QUADRANT MARKED MESH PATTERNS IN 132-AVOIDING PERMUTATIONS III

Transcription

1 #A39 INTEGERS 5 (05) QUADRANT MARKED MESH PATTERNS IN 3-AVOIDING PERMUTATIONS III Sergey Kitaev Department of Computer and Information Sciences, University of Strathclyde, Glasgow, United Kingdom sergey.kitaev@cis.strath.ac.uk Je rey Remmel Department of Mathematics, University of California, San Diego, California jremmel@ucsd.edu Mark Tiefenbruck Center for Communications Research, La Jolla, San Diego, California mark@tiefenbruck.org Received: 3/4/3, Revised: 4/5/5, Accepted: 9/9/5, Published: 0/3/5 Abstract Given a permutation =... n in the symmetric group S n, we say that i matches the marked mesh pattern MMP(a, b, c, d) in if there are at least a points to the right of i in which are greater than i, at least b points to the left of i in which are greater than i, at least c points to the left of i in which are smaller than i, and at least d points to the right of i in which are smaller than i. This paper is continuation of the systematic study of the distributions of quadrant marked mesh patterns in 3-avoiding permutations started by the present authors, where we studied the distribution of the number of matches of MMP(a, b, c, d) in 3-avoiding permutations where at most two elements of a, b, c, d are greater than zero and the remaining elements are zero. In this paper, we study the distribution of the number of matches of MMP(a, b, c, d) in 3-avoiding permutations where at least three of a, b, c, d are greater than zero. We provide explicit recurrence relations to enumerate our objects which can be used to give closed forms for the generating functions associated with such distributions. In many cases, we provide combinatorial explanations of the coe cients that appear in our generating functions.. Introduction The notion of mesh patterns was introduced by Brändén and Claesson [] to provide explicit expansions for certain permutation statistics as, possibly infinite, linear

2 INTEGERS: 5 (05) combinations of (classical) permutation patterns. This notion was further studied in [, 3, 5, 6, 9, ]. Kitaev and Remmel [6] initiated the systematic study of the distributions of quadrant marked mesh patterns on permutations. The study was extended to 3-avoiding permutations by Kitaev, Remmel and Tiefenbruck in [9, 0], and the present paper continues this line of research. Kitaev and Remmel also studied the distributions of quadrant marked mesh patterns in up-down and down-up permutations [7, 8]. Let =... n be a permutation written in one-line notation. Then we will consider the graph of, G( ), to be the set of points (i, i) for i =,..., n. For example, the graph of the permutation = is pictured in Figure. Then if we draw a coordinate system centered at a point (i, i), we will be interested in the points that lie in the four quadrants I, II, III, and IV of that coordinate system as pictured in Figure. Let N = {,,...} denote the positive integers. For any a, b, c, d {0} [ N and any =... n S n, the set of all permutations of length n, we say that i matches the quadrant marked mesh pattern MMP(a, b, c, d) in if, in G( ) relative to the coordinate system which has the point (i, i) as its origin, there are at least a points in quadrant I, at least b points in quadrant II, at least c points in quadrant III, and at least d points in quadrant IV. For example, if = , the point 4 = 5 matches the marked mesh pattern MMP(,,, ) since, in G( ) relative to the coordinate system with the origin at (4, 5), there are 3 points in quadrant I, point in quadrant II, points in quadrant III, and points in quadrant IV. Note that if a coordinate in MMP(a, b, c, d) is 0, then there is no condition imposed on the points in the corresponding quadrant. In addition, we considered patterns MMP(a, b, c, d) where a, b, c, d {;}[{0}[N. Here when a coordinate of MMP(a, b, c, d) is the empty set, then for i to match MMP(a, b, c, d) in =... n S n, it must be the case that there are no points in G( ) relative to the coordinate system with the origin at (i, i) in the corresponding quadrant. For example, if = , the point 3 = matches the marked mesh pattern MMP(4,, ;, ;) since in G( ) relative to the coordinate system with the origin at (3, ), there are 6 points in quadrant I, points in quadrant II, no points in quadrants III and IV. We let mmp (a,b,c,d) ( ) denote the number of i such that i matches MMP(a, b, c, d) in. Note how the (two-dimensional) notation of Úlfarsson [] for marked mesh patterns corresponds to our (one-line) notation for quadrant marked mesh patterns. For example, Given a sequence w = w... w n of distinct integers, let red(w) be the permutation found by replacing the i-th smallest integer that appears in by i. For example, if = 754, then red( ) = 43. Given a permutation =... j in the symmetric group S j, we say that the pattern occurs in =... n S n provided there exist apple i < < i j apple n such that red( i... i j ) =. We say that a permutation

3 INTEGERS: 5 (05) II III I IV Figure : The graph of = k MMP(k,0,0,0) = MMP(0,0,k,0) = k a MMP(0,a,b,c) = MMP(0,0, 0,k) = b c k Figure : Úlfarsson notation for quadrant marked mesh patterns. avoids the pattern if does not occur in. Let S n ( ) denote the set of permutations in S n which avoid. In the theory of permutation patterns, is called a classical pattern. See [4] for a comprehensive introduction to patterns in permutations. It has been a rather popular direction of research in the literature on permutation patterns to study permutations avoiding a 3-letter pattern subject to extra restrictions (see [4, Subsection 6..5]). In [9], we started the study of the generating functions Q (a,b,c,d) 3 (t, x) = + X Q (a,b,c,d) n,3 (x)t n n where for any a, b, c, d {;} [ N, Q (a,b,c,d) n,3 (x) = X S n(3) x mmp(a,b,c,d) ( ). For any a, b, c, d, we will write Q (a,b,c,d) n,3 (x) x k for the coe cient of x k in Q (a,b,c,d) n,3 (x). For any fixed (a, b, c, d), we know that Q (a,b,c,d) n,3 () is the number of 3-avoiding permutations in S n which is the nth Catalan number C n = n+ coe n n. Thus the cients in the polynomial Q (a,b,c,d) n,3 (x) represent a refinement of the nth Catalan number. It is then a natural question to ask whether (i) we can give explicit formulas

