排列形表─貝爾形表

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(1)國立高雄大學應用數學系碩士論文. 排列形表─貝爾形表 Bell Permutation Tableaux. 研究生：何澤初撰指導教授：游森棚. 中華民國一百零一年七月.

(2) 致謝這本論文的完成，首先要感謝的就是我的指導教授游森棚教授，他在我大學就讀應用數學系的期間，開啟了我對組合學的興趣，也感謝他願意在我碩士兩年的期間，用心的指導我；除了論文方面的要求，對我的其他方面要求也很高，擔任教官的線性代數助教讓我對線性代數的認識又更深了一層，也從教官身上學到如何帶學生，以及對自己高標準的要求；教官說過：沒見過好的，不知道可以有多好，感謝教官讓我看到努力可以做到什麼程度，看到對研究有熱情可以做到什麼地步，讓我在學術上與生活上都有深刻的體驗與成長。也要感謝學姊顏珮嵐博士，撥時間與我共同討論問題、還有大力的幫助我撰寫論文以及訂正寫法，真的十分的感謝她。再來要感謝傅東山教授以及潘業忠教授在我論文撰寫期間不吝給予建議。然後感謝口試委員嚴志弘教授，用心的看我的論文，不但詳細的告訴我如何修改，還寄範本讓我知道自己的格式不對在哪裡。還要感謝我的好朋友兼好同學蔡維迦，除了修課時一起討論問題，還有一起練習口試跟互相督促論文、一起睡研究室、一起三點起來唸書趕報告，還有一同上山下海的；以及一同修課的立志學長跟建太學長，大家一起討論，解開、困住在難題上的時光是難忘的。感謝佳芳學姐跟基祥學長在我Latex遇上困難的時候，願意在百忙之餘提供協助；也要感謝大學同學邱昇昱、許耀中，除了讓我問Latex跟英文文法的膚淺問題，還和凃雅婷同學跟曾品源同學，一起撥空聽我練習口試、提出建議，真的是十分的感謝他們的協助。而在高大的這幾年，感謝應用數學系，系辦公室的千惠姐以及雅鳳姐提供行政方面、各類跑流程的協助，感謝劉欽岳教授、張惠蘭教授對我的論文題目提出指導，還有系上各位教授以及學弟妹。最後我要感謝我的父親跟母親，生我養我、願意提供我資源讓我能夠唸書唸到碩士畢業，以及我的兩位弟弟，在我無法抽身的時候，代替我打點、照料家中的一切，由衷的感謝他們。無論幫助的程度大或者小，如果沒有他們的幫助，我想我的論文是無法如期完成的，再次的感謝幫助我的每個教授、老師、學長姐、朋友、家人。. 何澤初謹致於國立高雄大學應用數學系碩士班民國一○一年七月.

(3) Bell permutation tableaux. Tze-Chu Ho Advisor : Sen-Peng Eu. Department of Applied Mathematics National University of Kaohsiung Kaohsiung, Taiwan 811, R.O.C. June 2012.

(4) Contents. Z`. ii. Abstract. iii. 1 Introduction. 1. 1.1. Permutation tableaux . . . . . . . . . . . . . . . . . . . . . . . . . . .. 1. 1.2. Bell tableaux and generalized pattern avoidance . . . . . . . . . . . .. 2. 1.3. The p, q-Stirling number and statistics on set partitions . . . . . . . .. 4. 1.4. The p, q-Stirling number and the Bell tableaux . . . . . . . . . . . . .. 6. 1.5. The p, q-Stirling number and Leroux’s 01-tableaux . . . . . . . . . . .. 8. 1.6. Enumeration of the intersections of six subclasses . . . . . . . . . . .. 9. 2 Characterizations of L-Bell, R-Bell, B-Bell tableaux. 12. 2.1. Corteel and Nadeau’s bijection . . . . . . . . . . . . . . . . . . . . . . 12. 2.2. Characterizations of L-Bell tableaux . . . . . . . . . . . . . . . . . . 13. 2.3. Characterizations of R-Bell tableaux . . . . . . . . . . . . . . . . . . 19. 2.4. Characterizations of B-Bell tableaux . . . . . . . . . . . . . . . . . . 21. 3 p, q-Statistics on Bell tableaux. 22. 3.1. q-statistics on L-Bell and R-Bell Tableaux . . . . . . . . . . . . . . . 22. 3.2. p, q-statistics on L0 -Bell, R0 -Bell, and B-Bell Tableaux . . . . . . . . . 27. 3.3. p, q-statistics on L-Bell Tableaux . . . . . . . . . . . . . . . . . . . . 30. 4 Cardinality of intersections of various subclasses of tableaux. 33. 4.1. The Catalan numbers and the Narayana numbers . . . . . . . . . . . 33. 4.2. The Bessel numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37. 4.3. The Motzkin numbers . . . . . . . . . . . . . . . . . . . . . . . . . . 38. 4.4. The Binomial numbers . . . . . . . . . . . . . . . . . . . . . . . . . . 41. 4.5. The number of matchings in Kn . . . . . . . . . . . . . . . . . . . . . 42. 5 Conclusion Remarks. 43. i.

(5) 排列形表─貝爾形表指導教授：游森棚教授國立高雄大學應用數學系. 學生：何澤初國立高雄大學應用數學系. 摘要. Postnikov 在一篇計算全正 Grassmannian 格數量的文章中，介紹了排列形表的概念。長度為 n 的排列形表數量有 n! 個是已知的。Corteel 和 Nadeau 給了一個排列形表與排列之間之間的雙射 φ ，並介紹了兩種排列形表的子集，數量都是貝爾數的 L-貝爾以及 R-貝爾形表。Chen 跟 Liu 刻畫了 L-貝爾形表在 phi 之下對應的排列。在本篇文章中，我們介紹三種全新的排列形表的子集，分別為 L’-貝爾、R’-貝爾以及 B-貝爾，並証明其數量也是貝爾數。我們也給出了在 φ 之下刻畫 R-貝爾和 B-貝爾的排列迴避模式。我們還介紹了這五個排列形表子集上的統計量，並証明分布與 Wach.White 在 Leroux 的 01-形表上的 p,q-斯特林數相同。同時，也給出了不同於 Corteel 和 Nadeau 且能保持 q 統計量的 L-貝爾和 R-貝爾之間的雙射。我們還探討了這五個集合之間的交集。事實證明，出現了許多熟悉的經典數列。其中，我們證明同樣長度的 L-貝爾和 R-貝爾交集形表數量會是貝塞爾數。關鍵字：排列、排列形表、貝爾數、01-形表、p,q-斯特林數、L-貝爾、R-貝爾、B-貝爾、 L’-貝爾、R’-貝爾、貝塞爾數。. ii.

(6) ABSTRACT The concept of permutation tableaux was introduced by Postnikov in the context of enumeration of the totally positive Grassmannian cells. It is known that the number of permutation tableaux of length n is n!. Corteel and Nadeau gave a bijection φ between the set of permutation tableaux and the set of permutations, and introduced L-Bell and R-Bell tableaux, both counted by the Bell numbers, as two subclasses of the permutation tableaux. Chen and Liu then characterized the corresponding permutations of the L-Bell tableaux under the bijection φ. In this thesis we introduce three new subclasses of permutation tableaux, namely L0 -Bell, R0 -Bell, and B-Bell tableaux, and prove that they are also counted by the Bell numbers. We give characterizations of R-Bell and B-Bell tableaux in terms of pattern-avoiding permutations under the bijection φ. We also introduce statistics on these five subclasses of tableaux and prove that they are equidistributed with the p, q-Stirling numbers of Wach and White on the 01-tableaux of Leroux. Meanwhile a new bijection consistent with the above qstatistic, but different from Corteel and Nadeau’s, is given between L-Bell tableaux and R-Bell tableaux. We also investigate the cardinality of the intersection of two of these subclasses. It turns out that many familiar classical sequences appear. Among them we prove that the cardinality of the intersection of the L-Bell and R-Bell tableaux (of the same length) is a Bessel number. Keywords: Permutation, Permutation tableaux, Bell numbers, 01-tableaux, p, qstirling numbers, L-Bell, R-Bell, B-Bell, L0 -Bell, R0 -Bell, Bessel numbers.. iii.

(7) 1. Introduction. In this chapter we overview the main results of this thesis.. 1.1. Permutation tableaux. A permutation tableau T is a filling of a Ferrers diagram (with possibly empty rows) such that each cell of the diagram is filled with 0 or 1 and subject to the following conditions: 1. Each column contains at least one 1. 2. There is no 0 with a 1 above it (in the same column) and simultaneously with a 1 to its left (in the same row). The length (or size) of T is the sum of the number of columns and rows. Furthermore, it is known that the number of permutation tableaux of length n is n!, hence the name ‘permutation tableaux’. A 0 is restricted if there is a 1 above it in the same column. A row is unrestricted if it contains no restricted 0’s, otherwise it is restricted. A row is zero row if there is contains no 1 in that row, otherwise it is nonzero row. A 1 is superfluous if there is a 1 above it in the same column. We label the South-East border of the shape of the tableau from 1 to its length, going from top-right to bottom-left. For example, Figure 3 is a permutation tableau of length 9. The rows 1 and 7 are unrestricted and the rows 2, 3, 4 are restricted.. The notion of the permutation tableaux arose from the algebraic geometry, namely the enumeration of the totally positive Grassmannian cells [12, 19] and was introduced by Steingrímsson and Williams [15]. The permutation tableaux are also closely-related to the partially asymmetric exclusion process model (PASEP for short) in statistical physics [6] and are in bijection with the alternative tableaux introduced by Viennot [17].. 1.

