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Catalan number

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Recursive integer sequence

Not to be confused withCatalan's constant.

TheC₅ = 42noncrossing partitions of a 5-element set (below, the other 10 of the52 partitions)

TheCatalan numbers are asequence ofnatural numbers that occur in variouscounting problems, often involvingrecursively defined objects. They are named afterEugène Catalan, though they were previously discovered in the 1730s byMinggatu.

Then-th Catalan number can be expressed directly in terms of thecentral binomial coefficients by

C_{n}={\frac {1}{n+1}}{2n \choose n}={\frac {(2n)!}{(n+1)!\,n!}}\qquad {\text{for }}n\geq 0.

The first Catalan numbers forn = 0, 1, 2, 3, ... are

1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, ... (sequenceA000108 in theOEIS).

Properties

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An alternative expression forC_n is

C_{n}={2n \choose n}-{2n \choose n+1}

for

n\geq 0\,,

which is equivalent to the expression given above because ${\tbinom {2n}{n+1}}={\tfrac {n}{n+1}}{\tbinom {2n}{n}}$ . This expression shows thatC_n is aninteger, which is not immediately obvious from the first formula given. This expression forms the basis for aproof of the correctness of the formula.

Another alternative expression is

C_{n}={\frac {1}{2n+1}}{2n+1 \choose n}\,,

which can be directly interpreted in terms of thecycle lemma; see below.

The Catalan numbers satisfy therecurrence relations

C_{0}=1\quad {\text{and}}\quad C_{n}=\sum _{i=1}^{n}C_{i-1}C_{n-i}\quad {\text{for }}n>0

and

C_{0}=1\quad {\text{and}}\quad C_{n}={\frac {2(2n-1)}{n+1}}C_{n-1}\quad {\text{for }}n>0.

Asymptotically, the Catalan numbers grow as $C_{n}\sim {\frac {4^{n}}{n^{3/2}{\sqrt {\pi }}}}\,,$ in the sense that the quotient of then-th Catalan number and the expression on the right tends towards 1 asn approaches infinity.

This can be proved by using theasymptotic growth of the central binomial coefficients, byStirling's approximation for $n! {\displaystyle n!}$ , orvia generating functions.

The only Catalan numbersC_n that are odd are those for whichn = 2^k − 1; all others are even. The only prime Catalan numbers areC₂ = 2 andC₃ = 5.^[1] More generally, the multiplicity with which a primep dividesC_n can be determined by first expressingn + 1 in basep. Forp = 2, the multiplicity is the number of 1 bits, minus 1. Forp an odd prime, count all digits greater than(p + 1) / 2; also count digits equal to(p + 1) / 2 unless final; and count digits equal to(p − 1) / 2 if not final and the next digit is counted.^[2] The only known odd Catalan numbers that do not have last digit 5 areC₀ = 1,C₁ = 1,C₇ = 429,C₃₁,C₁₂₇ andC₂₅₅. The odd Catalan numbers,C_n forn = 2^k − 1, do not have last digit 5 ifn + 1 has a base 5 representation containing 0, 1 and 2 only, except in the least significant place, which could also be a 3.^[3]

The Catalan numbers have the integral representations^[4]^[5]

C_{n}={\frac {1}{2\pi }}\int _{0}^{4}x^{n}{\sqrt {\frac {4-x}{x}}}\,dx\,={\frac {2}{\pi }}4^{n}\int _{-1}^{1}t^{2n}{\sqrt {1-t^{2}}}\,dt.

which immediately yields $\sum _{n=0}^{\infty }{\frac {C_{n}}{4^{n}}}=2$ .

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 $2k+1$ is $C_{k}$ . Since the 1D random walk is recurrent, the probability that the walker eventually arrives at −1 is $\sum _{n=0}^{\infty }{\frac {C_{n}}{2^{2n+1}}}=1$ .

Applications in combinatorics

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There are many counting problems incombinatorics whose solution is given by the Catalan numbers. The bookEnumerative Combinatorics: Volume 2 by combinatorialistRichard P. Stanley contains a set of exercises which describe 66 different interpretations of the Catalan numbers. Following are some examples, with illustrations of the casesC₃ = 5 andC₄ = 14.

