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  • 1. The Catalan Numbers: Sequence of Integer Pai Sukanya Suksak December 16, 20101. IntroductionCatalan Numbers: Sequence of Integer Figure 1 The central binomial coefficients of Pascal’s TriangleAs shown in figure 1, the middle numbers in Pascal’s triangle are following; 1, 2, 6, 20, 70,…They can be divided by 1, 2, 3, 4, 5,…then, we obtain the sequence of following numbers 1, 1, 2, 5, 14,… ⎛ 2n ⎞Since each middle number is derided from the binomial coefficient, ⎜ ⎟ , then we ⎝ n⎠divide each number by (n + 1). These numbers are called Catalan numbers. The formulaof Catalan numbers will be presented in Theorem 1.[1]2. General Knowledge of the Catalan NumbersCatalan numbers, which appear in many counting problem such as monotonic paths,Dyck words, dividing polygons, and noncrossing partitions, form a sequence of asequence of natural numbers.Historical Information Catalan Numbers were previously discussed by the Chinese mathematician AntuMing in 1730, but his work was not well-known in the Western world because it wasintroduced only in China. Then, in 1751, the Swiss mathematician Leonhard Euler found
  • 2. Catalan numbers while he was studying the triangulations of convex polygons. Until1838, the Belgian mathematician Eugene Charles Catalan discovered Catalan numberswhile he was studying the sequence of parentheses. Even though Catalan numbers werediscovered long time ago, they were named just after Eugene Charles Catalan introducedthem to the Western world.[2] In fact, there are various ways to define the Catalan numbers, we are going tointroduce one of them. Kreher and Stinson (1956) define the Catalan numbers in term oftotally balanced sequence. [3]Definition 1. Let n be a positive integer, and let a = [a1, a2, …, a2n-1, a2n] ∈ (Ζ2)2n. Thesequence is said to be totally balanced sequence if the following two properties aresatisfied: a) a contains n 0s and n 1s, and b) for any I, 1 ≤ I ≤ 2n, it hold that | {j : 1 ≤ j ≤ I, ai = 0}| ≥ | {j : 1 ≤ j ≤ I, ai = 1} |. Now, we are ready to define the Catalan numbers using the defintion of the totallybalance sequence.Definition. Let Cn denote the set of all totally balanced sequences in (Ζ2)2n. Cn is referredto a Catalan family of order n. The Catalan number Cn is defined to beCn = |Cn|. By the definition of the Catalan numbers above, the first fourth terms of theCatalan families Cn and the Catalan numbers Cn in Table 1.Table 1 shows the Catalan families Cn and the Catalan numbers Cn for 1 ≤ n ≤ 4. N Cn Cn 1 01 1 2 0011 0101 2 3 000111 001011 001101 010011 010101 3 4 00001111 00010111 00011011 00011101 00100111 4 00101011 00101101 00110011 00110101 01000111 01001011 01001101 01010011 01010101 (Kreher and Stinson, 1956) In the first theorem, we are going to state a formula of the Catalan number that iscommonly used.3. Formula of the Catalan Number and Its ProofTheorem. For any integer n ≥ 1, the Catalan number Cn is given in term of binomialcoefficients by
  • 3. 1 ⎛ 2n ⎞ (2n)! Cn = ⎜ n ⎟ = (n + 1)!n! , for n ≥ 1. n + 1⎝ ⎠ (a) For instance, the first tenth Catalan numbers are C1 = 1, C2 = 2, C3 = 5, C4 =14, C5 = 42, C6 = 132, C7 = 429, C8 = 1430, C9 = 4862 and C10 = 16796.Proof. To derive the formula, we are going to prove by using the triangulationdefinition. We are going to apply Euler’s Triangulation problem. Consider the ways oftriangulating n-gon, where 3 ≤ n ≤ 5.[2]Figure 2 Triangulations of an n-gon, where 3 ≤ n ≤ 5. One way of triangulating a triangle Two ways of triangulating a square http://www.toulouse.ca/EdgeGuarding/MobileGuards.html Five ways of triangulating a pentagon Euler used an inductive argument to establish the following formula. 2 ⋅ 6 ⋅10 ⋅ ...⋅ (4n − 10) An = , n ≥ 3. ( n − 1)! However, we need to extend the formula of Euler that was published in 1761 toinclude the case n = 0, 1, and 2 because the original formula makes sense only for n ≥ 3.Then we have 2 ⋅ 6 ⋅10 ⋅ ...⋅ (4k + 2) Ak+3 = , k ≥ 0. ( k + 2 )!Then, let Cn = Ak+2. Thus, 2 ⋅ 6 ⋅10 ⋅ ...⋅ (4n − 2) Cn = , n ≥ 1. ( n + 1)!