(8) 1. 0. 1 1. 0. 0. 1 2. 0. 1. 1 3. 0. 0. 0 4 5. 1. 7. 6. 8 9 Figure 1: A permutation tableau of length 9.. Let Sn be the set of permutations of [n] and P Tn be the set of permutation tableaux of length n. In the literature there are three bijections between Sn and P Tn . The first one, Φ, is introduced by Steingrímsson and Williams [15], which is a simpler description of a map proposed by Postnikov [12]. More properties of permutation tableaux with respect to the bijection Φ are studied by Burstein [2]. Later in [5], Corteel and Nadeau proposed two other bijections φ and ξ between Sn and P Tn . In this thesis we will solely focus on the bijection φ, which will be introduced in the next chapter.. 1.2. Bell tableaux and generalized pattern avoidance. In [5], two special subclasses of permutation tableaux with the Bell numbers as the cardinality are introduced, namely the L-Bell tableaux and the R-Bell tableaux. A permutation tableau is L-Bell if all its topmost 1’s are also leftmost 1’s. Similarly a permutation tableau is R-Bell if all its topmost 1’s are also rightmost 1’s. In this thesis we will introduce another three subclasses of permutation tableaux. The first two subclasses are the L0 -Bell and R0 -Bell tableaux. A permutation tableau is L0 -Bell if all its leftmost 1’s are also topmost 1’s, while a permutation tableau is R0 -Bell if all its rightmost 1’s are also topmost 1’s. Although it seems that the L0 -Bell tableaux and the R0 -Bell tableaux are in some sense symmetric with the L-Bell tableaux and the R-Bell tableaux, their structures are quite different. The 2.

(9) third subclass is the B-Bell tableaux. We call a permutation tableau is B-Bell if its first row contains no 0’s. We will prove that these three subclasses are all counted by the Bell numbers. Let k < n, a pattern is a permutation σ ∈ Sk , and a permutation π = π1 π2 . . . πn ∈ Sn avoids σ if there is no subsequence in π whose letters are in the same relative order as the letters of σ. For example, π ∈ Sn avoids the pattern 132 if there are no 1 ≤ i < j < k ≤ n such that πi < πk < πj . We denote the subset of σ-avoiding permutations in Sn by Sn (σ). The concept of pattern avoidance is generalized by Babson and Steingrímsson [1]. In their version, there is a further requirement that some of the letters in a pattern must be adjacent. For example, a 1-32 pattern requires the letters corresponding to ‘3’ and ‘2’ must be adjacent; that is, π is 1-32 avoiding if there are no 1 ≤ i < j ≤ n such that πi < πj+1 < πj . The usual 132-avoiding in the last paragraph should be written as 1-3-2-avoiding in this generalized pattern avoidance notation. In this thesis we will use the generalized pattern avoidance notation. Let us go back to the permutation tableaux. One of the main result of Chen and Liu [4] is a characterization of the L-Bell tableaux in pattern-avoiding permutations, which answers a question proposed by Corteel and Nadeau on finding a statistic on permutation tableaux equidistributed with the number of occurrences of 32-1 [5]: Theorem 1. [4] Let T be a permutation tableaux of length n and π be the corresponding permutation under the bijection φ. Then T is L-Bell if and only if π ∈ Sn (3-21). Note that via the trivial identity Sn (3-21) = Sn (321, 41-32, 42-31) we can say that T is a L-Bell tableau of length n iff π ∈ Sn (321,41 -32,42-31). The reason for writing in this form is that it is parallel to our result on the R-Bell tableaux. Our first result is the following characterizations of the R-Bell tableaux and BBell tableaux in pattern-avoiding permutations under the bijection φ. For the sake of completeness, we also list the result on L-Bell tableaux. 3.

(10) Theorem 2. Let T be a permutation tableau of length n and π be the corresponding permutation under the bijection φ, then 1. T is a L-Bell tableau iff π ∈ Sn (321, 41-32, 42-31), 2. T is a R-Bell tableau iff π ∈ Sn (321, 31-42, 42-31), 3. T is a B-Bell tableau iff π ∈ Sn (32-41, 31-42) and π has no descents after the number ‘1’. The proofs will be shown in Chapter 2. Meanwhile, a new proof of (1) with our method will be given.. 1.3. The p, q-Stirling number and statistics on set partitions. The Stirling number of the second kind Sn,k is defined to be the number of set partitions of [n] into k (nonempty) blocks. By definition we have S0,0 = 1, Sn,k = 0 if k > n, and the recursion Sn,k = Sn−1,k−1 + kSn−1,k , if n ≥ k > 0. The Bell numbers Bn is the sum of Sn,k , for 0 ≤ k ≤ n. In the literature, there are several different q-analogues of Sn,k . The one we prefer here is the following. The q-Stirling numbers of the second kind Sq [n, k] is defined by    Sq [n − 1, k − 1] + [k]q Sq [n − 1, k], if 0 < k ≤ n; Sq [n, k] = 1, if n = k = 0;   0, otherwise,. (1). where [k]q =. 1 − qk = 1 + q + q 2 + · · · + q k−1 . 1−q. It can be generalized further to p, q-analogues. The p, q-Stirling numbers of the second kind Sp,q [n, k], introduced by Leroux [10], is defined by    Sp,q [n − 1, k − 1] + [k]p,q Sp,q [n − 1, k], if 0 < k ≤ n; Sp,q [n, k] = 1, if n = k = 0;   0, otherwise, 4. (2).

(11) where [k]p,q =. pk − q k = pk−1 + pk−2 q + · · · + q k−1 . p−q. Note that S1,q [n, k] = Sq [n, k] and S1,1 [n, k] = S1 [n, k] = Sn,k . We now introduce statistics on set partitions. Let Pn,k denote the set of all set partitions of [n] into k blocks. A set partition π ∈ Pn,k is in the standard form if π = B1 /B2 / · · · /Bk , where Bi are blocks of π, and min B1 < min B2 < · · · < min Bk . And π is in max-decreasing form if max B1 > max B2 > · · · > max Bk . Usually we represent a set partition in its standard form. A pair (Bi , b) is a hard inversion of π if b ∈ Bj , i < j, and max Bi > b. The statistic hinv(π) is the number of hard inversions of π. For example, hinv(π) = 4 if π = 126/35/4 since the hard inversions are (B1 , 3), (B1 , 5), (B1 , 4), and (B2 , 4). The hinv was first introduced by Wachs and White [18] in investigating the restricted growth functions (RG functions) and was discussed in [13]. They are called ‘hard’ because from its definition it is nontrivial to verify that it satisfies the recursion of q-Stirling numbers. One of Wachs and White’s result is the following: Theorem 3. [18] If n, k ∈ N, then Sq [n, k] =. X. q hinv(π) .. π∈Pn,k. Note that the summation runs over all π in standard form. In [14], Steingrímsson introduced eight new statistics (extened to ten in [9]) for set partitions. Here we list three statistics lcb, los and ros only that are used in this thesis. The opener of a block is its smallest element and the closer is its largest element. Let lcbi (π) be the number of blocks to the left of the block containing i such that their closer is bigger than i, losi (π) be the number of blocks to the left of the block containing i such that their opener is smaller than i, and rosi (π) be the number of 5.

(12) blocks to the right of the block containing i such that their opener is smaller than i. Then we set lcb(π) =. X. lcbi (π),. i∈[n]. los(π) =. X. losi (π),. i∈[n]. ros(π) =. X. rosi (π).. i∈[n]. For example, If σ = 178/25/346, then lcb(σ) = 7, los(σ) = 8, ros(σ) = 5 since we can calculate them in the following way. σ = 178/25/346, lcbi (σ) = 000/11/221, losi (σ) = 000/11/222, rosi (σ) = 022/01/000. Let π c be the complement of π, which is the partition obtained by replacing i with n + 1 − i. If we write σ c in max-decreasing form is 128/47/356, then los(σ c ) = 7 since σ c = 128/47/356, los(σ c ) = 000/11/122. By definition it is easy to see that the statistics hinv(π) and lcb(π) are identical. Also we have that hinv(π in standard form) = lcb(π in standard form) = los(π c in max-decreasing form).. 1.4. The p, q-Stirling number and the Bell tableaux. A natural question is to find statistics on L-Bell, R-Bell, L0 -Bell, R0 -Bell, and B-Bell tableaux respectively such that the distribution is Sq [n, k] or Sp,q [n, k] and to find statistic-preserving bijections between these subclasses. 6.

(13) In this thesis, for each of the four subclasses of Bell tableaux (except for the RBell tableaux), we find such corresponding p, q-statistics. For the R-Bell tableaux, we find a q-statistic. Firstly we find a q-statistic for L-Bell and R-Bell tableaux, respectively. By abuse of notation we also denote the desired statistic by hinv(T ). Let LBn be the set of L-Bell tableaux of length n and LBn,k be the subset with n − k nonzero rows. RBn and RBn,k is defined similarly. Theorem 4.. 1. For T ∈ LBn,k , set hinv(T ) as the sum of the number of 0’s for. which there is a topmost 1 above it (i.e. restricted 0’s)or to its left and the number of 1’s for which there are topmost 1’s above it (i.e. superfluous 1’s) and to its left simultaneously. 2. For T ∈ RBn,k , set hinv(T ) as the sum of the number of 0’s for which there is a topmost 1 above it (i.e. restricted 0’s) and twice the number of 1’s for which there are topmost 1’s above it (i.e. superfluous 1’s) and to its right simultaneously. Then we have Sq [n, k] =. X. q hinv(T ) =. T ∈LBn,k. X. q hinv(T ) .. T ∈RBn,k. For p, q-statistics, we have results on L0 -Bell, R0 -Bell, B-Bell, and L-Bell tableaux. Again by abuse of notation we denote the desired statistics by nin(T ) and inv(T ). Theorem 5.. 1. Let L0 Bn be the set of L0 -Bell tableaux of length n and L0 Bn,k. be the subset with k rows. For T ∈ L0 Bn,k , let nin(T ) (resp. inv(T )) be the number of cells above (resp. below) topmost 1’s in T . 2. Let R0 Bn be the set of R0 -Bell tableaux of length n and L0 Bn,k be the subset with k rows. For T ∈ R0 Bn,k , let nin(T ) (resp. inv(T )) be the number of cells above (resp. below) topmost 1’s in T . 3. Let BBn be the set of B-Bell tableaux of length n and BBn,k be the subset with k − 1 columns. For T ∈ BBn,k , let nin(T ) be the number of 1’s below topmost 7.

