一個卡特蘭等式的重新審視 - 政大學術集成

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(1)國立政治大學應用數學系碩士學位論文. 立. 政治大. ‧ 國. 學. 一個卡特蘭等式的重新審視 A Catalan Identity revisited. ‧. n. er. io. sit. y. Nat. al. Ch. engchi. i Un. v. 指導教授：李陽明博士研究生：李珮瑄撰中華民國 109 年 6 月. DOI:10.6814/NCCU202000719.

(2) 謝. 在本人的寫作過程中，首先最主要感謝的是我的指導老師，李陽明老師。在整個過程中他給了我很大的幫助，在論文題目制定時，他首先肯定了我的題目大方向，但是同時又協助確立目標，讓我在寫作時有了具體方. 政治大. 向。在論文提綱制定時，我的思路不是很清晰，經過老師的幫忙，讓我具. 立. 體寫作時思路頓時清晰。在完成初稿後，老師認真查看了我的內容，指出. ‧ 國. 學. 了我存在的很多問題。在此十分感謝李老師的細心指導，才能讓我順利完成畢業。感謝口試委員陳天進、蔡炎龍老師給予指正我論文中的錯誤，在. ‧. 此深表感謝！同時也感謝其他幫助和指導過我的老師和同學。最後要感謝在整個論文寫作過程中幫助過我的每一位人。. n. er. io. sit. y. Nat. al. Ch. engchi. i Un. v. i DOI:10.6814/NCCU202000719.

(3) 中文要. 本篇論文探討卡特蘭等式(n + 2)Cn+1 = (4n + 2)Cn 證明方式以往都以計算方式推導得出，當我參加劉映君的口試時，發現她使用組合方法來證明這個等式。當我在尋找論文的主題時，讀到李陽明老師的一篇論文 ‵‵. 政治大. T he Chung − F eller theorem revisited′′ ，發現 Dyck 路徑也可以作為卡特. 立. 蘭等式的組合證明，因此我們完成(n + 2)Cn+1 = (4n + 2)Cn 的組合證明。. ‧ 國. 學. 通過 Dyck 路徑證明卡特蘭等式可以得到以下優勢： 1. 子路徑 C 在切換過程中不會改變。. ‧. 2. 由於x1 中的 P 的子路徑 B 為空，因此在交換 Ad 和 Bu 部分後，生成新的缺陷必連接在原始子路徑 C 之後。. Nat. sit. y. 由於x2 中的 Q 的子路徑 A 為空，因此在 Bu 交換和 Ad 部分後，生成新的. al. er. io. 提升必連接在原始子路徑 C 之後。. n. 3. 在計算函數g1 (g2 ) 的反函數的過程中，缺陷（提升）恢復模式必遵循. Ch. “後進先出”或“先進後出”規則。. engchi. i Un. v. 關鍵字：卡特蘭等式、Dyck 路徑. ii DOI:10.6814/NCCU202000719.

(4) Abstract When we first prove the Catalan identity, (n + 2)Cn+1 = (4n + 2)Cn . We often prove it by calculation.When I participated in the oral examination of YingJun Liu’s essay, I found that she used a combinatorial proof to prove this identity. When I was looking for the subject of the thesis, I read a paper by professor Young-. 政治大. Ming Chen, ‵‵ T he Chung − F eller theorem revisited′′ , which found that Dyck. 立. paths could also be used as a combinatorial proof of the Catalan identity. Therefore,. ‧ 國. 學. we completed the combinatorial proof of (n + 2)Cn+1 = (4n + 2)Cn . Proving the Catalan identity through the Dick paths can reveal the following. ‧. advantages:. 1.The subpath C does not change during the process of switching of the portions. sit. y. Nat. Ad and Bu.. al. er. io. 2.Since the subpath B of P in x1 is empty, a new flaw generated after. v. n. switching of the portions Ad and Bu must be followed by the original subpath C.. Ch. i Un. Since the subpath A of Q in x2 is empty, a new lift generated after. engchi. switching of the portions Bu and Ad must be followed by the original subpath C. 3.In the process of computing the preimage of a function g1 (g2 ), the flaws (lifts) recovery mode follows the ‵‵ Last−in−F irst−out′′ or ‵‵ F irst−in−Last−out′′ .. Keywords: Catalan identity, Dyck path. iii DOI:10.6814/NCCU202000719.

(5) Contents 謝. i. 中文要. ii. Abstract. 立. Contents. iv. 14. er. n. 4 Summary. 3. sit. io. 3 Paths Start with Down-step. al. 1. y. ‧ 國 Nat. 2 Paths Start with Up-step. v. ‧. 1 Introduction. iii. 學. List of Figures. 政治大. Ch. engchi. i Un. v. 25. Appendix A examples of Catalan identity. 26. A.1 (n + 2)Cn+1 = (4n + 2)Cn . . . . . . . . . . . . . . . . . . . . . . . . . . . . Bibliography. 26 30. DOI:10.6814/NCCU202000719.