Lattice of the 14 Dyck words of length 8 –( and) interpreted asup anddown

C_n is the number ofDyck words^[6] of length2n. A Dyck word is astring consisting ofn X's andn Y's such that no initial segment of the string has more Y's than X's. For example, the following are the Dyck words up to length 6:

- XY
- XXYY
- XYXY
- XXXYYY
- XYXXYY
- XYXYXY
- XXYYXY
- XXYXYY

Re-interpreting the symbol X as an opening parenthesis and Y as a closing parenthesis,C_n counts the number of expressions containingn pairs of parentheses which are correctly matched. For instance, forn = 3 these are

((()))
(()())
(())()
()(())
()()()

C_n is the number of different waysn + 1 factors can be completely parenthesized, i.e. the number of ways ofassociatingn applications of abinary operator (as in thematrix chain multiplication problem). Forn = 3, for example, we have the following five different complete parenthesizations of four factors:

((ab)c)d
(a(bc))d
(ab)(cd)
a((bc)d)
a(b(cd))

Successive applications of a binary operator can be represented in terms of afull binary tree, by labeling each leafa,b,c,d. It follows thatC_n is the number of full binarytrees withn + 1 leaves, or, equivalently, with a total ofn internal nodes:

Theassociahedron of order 4 with the C₄=14 full binary trees with 5 leaves

C_n is the number of non-isomorphicordered (or plane) trees withn + 1 vertices.^[7] Seeencoding ordered trees as binary trees. For example,C_n is the number of possibleparse trees for a sentence (assuming binary branching), in natural language processing.
C_n is the number of monotoniclattice paths along the edges of a grid withn ×n square cells, which do not pass above the diagonal. A monotonic path is one which starts in the lower left corner, finishes in the upper right corner, and consists entirely of edges pointing rightwards or upwards. Counting such paths is equivalent to counting Dyck words: X stands for "move right" and Y stands for "move up".

The following diagrams show the casen = 4:

This can be represented by listing the Catalan elements by column height:^[8]

- [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]

The dark triangle is the root node, the light triangles correspond to internal nodes of the binary trees, and the green bars are the leaves.

Aconvex polygon withn + 2 sides can be cut intotriangles by connecting vertices with non-crossingline segments (a form ofpolygon triangulation). The number of triangles formed isn and the number of different ways that this can be achieved isC_n. The following hexagons illustrate the casen = 4:

C_n is the number ofstack-sortablepermutations of{1, ...,n}. A permutationw is calledstack-sortable ifS(w) = (1, ...,n), whereS(w) is defined recursively as follows: writew =unv wheren is the largest element inw andu andv are shorter sequences, and setS(w) =S(u)S(v)n, withS being the identity for one-element sequences.
C_n is the number of permutations of{1, ...,n} that avoid thepermutation pattern 123 (or, alternatively, any of the other patterns of length 3); that is, the number of permutations with no three-term increasing subsequence. Forn = 3, these permutations are 132, 213, 231, 312 and 321. Forn = 4, they are 1432, 2143, 2413, 2431, 3142, 3214, 3241, 3412, 3421, 4132, 4213, 4231, 4312 and 4321.
C_n is the number ofnoncrossing partitions of the set{1, ...,n}.A fortiori,C_n never exceeds then-thBell number.C_n is also the number of noncrossing partitions of the set{1, ..., 2n} in which every block is of size 2.
C_n is the number of ways to tile a stairstep shape of heightn withn rectangles. Cutting across the anti-diagonal and looking at only the edges gives full binary trees. The following figure illustrates the casen = 4:

C_n is the number of ways to form a "mountain range" withn upstrokes andn downstrokes that all stay above a horizontal line. The mountain range interpretation is that the mountains will never go below the horizon.

Mountain Ranges
$n=0:$	*	1 way
$n=1:$	/\	1 way
$n=2:$	/\ /\/\,/\	2 ways
$n=3:$	/\ /\/\/\/\/\ /\/\/\,/\/\,/\/\,/\,/\	5 ways

C_n is the number ofstandard Young tableaux whose diagram is a 2-by-n rectangle. In other words, it is the number of ways the numbers1, 2, ..., 2n can be arranged in a 2-by-n rectangle so that each row and each column is increasing. As such, the formula can be derived as a special case of thehook-length formula.