4 INTEGERS: 5 (05) 4 for the coe cients that appear in Q (a,b,c,d) n,3 (x) or (ii) whether such coe cients count other interesting classes of combinatorial objects. Of course, there is an obvious answer to question (ii). That is, if one has a bijection from S n (3) to other classes of combinatorial objects which are counted by the Catalan numbers such as Dyck paths or binary trees, then one can use that bijection to give an interpretation of the pattern MMP(a, b, c, d) in the other setting. We shall see that in many cases, there are interesting connections with the coe cients that arise in our polynomials Q (a,b,c,d) n,3 (x) and other sets of combinatorial objects that do not just arise by such bijections. In particular, it is natural to try to understand Q (a,b,c,d) n,3 (0) which equals the number of S n (3) that have no occurrences of the pattern MMP(a, b, c, d) as well as the coe cient of the highest power of x that occurs in Q (a,b,c,d) n,3 (x) since that coe cient equals the number of S n (3) that have the maximum possible number of occurrences of the pattern MMP(a, b, c, d). We shall see that in many cases, Q (a,b,c,d) n,3 (x) x and Q (a,b,c,d) n,3 (x) x, the number of S n (3) with exactly one occurrence and two occurrences, respectively, of the pattern MMP(a, b, c, d) also have interesting combinatorics associated with them. There are many more interesting questions of this type that can be pursued, but due to space considerations, we shall mostly restrict ourselves to trying to understand the four coe cients in Q (a,b,c,d) n,3 (x) described above. We should note, however, that there is a uniform way to compute generating functions of the form F (a,b,c,d) k (t) = X Q (a,b,c,d) n,3 (x) x kt n. n 0 That is, F (a,b,c,d) k (t) is just the result of setting x = 0 in the generating k k Q(a,b,c,d) 3 (t, x). Due to space considerations, we will not pursue the study of the functions F (a,b,c,d) k (t) for k in this paper. There is one obvious symmetry for such generating functions which is induced by the fact that if S n (3), then S n (3). That is, the following lemma was proved in [9]. Lemma. ([9]) For any a, b, c, d N [ {0} [ {;}, Q (a,b,c,d) n,3 (x) = Q (a,d,c,b) n,3 (x). In [9], we studied the generating functions Q (k,0,0,0) 3 (t, x), Q (0,k,0,0) 3 (t, x), and Q (0,0,k,0) 3 (t, x), where k can be either the empty set or a positive integer, as well as the generating functions Q (k,0,;,0) 3 (t, x) and Q (;,0,k,0) 3 (t, x). In [0], we studied Q (k,0,`,0) n,3 (t, x), Q (k,0,0,`) n,3 (t, x), Q (0,k,`,0) n,3 (t, x), and Q (0,k,0,`) n,3 (t, x), where k, `. We also showed that sequences of the form (Q (a,b,c,d) n,3 (x) x r) n s count a variety of combinatorial objects that appear in the On-line Encyclopedia of Integer Sequences

5 INTEGERS: 5 (05) 5 (OEIS) []. Thus, our results gave new combinatorial interpretations of certain classical sequences such as the Fine numbers and the Fibonacci numbers as well as provided certain sequences that appear in the OEIS with a combinatorial interpretation where none had existed before. Another particular result of our studies in [9] is enumeration of permutations avoiding simultaneously the patterns 3 and 34, while in [0], we made a link to the Pell numbers. The main goal of this paper is to continue the study of Q (a,b,c,d) 3 (t, x) and combinatorial interpretations of sequences of the form (Q (a,b,c,d) n,3 (x) x r) n s in the case where a, b, c, d N and at least three of these parameters are non-zero. Next we list the key results from [9] and [0] which we need in this paper. Theorem. ([9, Theorem 3.]) and, for k, Hence and, for k, Q (0,0,0,0) 3 (t, x) = C(xt) = p 4xt xt Q (k,0,0,0) 3 (t, x) = Q (k,0,0,0) 3 (t, 0) = (k,0,0,0) tq 3 (t, x). Q (,0,0,0) 3 (t, 0) = t Theorem. ([9, Theorem 4.]) For k, Q (0,0,k,0) 3 (t, x) = + (tx t)(p k j=0 C jt j ) and = (k,0,0,0) tq 3 (t, 0). q ( + (tx t)( P k j=0 C jt j )) 4tx tx + (tx t)( P q k j=0 C jt j ) + ( + (tx t)( P k j=0 C jt j )) 4tx Q (0,0,k,0) 3 (t, 0) = t(c 0 + C t + + C k t k ). Theorem 3. ([0, Theorem 5]) For all k, `, Q (k,0,`,0) 3 (t, x) = (k,0,`,0) tq 3 (t, x). ()

6 INTEGERS: 5 (05) 6 Theorem 4. ([0, Theorem ]) For all k, `, Q (k,0,0,`) 3 (t, x) = X ` C`t` + j=0 C j t j (k,0,0,0) tq (k,0,0,` j) 3 (t, x) + t(q 3 (t, x) (k,0,0,0) tq 3 (t, x) Theorem 5. ([0, Theorem 4]) For all k, `, `Xj s=0 C s t s )). () Q (0,k,`,0) 3 (t, x) = kx C k t k + C j t j ( j=0 tq (0,0,`,0) (0,k i,`,0) 3 (t, x) + t(q 3 (t, x) tq (0,0,`,0) 3 (t, x) kxi s=0 C s t s )). (3) Theorem 6. ([0, Theorem 8]) For all k, `, where Q (0,k,0,`) 3 (t, x) = k,`(t, x), (4) t k+` X k+` X k,`(t, x) = C j t j C j t j+ + j=0 j=0 j=0 0 kx C j t j (0,k Q 3 j,0,`) (t, x) t Q (0,k,0,0) 3 (t, x)! kx C u t u u=0 0 X` C j t j (0,k,0,` j) Q 3 (t, x) j= k+` Xj s=0 Q (0,0,0,`) 3 (t, x) k+` Xj w=0 C s t s! A +! X` C v t v + v=0 C w t w! A. As it was pointed out in [9], avoidance of a marked mesh pattern without quadrants containing the empty set can always be expressed in terms of multi-avoidance of (possibly many) classical patterns. Thus, among our results we will re-derive several known facts in permutation patterns theory. However, our main goals are more ambitious aimed at finding distributions in question.