(6) List of Figures 2.1. f1 (BuAdC) = AdBuC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 3. 2.2. BuAdC and B ′ uA′ dC ′ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 4. 2.3. f1 (BuAdC) = AdBuC and f1 (B ′ uA′ dC ′ ) = A′ dB ′ uC ′ . . . . . . . . . . . .. 4. 2.4. BuAdC and B ′ uA′ dC ′ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 5. 2.6. 治政 f (BuAdC) = AdBuC and f (B uA dC ) = A大 dB uC . . . . . . . . . . . . 立 BuAdC and B uA dC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 2.7. f1 (BuAdC) = AdBuC and f1 (B ′ uA′ dC ′ ) = A′ dB ′ uC ′ . . . . . . . . . . . .. 6. 2.8. BuAdC and B ′ uA′ dC ′ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 7. 2.9. f1 (BuAdC) = AdBuC and f1 (B ′ uA′ dC ′ ) = A′ dB ′ uC ′ . . . . . . . . . . . .. 7. 2.10 the new f law will be connected to original f laws behind . . . . . . . . . . .. 8. 1. 1. ′. ′. ′. ′. ′. ′. ′. ′. ′. 學. ‧. ‧ 國. 2.5. 5. Nat. y. 6. sit. 2.11 Rising f rom; F alling to; F alling f rom; Rising to . . . . . . . . . . . . . . P reimage under f1. n. al. er. io. 2.12 AdBuC −−−−−−−−−−−→ BuAdC . . . . . . . . . . . . . . . . . . . . . . . .. i Un. v. 9 9. ⃗uS) = R • S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.13 h1 (Rd⃗. 13. engchi. Ch. 3.1. f2 (AdBuC) = BuAdC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 14. 3.2. AdBuC and A′ dB ′ uC ′ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 15. 3.3. f2 (AdBuC) = BuAdC and f2 (A′ dB ′ uC ′ ) = B ′ uA′ dC ′ . . . . . . . . . . . .. 15. 3.4. AdBuC and A′ dB ′ uC ′ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 16. 3.5. f2 (AdBuC) = BuAdC and f2 (A′ dB ′ uC ′ ) = B ′ uA′ dC ′ . . . . . . . . . . . .. 16. 3.6. AdBuC and A′ dB ′ uC ′ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 17. 3.7. f2 (AdBuC) = BuAdC and f2 (A′ dB ′ uC ′ ) = B ′ uA′ dC ′ . . . . . . . . . . . .. 17. 3.8. AdBuC and A′ dB ′ uC ′ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 18. 3.9. f2 (AdBuC) = BuAdC and f2 (A′ dB ′ uC ′ ) = B ′ uA′ dC ′ . . . . . . . . . . . .. 18. 3.10 the new lif t will be connected to original lif t behind . . . . . . . . . . . .. 19. DOI:10.6814/NCCU202000719.

(7) 3.11 Rising f rom; F alling to; F alling f rom; Rising to . . . . . . . . . . . . . . P reimage under f2. 20. 3.12 BuAdC −−−−−−−−−−−→ AdBuC . . . . . . . . . . . . . . . . . . . . . . . .. 20. ⃗ =R•S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.13 h2 (R⃗udS). 24. 立. 政治大. ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. i Un. v. DOI:10.6814/NCCU202000719.

(8) Chapter 1 Introduction • When we first prove the Catalan identity, (n + 2)Cn+1 = (4n + 2)Cn . We often prove it. 政治大. by calculation. The proof method is as follows: 2n+2 (n+2)Cn+1. n. er. io. sit. y. ‧. Nat. n+2 (2n+2)! (n+1)!(n+1)! = (2n+1)(2n)!(2n+2) (n+1)n!n!(n+1) = 2(2n+1)(2n)!(2n+2) (n+1)n!n!(2n+2) = (4n+2)(2n)! (n+1)n!n! a l (4n+2)Cn2n i v+ = C h n+1 = (4n n i U e. 學. ‧ 國. (n + 2)Cn+1 =立 =. ngch. 2)Cn. When I participated in the oral examination of Ying-Jun Liu’s essay, I found that she used a combinatorial of proofs to prove this identity. When I was looking for the subject of the thesis, I read a paper by teacher Young-Ming Chen, ”The Chung-Feller theorem revisited”, which found that Dyck paths could also be used as a combinatorial proof of the Catalan identity. Therefore, we complete the combinatorial proof of (n + 2)Cn+1 = (4n + 2)Cn . [5] [7] [1]. 1 DOI:10.6814/NCCU202000719.

(9) Definition 1.0.1. An up-step is denoted by u = (1, 1). A down-step is denoted by d = (1, −1). A step is either an up-step(u) or a down-step(d). A path consists of consecutive steps. A subpath is some of consecutive steps of a path. Definition 1.0.2. A path is that all up-steps and down-steps are above the x-axis. Definition 1.0.3. A totally bad path is that all up-steps and down-steps are below the x-axis. Definition 1.0.4. A flaw is a down-step below the x-axis. Definition 1.0.5. A lift is a up-step above the x-axis.. 政治大. Definition 1.0.6. In An n-Dyck path, Cn is the number of good paths from (0, 0) to (2n, 0). 立. Definition 1.0.7. An n-Dyck path is a path from (0, 0) to (2n, 0) with n up-steps. ‧ 國. 學. and n down-steps.. An n-Dyck path with k flaws if it has k down-steps below the x-axis.. ‧. sit. Nat. down-steps in R is denoted by |R| = r,0≤r≤2n.. y. Definition 1.0.8. Let R is a subpath of n-Dyck path. The number of up-steps and. n. al. er. io. Definition 1.0.9. The set Dn,k consists of all n-Dyck paths with k flaws, 0≤k≤n.. Ch. i Un. v. For more details ,we refer to [4] [3] [2] [10] [9] [6] [8] [11]. engchi. 2 DOI:10.6814/NCCU202000719.

(10) Chapter 2 Paths Start with Up-step Definition 2.0.1. Define a function f1 from Dn+1,k into Dn+1,k+1 by the following:. 政治大. 1. The set Dn+1,k consists of all n + 1-Dyck paths with k flaws, 0≤k≤n. Each path in Dn+1,k. 立. can be factorized into BuAdC, where B is a subpath all bellow the x-xais, say with k1. ‧ 國. 學. flaws, 0≤k1 ≤k, u is the first up step above the x-xais, A is a subpath all above the x-xais, d is the first step to contact the x-axis after A and above the x-axis, say uAd with 0 flaws,. ‧. and C is the remaining path with k − k1 flaws, 0≤k≤n.. sit. y. Nat. 2. The set Dn+1,k+1 consists of all n + 1-Dyck path with k + 1 flaws, 0≤k≤n. Each path in. io. flaws.. al. iv n C A、B 、C may be empty, and they same number of up-steps h ehave n gthe chi U n. Note:. er. Dn+1,k+1 can be factorized into AdBuC, where A,dBu, and C have 0, k1 + 1, and k − k1. and down-steps.. i.e. f1 (BuAdC) = AdBuC. A u. d. C. A. C. f1 u. d B. x-axis. B Figure 2.1: f1 (BuAdC) = AdBuC. 3 DOI:10.6814/NCCU202000719.