123   124   125   134   135456   356   346   256   246

$C_{n}$ is the number of lengthn sequences that start with $1 {\displaystyle 1}$ , and can increase by either $0 {\displaystyle 0}$ or $1 {\displaystyle 1}$ , or decrease by any number (to at least $1 {\displaystyle 1}$ ). For $n=4$ these are $1234,1233,1232,1231,1223,1222,1221,1212,1211,1123,1122,1121,1112,1111$ . From a Dyck path, start a counter at0. An X increases the counter by1 and a Y decreases it by1. Record the values at only the X's. Compared to the similar representation of theBell numbers, only $1213 {\displaystyle 1213}$ is missing.

Proof of the formula

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There are several ways of explaining why the formula

C_{n}={\frac {1}{n+1}}{2n \choose n}

solves the combinatorial problems listed above. The first proof below uses agenerating function. The other proofs are examples ofbijective proofs; they involve literally counting a collection of some kind of object to arrive at the correct formula.

First proof

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We first observe that all of the combinatorial problems listed above satisfySegner's^[9]recurrence relation

C_{0}=1\quad {\text{and}}\quad C_{n+1}=\sum _{i=0}^{n}C_{i}\,C_{n-i}\quad {\text{for }}n\geq 0.

For example, every Dyck wordw of length ≥ 2 can be written in a unique way in the form

w = Xw₁Yw₂

with (possibly empty) Dyck wordsw₁ andw₂.

Thegenerating function for the Catalan numbers is defined by

c(x)=\sum _{n=0}^{\infty }C_{n}x^{n}.

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 intopower series. On the one hand, the recurrence relation uniquely determines the Catalan numbers; on the other hand, interpretingxc² −c + 1 = 0 as aquadratic equation ofc and using thequadratic formula, the generating function relation can be algebraically solved to yield two solution possibilities

c(x)={\frac {1+{\sqrt {1-4x}}}{2x}}

c(x)={\frac {1-{\sqrt {1-4x}}}{2x}}

From the two possibilities, the second must be chosen because only the second gives

C_{0}=\lim _{x\to 0}c(x)=1

The square root term can be expanded as a power series using thebinomial series

${\begin{aligned}1-{\sqrt {1-4x}}&=-\sum _{n=1}^{\infty }{\binom {1/2}{n}}(-4x)^{n}=-\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}(2n-3)!!}{2^{n}n!}}(-4x)^{n}\\&=-\sum _{n=0}^{\infty }{\frac {(-1)^{n}(2n-1)!!}{2^{n+1}(n+1)!}}(-4x)^{n+1}=\sum _{n=0}^{\infty }{\frac {2^{n+1}(2n-1)!!}{(n+1)!}}x^{n+1}\\&=\sum _{n=0}^{\infty }{\frac {2(2n)!}{(n+1)!n!}}x^{n+1}=\sum _{n=0}^{\infty }{\frac {2}{n+1}}{\binom {2n}{n}}x^{n+1}\,.\end{aligned}}$ Thus, $c(x)={\frac {1-{\sqrt {1-4x}}}{2x}}=\sum _{n=0}^{\infty }{\frac {1}{n+1}}{\binom {2n}{n}}x^{n}\,.$

Second proof

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Figure 1. The invalid portion of the path (dotted red) is flipped (solid red). Bad paths (after the flip) reach(n − 1,n + 1) instead of(n,n).

Call a bad path one that starts at $(x,y)=(0,0)$ , ends at $(n,n)$ , is monotonic, and contains a point above the $y=x$ line. We count the number of bad paths by establishing a bijection with paths that start at $(0,0)$ , end at $(n-1,n+1)$ , and are monotonic.