7 INTEGERS: 5 (05) 7. Q (k,0,m,`) n,3 (x) = Q (k,`,m,0) n,3 (x) where k, `, m By Lemma, we know that Q (k,0,m,`) n,3 (x) = Q (k,`,m,0) n,3 (x). Thus, we will only consider Q (k,`,m,0) n,3 (x) in this section. Throughout this paper, we shall classify the 3-avoiding permutations =... n by the position of n in. That is, let S n (i) (3) denote the set of S n (3) such that i = n. Clearly each S n (i) (3) has the structure pictured in Figure 3. That is, in the graph of, the elements to the left of n, A i ( ), have the structure of a 3-avoiding permutation, the elements to the right of n, B i ( ), have the structure of a 3-avoiding permutation, and all the elements in A i ( ) lie above all the elements in B i ( ). It is well-known that the number of 3-avoiding permutations in S n is the Catalan number C n = n n+ n and the generating function for the C n s is given by C(t) = X n 0 C n t n = p 4t t = + p 4t. n A (σ) i (σ) B i i n Figure 3: The structure of 3-avoiding permutations. Suppose that n `. It is clear that n cannot match the pattern MMP(k, `, m, 0) for k in any S n (3). For apple i apple n, it is easy to see that as we sum over all the permutations in S n (i) (3), our choices for the structure for A i ( ) will contribute a factor of Q (k,`,m,0) i,3 (x) to Q (k,`,m,0) n,3 (x). Similarly, our choices for the structure for B i ( ) will contribute a factor of Q (k,` i,m,0) n i,3 (x) to Q (k,`,m,0) n,3 (x) if i < ` since... i will automatically be in the second quadrant relative to the coordinate system with the origin at (s, s) for any s > i. However if i `, then our choices for the structure for B i ( ) will contribute a factor of Q (k,0,m,0) n i,3 (x) to Q (k,`,m,0) n,3 (x). It follows that for n `, X` Q (k,`,m,0) n,3 (x) = Q (k,`,m,0) (k,` i,m,0) i,3 (x)q i= n i,3 (x) + Q (k,`,m,0) i,3 (x)q (k,0,m,0) n i,3 (x). i=`

8 INTEGERS: 5 (05) 8 Note that for i < `, Q (k,`,m,0) i,3 (x) = C i. Thus, for n `, X` Q (k,`,m,0) (k,` i,m,0) n,3 (x) = C i Q n i,3 (x) + i= Q (k,`,m,0) i,3 (x)q (k,0,m,0) n i,3 i=` (x). (5) Multiplying both sides of (5) by t n and summing for n k, `, `, we see that for X` X` X 3 (t, x) = C j t j + C i t i Q (k,`,m,0) j=0 t X n ` = ` i= i= X X` C j t j + j=0 tq (k,0,m,0) s=0 u ` i Q (k,` i,m,0) u,3 (x)t u + Q (k,`,m,0) i,3 (x)t i Q (k,0,m,0) n i,3 (x)tn i i= C i t i (k,` i,m,0) (Q 3 (t, x) (k,`,m,0) 3 (t, x)(q 3 (t, x) s=0 (k,`,m,0) `Xi j=0 X` C s t s ) C j t j )+ = C` t` + tq (k,0,m,0) 3 (t, x)q 3 (t, x)+ X` `X s C s t s (k,` s,m,0) ( + tq 3 (t, x) tq (k,0,m,0) 3 (t, x) t Thus, we have the following theorem. Theorem 7. For all k, `, m, Q (k,`,m,0) 3 (t, x) = C` t` + tq (k,0,m,0) (k,`,m,0) 3 (t, x)q 3 (t, x)+ s=0 j=0 C j t j ). X` `X s C s t s (k,` s,m,0) ( + tq 3 (t, x) tq (k,0,m,0) 3 (t, x) t C j t j ). (6) Note that since we can compute Q (k,0,m,0) 3 (t, x) by Theorem 3 and Q (0,`,m,0) 3 (t, x) by Theorem 5, we can use (6) to compute Q (k,`,m,0) 3 (t, x) for any k, `, m... Explicit formulas for Q (k,`,m,0) n,3 (x) x r It follows from Theorem 7 that Q (k,,m,0) 3 (t, x) = + tq (k,0,m,0) (k,,m,0) 3 (t, x)q 3 (t, x) and (7) Q (k,,m,0) 3 (t, x) = + tq (k,0,m,0) (k,,m,0) 3 (t, x)(q 3 (t, x) ) + tq (k,,m,0) 3 (t, x). j=0

9 INTEGERS: 5 (05) 9 Note that it follows from Theorems 3 and 5 that Q (,,,0) 3 (t, 0) = + tq (,0,,0) 3 (t, 0)Q (0,,,0) 3 (t, 0) = + t t t t t = 3t + t + t 3 ( t). Thus, the generating function of the sequence {Q (,,,0) t n,3 (0)} n is t which is the generating function of the sequence A04563 in the OEIS. The n-th term a n of this sequence has many combinatorial interpretations including the number of s in all partitions of n + and the number of 3-avoiding permutations of S n+ which contain exactly one occurrence of the pattern 3. We note that for a permutation to avoid the pattern MMP(,,, 0), it must simultaneously avoid the patterns 34, 43, 34, and 43. Thus, the number of permutations S n (3) which avoid MMP(,,, 0) is the number of permutations in S n that simultaneously avoid the patterns 3, 34, and 43. Problem. Find simple bijections between the set of permutations S n (3) which avoid MMP(,,, 0) and the other combinatorial interpretations of the sequence A04563 in the OEIS. Note that it follows from Theorem 3 and our previous results that Q (,,,0) 3 (t, 0) = + tq (,0,,0) 3 (t, 0)Q (,,,0) 3 (t, 0) t = + t 3t + t 3t + t + t 3 ( t) = 4t + 4t + t 4 5t + 7t t 3. The sequence (Q (,,,0) n,3 (0)) n is the sequence A4586 in the OIES which has the generating function 3t+t +t 3 ( 3t+t )( t). That is, 4t+4t +t 4 5t+7t t = t( 3t+t +t 3 ) 3 ( 3t+t )( t). This sequence has no listed combinatorial interpretation so we have found a combinatorial interpretation of this sequence. Similarly, Q (3,,,0) 3 (t, 0) = + tq (3,0,,0) 3 (t, 0)Q (,,,0) 3 (t, 0) 3t + t = + t 4t + 3t 4t + 4t + t 4 5t + 7t t 3 = 5t + 7t t 3 + t 5 6t + t 6t 3.