(14) 1’s (i.e. superfluous 1’s) and inv(T ) be the number of 0’s below topmost 1’s (i.e. restricted 0’s) in T . 4. Let LBn,k be the set of L-Bell tableaux of length n with n − k nonzero rows. For T ∈ LBn,k , let a star row be a row that has a leftmost 1 but no topmost 1’s. Set nin(T ) as the number of pairs x and y, where x is a leftmost 1, and y is an unrestricted 0 that is not in a star row and above x. Let inv(T ) be the cardinality of A ∪ B, where A is the set of cells that is not in a star row and below a topmost 1, and B is the set of pairs u and v in the same column, where u is a cell in A and v is a leftmost 1 below u. Then we have Sp,q [n, k] =. X. pnin(T ) q inv(T ) =. X. pnin(T ) q inv(T ). T ∈R0 Bn,k. T ∈L0 Bn,k. =. X. pnin(T ) q inv(T ) =. T ∈BBn,k. X. pnin(T ) q inv(T ) .. T ∈LBn,k. Note that, the p, q-statistics on L-Bell tableaux in Theorem 5 is different from the one in Theorem 4 by letting p = 1.. 1.5. The p, q-Stirling number and Leroux’s 01-tableaux. There is another set of tableaux (though they are not a subclass of permutation tableaux), introduced by Leroux [10], which are also counted by the Bell numbers. A 01-tableau is a 01-filling of a Ferrers diagram (with possibly empty rows) such that there is exactly one 1 per column. The length (or size) of a 01-tableaux is again the sum of the number of columns and rows. For example, Figure 2 illustrates a 01-tableau of length 9. Let Ln be the set of 01-tableaux of length n and Ln,k be the subset with k rows. There is a simple bijection between Ln and the set partition of [n]. Given a 01-tableaux of length n with k rows, its corresponding set partition of [n] with k blocks. A block consisting of i, the label of row, and every column label of 1’s in that row i. Therefore, a zero row forms a single block which means the block contains only one element. For example the 01-tableaux in Figure 2 corresponds to 8.

(15) 0. 0. 1. 0. 1. 1. 1. 0. 1. 0. 0. 0. 0. 0 32 4 5. 9. 8. 7. 6. 1. 1. Figure 2: A 01-tableau of length 9.. the set partition 1247/3689/5. By this bijection the cardinality of Ln is again the Bell numbers, and the cardinality of Ln,k is Sn,k . Let the statistic nin(T ) (resp. inv(T )) of a 01-tableau be the number of 0’s above (resp. below) 1’s. For example, the tableau T of Figure 2 has nin(T ) = 5 and inv(T ) = 4. Leroux [10] proved the following: Theorem 6. [10] Let Sp,q [n, k] be as above, then we have X. pnin(T ) q inv(T ) = Sp,q [n, k].. T ∈Ln,k. Later in [11], Médicis and Leroux found the explicit formula of Sq [n, k]. Theorem 7. [11] Let Sq [n, k] be as above, then we have µ ¶· ¸ n X j 1 n−j n (−1) . Sq [n, k] = n−k (q − 1) j k q j=k. 1.6. Enumeration of the intersections of six subclasses. Now we have six subclasses of tableaux, LBn , RBn , BBn , L0 Bn , R0 Bn , and Ln , and each of them is enumerated by the Bell numbers. It is natural to ask the cardinality if we take the intersection of some of them. ¡¢ It turns out that among the nontrivial 62 intersections of two of these six sub¡¢ classes, then we get the Table 1, and 63 intersections of three of these subclasses with Table 2 and Table 3, almost all of them are enumerated by some classical numbers. We collect our results in Table 1, 2 and 3. One of them has an asterisk in front of it, which means a conjecture that we still have no proof.. 9.

(16) Intersection of tableaux. The cardinality / The distribution (w.r.t. some certain statistic). Ln ∩ P Tn , Ln ∩ L0 Bn , Ln ∩ R0 Bn. Catalan numbers Cn / Narayana numbers. LBn ∩ RBn , *L0 Bn ∩ R0 Bn. Bessel numbers Besn / Number of blocks. Ln ∩ LBn , Ln ∩ RBn , RBn ∩ L0 Bn , LBn ∩ R0 Bn. Motzkin numbers Mn / Number of horizontal steps. BBn ∩ LBn , BBn ∩ RBn , BBn ∩ Ln , BBn ∩ L0 Bn , BBn ∩ R0 Bn. 2n−1 / Binomial numbers. LBn ∩ L0 Bn , RBn ∩ R0 Bn. Number of matchings in Kn / Number of matchings of size k in Kn. Table 1: The intersection of two these six subclasses Intersection of tableaux. The cardinality / The distribution (w.r.t. some certain statistic). Ln ∩ L0 Bn ∩ R0 B. Catalan numbers Cn / Narayana numbers. Ln ∩ LBn ∩ RBn , Ln ∩ LBn ∩ L0 Bn , Ln ∩ LBn ∩ R0 Bn , Ln ∩ RBn ∩ L0 Bn , Ln ∩ RBn ∩ R0 Bn , LBn ∩ L0 Bn ∩ R0 Bn , LBn ∩ RBn ∩ L0 Bn , RBn ∩ LBn ∩ R0 Bn , RBn ∩ R0 Bn ∩ L0 Bn. Motzkin numbers Mn / Number of horizontal steps. Table 2: The intersection of three of these six subclasses (I) Here we select one of our results as an example. We say two blocks Bi and Bj overlap in a set partition if min(Bi ) < min(Bj ) < max(Bi ) < max(Bj ). 10.

(17) A set partition is called non-overlapping if no pair of blocks overlaps. For example, σ = 127/36/45 is non-overlapping. Let Besn be the number of non-overlapping set partitions of [n] and it is known {Besn }n≥0 := {1, 1, 2, 5, 14, 43, 143, 509, . . . }, which is called the Bessel numbers [8]. One of our results shows that the number of tableaux which is simultaneously L-Bell and R-Bell is a Bessel number. Note that the distribution {1, n − 1} means the set is divided into two subsets, ¡¢ one with cardinality 1, another with cardinality n − 1. Since the proofs of 63 intersections of three subclasses are mostly trivial, we omit them in the thesis. Intersection of tableaux. The cardinality / The distribution (w.r.t. some certain statistic). BBn ∩ LBn ∩ Ln , BBn ∩ RBn ∩ Ln , BBn ∩ RBn ∩ L0 Bn , BBn ∩ LBn ∩ R0 Bn , BBn ∩ RBn ∩ R0 Bn , BBn ∩ RBn ∩ L0 Bn. n / {1, n − 1}. BBn ∩ L0 Bn ∩ Ln , BBn ∩ RBn ∩ LBn , BBn ∩ R0 Bn ∩ Ln , BBn ∩ R0 Bn ∩ L0 Bn. 2n−1 / Binomial numbers. Table 3: The intersection of three of these six subclasses (II) The rest of the thesis is organized as follows. In Chapter 2 we will review the bijection φ between the P Tn and Sn and give characterizations of L-Bell, R-Bell, BBell tableaux in terms of pattern avoidance permutations, respectively. In Chapter 3, we will introduce statistics nin and inv on L-Bell, R-Bell, B-Bell, L0 -Bell, and R0 -Bell tableaux respectively. In Chapter 4, we will give some enumerative results on the cardinality of the intersections between these six subclasses of tableaux. In the last chapter, we propose some problems.. 11.

(18) 2. Characterizations of L-Bell, R-Bell, B-Bell tableaux. 2.1. Corteel and Nadeau’s bijection. In this section, we review Corteel and Nadeau’s bijection φ between P Tn and Sn [5] and introduce the notation used in our proofs. For a tableau T , the first step of φ is initialize the permutation σ to the list of the labels of the unrestricted rows in increasing order. Then for each column, starting from the left and proceeding to the right, if the column is labeled by j and if (i, j) is the topmost 1 of the column then we add j immediately to the left of i in the permutation σ. Moreover if column j contains rightmost restricted 0’s in rows i1 , i2 , . . . , ik then we add i1 , i2 , . . . , ik in increasing order to the left of j in the permutation σ. Note that a column in T is a descent in φ(T ). Now we introduce the notations used in our proofs. Let c1 , c2 , . . . , cs be the columns of T with c1 > c2 > · · · > cs and r1 , r2 , . . . , rs be the corresponding rows of c1 , c2 , . . . , cs , where cells (rp , cp ) are topmost 1’s in T , for all 1 ≤ p ≤ s. Let R0 denote the set of its unrestricted rows and Rp = {r | cell (r, cp ) is a rightmost restricted 0}. Moreover, Let Listp denote the list obtained after the (p + 1)-th step; i.e., Listp is the list we obtain after inserting a sub-words Rp cp to the left of rp , where Rp represents that the elements in it are in increasing order.. For example, we consider the tableau T in Figure 3. The unrestricted rows are rows 1, 7 and 9. The rightmost restricted zeros are in cells (2, 8), (3, 8) and (4, 5). Step 1:R0 = {1, 7, 9}. ⇒ List0 = (1, 7, 9). Step 2:c1 = 8, r1 = 1, R1 = {2, 3}. ⇒ List1 = (2, 3, 8, 1, 7, 9). Step 3:c2 = 6, r2 = 3, R2 = ∅. ⇒ List2 = (2, 6, 3, 8, 1, 7, 9). Step 4:c3 = 5, r3 = 1, R3 = {4}. ⇒ List3 = (2, 6, 3, 8, 4, 5, 1, 7, 9). then the permutation we obtain is (2, 6, 3, 8, 4, 5, 1, 7, 9).. 12.