(11) Theorem 2.0.1. Define f1 :Dn+1,k →Dn+1,k+1 by f1 (BuAdC) = AdBuC. The function f1 is one-to-one. Proof. Let |B| = 2i, 0≤i≤n; |u| = 1; |A| = 2j, 0≤j≤n − i; |d| = 1; |C| = 2(n − i − j − 1). Claim: f1 one-to-one. (i.e.f1 (BuAdC) = f1 (B ′ uA′ dC ′ ) ⇒ AdBuC = A′ dB ′ uC ′ ) Suppose |B ′ | = 2r, 0≤r≤n; |u| = 1; |A′ | = 2s, 0≤s≤n−r; |d| = 1; |C ′ | = 2(n−r−s−1). Claim 1: |A| = |A′ | case 1: 2j < 2s. d. B′. Figure 2.2: BuAdC and B ′ uA′ dC ′. n. T he (2j+1)st step. C′. sit er. A′. io. al C Ch u. d. d. y. Nat. A. u. ‧. ‧ 國. 立. 政治大. 學. B. C. T he (2j+1)st step. A. T he (2j+1)st step. A u. ′. n U engchi. B. T he (2j+1)st step v i. C′. u. d B′. Figure 2.3: f1 (BuAdC) = AdBuC and f1 (B ′ uA′ dC ′ ) = A′ dB ′ uC ′. To see Figure 2.3 when we start on (0, 0) to walk along the two path AdBuC and A′ dB ′ uC ′ . The (2j + 1)st step of AdBuC is below the x-axis, but the (2j + 1)st step of A′ dB ′ uC ′ is still above the x-axis. This is a contradiction as two paths are the same.. 4 DOI:10.6814/NCCU202000719.

(12) case 2: 2j > 2s. A. A′. T he (2s+1)st step. u. d. C. u. T he (2s+1)st step. d. C′. B′. B. Figure 2.4: BuAdC and B ′ uA′ dC ′. A. 立. T he (2s+1)st step. u. ′. C′. T he (2s+1)st step. ‧ 國. B. u. d. 學. d. C. 政治大 A B′. ‧. Figure 2.5: f1 (BuAdC) = AdBuC and f1 (B ′ uA′ dC ′ ) = A′ dB ′ uC ′. sit. y. Nat. er. io. To see Figure 2.5 when we start on (0, 0) to walk along the path the two path AdBuC and. al. n. iv n C of A′ dB ′ uC ′ is below the x-axis.hThis e nisga ccontradiction h i U as two paths are the same. Thus, A′ dB ′ uC ′ . The (2s + 1)st step of AdBuC is still above the x-axis, but the (2s + 1)st step. we have proof that j = s. ∴ |A| = |A′ | and A = A′ .. 5 DOI:10.6814/NCCU202000719.

(13) Claim 2: |B| = |B ′ | recall:Let |B| = 2i, 0≤i≤n; |u| = 1; |A| = 2j, 0≤j≤n − i; |d| = 1; |C| = 2(n − i − j − 1). Suppose |B ′ | = 2r, 0≤r≤n; |u| = 1; |A′ | = 2s, 0≤s≤n − r; |d| = 1; |C ′ | = 2(n − r − s − 1). case 1: 2i < 2r. T he (2j+2i+2)nd step. A′. A u. C. d. u. C′. T he (2j+2i+2)nd step. 政治大 B. B. d. ′. 立. ‧. ‧ 國. 學. Figure 2.6: BuAdC and B ′ uA′ dC ′. n B. Ch. u. d. y = −1. al. C′. sit. io. u. d. A′. C. er. A. y. T he (2j+2i+2)nd step. Nat. T he (2j+2i+2)nd step. engchi. i Un. v. y = −1. B′. Figure 2.7: f1 (BuAdC) = AdBuC and f1 (B ′ uA′ dC ′ ) = A′ dB ′ uC ′. To see Figure 2.7 when we start on (2j + 1, −1) to walk along the path B and B ′ . The (2j + 2i + 2)nd step of Bu is above y = −1, but the (2j + 2i + 2)nd step of B ′ is still below y = −1. This is a contradiction as two paths are the same.. 6 DOI:10.6814/NCCU202000719.

(14) case 2: 2i > 2r. T he (2j+2r+2)nd step. A′. A u. d. C. u. T he (2j+2r+2)nd step. d. C′. B′. B. Figure 2.8: BuAdC and B ′ uA′ dC ′. 政治大 nd T he (2j+2r+2) step. 立. A. A′. ‧ 國. d. u. C′. 學. C. u. d. y = −1. T he (2j+2r+2)nd step. B′. ‧. B. y = −1. er. io. sit. y. Nat. Figure 2.9: f1 (BuAdC) = AdBuC and f1 (B ′ uA′ dC ′ ) = A′ dB ′ uC ′. al. iv n C nd (2j + 2r + 2)nd step of B ′ u is h above e nyg=c−1, h ibutUthe (2j + 2r + 2) step of B is still n. To see Figure 2.9 when we star on (2j + 1, −1) to walk along the path B and B ′ . The. below y = −1. This is a contradiction as two paths are the same. Thus, we have proof. that j = s. ∴ |B| = |B ′ | and B = B ′ . Since A = A′ and B = B ′ , ∵ AdBuC = A′ dB ′ uC ′ ⇒ C = C ′ ∴ BuAdC = B ′ uA′ dC ′ Therefore f1 is one-to-one.. 7 DOI:10.6814/NCCU202000719.

(15) Lemma 2.0.2. If P is a path in Dn+1,k , then f1 (P ) has k + 1 flaws. Moreover, the (k + 1)st flaw will be connected to original flaws behind. Note: A down step below the x-axis is called flaw. The flaws are counted from right to left and bottome-up. Proof. The function f1 from Dn+1,k into Dn+1,k+1 . Suppose P has k flaws in Dn+1,k , and P = BuAdC, where B, uAd, C have k − j flaws, 0 flaw, j flaws, respectively. The k th flaw is in subpath B.Then f1 (P ) has k + 1 flaws on Dn+1,k+1 , and f1 (P ) = AdBuC, where A, dBu, C have 0 flaw, k − j + 1 flaws, j flaws, respectively.. A u. ‧ 國. d. d. C u. 學. x-axis. B. A. x-axis. ‧. d. u. C. C. io. sit. y. Nat. Figure 2.10: the new f law will be connected to original f laws behind. n. al. er. u. B A. 立. C. d. 政治大 A. Ch. engchi. i Un. v. Notice that the path f1 (P ), d is connected to the k th flaw in subpath B behind. d becomes the (k + 1)st flaw. Therefore, no matter how many times we use f1 , d is conneted to the k th flaw in subpath B behind. d is still the (k + 1)st flaw.. 8 DOI:10.6814/NCCU202000719.