10 INTEGERS: 5 (05) 0 Q (,,,0) 3 (t, 0) = + tq (,0,,0) 3 (t, 0)Q (0,,,0) 3 (t, 0) = + t t t t t t t t t = 3t + 3t3 + 3t 4 + t 5 ( t t ). Q (,,,0) 3 (t, 0) = + tq (,0,,0) 3 (t, 0)Q (,,,0) 3 (t, 0) t t = + t 3t + t 3 3t + 3t3 + 3t 4 + t 5 ( t t ) = 4t + t + 4t 3 + t 4 + t 5 + t 6 ( t t )( 3t + t 3. ) Using (7) and Theorem 3, we have computed the following. Q (,,,0) 3 (t, x) = + t + t + 5t 3 + ( + x)t x + x t x + 8x + x 3 t x + 97x + 6x 3 + x 4 t x + 408x + 84x x 4 + x 5 t x + 479x + 958x x x 5 + x 6 t 9 +. Q (,,,0) 3 (t, x) = + t + t + 5t 3 + 4t 4 + (38 + 4x)t x + 4x t x + 34x + 4x 3 t x + 00x + 4x 3 + 4x 4 t x + 964x + 304x x 4 + 4x 5 t 9 +. Q (,,3,0) 3 (t, x) = + t + t + 5t 3 + 4t 4 + 4t 5 + ( + 0x)t x + 0x t x + 88x + 0x 3 t x + 5x + 08x 3 + 0x 4 t 9 +. We can explain the highest and second highest coe That is, we have the following theorem. cients of x in these series. Theorem 8. (i) For all m and n 3+m, the highest power of x that occurs in Q (,,m,0) n,3 (x) is x n m which appears with a coe cient of C m. (ii) For n 5, Q (,,,0) n,3 (x) x n 4 = 6 + n. (iii) For m and n 4 + m, Q (,,m,0) n,3 (x) x n 3 m = C m+ + 8C m + 4C m (n m 4).

11 INTEGERS: 5 (05) Proof. It is easy to see that for the maximum number of MMP(,, m, 0)-matches in a S n (3), the permutation must be of the form (n ) (m + )... (n )n or n (m + )... (n )(n ) where S m (3). Thus, the highest power of x occurring in Q (,,m,0) n,3 (x) is x n m which occurs with a coe cient of C m. For parts (ii) and (iii), by (5) we have the recursion that Q (,,m,0) n,3 (x) = Q (0,,m,0) i,3 (x)q(,0,m,0) n i,3 i= (x). (8) We proved in [0, Theorem 5] and [0, Theorem 6] that for n m + the highest power of x which occurs in either Q (0,,m,0) n,3 (x) or Q (,0,m,0) n,3 (x) is x n m and Q (0,,m,0) n,3 (x) x n m = Q (,0,m,0) n,3 (x) x n m = C m. It is then easy to check that the highest power of x in Q (0,,m,0) is less than x n 3 m for i = 3,..., n. We also proved in [0, Theorems 9,0, and 5] that Q (,0,,0) n,3 (x) x n 3 =Q (0,,,0) n n,3 (x) x n 3 = + for n 4 and Q (,0,m,0) n,3 (x) x n m =Q (0,,m,0) n,3 (x) x n m i,3 (x)q(,0,m,0) n i,3 (x) =C m+ + C m + C m (n m) for n 3 + m and m. For m =, we are left with 4 cases to consider in the recursion (8). Case. i =. In this case, Q (0,,,0) i,3 (x)q(,0,,0) n i,3 (x) x n 4 = Q(,0,,0) n,3 (x) x n 4 and Q (,0,,0) n n,3 (x) x n 4 = + for n 5. Case. i =. In this case, Q (0,,,0) i,3 (x)q(,0,,0) n i,3 (x) x n 4 = Q(,0,,0) n,3 (x) x n 4 and Q (,0,,0) n,3 (x) xn 4 = for n 5. Case 3. i = n. In this case, Q (0,,,0) i,3 (x)q(,0,,0) n i,3 (x) x n 4 = Q(0,,,0) n,3 (x) x n 4 and Q (0,,,0) n,3 (x) xn 4 = for n 5. Case 4. i = n. In this case, Q (0,,,0) i,3 (x)q(,0,,0) n i,3 (x) x n 4 = Q(0,,,0) n,3 (x) x n 4 and Q (0,,,0) n n,3 (x) x n 4 = + for n 5. Thus, Q (,,,0) n,3 (x) x n 4 = 6 + n for n 5.

12 INTEGERS: 5 (05) Next we consider the case when m. Again we have 4 cases. Case. i =. We have Q (0,,m,0) i,3 (x)q(,0,m,0) n i,3 (x) x n 3 m = Q(,0,m,0) n,3 (x) x n 3 m, and Q (,0,m,0) n,3 (x) x n 3 ` = C m+ + 3C m + C m (n 4 m) for n 4 + m. Case. i =. We have Q (0,,m,0) i,3 (x)q(,0,m,0) n i,3 (x) x n 3 m = Q(,0,m,0) n,3 (x) x n 3 m, and Q (,0,m,0) n,3 (x) x n 3 m = C m for n 4 + m. Case 3. i = n and. Here, Q (0,,m,0) i,3 (x)q(,0,m,0) n i,3 (x) x n 3 m = Q(0,,m,0) n,3 (x) x n 3 m, Q (0,,m,0) n,3 (x) x n 3 m = C m for n 4 + m. Case 4. i = n. We have Q (0,,m,0) i,3 (x)q(,0,m,0) n i,3 (x) x n 3 m = Q(0,,m,0) n,3 (x) x n 3 m, and Q (0,,m,0) n,3 (x) x n 3 m = C m+ + 3C m + C m (n 4 m) for n 4 + m. Thus, for n 4 + m, Q (,,m,0) n,3 (x) x n 3 m = C m+ + 8C m + 4C m (n 4 m). Thus, when m =, we obtain that and, for m = 3, we obtain that which agrees with our computed series. Q (,,,0) n,3 (x) x n 5 = 6 + 8(n 6) for n 6 Q (,,3,0) n,3 (x) x n 6 = (n 7)for n 7 k One can also find the coe cient of the highest power of x in Q (k,,,0) n (x) for and Q (,`,,0) n (x) for `. Theorem 9. (i) For all k and n 3 + k, the highest power of x that occurs in Q (k,,,0) n,3 (x) is x n k which appears with a coe cient of k +. (ii) For all ` and n 3 + `, the highest power of x that occurs in Q (,`,,0) n,3 (x) is x n ` which appears with a coe cient of C`+.