  • 4. The expression of Cn can be rewritten as (4n − 2) ⋅ 2 ⋅ 6 ⋅10 ⋅ ...⋅ (4n − 6) Cn = ( n + 1) n! (4n − 2) ⋅ Cn −1 = ( n + 1)When n = 1, we will have C0 = C1 = 1. Thus, we can define C0 = 1 and Cn recursively asfollowing; (4n − 2) ⋅ Cn −1 C0 = 1 and Cn = ( n + 1)To show the recursive formula implying the formula (a) of the Catalan number, it can beprocessed algebraically. (4n − 2) ⋅ Cn −1 Cn = ( n + 1) (4n − 2) ⋅ (4n − 6) ⋅ Cn − 2 = ( n + 1) n (4n − 2) ⋅ (4n − 6) ⋅ ( 4n − 10 ) ⋅ Cn − 3 = ( n + 1) n(n − 1) (4n − 2) ⋅ (4n − 6) ⋅ ( 4n − 10 ) ⋅... ⋅ 6 ⋅ 2 ⋅ C0 = ( n + 1) n(n − 1) ⋅ ⋅ ⋅ 3 ⋅ 2 ⎡(2n − 1) ⋅ (2n − 3) ⋅ ( 2n − 5 ) ⋅... ⋅ 3 ⋅1⎤ ⋅ 2 n = ⎣ ⎦ ( n + 1)! 2 n (2n)! = 2 n n!( n + 1)! (2n)! = n!( n + 1)! 1 ⎛ 2n ⎞ =  n + 1⎜ n ⎟ ⎝ ⎠ Thus, we obtain the formula (a) of the Catalan number by using the recursionrelation as we wish.
  • 5. 4. Properties of the Catalan numbers 4.1 Parity of Catalan Numbers Definition. Let r be the remainder, then the parity of an integer is defined as its attribute of being even or odd, which means 1) let b = 2 and a be an integer, if the remainder, r, such that r = 0, the integer a is called even and has the form a = 2q; 2) if the remainder r, such that r = 1, the integer a is called odd and has the form a = 2q + 1. Two given integers, s and t, have the same parity, if they are both even, or both odd; but if one of them is even, and the other is odd, then s and t are said to be of different parity. [4] Table 2 The first 18 Catalan Numbers (Koshy and Salmassi, 2006)n 0 1 2 3 4 5 6 7 8Cn 1 1 2 5 14 42 132 429 1430n 9 10 11 12 13 14 15 16 17Cn 4862 16796 58786 208012 742900 2674440 9694845 35357670 129644790 As shown in Table 2, for n ≤ 17, the Catalan numbers, Cn, are odd when n = 0, 1, 3, 7, and 15. All of those numbers that have the same form which is 2m – 1 when m > 0 (i.e. the 2nd and 3rd Catalan numbers) are prime. The numbers of the form 2m – 1 where m is an integer are known as Mersenne numbers.[4] In order to prove the next theorem, we are first going to introduce one of the generalized Catalan numbers formulas. [5] Lemma. For any p ≥ 2, the generalized Catalan number Cp (n) , such that 1 ⎛ pn ⎞ C p (n) = , can be defined recursively by ( p − 1)(n + 1) ⎜ n ⎟ ⎝ ⎠ ( p −1)n +1 p ⎛ ( p − 1)(n − k) + 1⎞ Cp (0) = 1, and C p (n) = ∑ (−1)k −1 ⎜ ⎝ k ⎟ C p (n − k) , ⎠ n ≥ 1. k =1 In particular, p = 2 corresponds to the usual Catalan numbers C2(n) satisfy the linear recursion. i.e., n +1 2 ⎛ (n − k) + 1⎞ C2 (n) = ∑ (−1)k −1 ⎜ ⎟ C2 (n − k) , n ≥ 1. k =1 ⎝ k ⎠ pk − 1 Theorem. The prime p | Cp(n) if and only if n ≠ for all integers k ≥ 0. In p −1 particular, C2(n) is odd if and only if n is a Mersenne number 2k – 1 where some integers k. [5]
