck
infty | |
\sum | |
n=0 |
Cn | |
22n+1 |
=1
There are many counting problems in combinatorics whose solution is given by the Catalan numbers. The book Enumerative Combinatorics: Volume 2 by combinatorialist Richard P. Stanley contains a set of exercises which describe 66 different interpretations of the Catalan numbers. Following are some examples, with illustrations of the cases and .
XY
XXYY XYXY
XXXYYY XYXXYY XYXYXY XXYYXY XXYXYY
((ab)c)d (a(bc))d (ab)(cd) a((bc)d) a(b(cd))
The following diagrams show the case :This can be represented by listing the Catalan elements by column height:[6]
[0,0,0,0] [0,0,0,1] [0,0,0,2] [0,0,1,1]
[0,1,1,1] [0,0,1,2] [0,0,0,3] [0,1,1,2] [0,0,2,2] [0,0,1,3]
[0,0,2,3] [0,1,1,3] [0,1,2,2] [0,1,2,3]
n=0: | 1 way | ||
n=1: | /\ | 1 way | |
n=2: | /\ /\/\,/\ | 2 ways | |
n=3: | /\ /\/\/\/\/\ /\/\/\,/\/\,/\/\,/\,/\ | 5 ways |
123 124 125 134 135 456 356 346 256 246
Cn
1
0
1
1
n=4
1234,1233,1232,1231,1223,1222,1221,1212,1211,1123,1122,1121,1112,1111
1213
There are several ways of explaining why the formula
Cn=
1 | |
n+1 |
{2n\choosen}
We first observe that all of the combinatorial problems listed above satisfy Segner's[7] recurrence relation
C0=1 and Cn+1
n | |
=\sum | |
i=0 |
CiCn-i forn\ge0.
For example, every Dyck word of length ≥ 2 can be written in a unique way in the form
with (possibly empty) Dyck words and .
The generating function for the Catalan numbers is defined by
infty | |
c(x)=\sum | |
n=0 |
Cnxn.
The recurrence relation given above can then be summarized in generating function form by the relation
c(x)=1+xc(x)2;
in other words, this equation follows from the recurrence relation by expanding both sides into power series. On the one hand, the recurrence relation uniquely determines the Catalan numbers; on the other hand, interpreting as a quadratic equation of and using the quadratic formula, the generating function relation can be algebraically solved to yield two solution possibilities
c(x)=
1+\sqrt{1-4x | |
c(x)=
1-\sqrt{1-4x | |
From the two possibilities, the second must be chosen because only the second gives
C0=\limxc(x)=1
The square root term can be expanded as a power series using the binomial series
Thus,
We count the number of paths which start and end on the diagonal of an grid. All such paths have right and up steps. Since we can choose which of the steps are up or right, there are in total
\tbinom{2n}{n}
The part of the path after the higher diagonal is then flipped about that diagonal, as illustrated with the red dotted line. This swaps all the right steps to up steps and vice versa. In the section of the path that is not reflected, there is one more up step than right steps, so therefore the remaining section of the bad path has one more right step than up steps. When this portion of the path is reflected, it will have one more up step than right steps.
Since there are still steps, there are now up steps and right steps. So, instead of reaching, all bad paths after reflection end at . Because every monotonic path in the grid meets the higher diagonal, and because the reflection process is reversible, the reflection is therefore a bijection between bad paths in the original grid and monotonic paths in the new grid.
The number of bad paths is therefore:
{n-1+n+1\choosen-1}={2n\choosen-1}={2n\choosen+1}
and the number of Catalan paths (i.e. good paths) is obtained by removing the number of bad paths from the total number of monotonic paths of the original grid,
Cn={2n\choosen}-{2n\choosen+1}=
1 | |
n+1 |
{2n\choosen}.
In terms of Dyck words, we start with a (non-Dyck) sequence of X's and Y's and interchange all X's and Y's after the first Y that violates the Dyck condition. After this Y, note that there is exactly one more Y than there are Xs.