(19) 1. 0. 1 1. 0. 0. 1 2. 0. 1. 1 3. 0. 0. 0 4 5. 1. 7. 6. 8 9 Figure 3: A tableau corresponds to the permutation (2, 6, 3, 8, 4, 5, 1, 7, 9).. 2.2. Characterizations of L-Bell tableaux. Let φ(LBn ) be the set of permutations under Corteel and Nadeau’s bijection applying on the set LBn . We use a result from [5] in order to characterize φ(LBn ). Proposition 1. [5] There exists a bijection between L-Bell tableaux of length n such that the sum of the number of columns and the number of zero rows is k and set partitions of [n] with k blocks. In the following we prove three lemmas to show that φ(LBn ) ⊆ Sn (3-21) and construct a bijection h between Sn (3-21) and set partitions of [n]. Finally, we apply Proposition 1 to prove the result. Lemma 1. Let T ∈ LBn , c1 , c2 , . . . , cs are the columns of T with c1 > c2 > · · · > cs and r1 , r2 , · · · , rs be the corresponding rows where cell (ri , ci ) is a topmost 1, for all 1 ≤ i ≤ s. Then for a column cl , 1. None of the cells to the right of (rl , cl ) is a restricted 0, and all of the cells to the left of (rl , cl ) are 0’s. 2. If rl is an unrestricted row, then rt > rl , for all t < l. That is, the topmost 1 (rt , ct ) is in the lower left corner of the topmost 1 (rl , cl ), for all t < l Proof.. 1. A restricted 0 to the right of (rl , cl ) will lead to a exist of a topmost. 1 above it in T , which contradicts to the definition of permutation tableaux. The cells to the left of (rl , cl ) are all 0’s since T ∈ LBn . 13.

(20) 2. If rl is an unrestricted row, i.e., none of the cells to the left of (rl , cl ) is a restricted 0, then the cells in the upper left of it are not topmost 1’s. Thus rt > rl , for all t < l.. Lemma 2. Let T ∈ LBn and follow the notations above. R0 is the set of unrestricted rows in T and Rp ={r | cell (r, cp ) is a rightmost restricted 0 }. Then 1. If rq ∈ Rp for some p < q, then cq is to the left of cp in σ = φ(T ). 2. If there exist 1 ≤ k1 , k2 , . . . , km ≤ s such that: (a) k1 > k2 > · · · > km , (b) rkt−1 ∈ Rkt , for all 2 ≤ t ≤ m, and (c) rkm ∈ R0 , Moreover, if there exists rw ∈ R0 such that rw is to the left of rkm in List0 , then ck1 , ck2 , . . . , ckm in σ are between rw and max Rkm . Proof.. 1. If rq ∈ Rp for some p < q, then rq > rp and we insert Rp cp into the list. before we insert Rq cq . Since rq ∈ Rp , we insert Rq cq to the left of rq . Hence cq is to the left of cp in σ. 2. According to (1), the order of ck1 , ck2 , . . . , ckm in σ is ck1 , ck2 , . . . , ckm . Suppose that rw is already to the left of rkm in List0 . Thus after we insert Rkm ckm into the list, Rkm ckm are between rw and rkm in Listkm . Moreover, after we insert Rkm−1 ckm−1 into the list, ckm−1 is between rw and max Rkm (say rw0 ) in Listkm−1 since rkm−1 ∈ Rkm . That is, ck1 , ck2 , . . . , ckm in σ are between rw and rw0 , see Figure 4 as an illustion.. According to Lemma 2, we know that if rk is a restricted row, then there exists q < k such that rk ∈ Rq . Furthermore, if rq is also a restricted row, then rq ∈ Ru for some u < q. In other words, for a restricted row rk , we can find rk = rk1 , rk2 , . . . , rkm 14.

(21) ... List0. rw. Listkm. Listkm−1 rw. r w rkm R0. .... ... r ... c r km−1 rw0 km k m R km. .... ... Rkm−1. ckm−1 rkm−1 rw0. c km r k m. Figure 4: An illustration of the insertion steps.. satisfy (a), (b), and (c) in Lemma 2. Then we call rkm = rk . That is, for a restricted row rk1 = rk , we can find a zigzag path (rk1 , ck1 ), (rk1 , ck2 ), (rk2 , ck2 ), (rk2 , ck3 ), . . . , (rkm−1 , ckm ), (rkm , ckm ) such that the path passes through topmost 1’s and rightmost restricted 0’s alternately, see Figure 5.. rk3. 1 0. c k3. r k2. 1 0. 1. c k2. c k1. r k1. Figure 5: A zigzag path in LB.. In addition, the order of ck1 , ck2 , . . . , ckm in σ is ck1 , ck2 , . . . , ckm , where k1 > k2 > · · · > km . Thus, those descents in σ are in increasing order. Lemma 3. Let T ∈ LBn , two columns cp , cq with p < q and cp , cq are the corresponding columns. Then the behavior of these two paths are either disjoint or they both pass through rightmost restricted 0’s in some column cj where 1 ≤ j ≤ p and then pass through the same cells after that (See Figure 6). Moreover, for the latter 15.

(22) case, we have cq is to the left of cp in σ.. 1. 1 1 0. or 1. 0. cq¯. cp¯. cp. cq cp¯ = cq¯ cj. cp. cq. Figure 6: The behavior of two zigzag paths in LB.. Proof. Suppose that these two paths are intersecting at some cell and then separate after that. See figure 7. That is, the path starting at cq passes cells (rqt , cqt ), (rqt , cqt+1 ) and the path starting at cp passes cells (rpl−1 , cpl ), (rpl , cpl ) where cqt+1 > cpl > cqt , rpl < rqt < rpl−1 . Noth that (rqt , cqt ), (rpl , cpl ) are topmost 1’s and (rqt , cqt+1 ), (rpl−1 , cpl ) are rightmost restricted 0’s.. 1 0. rp l. 0. rqt. 1. rpl−1. 0 cq. cp cqt+1. cpl. c qt. Figure 7: An intersection of two zigzag paths in LB. Since T ∈ LB, cell (rqt , cqt ) is a leftmost 1. Thus the intersection cell (rqt , cpl ) is a 0. Furthermore, it is a restricted 0 as cell (rpl , cpl ) is a topmost 1 above it. This leads to the fact that the cell (rqt , cqt+1 ) which is to the left of cell (rqt , cpl ) is not a rightmost restricted 0, a contradiction. 16.

(23) Since T ∈ LB, all topmost 1’s belong to distinct rows. Thus two paths intersect only if they both pass through rightmost restricted 0’s in some column cj , where 1 ≤ j ≤ p. Then by the behavior of path, they merge together after that. Moreover, if the path starting at cp (respectively cq ) passes through the rightmost restricted zero (rp0 , cj ) (respectively (rq0 , cj )), then rq0 < rp0 and rq0 , rp0 ∈ Rj . Thus rq0 is to the left of rp0 in listj . Hence cq is to the left of cp in σ. proof of Theorem 2(1). Now we are ready to prove the first part of Theorem 2(1), i.e., φ(LBn ) ⊆ Sn (3-21). For a tableau T ∈ LBn , we show that the descents in σ = φ(T ) are in increasing order. That is, for any two columns cp , cq with p < q, cq is to the left to cp in σ. We show that for any two columns cp , cq with p < q, cq is to the left of cp in σ. Hence the descents in σ are in increasing order. we consider four cases: Case 1. rp , rq are both unrestricted rows. Since rq is unrestricted row, by Lemma 1(2), rq < rp . Thus rq is to the left of rq in the list List0 . Hence, cq is to the left of cp in σ. Case 2. rp is an unrestricted row, while rq is a restricted row. Subcase 1. rq < rp . Then there exists a zigzag path starting from (rq , cq ). By Lemma 1(2), the path ends at cell (rq , cq ), where cq < cp and rq < rq . Since rq and rp are both unrestricted rows and rq < rp , we have rq is to the left of rq in list List0 . By Lemma 2(2), cq is to the left of cq . Hence, it is to the left of cp in σ. Subcase 2. rq > rp . By Lemma 1(2), the path starting from (rq , cq ) ends in either the upper right or the lower left areas of cell (rp , cp ). If it ends in the upper right area of (rp , cp ), then we have cq is to the left of cp in σ by the similar discussion in Case 1. Suppose it passes cell (rq0 , cp ) for some rq0 > rp , which means (rq0 , cp ) is a rightmost restricted 0. Then it ends at (rp , cp ). By Lemma 3, cq is to the left of cp in σ. Case 3. rp is a restricted row, while rq is an unrestricted row. By Lemma 1(2), rq < rp and the path starting from (rp , cp ) ends at a cell below the row rq , say (rp , cp ). Then rp , rq ∈ R0 and rq < rp . Hence, rq is to the left of rp . By Lemma 2(2), cp in σ is between rq and max Rp . Thus we still have the result that cq is to the left 17.