For a given bad path, construct a reflected path as follows. Let $P {\displaystyle P}$ be the first point on the bad path intersecting the line $y=x+1$ . The bad path from $(0,0)$ to $P {\displaystyle P}$ is the beginning of the reflected path. The part of the bad path from $P {\displaystyle P}$ to $(n,n)$ reflected across the line $y=x+1$ is the rest of the reflected path. See the illustration for an example. The black line is the points shared between the two paths, the dotted red line is the rest of the bad path, the solid red line is the rest of the reflected path.

This is a bijection because every monotonic path from $(0,0)$ to $(n-1,n+1)$ is constructable from a bad path, and every reflected path is uniquely invertible by finding the unique point $P {\displaystyle P}$ , which must exist because every such path must intersect $y=x+1$ .

The number of steps in the reflected path is $(n-1)+(n+1)=2n$ . The number of upward steps is $n+1$ because the path is monotonic and starts at $y=0$ and ends at $y=n+1$ .

The number of reflected paths can be counted in the usual way, by counting how many way upward steps may be distributed among total steps, which is

{2n \choose n+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,

C_{n}={2n \choose n}-{2n \choose n+1}={\frac {1}{n+1}}{2n \choose n}.

This proof can be restated in terms of Dyck words. We start with a (non-Dyck) sequence ofn X's andn Y's and interchange all X's and Y's after the first Y that violates the Dyck condition.

Third proof

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This bijective proof provides a natural explanation for the termn + 1 appearing in the denominator of the formula for C_n. A generalized version of this proof can be found in a paper of Rukavicka Josef (2011).^[10]

Given a monotonic path, theexceedance of the path is defined to be the number ofvertical 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 is1 less than the one we started with.

Starting from the bottom left, follow the path until it first travels above the diagonal.
Continue to follow the path until ittouches the diagonal again. Denote byX the first such edge that is reached.
Swap the portion of the path occurring beforeX with the portion occurring afterX.

In Figure 3, the black dot indicates the point where the path first crosses the diagonal. The black edge isX, 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.

Figure 3. The green and red portions are being exchanged.

The exceedance has dropped from3 to2. In fact, the algorithm causes the exceedance to decrease by1 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.

Figure 4. All monotonic paths in a 3×3 grid, illustrating the exceedance-decreasing algorithm.

It can be seen that this process isreversible: given any pathP whose exceedance is less thann, there is exactly one path which yieldsP when the algorithm is applied to it. Indeed, the (black) edgeX, which originally was the first horizontal step ending on the diagonal, has become thelast horizontal stepstarting on the diagonal. Alternatively, reverse the original algorithm to look for the first edge that passesbelow the diagonal.

This implies that the number of paths of exceedancen is equal to the number of paths of exceedancen − 1, which is equal to the number of paths of exceedancen − 2, and so on, down to zero. In other words, we have split up the set ofall monotonic paths inton + 1 equally sized classes, corresponding to the possible exceedances between 0 andn. Since there are $\textstyle {2n \choose n}$ monotonic paths, we obtain the desired formula $\textstyle C_{n}={\frac {1}{n+1}}{2n \choose n}.$

Figure 4 illustrates the situation for n = 3. 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 C₃ = 5, and the last column displays all paths no higher than the diagonal.

Using Dyck words, start with a sequence from $\textstyle {\binom {2n}{n}}$ . Let $X_{d}$ be the firstX that brings an initial subsequence to equality, and configure the sequence as $(F)X_{d}(L)$ . The new sequence is $L X F {\displaystyle LXF}$ .

Fourth proof

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This proof uses the triangulation definition of Catalan numbers to establish a relation betweenC_n andC_n+1.

Given a polygonP withn + 2 sides and a triangulation, mark one of its sides as the base, and also orient one of its2n + 1 total edges. There are(4n + 2)C_n such marked triangulations for a given base.

Given a polygonQ withn + 3 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(n + 2)C_{n + 1} such marked triangulations for a given base.

There is a simple bijection between these two marked triangulations: We can either collapse the triangle inQ whose side is marked (in two ways, and subtract the two that cannot collapse the base), or, in reverse, expand the oriented edge inP to a triangle and mark its new side.