(16) Note: Rising from. Falling to. x-axis. x-axis. x-axis. x-axis. Falling from. Rising to. Figure 2.11: Rising f rom; F alling to; F alling f rom; Rising to. 政治大 , then f (Q) has k flaws. −1 1. Lemma 2.0.3. If Q is a path in Dn+1,k+1. 立. Moreover,the (k + 1)st flaw in Q will be restored.. ‧ 國. 學. i.e. The (k + 1)st flaw d and the first up-step u rising to the x-axis on the right side of d in. ‧. Dn+1,k+1 are restored to the first down-step d falling to the x-axis and the first up-step u. y. sit. n. al. er. io. Show by formula:. Nat. rising from the x-axis in Dn+1,k .. A y=-1. C d. Empty area. u. Ch. engchi. P reimage under f1 − −−−−−−−−−−−−−− →. iv n Uy=1 u. A Empty area. d. C x-axis. B. B. P reimage under f1. Figure 2.12: AdBuC −−−−−−−−−−−→ BuAdC. In Q, the (k + 1)st flaw d and the first up-step u rising to the x-axis on the right side of d in Dn+1,k+1 , we can observe that there is an empty area enclosed by the u, d, the x-axis, and the horizontal line y = −1. After switching two portions Ad and Bu, another empty area is enclosed by the u is the first up-step rising from the x-axis, and d is the first down-step falling to the x-axis, and the horizontal line y = 1 in f1−1 (Q). And the remaining segments which are behind u is the fixed subpath C. 9 DOI:10.6814/NCCU202000719.

(17) Theorem 2.0.2. Define f1 :Dn+1,k →Dn+1,k+1 by f1 (BuAdC) = AdBuC. The function f1 is onto. Proof. Claim: f1 is onto. (f1 :Dn+1,k → Dn+1,k+1 ) For any path Q = AdBuC in Dn+1,k+1 which 1≤k +1≤n+1. We choose the (k +1)st flaw d and choose the first up-step u rising to the x-axis on the right side of d. We switch the portions Ad and Bu then we can restore the (k+1)st flaw and get a new path P = BuAdC in Dn+1,k , by Lemma 2.2、Lemma 2.3. Where Q has at least one flaw and P has at most n flaws. In fact, if Q has k + 1 flaws then P has k flaws. So every Q in Dn+1,k+1 , we can find a path P in Dn+1,k , such that f1 (P ) = Q. Therefore f1 is onto.. 政治大. Hence, f1 is one-to-one and onto by Theorem 2.1 and Theorem 2.2.. 立. Note: Let f1−1 be the inverse function of f1 .. ‧ 國. 學. Definition 2.0.4. The set X1 consists of all paths in n + 1-Dyck paths with k flaws.Each − → → path in X can be factorized into B − u A d C. The set Y consists of all 1. 1. ‧. n + 1-Dyck paths which are totally bad paths.. Define a function g1 from X1 into Y1 by the following:. y. Nat. n+1−k times. n. al. er. io. sit. 1. g1 (P ) = f1n+1−k (P ), where P in X1 and f1n+1−k = f1 ◦ f1 ◦ · · · ◦ f1 {z } |. i Un. v. → 2. The first up-step rising from x-axis denote by − u and the first down-step falling to − → x-axis denote by d .. Ch. engchi. → Lemma 2.0.5. Suppose that P in X1 . In f1 (P ), the first up-step − u rising from the x-axis of P − → − →→ − →→ connectswith the first down-step d falling to the x-axis of P to be d − u and d − u − → − → → → is below the x-axis. Let P = − u A d C, then f (P ) = A d − u C. 1. − → → Proof. We may assume that P = B − u A d C is any path of Dn+1,k . Since the subpath B is − → − →→ − → → → empty, then P = − u A d C ∈ X1 and f1 (P ) = A d − u C.We know − u A d has 0 flaw, and − →→ C must have k flaws in P . Then A, d − u , and C have 0 flaw, 1 flaws, and k flaws in − →→ f1 (P ), respectively. Therefore, d − u is below the x-axis.. Note: In f1i (P ), 1≤i≤n + 1 − k, the segments below the x-axis will be always below the x-axis. 10 DOI:10.6814/NCCU202000719.

(18) Theorem 2.0.3. The function g1 :X1 →Y1 is one-to-one. Proof. Suppose that g1 (P ) = g1 (Q), where P has k flaws and Q has h flaws in X1 . To prove that P = Q. By definition g1 (P ) = f1n+1−k (P ), g1 (Q) = f1n+1−h (Q) case 1: k < h f1n+1−k (P ) = f1n+1−h (Q) ⇒ f1 (f1n−k (P )) = f1 (f1n−h (Q)) ∵ f1 is one-to-one. ∴ f1n−k (P ) = f1n−h (Q). For the same reason, the following can be obtained f1 (f1n−k−1 (P )) = f1 (f1n−h−1 (Q)) ⇒ f1n−k−1 (P ) = f1n−h−1 (Q) Since f1 is one-to-one, use this method n + 1 − h times f1 (f1h−k (P )) = f1 (Q) ⇒ f1h−k (P ) = Q − →→ − → → The d − u of f h−k (P ) are below the x-axis, but the − u and d of Q are above the x-axis 1. 立. by Lemma 2.5.. 政治大. case 2: k > h. f1n+1−k (P ) = f1n+1−h (Q) ⇒ f1 (f1n−k (P )) = f1 (f1n−h (Q)). y. ∴ f1n−k (P ) = f1n−h (Q). Nat. ∵ f1 is one-to-one. ‧. ‧ 國. 學. This is a contradiction.. sit. For the same reason, the following can be obtained. er. io. f1 (f1n−k−1 (P )) = f1 (f1n−h−1 (Q)) ⇒ f1n−k−1 (P ) = f1n−h−1 (Q). al. iv n C k−h ⇒ P = fh1 e(Q) n g c h−→i −→U k−h. n. Since f1 is one-to-one, use this method n + 1 − k times f1 (f1k−h (Q)). f1 (P ) = − → → The − u and d of P are above the x-axis, but d u of f1. (Q) are below the x-axis by. Lemma 2.5. This is a contradiction. case 3: k = h f1n+1−k (P ) = f1n+1−h (Q) ⇒ f1 (f1n−k (P )) = f1 (f1n−h (Q)) ∵ f1 is one-to-one. ∴ f1n−k (P ) = f1n−h (Q). Use this method n + 1 − h times, we have f1 (P ) = f1 (Q) ⇒ P = Q Therefore, g1 is one-to-one.. 11 DOI:10.6814/NCCU202000719.