(24) of cp in σ. Case 4. rp , rq are both restricted rows. Subcase 1. rp < rq and these two paths end at the same cell, i.e., rp = rq . Since (rp , cp ) is a topmost 1, there is no restricted 0 above it in column cp . Thus, these two paths intersect at cells in column ct for some t ≤ p according to Lemma 3. We prove that these two path will not intersect at the left area of column cp . Assume that t < p, then we call cell (rm , cm ) as the last cell, when the path starting at (rq , cq ) passes before entering the left area of column cp , i.e., p < m < q and rp < rm < rq . Then the cell (rm , cp ) is a restricted 0 by Lemma 1(1). This implies that these two paths do not intersect at cells in column ct , a contradiction. Hence, t = p. That is, rm ∈ Rp and rm is to the left of cp in Listp . By the similar discussion in Lemma 2(2), cq is to the left of rm in σ, thus cq is to the left of cp in σ. Subcase 2. rp < rq and these two paths are disjoint. By the similar discussion in the previous case, we know that the path starting at cell (rq , cq ) do not pass through cells in the lower left area of (rp , cp ). It neither ends at cells in the lower right area of (rp , cp ) according to Lemma 1(2). Thus it passes through the upper right area of (rp , cp ) and (rq , cq ) is in the upper right area of (rp , cp ) in the end. This implies that rq is to the left of rp in List0 . By Lemma 2(2), cp is between rq and max Rp . Then we have that cq is to the left of cp in σ. Subcase 3. rp > rq . By the similar discussion, we have (rq , cq ) is in the upper right area of (rp , cp ) if these two paths are disjoint; otherwise, they intersect at cells in column ct for some t < p. For the former case, we have cq is to the left of cp in σ, hence cq is to the left of cp in σ. For the latter case, by Lemma 3, let (rq0 , ct ) and (rp0 , ct ) are rightmost restricted 0’s, where these two paths pass through column ct , then rq0 < rp0 and rq0 , rp0 ∈ Rt . We have that rq0 is to the left of rp0 in Listt . By the similar discussion in Lemma 2(2), cq is to the left of cp in σ. The second part of the proof is to find a bijection h from Sn (3-21) to set partitions. For a permutation σ in Sn (3-21), its descents are in increasing order. We obtain the set partition π = h(σ) by cutting σ in front of each left-to-right maximum into blocks. Note that, a left-to-right maximum that is not a descent forms a single 18.

(25) block, while a left-to-right maximum that is a descent forms a block of size greater than 1. For a set partition π, we arrange the blocks in increasing order of their maximum. In each block, we put its maximum in the first place, and then arrange the other elements in increasing order. Finally, we delete the blocks to obtain a permutation which has its descents in increasing order.. 2.3. Characterizations of R-Bell tableaux. Let φ(RBn ) be the set of permutations under Corteel and Nadeau’s bijection applying on the set RBn . proof of Theorem 2(2). Let T ∈ RBn and c1 , c2 , . . . , cs be the columns of T with c1 > c2 > · · · > cs . Let r1 , r2 , . . . , rs be the rows of T such that cell (rl , cl ) is a topmost 1, for all 1 ≤ l ≤ s. Then we have rl 6= rm , for all l 6= m. From R-Bell tableaux to permutations: we prove φ(T ) = σ ∈ Sn (321) first and then use this property to show that σ ∈ Sn (31-42, 42-31). Since a descent in σ is exactly a column in T , it is sufficient to say that there are no consecutive columns in σ. In bijection φ, the first step is to write down the unrestricted rows in increasing order, and then insert a sub-words Rl cl to the left of rl at each step, where Rl represents the rows in which the rightmost restricted 0’s occur in column cl in increasing order. Since rl 6= rm , for all m, we know that cm will not appear between cl and rl in σ. In other words, we have cl rl in σ for all l and hence σ ∈ Sn (321). Now we have an observation on R-Bell that if cell (i, j) is a topmost 1 in T , then j is a descent in σ and we have ji in σ, and vice versa. Now we consider two descents cp and cq in σ with cp > cq . Claim : For cp > cq , either rp < rq or rp > cq . Suppose that rp > rq and rp < cq , then (rq , cq ) and (rp , cp ) are topmost 1’s. Thus cell (rp , cq ) is a 1 since T is a permutation tableau. This implies that cell (rp , cp ) is a topmost 1 but not a rightmost 1, a contradiction. By Claim, we have either rp < rq or rp > cq . Then the possible patterns occurring in the former case are 41-32 or 32-41, and the possible patterns occurring 19.

(26) in the latter case are 43-21 or 21-43. That is to say, σ ∈ Sn (31-42, 42-31). From permutations to R-Bell tableaux: Suppose σ ∈ Sn (321,31-42,42 -31) and T =φ−1 (σ)∈ / RB, then there exists a topmost 1 which is not a rightmost 1. We call it as an illegal topmost 1. Now we take a column cl such that (rl , cl ) is the rightmost illegal topmost 1, and let (rl , ck ) be the rightmost 1 in row rl . Case 1 : The cell (rl , ck ) is not a topmost 1. Then rk < rl . By the choice of cl , there are all 0’s to the right of (rl , ck ) in row rl and to the right of (rk , ck ) in row rk ; furthermore, rt is neither rl nor rk for a column ct with cl > ct > ck . That is to say, we still have cl rl at each time we insert Rt ct . After the step we insert Rk ck to the right of rk , we have either · · · cl rl · · · ck rk · · · or · · · ck rk · · · cl rl · · · . Now either a pattern 42-31 or 31-42 occurs. We continue to insert Rq cq into it, for every cq < ck . By the choice of ck , we have rq 6= rl , rk . Thus, we still have either · · · cl rl · · · ck rk · · · or · · · ck rk · · · cl rl · · · . This results in the occurrences of the pattern being either 31-42 or 42-31 in σ, a contradiction. Case 2: The cell (rl , ck ) is a topmost 1. Then by the choice of cl , we have rt 6= rl , for all cl > ct > ck . By the similar discussion above, we have · · · cl Rk ck rl · · · after the step we insert Rk ck . If Rk = ∅, then we still have · · · cl ck rl · · · in σ, since rq 6= rl , for all cq < ck . That is a 321 pattern, a contradiction. Hence Rk 6= ∅. Suppose that rk1 is the smallest label in Rk , then we have · · · cl rk1 · · · after we insert Rk ck , and rk1 > rl . Now we consider the cells to the right of (rk1 , ck ). Subcase 2.1: There is no topmost 1 to the right of (rk1 , ck ). Hence for every cq < ck , we have rq 6= rk1 . Thus, we have · · · cl rk1 · · · ck rl · · · in σ. This is a pattern 42-31 occurs in σ, a contradiction. Subcase 2.2: There exists a topmost 1 to the right of (rk1 , ck ), say (rk1 , cq ). Then by the choice of cl , (rk1 , cq ) is the only one topmost 1 to the right of (rk1 , cq ). After each step we insert Rm cm with ck > cm > cq , we have cl rk1 and ck rl by the similar discussion in Subcase 2.1. For column cq , we insert Rq cq to the left of rk1 since rk1 ∈ Rk and rq = rk1 . Since there is no topmost 1 to the right of (rk1 , cq ), we still have cq rk1 in σ. Then we continue inserting the remaining columns; in the end, we consider the label next to cl in σ, say v. Thus v must be a row i, otherwise σ has 20.

(27) a pattern 321, a contradiction. That is, v ∈ Rj for some column cj with cj < cq and v > rj > rk1 . Thus · · · cl v · · · cq rk1 · · · is a pattern 42-31 in σ, a contradiction.. 2.4. Characterizations of B-Bell tableaux. Let φ(BBn ) be the set of permutations under Corteel and Nadeau’s bijection applying on the set BBn . proof of Theorem 2(3). Let T ∈ BBn with columns c1 , c2 , . . . , cs , where c1 > c2 > · · · > cs . Since the first row of T contains no 0’s, row 1 is an unrestricted row. Moreover, (1, cl ) are topmost 1’s for all 1 ≤ l ≤ s. The first step of φ is to write down the unrestricted rows in increasing order, thus the number ‘1’ is the first label in the list. Then insert Rl cl to the list. Since (1, cl ) are topmost 1’s for all 1 ≤ l ≤ s, we obtain σ = R1 c1 R2 c2 · · · Rs cs R0 . Note that, R0 is the unrestricted rows of T in increasing order, thus there is no descent after ‘1’ in σ. Furthermore, the descents are decreasing, hence σ avoids patterns 31-42 and 32-41. Conversely, suppose that there exists a permutation σ ∈ Sn (31-42, 32-41) and there is no descent after the number ‘1’ such that φ−1 (σ) = T ∈ / BBn ; i.e., there exists a column cl such that the cell rl 6= 1. If rl is an unrestricted row, then we have · · · 1 · · · Rl cl rl · · · in Listl . Thus cl will be a descent after the number ‘1’, a contradiction. If rl is a restricted row. Let rl ∈ Rw for some cw > cl . Thus we have · · · rl · · · cw rw · · · in Listl−1 . Hence · · · cl rl · · · cw rw · · · in Listl . This pair of descents cl , cw forms either a 31-42 or a 32-41 pattern in σ, a contradiction. We also obtain a bijection between set partitions and the set of Sn (31-42,32-41) in which there is no descent after the number ‘1’. Theorem 8. There is a bijection between set partitions of [n] and the set of permutations in Sn (31-42, 32-41) in which there is no descent after the number ‘1’. Proof. For every block in a set partition of [n], we arrange the elements in it in increasing order. Then we arrange the blocks that do not contain 1 in increasing order by its largest number. Finally, we put the block that contains 1 in the last. 21.