Thus

(4n+2)C_{n}=(n+2)C_{n+1}

Write $\textstyle {\frac {4n-2}{n+1}}C_{n-1}=C_{n}.$

Because

(2n)!=(2n)!!(2n-1)!!=2^{n}n!(2n-1)!!

we have

{\frac {(2n)!}{n!}}=2^{n}(2n-1)!!=(4n-2)!!!!.

Applying the recursion with $C_{0}=1$ gives the result.

Fifth proof

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This proof is based on theDyck words interpretation of the Catalan numbers, so $C_{n}$ is the number of ways to correctly matchn pairs of brackets. We denote a (possibly empty) correct string withc and its inverse withc'. Since anyc can be uniquely decomposed into $c=(c_{1})c_{2}$ , summing over the possible lengths of $c_{1}$ immediately gives the recursive definition

C_{0}=1\quad {\text{and}}\quad C_{n+1}=\sum _{i=0}^{n}C_{i}\,C_{n-i}\quad {\text{for }}n\geq 0

Letb be a balanced string of length2n, i.e.b contains an equal number of $($ and $)$ , so $\textstyle B_{n}={2n \choose n}$ . A balanced string can also be uniquely decomposed into either $(c)b$ or $)c'(b$ , so

B_{n+1}=2\sum _{i=0}^{n}B_{i}C_{n-i}.

Any incorrect (non-Catalan) balanced string starts with $c)$ , and the remaining string has one more $($ than $)$ , so

B_{n+1}-C_{n+1}=\sum _{i=0}^{n}{2i+1 \choose i}C_{n-i}

Also, from the definitions, we have:

B_{n+1}-C_{n+1}=2\sum _{i=0}^{n}B_{i}C_{n-i}-\sum _{i=0}^{n}C_{i}\,C_{n-i}=\sum _{i=0}^{n}(2B_{i}-C_{i})C_{n-i}.

Therefore, as this is true for alln,

2B_{i}-C_{i}={\binom {2i+1}{i}}

C_{i}=2B_{i}-{\binom {2i+1}{i}}

C_{i}=2{\binom {2i}{i}}-{\binom {2i+1}{i}}

C_{i}={\frac {1}{i+1}}{\binom {2i}{i}}

Sixth proof

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This proof is based on theDyck words interpretation of the Catalan numbers and uses thecycle lemma of Dvoretzky and Motzkin.^[11]^[12]

We call a sequence of X's and Y'sdominating if, reading from left to right, the number of X's is always strictly greater than the number of Y's. The cycle lemma^[13] states that any sequence of $m {\displaystyle m}$ X's and $n {\displaystyle n}$ Y's, where $m>n$ , has precisely $m-n$ dominatingcircular shifts. To see this, arrange the given sequence of $m+n$ X's and Y's in a circle. Repeatedly removing XY pairs leaves exactly $m-n$ X's. Each of these X's was the start of a dominating circular shift before anything was removed. For example, consider ${\mathit {XXYXY}}$ . This sequence is dominating, but none of its circular shifts ${\mathit {XYXYX}}$ , ${\mathit {YXYXX}}$ , ${\mathit {XYXXY}}$ and ${\mathit {YXXYX}}$ are.

A string is a Dyck word of $n {\displaystyle n}$ X's and $n {\displaystyle n}$ Y's if and only if prepending an X to the Dyck word gives a dominating sequence with $n+1$ X's and $n {\displaystyle n}$ Y's, so we can count the former by instead counting the latter. In particular, when $m=n+1$ , there is exactly one dominating circular shift. There are $\textstyle {2n+1 \choose n}$ sequences with exactly $n+1$ X's and $n {\displaystyle n}$ Y's. For each of these, only one of the $2n+1$ circular shifts is dominating. Therefore there are $\textstyle {\frac {1}{2n+1}}{2n+1 \choose n}=C_{n}$ distinct sequences of $n+1$ X's and $n {\displaystyle n}$ Y's that are dominating, each of which corresponds to exactly one Dyck word.

Hankel matrix

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Then ×nHankel matrix whose(i,j) entry is the Catalan numberC_i+j−2 hasdeterminant 1, regardless of the value ofn. For example, forn = 4 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(i,j) entry is filled with the Catalan numberC_i+j−1 then the determinant is still 1, regardless of the value ofn.For example, forn = 4 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 then ×n submatrix starting at2 has determinantn + 1.