This bijective proof provides a natural explanation for the term appearing in the denominator of the formula for . A generalized version of this proof can be found in a paper of Rukavicka Josef (2011).[8]
Given a monotonic path, the exceedance of the path is defined to be the number of vertical edges above the diagonal. For example, in Figure 2, the edges above the diagonal are marked in red, so the exceedance of this path is 5.
Given a monotonic path whose exceedance is not zero, we apply the following algorithm to construct a new path whose exceedance is less than the one we started with.
In Figure 3, the black dot indicates the point where the path first crosses the diagonal. The black edge is, and we place the last lattice point of the red portion in the top-right corner, and the first lattice point of the green portion in the bottom-left corner, and place X accordingly, to make a new path, shown in the second diagram.
The exceedance has dropped from to . In fact, the algorithm causes the exceedance to decrease by for any path that we feed it, because the first vertical step starting on the diagonal (at the point marked with a black dot) is the only vertical edge that changes from being above the diagonal to being below it when we apply the algorithm - all the other vertical edges stay on the same side of the diagonal.
It can be seen that this process is reversible: given any path whose exceedance is less than, there is exactly one path which yields when the algorithm is applied to it. Indeed, the (black) edge, which originally was the first horizontal step ending on the diagonal, has become the last horizontal step starting on the diagonal. Alternatively, reverse the original algorithm to look for the first edge that passes below the diagonal.
This implies that the number of paths of exceedance is equal to the number of paths of exceedance, which is equal to the number of paths of exceedance, and so on, down to zero. In other words, we have split up the set of all monotonic paths into equally sized classes, corresponding to the possible exceedances between 0 and . Since there are
style{2n\choosen}
styleCn=
1 | |
n+1 |
{2n\choosen}.
Figure 4 illustrates the situation for . Each of the 20 possible monotonic paths appears somewhere in the table. The first column shows all paths of exceedance three, which lie entirely above the diagonal. The columns to the right show the result of successive applications of the algorithm, with the exceedance decreasing one unit at a time. There are five rows, that is , and the last column displays all paths no higher than the diagonal.
Using Dyck words, start with a sequence from
style\binom{2n}{n}
Xd
(F)Xd(L)
LXF
This proof uses the triangulation definition of Catalan numbers to establish a relation between and .
Given a polygon with sides and a triangulation, mark one of its sides as the base, and also orient one of its total edges. There are such marked triangulations for a given base.
Given a polygon with sides and a (different) triangulation, again mark one of its sides as the base. Mark one of the sides other than the base side (and not an inner triangle edge). There are such marked triangulations for a given base.
There is a simple bijection between these two marked triangulations: We can either collapse the triangle in whose side is marked (in two ways, and subtract the two that cannot collapse the base), or, in reverse, expand the oriented edge in to a triangle and mark its new side.
Thus
(4n+2)Cn=(n+2)Cn+1
Write
| ||||
n-1 |
=Cn.
Because
(2n)!=(2n)!!(2n-1)!!=2nn!(2n-1)!!
we have
(2n)! | |
n! |
=2n(2n-1)!!=(4n-2)!!!!.
Applying the recursion with
C0=1
This proof is based on the Dyck words interpretation of the Catalan numbers, so
Cn
c=(c1)c2
c1
C0=1 and Cn+1=
n | |
\sum | |
i=0 |
CiCn-i forn\ge0
Let be a balanced string of length, i.e. contains an equal number of
(
)
styleBn={2n\choosen}
(c)b
)c'(b
Bn+1=
n | |
2\sum | |
i=0 |
BiCn-i.
Any incorrect (non-Catalan) balanced string starts with
c)
(
)
Bn+1-Cn+1=
n | |
\sum | |
i=0 |
{2i+1\choosei}Cn-i
Also, from the definitions, we have:
Bn+1-Cn+1=
n | |
2\sum | |
i=0 |
BiCn-i-
n | |
\sum | |
i=0 |
CiCn-i=
n | |
\sum | |
i=0 |
(2Bi-Ci)Cn-i.