(28) Thus after deleting the blocks, we get a permutation whose descents are increasing and in which there is no descent after the number ‘1’. The reverse is easy.. We form a set partition by cutting a permutation in. Sn (31-42, 32-41) into blocks after each descent. By Theorems 2(3) and 8, we know that the B-Bell tableaux are counted by the Bell numbers.. 3. p, q-Statistics on Bell tableaux. 3.1. q-statistics on L-Bell and R-Bell Tableaux. We called the bijection in Proposition 1 f and let T be a L-Bell tableau whose sum of the number of columns and the number of zero rows is k. Then f (T ) is a set partition of [n] with k blocks formed by the following: 1. For every column j of T , construct a block of f (T ) that contains j and row i whenever its leftmost 1 is in column j, 2. Every zero row forms a block of size 1. Note that the largest element in each block of f (T ) is a column of T . For a L-Bell tableau T , let hinv(T ) be the sum of the number of the restricted 0’s, the number of 0’s for which there is a topmost 1 to the left of it and the number of the superfluous 1’s for which there is a topmost 1 to the left of it (stars in Figure 8). For example, the L-Bell tableau T in Figure 8 has hinv(T ) = 6. The image of T under f is π = 178/25/346. The complement of π, π c = 128/356/47, also has hinv(π c ) = 6. Now we are ready to prove our first result in this section: Proof of Theorem 4 (1). Suppose that T is a L-Bell tableau of length n with n − k nonzero rows, then f (T ) is a set partition of [n] with k blocks. As we have mentioned. 22.

(29) 1 0 0 0 1. 0 0 1 1. 0 1 1 1. 6 7. 5. 1 ? ? 1 ? 1 2 ? 1 ? 3 ? 4. 1 2 3 4. 6 7. 8. 5. 8. Figure 8: A L-Bell tableau of length 8 and its hinv.. in section 1.3, for a set partition π, hinv(π in standard form) = lcb(π in standard form) = los(π c in max-decreasing form). We prove this theorem by showing that hinv(T ) = los(f (T ) in max-decreasing form), thus hinv(T ) = lcb(f (T )c in standard form) = hinv(f (T )c in standard form). By applying Theorem 3 we complete the proof. Now we show that hinv(T ) = los(f (T ) in the max-decreasing form). We first classify the hinv(T ) into four types: 1. For a superfluous 1’s in cell (i, j) that has topmost 1’s to its left, it contributes once to hinv(T ). For example, (3, 5) in Figure 8. Let the cell (i0 , j) be the topmost 1 above it and (i, j 0 ) be a topmost 1 to its left, where i0 < i < j < j 0 . Since T is a L-Bell tableau, they are also leftmost 1’s. Thus in f (T ), i and j 0 are in the same block. Besides, i and j 0 are the opener and closer since (i, j 0 ) is a topmost 1. Hence, we have the block containing i, j 0 as its opener and closer is to the left of the block containing i0 , j as its opener and closer in f (T ). Therefore the opener i contributes once to losj (f (T )), where j is a closer. 2. For a 0 in cell (i, j) that has a topmost 1 to its left, it contributes once to hinv(T ). For example, (1, 6) and (1, 5) in Figure 8. Let the cell (i, j 0 ) be 23.

(30) a topmost 1 to its left and (i0 , j) be the topmost 1 in column j, where i < i0 < j < j 0 . By the similar discussion in previous case, we have the block containing i, j 0 as its opener and closer is to the left of the block containing i0 , j as its opener and closer in f (T ). Therefore the opener i contributes once to losj (f (T )), where j is a closer. 3. For a restricted 0’s in cell (i, j) that has topmost 1’s to its right, it contributes once to hinv(T ). For example, (2, 8) and (3, 8) in Figure 8. Let the cell (i0 , j) be the topmost 1 above it and (i, j 0 ) be a topmost 1 to its right, where i0 < i < j 0 < j. Similarly, we have the block containing i0 , j as its opener and closer is to the left of the block containing i, j 0 as its opener and closer in f (T ). We regard the contribution made by (i, j) in T as the contribution made by the opener i0 to losi (f (T )). Therefore the opener i0 contributes once to losi (f (T )), where i is an opener. 4. For a restricted 0’s in cell (i, j) that has is no topmost 1 to its right, it contributes once to hinv(T ). For example, (4, 8) in Figure 8. Let the cell (i0 , j) be the topmost 1 above it and (i, j 0 ) be a leftmost 1 in row i, where i0 < i < j 0 < j. Similarly, we have the block containing i0 , j as its opener and closer is to the left of the block containing i, j 0 , where i is neither an opener nor a closer and j 0 is a closer in f (T ). We regard the contribution made by (i, j) in T as the contribution made by the opener i0 to losi (f (T )). Therefore the opener i0 contributes once to losi (f (T )), where i is neither an opener nor a closer. Thus hinv(T ) ≤ los(f (T ) in the max-decreasing form). Besides, it is easy to see from above that every pair of (i, j), where i contributes to losj (f (T )), is counted exactly once. Hence, we have hinv(T ) = los(f (T ) in the max-decreasing form). Now we turn to the q-statistic on R-Bell tableaux. For a R-Bell tableau T , let hinv(T ) be the sum of the followings: 1. the number of the restricted 0’s (diamonds in Figure 9), and 24.

(31) 2. twice the number of the superfluous 1’s for which there is a topmost 1 to the right of it (stars in Figure 9). For example the R-Bell tableau T in Figure 9 has hinv(T ) = 6. Note that the 1 in cell (2, 8) contributes twice hinv(T ) as well as the cell (3, 6) does.. 1 1 0 0 1 8. 0 1 1 1. 0 0 1 1. 6 7. 5. 1. 1 1. 2 3. . 4. . 1. 1. 1. 2. ?. 1. 1 ?. 3. 2. 1. 4 6 7. 4. 5. 8. 3. 6 7. 5. 8. Figure 9: A R-Bell tableau of size 8.. We prove the distribution of statistic hinv on R-Bell tableaux by providing a bijection between LBn,k and RBn,k . Although Corteel and Nadeau already gave a bijection between LBn,k and RBn,k in [5], the statistic hinv between the corresponding tableaux vary and the bijection is tedious and use PT-words as its medium. The one we give here is easier than Corteel and Nadeau’s and is a statistic-preserving map for hinv. Now we introduce the bijection between LBn,k and RBn,k , say θ. From L-Bell tableaux to R-Bell tableaux.. The shapes of a L-Bell tableau T. and its corresponding R-Bell tableau T 0 are the same, by [5]. In each row, the topmost 1’s in T remain topmost 1’s in T 0 (but the positions may change). We do the bijection row by row from the bottom to the top. For each row, we fill the cells in T 0 with 0’s whenever there is a topmost 1 below it in the same column. Then 1. if there is no topmost 1 in the row, we fill the empty cells in T 0 from the left to the right with the restricted 0’s and the superfluous 1’s in the same order in T ; 2. otherwise, we write down the restricted 0’s and the superfluous 1’s in T to form a list, and we transform 0’s to 1’s and 1’s to 0’s in it, respectively; then 25.

(32) we fill the empty cells in T 0 from the left to the right with elements in reversing order of the list. Finally, we fill the remaining empty cell in T 0 with topmost 1’s to complete this row. See Figure 10 as an example.. 1 0 0 0 1. 0 0 1 1. 0 1 1 1. 6 7. 5. 0 0 0 1 1 0 1 1 1. 1 2 3 4. 1. 0 1 1 1. 8. 0 1 1 0 1 0 1 1. 0 0 1 1. 1 1 0 0 1. 0 1 1 1. 0 0 1 1. 6 7. 5. 1 2 3 4. 8. Figure 10: A bijection from L-Bell tableau to R-Bell tableau.. Note that, we put the restricted 0’s to the left of 1’s in each row operation, hence a topmost 1 in T becomes a rightmost 1 in the resulting tableau T 0 . Thus T 0 is a R-Bell tableau. From R-Bell tableaux to L-Bell tableaux. We apply the inverse operation in each row operation. Similarly, the topmost 1’s in R-Bell tableaux T 0 remain topmost 1’s in L-Bell tableaux T in each row (but the positions may change). We apply the inverse operation row by row from the bottom to the top. For each row, we fill the cells in T with 0’s whenever there is a topmost 1 below it in the same column. Then 1. if there is no topmost 1 in the row, we fill the empty cells in T from the left to the right with the restricted 0’s and the superfluous 1’s in the same order in T 0 ; 2. otherwise, we write down the restricted 0’s and the superfluous 1’s in the reversing order in T 0 to form a list, and we transform 0’s to 1’s and 1’s to 0’s in it, respectively; then we insert the topmost 1 to the list between 0’s and 1’s. Finally, we fill the empty cells in T from the left to the right with elements in the order of the list.. 26.

(33) proof of Theorem 4 (2). Now we prove that the statistics hinv(T ) and hinv(T 0 ) are equal. If we regard a contribution made by a 0 of type(2) as a contribution made by a 0 of type(3) in the proof of 4(1), then hinv(T ) of a L-Bell tableau T can be classified into three types: 1. For a superfluous 1 which has topmost 1’s to the left of it contributes once to hinv(T ). 2. For a restricted 0 which has topmost 1’s to the right of it contributes twice to hinv(T ). 3. For a restricted 0 which has no topmost 1 to its right contributes once to hinv(T ). On the other hand, for the statistics hinv(T 0 ) of the corresponding R-Bell tableau T 0 , we also classify them into three types: 1. For a restricted 0 which has topmost 1’s to the right of it contributes once to hinv(T 0 ). 2. For a superfluous 1 which has topmost 1’s to the right of it contributes twice to hinv(T 0 ). 3. For a restricted 0 which has no topmost 1 to its right contributes once to hinv(T 0 ). For example, (3, 8), (2, 8), and (4, 8) are of type(1), (2), and (3), respectively in Figure 9. It is easy to see that the bijection θ maps elements of types (i) in T to type (i) in T 0 , where i = 1, 2, 3, hence the hinv is invariant under θ.. 3.2. p, q-statistics on L0 -Bell, R0 -Bell, and B-Bell Tableaux. Let L0 Bn,k be the set of L0 -Bell tableaux with k rows. There is a natural bijection g between L0 Bn,k and set partition of [n] with k blocks. For a L0 -Bell tableau with k 27.