(19) Theorem 2.0.4. g1 :X1 →Y1 is onto, where Y1 is totally bad path of n + 1-Dyck and g1 (P ) = f1n+1−k (P ). − →→ Proof. Give Q ∈ Y1 , Q has n + 1 flaw, and d − u is the (k + 1)st flaw in Q. −(n+1−k). Define: The preimage of the function g1−1 = f1. = f1−1 ◦ f1−1 ◦ · · · ◦ f1−1 | {z }. Since f1−1 (Q) is the preimage of Q under f1 and has n flaws.. n+1−k times. −(n−k). Using this way for n − k times, we get the path f1 (Q) which has k + 1 flaws. − → −(n−k) → In f1 (Q), d − u is still under the x-axis and is the (k + 1)st flaw.We use this way − → −(n+1−k) → again, we get the path f (Q) which has k flaws and − u A d is above the x-axis, as 1. f1 is onto and by Lemma 2.5. −(n+1−k). So we have f1. −(n+1−k). (Q) = f1. (f1n+1−k (P )) = P ,. 政治大 (Q) ⇒ g (P ) = f (P ) 立. where P has k flaws,P ∈ X1 . −(n+1−k). Let P = f1. 1. n+1−k 1. −(n+1−k). )(Q)). 學. ‧ 國. = f1n+1−k ((f1 =Q. Thus g1 is onto.. ‧ sit. y. Nat. n. al. er. io. Hence, g1 is one-to-one and onto by Theorem 2.3 and Theorem 2.4.. Ch. engchi. i Un. v. 12 DOI:10.6814/NCCU202000719.

(20) Definition 2.0.6. The set Z1 contains the totally bad path for all n-Dyck paths which − →→ replaces d − u in Y with a dot mark, and all paths in Y are n + 1-Dyck 1. 1. paths which are totally bad path. Let h1 be the function from Y1 into Z1 . − →→ − →→ i.e. Q = R d − u S is (2n + 2, 0) path, where R, d − u , and S are all totally bad paths. h1 (Q) = − →− h (R d → u S) = R • S 1. Theorem 2.0.5. h1 is one-to-one and onto. Proof. It is clear that h1 is one-to-one. Given Q′ = R • S ∈ Z1 − →→ − →→ We can change • into d − u . Thus R • S ⇒ R d − u S ∈ Y1 .. 政治大. Therefore, h1 is one-to-one and onto.. ‧. R. S. io. sit. y. S. x-axis. ⃗uS) = R • S Figure 2.13: h1 (Rd⃗. n. al. er. ⃗u. h1. Nat. d⃗. 學. R. ‧ 國. 立. Ch. engchi. i Un. v. 13 DOI:10.6814/NCCU202000719.

(21) Chapter 3 Paths Start with Down-step Definition 3.0.1. Define a function f2 from Dn+1,k+1 into Dn+1,k by the following:. 治政大k + 1 flaws, 0≤k≤n. Each path in consists of all n + 1-Dyck paths with 立 can be factorized into AdBuC, where A is a subpath all above the x-xais, d. 1. The set Dn+1,k+1. 學. ‧ 國. Dn+1,k+1. is the first down step below the x-xais, B is a subpath all bellow the x-xais, say with k2 flaws, 0≤k2 ≤k, u is the first up step contact the x-axis after B and below the x-axis, say. ‧. uBd with k2 + 1 flaws, and C is the remaining path with k − k2 flaws, 0≤k≤n.. sit. y. Nat. 2. The set Dn+1,k consists of all n + 1-Dyck path with k flaws, 0≤k≤n. Each path in Dn+1,k can. n. al. er. io. be factorized into BuAdC, where B,dAu, and C have k2 , 0, and k − k2 flaws.. i Un. v. Note: A、B 、C may be empty, and they have the same number of up-steps and down-steps.. Ch. engchi. i.e. f2 (AdBuC) = BuAdC. A. A u. f2 u. d B. C. d. C x-axis. B. Figure 3.1: f2 (AdBuC) = BuAdC. 14 DOI:10.6814/NCCU202000719.

(22) Theorem 3.0.1. Define f2 :Dn+1,k+1 →Dn+1,k by f2 (AdBuC) = BuAdC. The function f2 is one-to-one. Proof. Let |B| = 2i, 0≤i≤n; |u| = 1; |A| = 2j, 0≤j≤n − i; |d| = 1; |C| = 2(n − i − j − 1). Claim: f2 one-to-one. (i.e.f2 (AdBuC) = f2 (A′ dB ′ uC ′ ) ⇒ BuAdC = B ′ uA′ dC ′ ) Suppose |B ′ | = 2r, 0≤r≤n; |u| = 1; |A′ | = 2s, 0≤s≤n−r; |d| = 1; |C ′ | = 2(n−r−s−1). Claim 1: |B| = |B ′ | case 1: 2i < 2r. T he (2i+1)st step. T he (2i+1)st step. A. C u. d. 立. 政治 A大 ′. y. sit er. al C Ch. ‧ 國. n. B. io. T he (2i+1)st step. u. d. B′. ‧. Figure 3.2: AdBuC and A′ dB ′ uC ′. Nat A. u. d. 學. B. C′. n U engchi ′. iv. u. A′ d. C′. T he (2i+1)st step. B. Figure 3.3: f2 (AdBuC) = BuAdC and f2 (A′ dB ′ uC ′ ) = B ′ uA′ dC ′. To see Figure 3.3 when we start on (0, 0) to walk along the two path BuAdC and B ′ uA′ dC ′ . The (2i + 1)st step of BuAdC is above the x-axis, but the (2i + 1)st step of B ′ uA′ dC ′ is still below the x-axis. This is a contradiction as two paths are the same.. 15 DOI:10.6814/NCCU202000719.