13 INTEGERS: 5 (05) 3 Proof. For (i), it is easy to see that the permutations in S n (3) which have the most occurrences of MMP(k,,, 0) start with n j for some 0 apple j apple k followed by an increasing sequence which will have n k elements matching MMP(k,,, 0). For (ii), it is easy to see that the way to construct a permutation of S n (3) which has the maximum number of occurrences of MMP(, `,, 0) is to start with a rearrangement =... `+ of {n `, n `+,..., n} such that red( ) S`+ (3) and then let =... `... (n ` ) `+, which will have n ` elements that match MMP(, `,, 0). One can also find a formula for the second highest coe for any k. Theorem 0. For any k and n 4 + k, Q (k,,,0) k + n,3 (x) x n k 3 = 3 + k + + (k + ) cient of x in Q (k,,,0) n (x) n k. (9) Proof. We proceed by induction on k. Note that our formula reduces to part (ii) of Theorem 8 when k =. Thus assume that k > and our formula holds for k. By recursion (5), we see that Q (k,,,0) (k,,,0) n,3 (x) = Q (x). (0) i= i,3 (x)q (k,0,,0) n i,3 From [0, Theorem 6], the highest power of x that can occur in Q (k,0,,0) n,3 (x) is x n k, which occurs with a coe cient of for n k + 3. Similarly, from Theorem 9, we know that the highest power of x that can occur in Q (k,,,0) n,3 (x) is x n k which occurs with a coe cient of k+. One can then easily check that for n k+4, there are only four terms on the right-hand side of (0) that can contribute to the coe cient of Q (k,,,0) n,3 (x) x n k 3. (k,,,0) Case. i =. In this case, Q i,3 (x)q (k,0,,0) n i,3 (x) x n k = 3 Q(k,0,,0) n,3 (x) x n k 3 and by [0, Theorem 9], we know that Q (k,0,,0) n,3 (x) x n k 3 = k + n k for n k + 4. (k,,,0) Case. i =. In this case, Q i,3 (x)q (k,0,,0) n i,3 (x) x n k = 3 Q(k,0,,0) n,3 (x) x n k 3 and by [0, Theorem 6], we know that Q (k,0,,0) n,3 (x) xn k 3 = for n k + 4. (k,,,0) Case 3. i = n. Here, Q i,3 (x)q (k,0,,0) n i,3 (x),,,0) xn 4 = Q(k n,3 (x) x n k 3, and by Theorem 9, we know that Q (k,,,0) n,3 (x) x n k 3 = k for n k + 4.

14 INTEGERS: 5 (05) 4 (k,,,0) Case 4. i = n. We have Q i,3 (x)q (k,0,,0) n i,3 (x),,,0) xn 4 = Q(k n,3 (x) x n k 3, and by induction, and Q (0,,,0) n,3 (x) x n 4 = 3 k n k + (k ) + + k for n k + 4. Summing up these four cases, we find that for n k + 4, Q (k,,,0) k + n k n,3 (x) x n k 3 = 3 + k + + (k + ). For example, for k = and k = 3, we obtain that Q (,,,0) n 3 n,3 (x) x n 5 = for n 6 Q (3,,,0) n 4 n,3 (x) x n 6 = for n 7, which agrees with the expansions of the series for Q (,,,0) 3 (t, x) and Q (3,,,0) 3 (t, x) given below. One can ask whether there is a similar formula for the second highest coe cient of x in Q (,`,,0) n,3 (x) as a function of `. We conjecture that it is possible to find such a formula but it will be more complicated because it is no longer the case that a fixed number of terms in the recursion (5) contribute to the coe cient of the second highest power of x that occurs in Q (,`,,0) n,3 (x). That is, as ` grows, the number of terms in the recursion (5) that contribute to the coe cient of the second highest power of x that occurs in Q (,`,,0) n,3 (x) grows. We can, however, give an explicit formula in the case where ` =. Theorem. For n 6, Q (,,,0) n,3 (x) x n 5 = Proof. In this case, the recursion (5) becomes Q (,,,0) n n,3 (x) = Q (,,,0) n,3 (x) + X i= n 3 Q (0,,,0) i. (),3 (x)q(,0,,0) n i,3 (x). () One can then easily check that for n 6, there are only five terms on the righthand side of () that can contribute to the coe cient of Q (,,,0) n,3 (x) x n 5. Case. In part (ii) of Theorem 8, we proved that Q (,,,0) n 3 n,3 (x) x n 5 = 6 + for n 6.