(34) rows, we can map each column to the row where the topmost 1 belongs. Thus the L0 -Bell tableaux with k rows are counted by the Stirling number of the second kind Sn,k , and the cardinality of it is counted by the Bell numbers. proof of Theorem 5(1). It is easy to see that the sum satisfies the same boundary conditions as Sp,q [n, k]. To verify the recursion, we consider T ∈ L0 Bn,k and the permutation tableau T 0 obtained from T by deleting the label n in the shape. If n is a row of T , then it is an empty row. Thus T 0 ∈ L0 Bn−1,k−1 and nin(T ) = nin(T 0 ), inv(T ) = inv(T 0 ). Suppose, on the contrary, that n is a column of T . Since T is a L0 -Bell tableau, there is exactly one 1 in column n, say cell (i, n) for some 1 ≤ i ≤ k. Then nin(T ) = nin(T 0 ) + i − 1, and inv(T ) = inv(T 0 ) + k − i. Thus. X. X. pnin(T ) q inv(T ) =. 0. pnin(T ) q inv(T. 0). T 0 ∈L0 Bn−1,k−1. T ∈L0 Bn,k. +. k X. X. pnin(T. 0 )+i−1. q inv(T. 0 )+k−i. i=1 T 0 ∈L0 Bn−1,k. =. X. 0. pnin(T ) q inv(T. 0). T 0 ∈L0 Bn−1,k−1. X. + [k]p,q. 0. 0. pnin(T ) q inv(T ) .. T 0 ∈L0 Bn−1,k. In fact, there is an obvious bijection between Ln,k and L0 Bn,k by simply replacing illegal 0’s (0’s which have a 1 to the left of it and a 1 above it simultaneously) with the superfluous 1’s. Therefore, the proofs for Ln,k and L0 Bn,k are the same. However, for R0 -Bell tableaux, it is not obvious to verify that the statistics satisfy the recursions that define the p, q-Stirling numbers. We give a bijection θ0 between L0 Bn,k and R0 Bn,k which is in some sense similar to the bijection θ we gave from L-Bell tableaux to R-Bell tableaux, and is statistic-preserving on nin and inv. This also proves that R0 -Bell tableaux with k rows are counted by the Stirling number of the second kind Sn,k , and the cardinality of it is counted by the Bell numbers. 28.

(35) From L0 -Bell tableaux to R0 -Bell tableaux. The shapes of a L0 -Bell tableau T and its corresponding R0 -Bell tableau T 0 are the same. In each row, the topmost 1’s in T remain topmost 1’s in T 0 (but the positions may change). We do the bijection row by row from the bottom to the top. For each row, we fill the cells in T 0 with 0’s whenever there is a topmost 1 below it in the same column. Then from the left to the right, we fill 0’s as many as the number of the restricted 0’s T has in the same row. Finally, we fill the remaining empty cells in T 0 with topmost 1’s and the superfluous 1’s in reversing order of T to complete this row. See Figure 11 as an example. 1. 0. 0. 0. 0. 0. 0. 1. 1. 0. 0. 0. 0. 0. 0. 1. 0. 0. 0. 0. 1. 0. 0. 2. 0. 0. 0. 0. 1. 0. 0. 2. 0. 1. 0. 0. 1. 1. 1. 3. 0. 1. 1. 0. 1. 0. 3. 0. 0. 0. 1. 1. 1. 0. 0. 1. 0. 0. 0. 1. 1. 1 7 8. 1 6. 1 4 5. 7 8. 4 5. 6. 0 0 1 1 12 11 10 9. 12 11 10 9. Figure 11: A L0 -Bell tableau and its corresponding R0 -Bell tableau. Since we put the restricted 0’s to the left of 1’s in each row, the resulting tableau T 0 is still a permutation tableau. Moreover, a leftmost 1 in T becomes a rightmost 1 in T 0 after we reverse the order of 1’s, hence T 0 is a R0 -Bell tableau. From R0 -Bell tableaux to L0 -Bell tableaux. We do the same steps as what we do from L0 Bn,k to R0 Bn,k . proof of Theorem 5(2). We prove it by showing that the statistics are invariant under the bijection θ0 we gave between L0 Bn,k and R0 Bn,k . We consider the statistics nin and inv among these two permutation tableaux. For each topmost 1 in each row, we just shift its position in the same row, and this will not alter the number of cells above it. Thus the statistic nin is unchanged. Furthermore, the number of cells below topmost 1’s plus the number of cells above topmost 1’s equals to the number of total cells minus the number of topmost 1’s. Thus the statistic inv is. 29.

(36) unchanged as well as nin. Then X. pnin(T ) q inv(T ) =. T ∈R0 Bn,k. X. pnin(T ) q inv(T ) = Sp,q [n, k].. T ∈L0 Bn,k. proof of Theorem 5(3). It is easy to see that the sum satisfy the same boundary conditions as Sp,q [n, k]. Now we consider T ∈ BBn,k and the permutation tableau obtained from T by contracting the label 2 in the shape (deleting label 2 and merge label 1 and 3). Note that, T 0 is still a B-Bell tableau. If label 2 is a column, then T 0 is nothing but a tableau resulted from T by deleting the upper rightmost cell. Thus nin(T 0 ) = nin(T ), and inv(T 0 ) = inv(T ). On the other hand, that label 2 is a row in T . Then the behavior in row 2 is the following: from the left to the right, there are i consecutive 0’s and k − 1 − i consecutive 1’s for some 0 ≤ i ≤ k − 1. Thus nin(T ) = nin(T 0 ) + k − 1 − i, and inv(T ) = inv(T 0 ) + i. Then X. X. pnin(T ) q inv(T ) =. 0. pnin(T ) q inv(T. 0). T 0 ∈BBn−1,k−1. T ∈BBn,k. +. k−1 X. X. pnin(T. 0 )+k−1−i. q inv(T. 0 )+i. i=0 T 0 ∈BBn−1,k. =. X. 0. pnin(T ) q inv(T. 0). T 0 ∈BBn−1,k−1. X. + [k]p,q. 0. 0. pnin(T ) q inv(T ) .. T 0 ∈BBn−1,k. 3.3. p, q-statistics on L-Bell Tableaux. So far we have Corteel’s bijection f between set partitions and L-Bell tableaux, and a natural bijection g between set partitions and L0 -Bell tableaux (see Chapter 3.1 and 3.2). We use these two bijections to find our p, q-statistics on L-Bell Tableaux. Let us review the p, q-statistics on L-Bell Tableaux. Let LBn,k be the set of L-Bell tableaux of length n with n − k nonzero rows. For T ∈ LBn,k , let a star row 30.

(37) be a row that has a leftmost 1 but no topmost 1’s. Set nin(T ) as the number of pairs x and y, where x is a leftmost 1, and y is an unrestricted 0 that is not in a star row and above x. Let inv(T ) be the cardinality of A ∪ B, where A is the set of cells that is not in a star row and below a topmost 1, and B is the set of pairs u and v in the same column, where u is a cell in A and v is a leftmost 1 below u. For example, the set partition π = 1, 5, 8, 10/2, 3, 6/4, 9/7 and its corresponding L0 -Bell and L-Bell tableaux are shown in Figure 12. The left L0 -Bell tableau T 0 has nin(T 0 ) = 4 and inv(T 0 ) = 10. For the right L-Bell tableau T , its star rows are 3, 5, and 8. Then the pairs x and y are ((2, 6), (1, 6)), ((3, 6), (1, 6)), ((4, 9), (1, 9)), and ((4, 9), (2, 9)). The cells in A of T are labelled by •’s, and the pairs in B of T are ((2, 10), (5, 10)), ((2, 10), (8, 10)), ((4, 10), (5, 10)), ((4, 10), (8, 10)), and ((7, 10), (8, 10)). Thus, nin(T ) = 4 and inv(T ) = 5 + 5 = 10. 1. 0. 1. 0. 1. 0. 1. 1. 0. 0. 0. 1. 1. 1. 2. 0. 0. 1. 1. 1. 1. 0. 0. 0. 6 7. 5. 10 9. 8. 0. 0. 1. 0. 1. 2. 0. 1. 3. •. 3 4. 0. 0• 1. 1• 4. 1. 1. 1. 5. 0 • 0 • 76 1 1 8 10 9. Figure 12: The corresponding L0 -Bell and L-Bell tableaux.. proof of Theorem 5(4). We prove this result by using set partitions as the intermediate step. Let π = B1 /B2 / · · · /Bk be a set partition in standard form, and T and T 0 be its corresponding L-Bell and L0 -Bell tableaux, respectively. We consider the L0 -Bell tableau T 0 together with π first. From the bijection g (mapping each column to the row where the topmost 1 belongs), we know that an opener in π represents a row in T 0 and an element which is not an opener represents a column in T 0 . That is, in each block Bt , there are |Bt | − 1 column labels. For a 31.