(23) case 2: 2i > 2r. T he (2r+1)st step. A′. A. C′. C u. d. u. d. T he (2r+1)st step. B′. B Figure 3.4: AdBuC and A′ dB ′ uC ′. A. 政治大. 立C. T he (2r+1)st step. d. T he (2r+1)st step. B. u. 學. ‧ 國. u. A′. B′. d. C′. ‧. al. er. io. sit. y. Nat. Figure 3.5: f2 (AdBuC) = BuAdC and f2 (A′ dB ′ uC ′ ) = B ′ uA′ dC ′. n. To see Figure 3.5 when we start on (0, 0) to walk along the path the two path BuAdC ′. ′. ′. Ch. i Un. v. and B uA dC . The (2r + 1)st step of BuAdC is below the x-axis, but the (2r + 1)st step. engchi. of B ′ uA′ dC ′ is still above the x-axis. This is a contradiction as two paths are the same. Thus, we have proof that i = r. ∴ |B| = |B ′ | and B = B ′ .. 16 DOI:10.6814/NCCU202000719.

(24) Claim 2: |A| = |A′ | recall:Let |B| = 2i, 0≤i≤n; |u| = 1; |A| = 2j, 0≤j≤n − i; |d| = 1; |C| = 2(n − i − j − 1). Suppose |B ′ | = 2r, 0≤r≤n; |u| = 1; |A′ | = 2s, 0≤s≤n − r; |d| = 1; |C ′ | = 2(n − r − s − 1). case 1: 2j < 2s. T he (2i+2j+2)nd step. A′. A. C′. C d. u. T he (2i+2j+2)nd step. u. d. B. 立. 政治大. B′. Figure 3.6: AdBuC and A′ dB ′ uC ′. ‧. ‧ 國. 學 T he (2i+2j+2)nd step. T he (2i+2j+2)nd step. y y=1. n. al. sit. C. d. io. B. u. Ch. i Un. B′. engchi. u. er. y=1. Nat. A. A′. C′ d. v. Figure 3.7: f2 (AdBuC) = BuAdC and f2 (A′ dB ′ uC ′ ) = B ′ uA′ dC ′. To see Figure 3.7 when we start on (2j + 1, 1) to walk along the path A and A′ . The (2i + 2j + 2)nd step of Ad is below y = 1, but the (2i + 2j + 2)nd step of A′ d is still above y = 1. This is a contradiction as two paths are the same.. 17 DOI:10.6814/NCCU202000719.

(25) case 2: 2j > 2s. T he (2i+2s+2)nd step. A. A′. C′. C u. d. u. d. T he (2i+2s+2)nd step. B. B′. Figure 3.8: AdBuC and A′ dB ′ uC ′. A. 政治 T he (2i+2s+2)nd step 大 A nd ′. 立. y=1. B. step. C. d. y=1. C′. u. 學. u. B′. d. ‧. ‧ 國. T he (2i+2s+2). al. er. io. sit. y. Nat. Figure 3.9: f2 (AdBuC) = BuAdC and f2 (A′ dB ′ uC ′ ) = B ′ uA′ dC ′. n. To see Figure 3.9 when we star on (2i + 1, 1) to walk along the path A and A′ . The (2i + 2s + 2). nd. Ch. i Un. v. step of Ad is above y = 1, but the (2i + 2s + 2)nd step of A′ d is still below. engchi. y = 1. This is a contradiction as two paths are the same. Thus, we have proof that j = s. ∴ |A| = |A′ | and A = A′ . Since B = B ′ and A = A′ , ∵ BuAdC = B ′ uA′ dC ′ ⇒ C = C ′ ∴ AdBuC = A′ dB ′ uC ′ Therefore f2 is one-to-one.. 18 DOI:10.6814/NCCU202000719.

(26) Lemma 3.0.2. If Q is a path in Dn+1,k+1 , then f2 (Q) has k flaws. Moreover, the k + 1st flaw will be restored and connected to original lift behind. Note: A up-step above the x-axis is called lift. The lifts are counted from right to left and top to bottom. Proof. The function f2 from Dn+1,k+1 into Dn+1,k . Suppose P has k + 1 flaws in Dn+1,k+1 , and Q = AdBuC, where A, dBu, C have 0 flaw, k + 1 − j flaws, j flaws, respectively. (i.e.A, dBu, C have n − k − i lifts, 0 lift, i lifts, respectively.) The (k + 1)th flaw is in subpath dBu.Then f2 (Q) has k flaws on Dn+1,k , and f2 (Q) = BuAdC, where B, uAb, C have k − j flaws, 0 flaw, j flaws, respectively.. 政治大. 立. A. ‧ 國 u. B. y. u. d. B. n. al. C. x-axis. er. B. io. d. Nat. C. u. x-axis. ‧. B. C. sit. d. d. 學. u. C. A. i Un. v. Figure 3.10: the new lif t will be connected to original lif t behind. Ch. engchi. Notice that the path f2 (Q), u is connected to the (n − k)th flaw in subpath A behind. u becomes the (n − k + 1)st lift. Therefore, no matter how many times we use f2 , u is conneted to the (n − k + 1)st lift in subpath A behind. u is still the (n − k + 1)st lift.. 19 DOI:10.6814/NCCU202000719.