\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.

History

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Catalan numbers in Mingantu's bookThe Quick Method for Obtaining the Precise Ratio of Division of a Circle volume III

The Catalan sequence was described in 1751 byLeonhard Euler, who was interested in the number of different ways of dividing a polygon into triangles. The sequence is named afterEugène Charles Catalan, who discovered the connection to parenthesized expressions during his exploration of theTowers of Hanoi puzzle. The reflection counting trick (second proof) for Dyck words was found byDésiré André in 1887.

The name “Catalan numbers” originated fromJohn Riordan.^[14]

In 1988, it came to light that the Catalan number sequence had been used inChina by the Mongolian mathematicianMingantu by 1730, when he started to write his bookGe 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.^[15]^[16] 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, Mingantu used the Catalan sequence to express series expansions of $\sin(2\alpha )$ and $\sin(4\alpha )$ in terms of $\sin(\alpha )$ .

Generalizations

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The Catalan numbers can be interpreted as a special case of theBertrand's ballot theorem. Specifically, $C_{n}$ is the number of ways for a candidate A withn + 1 votes to lead candidate B withn votes.

The two-parameter sequence of non-negative integers ${\frac {(2m)!(2n)!}{(m+n)!m!n!}}$ is a generalization of the Catalan numbers. These are namedsuper-Catalan numbers, perIra Gessel. These should not confused with theSchröder–Hipparchus numbers, which sometimes are also called super-Catalan numbers.

For $m=1$ , this is just two times the ordinary Catalan numbers, and for $m=n$ , the numbers have an easy combinatorial description.However, other combinatorial descriptions are only known^[17]for $m=2,3$ and $4 {\displaystyle 4}$ ,^[18]and it is an open problem to find a general combinatorial interpretation.

Sergey Fomin and Nathan Reading have given a generalized Catalan number associated to any finite crystallographicCoxeter group, namely the number of fully commutative elements of the group; in terms of the associatedroot system, it is the number of anti-chains (or order ideals) in the poset of positive roots. The classical Catalan number $C_{n}$ corresponds to the root system of type $A_{n}$ . The classical recurrence relation generalizes: the Catalan number of a Coxeter diagram is equal to the sum of the Catalan numbers of all its maximal proper sub-diagrams.^[19]

The Catalan numbers are a solution of a version of theHausdorff moment problem.^[20]

For coprime positive integersr ands, therational Catalan numbers ${\frac {1}{r+s}}{\binom {r+s}{r}}$ count the number of lattice paths with steps of unit length rightwards and upwards from(0,0) to(r,s) that never go above the linery =sx.^[21]

Catalan k-fold convolution

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The Catalank-fold convolution is:

\sum _{i_{1}+\cdots +i_{k}=n \atop i_{1},\ldots ,i_{k}\geq 0}C_{i_{1}}\cdots C_{i_{k}}={\dfrac {k}{2n+k}}{\binom {2n+k}{n}}