(38) topmost 1 in cell (i, j) and a cell (i0 , j) above it, we have i0 is an opener in π and i0 is to the left of the block containing j since i0 < i. Thus, for a column label j in Bt , the number of cells above the topmost 1 in cell (i, j) is losj (π) = t − 1. Hence 0. nin(T ) =. X. losj (π) =. k X. (t − 1)(|Bt | − 1).. t=1. j is not an opener in π. For a topmost 1 in cell (i, j) and a cell (i0 , j) below it, we have i0 is an opener in π and i0 is to the right of the block containing j since i0 > i. Thus, for a column label j, the number of cells below the topmost 1 cell (i, j) is the number of openers to its right which is smaller than j; that is, rosj (π). Thus inv(T 0 ) =. X. rosj (π) =. X. rosj (π) = ros(π).. j∈[n]. j is not an opener in π. Note that rosj (π) = 0 whenever j is an opener in π. Then we consider the L-Bell tableau T together with π. From the bijection f (mapping each row to the column where the leftmost 1 belongs), we know that the closer of a non-single block in π represents a column in T . An element which is not the closer of a non-single block represents a row in T , so does an element in a single block. The star rows are the elements which are neither openers nor closers of a block. For a pair x and y in the statistic nin(T ), say x = (i, j) and y = (i0 , j), let Bt be the block in π containing both i and j, and let Bs be the block containing i0 . Since y = (i0 , j) is not in a star row, i0 is an opener of Bs . If s > t, then i0 is bigger than the opener i00 of Bt . Thus (i00 , j) is a leftmost 1 above y = (i0 , j), which contradicts to the assumption that y = (i0 , j) is an unrestricted 0. Hence s < t. In other words, the number of y’s which is paired with x is the number of the openers to the left of the non-closer i; that is losi (π) = t − 1. Furthermore, there are |Bt | − 1 non-closers in each block Bt . Hence, nin(T ) =. X. k X losi (π) = (t − 1)(|Bt | − 1). t=1. i is not a closer in π. For a cell (i, j) in A, let (i0 , j) be the topmost 1 above it. Since i is not a star row, there is a topmost 1 in row i. Note that, i0 < i < j. Thus we have the block 32.

(39) of π which contains i as its opener is to the right of the block containing i0 and j as its opener and closer. For a column label j, the number of cells in A belong it is the number of openers to its right which is smaller than j; that is, rosj (π) where j is a closer of a non-single block. Now we consider a pair u and v in B. Since u is in A, let u be (i, j) and (i0 , j) be the topmost 1 above it. Let v be a leftmost 1 below u, say (i00 , j). Note that, i0 < i < i00 < j. Thus we have the block containing i as its opener is to the right of the block containing i0 , i00 , and j, where i0 and j are the opener and closer, respectively , and i00 is a star row. Thus the number of u’s that is paired with v equals to the number of opener i’s which is to the right of the star row i00 and smaller than it; that is, rosi00 (π) where i00 is neither an opener nor a closer of a block. Together with the result we get from A, inv(T ) =. X. rosi (π) =. X. rosi (π) = ros(π).. i∈[n]. i is not an opener in π. Note that rosi (π) = 0 whenever i is an opener in π.. 4. Cardinality of intersections of various subclasses of tableaux. Now we have six subclasses of tableaux counted by the Bell numbers and we can investigate the cardinality of intersections of some of them in this chapter. The results we will prove in this thesis are listed in the table below.. 4.1. The Catalan numbers and the Narayana numbers. A Catalan path (or Dyck path) of length 2n is a lattice paths from (0, 0) to (2n, 0) with steps (1, 1) and (1, −1) that never goes below the x-axis. The Catalan number Cn is the number of the Catalan paths of length 2n. For example, C3 = 5. Figure 13 lists these 5 Catalan paths. It is known that {Cn }n≥0 := {1, 1, 2, 5, 14, 42, 132, 429, . . . }. 33.

(40) Intersection of tableaux. The cardinality / The distribution (w.r.t. some certain statistic). Ln ∩ P Tn , Ln ∩ L0 Bn , Ln ∩ R0 Bn. Catalan numbers Cn / Narayana numbers. LBn ∩ RBn , *L0 Bn ∩ R0 Bn. Bessel numbers Besn / Number of blocks. Ln ∩ LBn , Ln ∩ RBn , RBn ∩ L0 Bn , LBn ∩ R0 Bn. Motzkin numbers Mn / Number of horizontal steps. BBn ∩ LBn , BBn ∩ RBn , BBn ∩ Ln , BBn ∩ L0 Bn , BBn ∩ R0 Bn. 2n−1 / Binomial numbers. LBn ∩ L0 Bn , RBn ∩ R0 Bn. Number of matchings in Kn / Number of matchings of size k in Kn. Table 4: The intersection of two of these six subclasses. Figure 13: Five Catalan paths of length 6.. Let U and D represent the steps (1, 1) and (1, −1) respectively. We encode a Catalan path with a word over U, D. For example, those 5 paths in Figure 13 are represented by U U U DDD, U U DDU D, U U DU DD, U DU U DD, U DU DU D, respectively. If δ is a non-empty Catalan path, then δ can be decomposed as δ = αU βD, where α and β are Catalan paths. The Narayana number N (n, k) is the number of the Catalan paths of length 2n with k valleys. By a valley in a path we mean a pair of consecutive DU steps. Theorem 9. The cardinality of the set of permutation tableaux in Ln is the n-th Catalan numbers. 34.

(41) We give two proofs of this result. proof I. We give a bijection from the set of permutation tableaux in Ln to the set of rooted plane tree with n edges. For a permutation tableau where each column contains exactly one 1, say T , we construct a rooted plane tree by the labels of the shape of T as follows: From label 1 to n, a row i represents a leftmost child and a column j represents other children having the common parent with row k where the cell (k, j) is 1. See Figure 14 for an example. 1. 1. 0. 0. 0. 0. 0. 0. 0. 1. 0. 0. 1. 1. 4 5. 9. 8. 7. 6. 1 3. 1. 1. 9 2 8. 2. 4. 3 5. 6. 7. Figure 14: A bijection between permutation tableaux in Ln and rooted plane tree. For a rooted plane tree with n edges, we traverse it in pre-order and label its edges. Then for a parent, the label of the leftmost child represents a row and the labels of other children represent a column which has a 1 in the row. We need to prove that the resulting tableau T is a permutation tableau where each column contains exactly one 1. From above we know that each column contains exactly one 1 in the row which is the leftmost child of their common parent. Moreover, suppose that there exist i < j < k < l such that cells (i, k) and (j, l) are 1’s and cell (j, k) is a 0. In the corresponding rooted plane tree, i and k have common parent and j and l have common parent, but the parents of i and j are distinct. Thus we have that i < k < j < l if i is not an ancestor of j; otherwise, i < j < l < k. Both of them lead to a contradiction. Hence, the resulting tableau is a permutation tableau in Ln . proof II. We give another bijection between the set of permutation tableaux in Ln and the set of lattice paths of length 2n. We call a cell is legal in a permutation tableau in Ln if there is no restricted 0 to the right side of it. For a permutation 35.

(42) tableau that each column contains exactly one 1, say T , we recurrently decompose T to construct a lattice path as U D with the following rules: 1. We map label 1 to a consecutive U D. 2. If the next label in tableau T is a row, then we insert a consecutive U D after the rightmost U ; if it is a column, and its 1 is located in the ith legal cell, then we insert a consecutive U D behind the ith D counting from the right.. 0. 0. 1. 1. 1. 2. 0. 0. 3. 5. 4. U U U DDU DU DD 123 4 5. Figure 15: A bijection between permutation tableaux in Ln and lattice path.. The inverse mapping from lattice path of length 2n to the set of permutation tableaux T in Ln is simple. Besides, in a lattice path we call the (k + 1)-th level of a U if U between the (x, k) and (x + 1, k + 1), for some x, k ∈ N ∪ {0}. For example, there is a Catalan path U U DU DD, the level of the third U is 2. Let the shape of T be defined by the U ’s in Catalan path. The k-th U represents a column label k in T if there is a D in front of it; otherwise, it is a row in T . If U is a column, and has level l in the Catalan path, then there is a 1 in the l-th legal cell in column U . See Figure 15 for an example. By the property of the lattice path, the number of permutation tableaux in Ln is the Catalan numbers. Furthermore, we map a column to a pair ‘DU ’ which means a valley in a Catalan path, therefore the number of columns satisfying the Narayana numbers N (n, k) with k columns on L n ∩ P Tn .. As we known in section 3.2 in chapter 3, there is an obvious bijection between Ln,k and L0 Bn,k by simply replacing illegal 0’s (0’s which have a 1 to the left of it and 36.

(43) a 1 above it simultaneously) with the superfluous 1’s, we have Ln ∩L0 Bn = Ln ∩P Tn . Since the difference between L0 Bn and R0 Bn is the position of the superfluous 1’s, we have Ln ∩ L0 Bn = Ln ∩ R0 Bn . Proposition 2. Ln ∩ L0 Bn = Ln ∩ P Tn Proposition 3. Ln ∩ R0 Bn = Ln ∩ L0 Bn. 4.2. The Bessel numbers. We say two blocks Bi and Bj overlap in a set partition if min(Bi ) < min(Bj ) < max(Bi ) < max(Bj ). A set partition is called non-overlapping if no pair of blocks overlaps. For example, σ = 127/36/45 is non-overlapping . Let Besn be the number of non-overlapping set partitions of [n] and it is known {Besn }n≥0 := {1, 1, 2, 5, 14, 43, 143, 509, . . . }, called the Bessel numbers [8]. Theorem 10. We have |LBn ∩ RBn | = Besn . proof of Theorem 10. We know from section 3. 3.1 that f is a bijection between LBn and set partitions of [n], and we prove this theorem by showing that T ∈ LBn \ RBn if and only if f (T ) is an overlapping set partition of [n]. Note that for a block γ of size greater than or equal to 2 in f (T ), max(γ) is a column of T and min(γ) is the row such that cell (min(γ), max(γ)) is a topmost 1 and cell (k, max(γ)) is leftmost 1 whenever k ∈ γ and k 6= min(γ). Since T ∈ LBn \ RBn , there exists a topmost 1 which is not a rightmost 1 in cell (i, j). Let cell (i, j 0 ) be the rightmost 1 in row i. According to T ∈ LBn , there is a topmost 1 above (i, j 0 ), say cell (i0 , j 0 ). Thus the block containing i0 and j 0 overlaps the block containing i and j and f (T ) is an overlapping set partition of [n]. 37.