15 INTEGERS: 5 (05) 5 Case. i =. In this case, Q (0,,,0) i,3 (x)q(,0,,0) n i,3 (x) x n 5 = Q(,0,,0) n,3 (x) x n 5 and by [0, Theorem 9], we know that Q (,0,,0) n,3 (x) x n 5 = + n 3 for n 6. Case 3. i = 3. In this case, Q (0,,,0) i,3 (x)q(,0,,0) n i,3 (x) x n 5 = Q(,0,,0) n 3,3 (x) x n 5 and by [0, Theorem 6], we know that Q (,0,,0) n,3 (x) xn 5 = for n 6. Case 4. i = n. In this case, Q (0,,,0) i,3 (x)q(,0,,0) n i,3 (x) x n 5 = Q(0,,,0) n,3 (x) x n 5 and by[0, Theorem 6 (i)], we know that Q (0,,,0) n,3 (x) xn 5 = for n 6. Case 5. i = n. In this case, Q (0,,,0) i,3 (x)q(,0,,0) n i,3 (x) x n 5 = Q(0,,,0) n,3 (x) x n 5 and by [0, Theorem 6 (ii)], Thus, for n 6, Q (0,,,0) n,3 (x) x n 5 = 6 + n 3 Q (,,,0) n,3 (x) x n 5 = for n 6. n 3. The formulas for the series Q (k,`,m,0) 3 (t, x) become increasingly complicated. For example, we have computed that Q (,,,0) 3 (t, x) = + Q (,,,0) 3 (t, x) = + Q (3,,,0) 3 (t, x) = + Q (,,,0) 3 (t, x) = + 4tx + t + x tx + p, ( + t( + x)) 4tx! 4tx t + p +t+x tx+ (+t( +x)) 4tx 0 B + tx +t+x tx+ p (+t( +x)) 4tx! 4tx t + ( +t+x tx+ p(+t( +x)) 4tx) tx p +t+x tx+ (+t( +x)) 4tx t tx p +t+x tx+ (+t( +x)) 4tx t +t(+t)( +x) p (+t(+t)( +x)) 4tx x, C A,,

16 INTEGERS: 5 (05) 6 Q (,,3,0) 3 (t, x) = + +t(+t+t )( +x) t p (+t(+t+t )( +x)) 4tx x, and Q (,,,0) 3 (t, x) = + t p 4t x ( t + tx 4x + t ( + x) t( + x)) ( + t + x tx + p + t ( + x) t( + x)). 3 We also have computed the following: Q (,,,0) 3 (t, x) = + t + t + 5t 3 + 4t 4 + (39 + 3x)t x + 3x t x + 3x + 3x 3 t x + 90x + 43x 3 + 3x 4 t x + 909x + 336x x 4 + 3x 5 t x x + 938x x x 5 + 3x 6 t 0 + ; Q (3,,,0) 3 (t, x) = + t + t + 5t 3 + 4t 4 + 4t 5 + (8 + 4x) t x + 4x t x + 47x + 4x 3 t x + 35x + 63x 3 + 4x 4 t x + 743x + 54x x 4 + 4x 5 t 0 + ; Q (,,,0) 3 (t, x) = + t + t + 5t 3 + 4t 4 + 4t 5 + (6 + 6x)t x + 6x t x + 59x + 6x 3 t x + 386x + 7x 3 + 6x 4 t x + 045x + 558x x 4 + 6x 5 t 0 + ; Q (,,3,0) 3 (t, x) = + t + t + 5t 3 + 4t 4 + 4t 5 + 3t 6 + (44 + 5x)t x + 5x t x + 5x + 5x 3 t x + 989x + 8x 3 + 5x 4 t 0 + ; Again one can easily explain the highest coe cient in Q (,,m,0) n,3 (x). That is, to have the maximum number of MMP(,, m, 0)-matches in a S n (3), the permutation must be of the form (n ) (m + )... (n 3)(n )n, (n ) (m + )... (n 3)(n )n, or n (m + )... (n 3)(n )(n ) where S m (3). Thus, the highest power of x occurring in Q (,,m,0) n,3 (x) is x n 3 m which occurs with a coe cient of 3C m.

17 INTEGERS: 5 (05) 7 We have computed the following: Q (,,,0) 3 (t, x) = + t + t + 5t 3 + 4t 4 + (37 + 5x)t x + 5x t x + 48x + 5x 3 t x + 77x + 68x 3 + 5x 4 t x + 48x + 508x x 4 + 5x 5 t 9 + ; Q (,,,0) 3 (t, x) = + t + t + 5t 3 + 4t 4 + 4t 5 + ( + 0x)t x + 0x t x + 9x + 0x 3 t x + 563x + x 3 + 0x 4 t 9 + ; Q (,,3,0) 3 (t, x) = + t + t + 5t 3 + 4t 4 + 4t 5 + 3t 6 + ( x)t x + 5x t x + 35x + 5x 3 t 9 +. Again, one can easily explain the highest coe cient in Q (,,m,0) n,3 (x). That is, to have the maximum number of MMP(,, m, 0)-matches in a S n (3), one must be of the form (n )(n ) (m + )... (n 3)n, (n )(n ) (m + )... (n 3)n, n(n ) (m + )... (n 3)(n ), n(n ) (m + )... (n 3)(n ), or (n )n (m + )... (n 3)(n ) where S m (3). Thus, the highest power of x occurring in Q (,,m,0) n,3 (x) is x n 3 m which occurs with a coe cient of 5C m. Finally, we have computed the following. Q (,,,0) 3 (t, x) = + t + t + 5t 3 + 4t 4 + 4t 5 + (3 + 9x)t 6 + ( x + 9x )t 7 + ( x + 96x + 9x 3 )t 8 + ( x + 609x + 3x 3 + 9x 4 )t 9 +. Q (,,,0) 3 (t, x) = + t + t + 5t 3 + 4t 4 + 4t 5 + 3t 6 + (4 + 8x)t 7 + ( x + 8x )t 8 + ( x + 83x + 8x 3 )t 9 + ( x + 9x + 9x 3 + 8x 4 )t 0 +. Q (,,3,0) 3 (t, x) = + t + t + 5t 3 + 4t 4 + 4t 5 + 3t t 7 + ( x)t 8 + ( x + 45x )t 9 + (48 + 6x + 47x + 45x 3 )t 0 + ( x + 349x + 56x x 4 )t +.