Notes

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^Koshy, Thomas; Salmassi, Mohammad (2006)."Parity and primality of Catalan numbers"(PDF).The College Mathematics Journal.37 (1):52–53.doi:10.2307/27646275.JSTOR 27646275. Archived fromthe original(PDF) on 2021-02-09. Retrieved2019-03-04.
^Sloane, N. J. A. (ed.)."Sequence A000108 (Catalan numbers)".TheOn-Line Encyclopedia of Integer Sequences. OEIS Foundation.
^"Catalan Number".
^Choi, Hayoung; Yeh, Yeong-Nan; Yoo, Seonguk (2020), "Catalan-like number sequences and Hausdorff moment sequences",Discrete Mathematics,343 (5): 111808, 11,arXiv:1809.07523,doi:10.1016/j.disc.2019.111808,MR 4052255,S2CID 214165563, Example 3.1
^Feng, Qi; Bai-Ni, Guo (2017), "Integral Representations of the Catalan Numbers and Their Applications",Mathematics,5 (3): 40,doi:10.3390/math5030040,Theorem 1
^Dyck paths
^Stanley p.221 example (e)
^Črepinšek, Matej; Mernik, Luka (2009)."An efficient representation for solving Catalan number related problems"(PDF).International Journal of Pure and Applied Mathematics.56 (4):589–604.
^A. de Segner, Enumeratio modorum, quibus figurae planae rectilineae per diagonales dividuntur in triangula.Novi commentarii academiae scientiarum Petropolitanae7 (1758/59) 203–209.
^Rukavicka Josef (2011),On Generalized Dyck Paths, Electronic Journal of Combinatoricsonline
^Dershowitz, Nachum; Zaks, Shmuel (1980), "Enumerations of ordered trees",Discrete Mathematics,31:9–28,doi:10.1016/0012-365x(80)90168-5,hdl:2027/uiuo.ark:/13960/t3kw6z60d
^Dvoretzky, Aryeh; Motzkin, Theodore (1947), "A problem of arrangements",Duke Mathematical Journal,14 (2):305–313,doi:10.1215/s0012-7094-47-01423-3
^Dershowitz, Nachum; Zaks, Shmuel (January 1990)."The Cycle Lemma and Some Applications"(PDF).European Journal of Combinatorics.11 (1):35–40.doi:10.1016/S0195-6698(13)80053-4.
^Stanley, Richard P. (2021). "Enumerative and Algebraic Combinatorics in the 1960's and 1970's".arXiv:2105.07884 [math.HO].
^Larcombe, Peter J."The 18th century Chinese discovery of the Catalan numbers"(PDF).
^"Ming Antu, the First Inventor of Catalan Numbers in the World". Archived fromthe original on 2020-01-31. Retrieved2014-06-24.
^Chen, Xin; Wang, Jane (2012). "The super Catalan numbers S(m, m + s) for s ≤ 4".arXiv:1208.4196 [math.CO].
^Gheorghiciuc, Irina; Orelowitz, Gidon (2020). "Super-Catalan Numbers of the Third and Fourth Kind".arXiv:2008.00133 [math.CO].
^Sergey Fomin and Nathan Reading, "Root systems and generalized associahedra", Geometric combinatorics, IAS/Park City Math. Ser.13,American Mathematical Society, Providence, RI, 2007, pp 63–131.arXiv:math/0505518
^Choi, Hayoung; Yeh, Yeong-Nan; Yoo, Seonguk (2020), "Catalan-like number sequences and Hausdorff moment sequences",Discrete Mathematics,343 (5): 111808, 11,arXiv:1809.07523,doi:10.1016/j.disc.2019.111808,MR 4052255,S2CID 214165563
^Krattenthaler, Christian (2015)."Lattice Path Enumeration"(PDF). InBóna, Miklós (ed.).Handbook of Enumerative Combinatorics. Discrete Mathematics and Its Applications (1 ed.). CRC Press. p. 598.ISBN 9780429170317.

References

[edit]

Stanley, Richard P. (2015),Catalan numbers. Cambridge University Press,ISBN 978-1-107-42774-7.
Conway andGuy (1996)The Book of Numbers. New York: Copernicus, pp. 96–106.
Gardner, Martin (1988),Time Travel and Other Mathematical Bewilderments, New York: W.H. Freeman and Company, pp. 253–266 (Ch. 20),Bibcode:1988ttom.book.....G,ISBN 0-7167-1924-X
Koshy, Thomas (2008),Catalan Numbers with Applications, Oxford University Press,ISBN 978-0-19-533454-8
Koshy, Thomas & Zhenguang Gao (2011) "Some divisibility properties of Catalan numbers",Mathematical Gazette 95:96–102.
Larcombe, P.J. (1999)."The 18th century Chinese discovery of the Catalan numbers"(PDF).Mathematical Spectrum.32:5–7.
Stanley, Richard P. (1999),Enumerative combinatorics. Vol. 2, Cambridge Studies in Advanced Mathematics, vol. 62,Cambridge University Press,ISBN 978-0-521-56069-6,MR 1676282
Egecioglu, Omer (2009),A Catalan–Hankel Determinant Evaluation(PDF)
Gheorghiciuc, Irina; Orelowitz, Gidon (2020),Super-Catalan Numbers of the Third and Fourth Kind,arXiv:2008.00133