  • 6. pk − 1Proof. We are going to prove that if the prime p | Cp(n) then n ≠ . p −1Let Cp(n) be the generalized Catalan number, for any p ≥ 2, i.e. ( p −1)n +1 p ⎛ ( p − 1)(n − k) + 1⎞ C p (n) = ∑ (−1)k −1 ⎜ ⎝ k ⎟ C p (n − k) . ⎠ k =1To prove the statement, we shall apply induction on n. For the base step, the result holdsfor n = 1, since Cp(1) = 1. Then, we assume n > 1 and that the result holds for all numberm, m < n. Let p r ≤ n ≤ p r +1 − 1 . We will consider the right-hand side of Cp(n). We will letit called modulo p. Then, we apply the induction hypothesis that for m < n, we will haveCp(m) is a multiple of p if (p – 1) m +1 is not a power of p. pN − 1For the term of modulo p, the summation of the term n – k is the form . However, p −1since p r ≤ n ≤ p r +1 , the only non-zero term modulo p, is the one corresponding to the p r +1 − 1index k for which (p – 1)(n – k) = pr – 1 if n ≤ (respectively, p −1 p r +1 − 1( p − 1)(n − k) = p r +1 − 1 if n > ). This term is, to within sign, p −1⎛ pr ⎞ ⎟ C ⎛ p − 1 ⎞ if n ≤ p − 1 (respectively, ( p − 1)(n − k) = p r +1 − 1 if r r +1⎜⎜ n − p − 1⎟ p ⎜ p − 1 ⎟ r⎜ ⎝ ⎠ p −1⎝ p −1 ⎟ ⎠ p r +1 − 1 ⎛ pr ⎞n> ). As the binomial coefficient ⎜ ⎟ is a multiple of p if and only if 0 < s < p −1 ⎝s ⎠ pr − 1pr, the above term is a multiple of p if and only if 0 < n − < p r if p −1 p r +1 − 1 p r +1 − 1n≤ (respectively, p r +1 < ( p − 1)n + 1 < p r + 2 if n > ). This is equivalent to p −1 p −1 p r +1 − 1 p r < ( p − 1)n + 1 < p r +1 if n ≤ (respectively, p r +1 < ( p − 1)n + 1 < p r + 2 if p −1 p r +1 − 1n> ). This implies that (p – 1)n + 1 is not a power of p. Therefore, we can p −1 pk − 1conclude that if the prime p | Cp(n) then n ≠ .  p −1
  • 7. 5. The Generating Function of Catalan NumbersThe linear recursion of generalized Catalan numbers, Cp(x), p ≥ 2, have been given in theprevious section. In this section, we will provide a proof of the generating function of theCatalan numberTheorem. Let C be the generating function of the second-ordered Catalan number, and 1 − 1 − 4xCn be the nth Catalan number, C(x) can be expressed as following; C(x) = .[6] 2xProof. We are going to show that the generating function of the Catalan number, C, 1 − 1 − 4xsatisfies the equation, C(x) = . 2xBased on the evident recurrence relation, we can express the nth Catalan number Cn asshown, Ck = ∑ Ci C j , k ≥ 1, and C0 = 1. i + j = k −1A sequence of numbers, a0, a1, a2,…, elements of the sequence can be formed the powerseries The generating function for the sequence is shown below. ∞ C(x) = ∑ Ck x k . k=0 = C0 + C1x + C2x2 + C3x3 +…then the recurrence relation expression shows that C satisfies xC2(x) – C(x) +1 = 0, C(0) = 1,so that C can be solved using Quadratic formula, i.e., 1 ± 1 − 4x C(x) = 2x In fact, we must choose the minus sign to obtain the positive coefficients of thepower of x in the generating function of C(x) because all Catalan numbers are alwayspositive. Moreover,Thus, the generating function of the Catalan number is 1 − 1 − 4x C(x) = (b). 2xTherefore, we have proved the formula of the generating function of the second-orderedCatalan numbers as we wish. [6]We generate the second-ordered Catalan numbers, since they are the usual one that wemainly discuss in this paper.