Therefore, as this is true for all,
2Bi-Ci=\binom{2i+1}{i}
Ci=2Bi-\binom{2i+1}{i}
Ci=2\binom{2i}{i}-\binom{2i+1}{i}
C | ||||
|
\binom{2i}{i}
This proof is based on the Dyck words interpretation of the Catalan numbers and uses the cycle lemma of Dvoretzky and Motzkin.
We call a sequence of X's and Y's dominating if, reading from left to right, the number of X's is always strictly greater than the number of Y's. The cycle lemma[9] states that any sequence of
m
n
m>n
m-n
m+n
m-n
XXYXY
XYXYX
YXYXX
XYXXY
YXXYX
A string is a Dyck word of
n
n
n+1
n
m=n+1
style{2n+1\choosen}
n+1
n
2n+1
|
\choosen}=Cn
n+1
n
The Hankel matrix whose entry is the Catalan number has determinant 1, regardless of the value of . For example, for we have
\det\begin{bmatrix}1&1&2&5\ 1&2&5&14\ 2&5&14&42\ 5&14&42&132\end{bmatrix}=1.
Moreover, if the indexing is "shifted" so that the entry is filled with the Catalan number then the determinant is still 1, regardless of the value of .For example, for we have
\det\begin{bmatrix}1&2&5&14\ 2&5&14&42\ 5&14&42&132\ 14&42&132&429\end{bmatrix}=1.
Taken together, these two conditions uniquely define the Catalan numbers.
Another feature unique to the Catalan–Hankel matrix is that the submatrix starting at has determinant .
\det\begin{bmatrix}2\end{bmatrix}=2
\det\begin{bmatrix}2&5\\5&14\end{bmatrix}=3
\det\begin{bmatrix}2&5&14\\5&14&42\ 14&42&132\end{bmatrix}=4
\det\begin{bmatrix}2&5&14&42\ 5&14&42&132\ 14&42&132&429\ 42&132&429&1430\end{bmatrix}=5
et cetera.
The Catalan sequence was described in the 18th century by Leonhard Euler, who was interested in the number of different ways of dividing a polygon into triangles. The sequence is named after Eugène Charles Catalan, who discovered the connection to parenthesized expressions during his exploration of the Towers of Hanoi puzzle. The reflection counting trick (second proof) for Dyck words was found by Désiré André in 1887.
The name “Catalan numbers” originated from John Riordan.[10]
In 1988, it came to light that the Catalan number sequence had been used in China by the Mongolian mathematician Mingantu by 1730.[11] [12] That is when he started to write his book Ge Yuan Mi Lu Jie Fa [The Quick Method for Obtaining the Precise Ratio of Division of a Circle], which was completed by his student Chen Jixin in 1774 but published sixty years later. Peter J. Larcombe (1999) sketched some of the features of the work of Mingantu, including the stimulus of Pierre Jartoux, who brought three infinite series to China early in the 1700s.
For instance, Ming used the Catalan sequence to express series expansions of
\sin(2\alpha)
\sin(4\alpha)
\sin(\alpha)
The Catalan numbers can be interpreted as a special case of the Bertrand's ballot theorem. Specifically,
Cn
The two-parameter sequence of non-negative integers
(2m)!(2n)! | |
(m+n)!m!n! |
For
m=1
m=n
m=2,3
4
Sergey Fomin and Nathan Reading have given a generalized Catalan number associated to any finite crystallographic Coxeter group, namely the number of fully commutative elements of the group; in terms of the associated root system, it is the number of anti-chains (or order ideals) in the poset of positive roots. The classical Catalan number
Cn
An
The Catalan numbers are a solution of a version of the Hausdorff moment problem.
The Catalan -fold convolution, where, is:[16]
\sum | |
i1+ … +im=n\atopi1,\ldots,im\ge0 |
C | |
i1 |
…
C | |
im |
=\begin{cases} \dfrac{m(n+1)(n+2) … (n+m/2-1)}{2(n+m/2+2)(n+m/2+3) … (n+m)}Cn+m/2,&meven,\\[5pt] \dfrac{m(n+1)(n+2) … (n+(m-1)/2)}{(n+(m+3)/2)(n+(m+3)/2+1) … (n+m)}Cn+(m-1)/2,&modd. \end{cases}