18 INTEGERS: 5 (05) 8 Again, one can easily explain the highest coe cient in Q (,,m,0) n,3 (x). That is, to have the maximum number of MMP(,, m, 0)-matches in a S n (3), one must be of the form n(n ) (m + )... (n 4)(n 3)(n ), (n )n (m + )... (n 4)(n 3)(n ), n(n ) (m + )... (n 4)(n 3)(n ), n(n 3) (m + )... (n 4)(n )(n ), (n )(n ) (m + )... (n 4)(n 3)n, (n )(n ) (m + )... (n 4)(n 3)n, (n )(n 3) (m + )... (n 4)(n )n, (n )(n 3) (m + )... (n 4)(n )n, or (n 3)(n ) (m + )... (n 4)(n )n where S m (3). Thus, the highest power of x occurring in Q (,,m,0) n,3 (x) is x n 4 m which occurs with a coe cient of 9C m. 3. Q (0,k,`,m) n,3 (x) where k, `, m Suppose that k, `, m and n k +m. It is clear that n cannot match the pattern MMP(0, k, `, m) for k, `, m in any S n (3). If =... n S n (3) and i = n, then we have three cases, depending on the value of i. Case. i < k. It is easy to see that as we sum over all the permutations in S n (i) (3), our choices for the structure for A i ( ) will contribute a factor of C i to Q (0,k,`,m) n,3 (x) since none of the elements j for j apple k can match MMP(0, k, `, m) in. Similarly, our choices for the structure for B i ( ) will contribute a factor of Q (0,k i,`,m) n i,3 (x) to Q (0,k,`,m) n,3 (x) since... i will automatically be in the second quadrant relative to the coordinate system with the origin at (s, s) for any s > i. Thus, the permutations in Case will contribute to Q (0,k,`,m) n,3 (x). kx (0,k i,`,m) C i Q n i,3 (x) i= Case. k apple i apple n m. It is easy to see that as we sum over all the permutations in S n (i) (3), our choices for the structure for A i ( ) will contribute a factor of Q (0,k,`,0) (x) to Q(0,k,`,m) n,3 (x) since the elements in B i ( ) will all be in i,3

19 INTEGERS: 5 (05) 9 the fourth quadrant relative to a coordinate system centered at (r, r) for r apple i in this case. Similarly, our choices for the structure for B i ( ) will contribute a factor of Q (0,0,`,m) n i,3 (x) to Q(0,k,`,m) n,3 (x) since... i will automatically be in the second quadrant relative to the coordinate system with the origin at (s, s) for any s > i. Thus, the permutations in Case will contribute to Q (0,k,`,m) n,3 (x). m i=k Q (0,k,`,0) i (x)q(0,0,`,m),3 n i,3 (x) Case 3. i n m +. It is easy to see that as we sum over all the permutations in S n (i) (3), our choices for the structure for A i ( ) will contribute a factor of (0,k,`,m (n i)) Q i,3 (x) to Q (0,k,`,m) n,3 (x) since the elements in B i ( ) will all be in the fourth quadrant relative to a coordinate system centered at (r, r) for r apple i in this case. Similarly, our choices for the structure for B i ( ) will contribute a factor of C n i to Q (0,k,`,m) n,3 (x) since the elements in B i ( ) do not have enough elements to the right to match MMP(0, k, `, m) in. Thus, the permutations in Case 3 will contribute (0,k,`,m (n i)) Q i,3 (x)c n i i=n m+ to Q (0,k,`,m) n,3 (x). Hence, for n k + m, kx m (0,k i,`,m) n,3 (x) = C i Q n i,3 (x) + Q (0,k,`,m) i= i=n m+ i=k Q (0,k,`,0) i (x)q(0,0,`,m),3 n i,3 (x)+ (0,k,`,m (n i)) Q i,3 (x)c n i. (3) Multiplying (3) by t n and summing, it is easy to compute that k+m X Q (0,k,`,m) 3 (t, x) = C p t p + p=0! kx k i+m X C i t i (0,k i,`,m) tq 3 (t, x) t C r t r + t i=0 mx j=0 Q (0,k,`,0) 3 (t, x)! kx C a t a a=0 r=0 Q (0,0,`,m) 3 (t, x) k+mx j C j t j (0,k,`,m j) tq 3 (t, x) t s=0 C s t s! mx b=0 C b t b! +. (4)

20 INTEGERS: 5 (05) 0 Note that the j = 0 term in the last sum is tq (0,k,`,m) 3 (t, x) t P k+m s=0 C s t s. Thus, taking the term tq (0,k,`,m) 3 (t, x) over to the other side and combining the sum t P k+m s=0 C s t s with the sum P k+m p=0 C p t p to obtain C k+m t k+m + ( t) P k+m p=0 C p t p and then dividing both sides by t will yield the following theorem. Theorem. k+m X Q (0,k,`,m) 3 (t, x) = C p t p + C k+m t k+m + t p=0 k t X C i t i (0,k i,`,m) Q 3 (t, x) t i=0! k t X Q (0,k,`,0) 3 (t, x) C a t a t a=0 m t X C j t j (0,k,`,m j) Q 3 (t, x) t j= k i+m X r=0 C r t r! + Q (0,0,`,m) 3 (t, x) k+mx j s=0 C s t s! mx b=0 C b t b! +. (5) Note that since we can compute Q (0,k,`,0) 3 (t, x) = Q (0,0,`,k) 3 (t, x) by Theorem 5, we can compute Q (0,k,`,m) 3 (t, x) for all k, `, m. 3.. Explicit formulas for Q (0,k,`,m) n,3 (x) x r It follows from Theorem that Q (0,,`,) 3 (t, x) = + t t + t t Q(0,,`,0) 3 (t, x)(q (0,0,`,) 3 (t, x) ) = t + t t Q(0,,`,0) 3 (t, x)(q (0,0,`,) 3 (t, x) ), Q (0,,`,) 3 (t, x) = + t + t t + t Q(0,,`,0) 3 (t, x)(q (0,0,`,) 3 (t, x) t ( + t))+ t 3 (t, x) t ), and Q (0,,`,) 3 (t, x) = + t + t + 5t3 t + t (Q(0,,`,) 3 (t, x) ( + t + t ))+ t t (Q(0,,`,0) 3 (t, x) )(Q (0,0,`,) 3 (t, x) ( + t))+ t t (Q(0,,`,) 3 (t, x) ( + t)). t