  • 8. 5. Interpretations of the Catalan Numbers There are connections among the combinatorial interpretations which are found inthe triangulation of n-gon with n+2 sides, the construction of the binary tree-diagramwith n+1 leaves, Dyck words with 2n length, and the monotonic paths with n × n squarecells.[7]A1) The Catalan Numbers and Triangulation of Polygon. The Catalan number Cn isthe number of different ways a polygon with n + 2 sides can be partition into triangle bydrawing nonintersecting diagonals. For example, the following pentagon as shown infigure 3 demonstrates the case n = 3. The 3rd Catalan number is 5, so there are 5 differentways to triangulate the pentagon. [7]Figure 3. Five different partitions of a convex pentagon into triangles. [7]Figure 4. The construction of the tree-diagram with n+1 leaves, corresponding to thepartitions. [7]Figure 5. The labeling of the branches of the tree-diagrams. [7]Figure 6. The codes with length 2n, derived from the labeled tree-diagrams. [7]
  • 9. Figure 7. The lattices paths which is called the monotonic path are obtained from thecodes. [7]A2) The Catalan Numbers and Tree Diagram. There are various applications in metrictrees. For example, the Catalan number, Cn, is the number of full binary trees with n + 1leaves when a full binary tree is defined as a tree whose every vertex has either twobranches (leaves) or no branch (leaf). [8] For instance, if we let a full binary tree with 3 + 1 = 4 leaves, then n = 3. The 3rdCatalan number, C3, is 5. Thus, there are five patterns of full binary trees with 4 leaves asshown in the Figure 5.A3) The Catalan Numbers and the Dyck Words. There is an application of the Catalannumber in Computer Science. In fact, the Catalan number, Cn, is the number of Dyckwords of the length 2n. [8] Dyck word is defined as “a string consisting of n X’s and n Y’ssuch that no initial segment of the string has more Y’s than X’s”. Then we construct monotonic path, which is a set of steps begining at the lowerleft corner, finish at the upper right corner of the grid with n×n square cells. Beside, eachstep stays below the diagonal, y = x. According to the codes. let R(right) stand for X andL(left) stand for Y. As been seen in Figure 6, given 3 × 3 square cells. Thus, there arefive different monotonic paths along the edge of the given grid. Thus, the Catalannumber, Cn, is the number of different monotonic paths that can occur in n x n gridsquare. In fact, monotonic path is similar to the ballot problem that we are going todiscuss.A4) The Catalan Numbers and the Ballot Problem (Hilton and Pederson). The ballot problem was introduced in the late nineteen century to determine theprobability of counting of votes when there are two candidates A, and B such that Areceives a votes, and B receives b votes with a > b. Originally, the problem was firstsolved by Bertrand, but it became famous in 1887 after the French mathematician DesireAndre gave the solution for the problem. We will introduce the term of p-good before weare continuing the discussion.Definition. A path from P to Q is p-good if it lies entirely below the line y = (p – 1)x.Thus, a bad path is defined as the complementary set of a set of good paths. [6] According to the linear recursion of generalized Catalan numbers, the usualCatalan numbers are defined when p = 2, so we will call 2-good paths for thisinterpretation. The method of solving the ballot problem is to count the number of 2-good pathsfrom (c,d) to (a,b), where (c,d), (a,b) are any two lattice points below the line y = x. [6]