External links

[edit]

Stanley, Richard P. (1998),Catalan addendum to Enumerative Combinatorics, Volume 2(PDF)
Weisstein, Eric W."Catalan Number".MathWorld.
Davis, Tom:Catalan numbers. Still more examples.
"Equivalence of Three Catalan Number Interpretations" from The Wolfram Demonstrations Project[1]
Learning materials related toPartition related number triangles at Wikiversity

Classes ofnatural numbers

Powers and related numbers
Achilles Power of 2 Power of 3 Power of 10 Square Cube Fourth power Fifth power Sixth power Seventh power Eighth power Perfect power Powerful Prime power

Of the forma × 2^b ± 1
Cullen Double Mersenne Fermat Mersenne Proth Thabit Woodall

Other polynomial numbers
Hilbert Idoneal Leyland Loeschian Lucky numbers of Euler

Recursively defined numbers
Fibonacci Jacobsthal Leonardo Lucas Narayana Padovan Pell Perrin Graham

Possessing a specific set of other numbers
Amenable Congruent Knödel Riesel Sierpiński

Expressible via specific sums
Nonhypotenuse Polite Practical Primary pseudoperfect Ulam Wolstenholme

Figurate numbers

2-dimensional

centered	Centered triangular Centered square Centered pentagonal Centered hexagonal Centered heptagonal Centered octagonal Centered nonagonal Centered decagonal Star
non-centered	Triangular Square Square triangular Pentagonal Hexagonal Heptagonal Octagonal Nonagonal Decagonal Dodecagonal

3-dimensional

centered	Centered tetrahedral Centered cube Centered octahedral Centered dodecahedral Centered icosahedral
non-centered	Tetrahedral Cubic Octahedral Dodecahedral Icosahedral Stella octangula
pyramidal	Square pyramidal

4-dimensional

non-centered	Pentatope Squared triangular Tesseractic

Combinatorial numbers
Bell Cake Catalan Dedekind Delannoy Euler Eulerian Fuss–Catalan Lah Lazy caterer's sequence Lobb Motzkin Narayana Ordered Bell Schröder Schröder–Hipparchus Stirling first Stirling second Telephone number Wedderburn–Etherington

Primes
Wieferich Wall–Sun–Sun Wolstenholme prime Wilson

Pseudoprimes
Carmichael number Catalan pseudoprime Elliptic pseudoprime Euler pseudoprime Euler–Jacobi pseudoprime Fermat pseudoprime Frobenius pseudoprime Lucas pseudoprime Lucas–Carmichael number Perrin pseudoprime Somer–Lucas pseudoprime Strong pseudoprime

Arithmetic functions anddynamics

Divisor functions	Abundant Almost perfect Arithmetic Betrothed Colossally abundant Deficient Descartes Hemiperfect Highly abundant Highly composite Hyperperfect Multiply perfect Perfect Practical Primitive abundant Quasiperfect Refactorable Semiperfect Sublime Superabundant Superior highly composite Superperfect
Prime omega functions	Almost prime Semiprime
Euler's totient function	Highly cototient Highly totient Noncototient Nontotient Perfect totient Sparsely totient
Aliquot sequences	Amicable Perfect Sociable Untouchable
Primorial	Euclid Fortunate

Otherprime factor ordivisor related numbers
Blum Cyclic Erdős–Nicolas Erdős–Woods Friendly Giuga Harmonic divisor Jordan–Pólya Lucas–Carmichael Pronic Regular Rough Smooth Sphenic Størmer Super-Poulet

Numeral system-dependent numbers

Arithmetic functions
anddynamics

Persistence
- Additive
- Multiplicative

Digit sum	Digit sum Digital root Self Sum-product
Digit product	Multiplicative digital root Sum-product
Coding-related	Meertens
Other	Dudeney Factorion Kaprekar Kaprekar's constant Keith Lychrel Narcissistic Perfect digit-to-digit invariant Perfect digital invariant Happy