(27) Note: Rising from. Falling to. x-axis x-axis x-axis Falling from. Rising to. x-axis. Figure 3.11: Rising f rom; F alling to; F alling f rom; Rising to. 治政 the (n − k + 1) lift in P will be dropped. 大立. Lemma 3.0.3. If P is a path in Dn+1,k , then f2−1 (P ) has k + 1 flaws.Moreover, st. ‧ 國. 學. i.e. The (n − k)th lift u and the first down-step d falling to the x-axis on the right side of u in Dn+1,k are dropped to the first up-step u rising to the x-axis and the first down-step d. Empty area. d. y. al. n. u. C. B. A P reimage under f2 − −−−−−−−−−−−−−− →. Ch. engchi. sit. io. y=1. er. A. Nat. Show by formula:. ‧. falling from the x-axis in Dn+1,k+1 .. i Un. v. y=-1. d. C Empty area. u. x-axis. B. P reimage under f2. Figure 3.12: BuAdC −−−−−−−−−−−→ AdBuC. In P , the (n − k + 1)st lift u and the first down-step d falling to the x-axis on the left side of u in Dn+1,k , we can observe that there is an empty area enclosed by the u, d, the x-axis, and the horizontal line y = 1. After switching two portions Bu and Ad, another empty area is enclosed by the d is the first down-step falling from the x-axis, and u is the first up-step rising to the x-axis, and the horizontal line y = −1 in f2−1 (P ). And the remaining segments which are behind d is the fixed subpath C.. 20 DOI:10.6814/NCCU202000719.

(28) Theorem 3.0.2. Define f2 :Dn+1,k+1 →Dn+1,k by f2 (AdBuC) = BuAdC. The function f2 is onto. Proof. Claim: f2 is onto. (f2 :Dn+1,k+1 → Dn+1,k ) For any path P = BuAdC in Dn+1,k which 0≤k≤n. We choose the (n − k + 1)st lift u and choose the first down-step d falling to the x-axis on the right side of u. We switch the portions Bu and Ad then we can drop the (n − k + 1)st lift (i.e. the (k + 1)st flaw) and get a new path Q = AdBuC in Dn+1,k+1 , by Lemma 3.2、Lemma 3.3. Where Q has at most n + 1 flaws and P has at least one flaw. In fact, if P has k flaws then Q has k + 1 flaws. So every P in Dn+1,k , we can find a path Q in Dn+1,k+1 , such that f2 (Q) = P . Therefore f2 is one-to one and onto.. 政治大. Hence, f2 is one-to-one and onto by Theorem 3.1 and Theorem 3.2.. 立. Note: Let f2−1 be the inverse function of f2 .. ‧ 國. 學. Definition 3.0.4. The set X2 consists of all paths in n + 1-Dyck paths with k + 1 flaws.Each − → → path in X can be factorized into A d B − u C. The set Y consists of all 2. 2. ‧. n + 1-Dyck paths which are good paths.. Nat. al. er. io. k+1 times. sit. 1. g2 (Q) = f2k+1 (Q) where Q in X2 , and f2k+1 = f2 ◦ f2 ◦ · · · ◦ f2 {z } |. y. Define a function g2 from X2 into Y2 by the following:. v. n. − → 2. The first down-step falling from x-axis denote by d and the first up-step rising to → x-axis denote by − u.. Ch. engchi. i Un. − → Lemma 3.0.5. Suppose that Q in X2 . In f2 (Q), the first down-step d falling from the → x-axis of Q connects with the first up-step − u rising to the x-axis of Q to. − → − → − → → − → → → → be − u d and − u d is above the x-axis. Let Q = d B − u C, then f2 (Q) = B − u d C.. − → → Proof. We may assume that Q = A d B − u C is any path of Dn+1,k+1 . Since the subpath A is − → − − → − → → → empty, then Q = d B → u C ∈ X and f (Q) = B − u d C.We know d B − u has k + 1 − j 2. 2. − → → flaws, and C must have j flaws in Q. Then B, − u d , and C have k − j flaws, 0 flaw, and − → → j flaws in f (Q), respectively. Therefore, − u d is above the x-axis. 2. Note: In f2i (Q), 1≤i≤K + 1, the segments above the x-axis will be always above the x-axis. 21 DOI:10.6814/NCCU202000719.

(29) Theorem 3.0.3. The function g2 :X2 →Y2 is one-to-one. Proof. Suppose that g2 (P ) = g2 (Q), where P has k flaws and Q has h flaws in X2 . To prove that P = Q. By definition g2 (P ) = f2k+1 (P ), g2 (Q) = f2h+1 (Q) case 1: k < h f2k+1 (P ) = f2h+1 (Q) ⇒ f2 (f2k (P )) = f2 (f2h (Q)) ∵ f2 is one-to-one. ∴ f2k (P ) = f2h (Q). For the same reason, the following can be obtained f2 (f2k−1 (P )) = f2 (f2h−1 (Q)) ⇒ f2k−1 (P ) = f2h−1 (Q) Since f2 is one-to-one, use this method k + 1 times f2 (P ) = f2 (f2h−k (Q)) ⇒ P = f2h−k (Q) − → − → → → The d and − u of P are below the x-axis, but the − u d of f2h−k (Q) are above the x-axis. 立. by Lemma 3.5.. 政治大. ‧ 國. 學. This is a contradiction. case 2: k > h. ∴ f2k (P ) = f2h (Q). y. Nat. ∵ f1 is one-to-one. ‧. f2k+1 (P ) = f2h+1 (Q) ⇒ f2 (f2k (P )) = f2 (f2h (Q)). n. al. i Un. Since f2 is one-to-one, use this method h + 1 times f2 (f2k−h (P )). = f2 (Q) ⇒. Ch. er. io. f2 (f2k−1 (P )) = f2 (f2h−1 (Q)) ⇒ f2k−1 (P ) = f2h−1 (Q). sit. For the same reason, the following can be obtained. v. e=nQg c h i −→. f2k−h (P ). − → → → The − u d of f2k−h (P ) are above the x-axis, but d and − u of Q are below the x-axis by Lemma 3.5. This is a contradiction. case3: k = h f2k+1 (P ) = f2h+1 (Q) ⇒ f2 (f2k (P )) = f2 (f2h (Q)) ∵ f2 is one-to-one. ∴ f2k (P ) = f2h (Q). Use this method h + 1 times, we have f2 (P ) = f2 (Q) ⇒ P = Q Therefore, g2 is one-to-one.. 22 DOI:10.6814/NCCU202000719.