  • 10. Theorem. Given the formula of the kth Catalan number, Ck, as following 1 ⎛ 2k ⎞ C0 = 1, and Ck = , k ≥ 1. k ⎜ k − 1⎟ ⎝ ⎠The kth Catalan numbers, Ck, is an expression of the number of good path from (0,-1) to(k, k-1), i.e., ⎛ 2k ⎞ ⎛ 2k ⎞ 1 ⎛ 2k ⎞ ⎜ k ⎟ − ⎜ k + 1⎟ = k ⎜ k − 1⎟ . ⎝ ⎠ ⎝ ⎠ ⎝ ⎠Proof. Let us assume that both good paths and bad paths exist. As shown in Figure 7., wehaved < c ≤ b < a. (Note that this condition is satisfied the ballot problem that the loserreceives at least 1 vote.). Let ℘ be bad path, PF, FQ be subpaths from P to Q, and ℘1 bethe path obtained from ℘1 by reflecting with the line y = x. Figure  7  The total number of paths from P(c,d) to Q(a,b) can be expressed as the binomialcoefficient. ⎛ (a + b) − (c + d)⎞ ⎜ ⎝ a−c ⎟; ⎠As we have seen from Figure 7, ℘ = ℘1℘2 . Then, if ℘ = ℘1 ℘2 , where ℘ is a path fromP(d, c) to Q(a,b) . Assume the rule ℘  ℘ be injective correspondence between the set ofbad paths from P to Q and the set of paths from P to Q. Thus, the number of bad paths from ⎛ (a + b) − (c + d)⎞P to Q is ⎜ ⎟ , and hence we have the number of good paths from P to Q as ⎝ a−d ⎠shown below ⎛ (a + b) − (c + d)⎞ ⎛ (a + b) − (c + d)⎞ ⎜ ⎝ a−c ⎟ −⎜ ⎠ ⎝ a−d ⎟. ⎠
  • 11. In particular, if the starting point of the path is (1,0) to ( a,b ) , the number of good paths isgiven as ⎛ a + b − 1⎞ ⎛ a + b − 1⎞ a − b ⎛ a + b ⎞ ⎜ a −1 ⎟ − ⎜ ⎝ ⎠ ⎝ ⎠ = a ⎟ a+b⎜ a ⎟ ⎝ ⎠ (1). a−bThus, is the probability that a path from (0,0) to ( a,b ) proceeds first to (1,0), then a+bcontinues as a good path to ( a,b ) . According to the solution of the ballot problem, if we letthe path starting from (0, -1) to ( k, k − 1) , we can rewrite (1) as ⎛ 2k ⎞ ⎛ 2k ⎞ 1 ⎛ 2k ⎞ ⎜ k ⎟ − ⎜ k + 1⎟ = k ⎜ k − 1⎟ . ⎝ ⎠ ⎝ ⎠ ⎝ ⎠Here, we obtain the formula of the kth Catalan number, Ck, as we wish. 1 ⎛ 2k ⎞ C0 = 1, and Ck = , k ≥ 1.  k ⎜ k − 1⎟ ⎝ ⎠References[1] Conway, J., Guy, R. (1996). “THE BOOK OF NUMBERS”. Copericus, New York.[2] Koshy, T. (2007). “ELEMENTARY NUMBER THEORY WITH APPLICATIONS”. Boston Academic Press, Massachusetts.[3] Kreher, D., Stinson, D. (1999). “COMBINATORIAL ALGORITHMS: GENERATION, ENUMERATION, ANS SEARCH”. CRC Press, Florida.[4] Koshy, T, & Salmassi, M. (2006). “Parity and Primality of Catalan Numbers”. The College Mathematics Journal 37, 1, pp. 52-53.[5] Sury, B. (2009). “Generalized Catalan Numbers: Linear Recursion and Divisibility”, Journal of Integer Sequences 12.[6] Hilton and J. Pedersen. (1991). “Catalan numbers, their generalization and their uses”, Math. Intelligencer 13, pp. 64–75.[7] Cofman, Judita (08/01/1997). "Catalan Numbers for the Classroom?". Elemente der Mathematik (0013-6018), 52 (3), p. 108.[8] Stanley, R. (1944). “ENUMERATIVE COMBENATORICS”. Wadsworth & Brooks/Cole Advanced Books & Software, California.  

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