(30) Theorem 3.0.4. g2 :X2 →Y2 is onto, where Y2 is good path of n + 1-Dyck and g2 (Q) = f2k+1 (Q). − → → Proof. Give P ∈ Y2 , P has 0 flaw, and − u d is the (n − k + 1)st lift in P . −(k+1). Definition:The preimage of the function g2 = f2. = f2−1 ◦ f2−1 ◦ · · · ◦ f2−1 | {z } k+1 times. Since f2−1 (P ) is the preimage of P under f2 and has 1 flaw.. Using this way for k times, we get the path f2−k (P ) which has k flaws. − → → u d is still above the x-axis and is the (n − k + 1)st lift.We use this way In f2−k (P ), − − → → −(k+1) again, we get the path f (p) which has k + 1 flaws and d B− u is under the x-axis, 2. as f2 is onto and by Lemma 3.5. −(k+1). So we have f2. −(k+1). (P ) = f2. (f2k+1 (Q)) = Q,. 政治大 (P ) ⇒ g (Q) = f (Q) 立. where Q has k + 1 flaws,Q ∈ X2 . −(k+1). Let Q = f2. k+1 2. 2. −(k+1). )(P )). 學. ‧ 國. = f2k+1 ((f2 =P. Thus g2 is onto.. ‧ sit. y. Nat. n. al. er. io. Hence, g2 is one-to-one and onto by Theorem 3.3 and Theorem 3.4.. Ch. engchi. i Un. v. 23 DOI:10.6814/NCCU202000719.

(31) Definition 3.0.6. The set Z2 contains the good path for all n-Dyck paths which replaces − → − → u d in Y with a dot mark, and all paths in Y are n + 1-Dyck paths which 2. 2. are good path. Let h2 be the function from Y2 into Z2 . − → − → → → i.e. P = R− u d S is (2n + 2, 0) path, where R, − u d , and S are all good paths. h2 (P ) = − → → h (R− u d S) = R • S 2. Theorem 3.0.5. h2 is one-to-one and onto. Proof. It’s clearly that h2 is one-to-one. Given P ′ = R • S ∈ Z2 − → − → → → We can change • into − u d . Thus R • S ⇒ R− u d S ∈ Y2 .. 政治大. Therefore, h2 is one-to-one and onto.. 立. R. S. S. h2. io. sit. Nat. y. ‧. x-axis. ⃗ =R•S Figure 3.13: h2 (R⃗udS). n. al. er. d⃗. ‧ 國. ⃗u. 學. R. Ch. engchi. i Un. v. We have completed the combinatorial proof of (n + 2)Cn+1 = (4n + 2)Cn .. 24 DOI:10.6814/NCCU202000719.

(32) Chapter 4 Summary • In this thesis, we prove the Catalan identity in combinatorial way.. 政治大. In Chapter 2, we give a bijective proof between ‵‵ P aths Start with U p − step′′ and ‵‵. 立. g1. h. 1 Dotted T otally Bad P aths′′ . Then we construct the functions in X1 −→ Y1 −→ Z1. ‧ 國. 學. that Paths Start with Up-step.. In Chapter 3, we give a bijective proof between ‵‵ P aths Start with Down − step′′ and ‵‵. g2. h. ‧. 2 Dotted Good P aths′′ .Then we construct the functions in X2 −→ Y2 −→ Z2 that Paths. Start with Down-step.. sit. y. Nat. n. al. er. io. Proving the Catalan identity through the Dyck paths can reveal the following advantages:. i Un. v. 1.The subpath C does not change during the process of switching of the portions Ad and Bu.. Ch. engchi. 2.Since the subpath B of P in x1 is empty, a new flaw generated after switching of the portions Ad and Bu must be followed by the original subpath C. Since the subpath A of Q in x2 is empty, a new lift generated after switching of the portions Bu and Ad must be followed by the original subpath C. 3.In the process of computing the preimage of a function g1 (g2 ), the flaws (lifts) recovery mode follows the ‵‵ Last − in − F irst − out′′ or ‵‵ F irst − in − Last − out′′ .. 25 DOI:10.6814/NCCU202000719.

(33) Appendix A examples of Catalan identity (n + 2)Cn+1 = (4n + 2)Cn 治. 立. • n=1. 政. 大. ‧. ‧ 國. 學. io. sit. y. Nat. n. al. er. A.1. Ch. engchi. i Un. v. 26 DOI:10.6814/NCCU202000719.

(34) • n=2. 立. 政治大. ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. i Un. v. 27 DOI:10.6814/NCCU202000719.

(35) • n=3. 立. 政治大. ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. i Un. v. 28 DOI:10.6814/NCCU202000719.

(36) 立. 政治大. ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. i Un. v. 29 DOI:10.6814/NCCU202000719.

(37) Bibliography [1] 劉映君. 一個卡特蘭等式的組合證明, 2017. [2] Ronald Alter. Some remarks and results on catalan numbers. 05 2019. [3] Ronald Alter and K.K Kubota. Prime and prime power divisibility of catalan numbers.. 政治大. Journal of Combinatorial Theory, Series A, 15(3):243 – 256, 1973.. 立. [4] Federico Ardila. Catalan numbers. The Mathematical Intelligencer, 38(2):4–5, Jun 2016.. ‧ 國. 學. [5] Young-Ming Chen. The chung–feller theorem revisited. Discrete Mathematics, 308:1328–. ‧. 1329, 04 2008.. [6] Ömer Eğecioğlu. A Catalan-Hankel determinant evaluation. In Proceedings of the Fortieth. y. Nat. sit. Southeastern International Conference on Combinatorics, Graph Theory and Computing,. n. al. er. io. volume 195, pages 49–63, 2009.. Ch. i Un. v. [7] R. Johnsonbaugh. Discrete Mathematics. Pearson/Prentice Hall, 2009.. engchi. [8] Thomas Koshy. Catalan numbers with applications. Oxford University Press, Oxford, 2009. [9] Tamás Lengyel. On divisibility properties of some differences of the central binomial coefficients and Catalan numbers. Integers, 13:Paper No. A10, 20, 2013. [10] Youngja Park and Sangwook Kim. Chung-Feller property of Schröder objects. Electron. J. Combin., 23(2):Paper 2.34, 14, 2016. [11] Matej Črepinšek and Luka Mernik. An efficient representation for solving Catalan number related problems. Int. J. Pure Appl. Math., 56(4):589–604, 2009.. 30 DOI:10.6814/NCCU202000719.

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