21 INTEGERS: 5 (05) Again the formula for Q (0,k,`,m) 3 (t, x) quickly becomes quite complicated. For example, + tx Q (0,,,) 3 (t, x) = + tx Q (0,,,) 3 (t, x) = Q (0,,3,) 3 (t, x) = + 0 p t+tx (+t( +x)) 4tx +t+x tx+ p (+t( +x)) 4tx t p t t +tx+t x p (+t(+t)( +x)) 4tx +t+t +x tx t x+ (+t(+t)( +x)) 4tx t,, and B t@ + C A x +t(+t+t )( +x) (+t(+t+t )( +x)) 4tx +t(+t+t )( +x) p t We used these formulas to compute the following: p (+t(+t+t )( +x)) 4tx x. Q (0,,,) 3 (t, x) = + t + t + 5t 3 + (3 + x)t 4 + (33 + 8x + x )t 5 + (8 + 39x + x + x 3 )t 6 + ( x + 70x + 5x 3 + x 4 )t 7 + ( x + 337x + x 3 + 0x 4 + x 5 )t 8 + ( x + 363x + 79x x 4 + 6x 5 + x 6 )t 9 + ( x + 489x x x x x 6 + x 7 )t 0 + ; Q (0,,,) 3 (t, x) = + t + t + 5t 3 + 4t 4 + (0 + x)t 5 + (3 + 7x + x )t 6 + (34 + 9x + x + x 3 )t 7 + ( x + 40x + 5x 3 + x 4 )t 8 + ( x + 745x + 00x 3 + 9x 4 + x 5 )t 9 + ( x x + 6x 3 + 7x x 5 + x 6 )t 0 + ; Q (0,,3,) 3 (t, x) = + t + t + 5t 3 + 4t 4 + 4t 5 + (7 + 5x)t 6 + ( x + 5x )t 7 + (5 + 46x + 54x + 5x 3 )t 8 + ( x + 359x + 64x 3 + 5x 4 )t 9 + ( x + 94x + 49x x 4 + 5x 5 )t 0 +. Our next theorem will explain the coe cient of the highest and second highest powers of x that appear in Q (0,,`,) n,3 (x) in these series. Theorem 3. (i) For n + k + ` + m, the highest power of x that occurs in Q (0,k,`,m) n,3 (x) is x n k ` m which appears with a coe cient of C k C`C m.

22 INTEGERS: 5 (05) (ii) For n 5, Q (0,,,) n,3 (x) x n 4 = 5 + n. (iii) For all ` and n 4 + `, Q (0,,`,) n,3 (x) x n 3 ` = C`+ + 6C` + C`(n 4 `). Proof. It is easy to see that the maximum number of matches of MMP(0, k, `, m) that are possible in a 3-avoiding permutation is a permutation of the form where is a rearrangement of {n k +,..., n} such that red( ) S k (3), is a rearrangement of {m +,..., m + `} such that red( ) S`(3), is the increasing sequence m + ` +, m + ` +,..., n k, and S m (3). Thus, the highest power in Q (0,k,`,m) n is x n k ` m which has a coe cient of C k C`C m. For parts (ii) and (iii), we note that it follows from (3) that Q (0,,`,) n,3 (x) = Q (0,,`,) n n,3 (x) + X i= Q (0,,`,0) i (x)q(0,0,`,),3 n i,3 (x). We proved in [0, Theorems 6 and 5] that the highest power of x that appears in Q (0,,`,0) n,3 (x) = Q (0,0,`,) n,3 (x) is x n ` which appears with a coe cient of C` for n ` +. This implies that the highest power of x appearing in Q (0,,`,0) (x)q(0,0,`,) i,3 n i,3 (x) is less than x n ` 3 for i = 3,..., n. Hence we have four cases to consider when we are computing Q (0,,,) n,3 (x) x n 4. Case. i =. In this case, Q (0,,,0) i,3 (x)q(0,0,,) n i,3 (x) x n 4 = Q(0,0,,) n,3 (x) x n 4 and we proved in [0, Theorems 9 and 5] that Q (0,0,,) n n,3 (x) x n 4 = Q(0,,,0) n,3 (x) x n 4 = + for n 5. Case. i =. In this case, Q (0,,,0) i,3 (x)q(0,0,,) n i,3 (x) x n 4 = Q(0,0,,) n,3 (x) x n 4 and we proved in [0, Theorems 6 and 5] that Q (0,0,,) n,3 (x) x n 4 = Q(0,,,0) n,3 (x) xn 4 = for n 5. Case 3. i = n. In this case, Q (0,,,0) i,3 (x)q(0,0,,) n i,3 (x) x n 4 = Q(0,,,0) n,3 (x) x n 4 and we proved in [0, Theorems 6 and 5] that Q (0,,,0) n,3 (x) xn 4 = for n 5. Case 4. Q (0,,,) n,3 (x) xn 4. By part (i), we know that Q(0,,,) n,3 (x) x n 4 = for n 5. Thus, Q (0,,,) n,3 (x) x n 4 = 5 + n for n 5. Again there are four cases to consider when computing Q (0,,`,) n,3 (x) x n 3 ` for `.

23 INTEGERS: 5 (05) 3 Case. i =. In this case, Q (0,,`,0) i,3 n i,3 (x) x n 3 ` = Q(0,0,`,) n,3 (x) x n 3 ` and we proved in [0, Theorems 0 and 5] that Q (0,0,`,) n,3 (x) x n 3 ` =Q(0,,`,0) n,3 (x) x n 3 ` =C`+ + 3C` + C`(n 4 `) for n 4 + `. Case. i =. In this case, Q (0,,`,0) i,3 n i,3 (x) x n 3 ` = Q(0,0,`,) n,3 (x) x n 3 ` and we proved in [0, Theorems 6 and 5] that Q (0,0,`,) n,3 (x) x n 3 ` = Q(0,,`,0) n,3 (x) x n 3 ` = C` for n 4 + `. Case 3. i = n. In this case, Q (0,,`,0) (x)q(0,0,`,) i,3 n i,3 (x) x = Q(0,,`,0) n 3 ` n,3 (x) x n 3 ` and we proved in [0, Theorems 6 and 5] that Q (0,,`,0) n,3 (x) x n 3 ` = C` for n 4 + `. Case 4. Q (0,,`,) n,3 (x) xn 3 `. By part (i), we know that Thus, Q (0,,`,) n,3 (x) x n 3 ` = C` for n 4 + `. Q (0,,`,) n,3 (x) x n 3 ` = C`+ + 6C` + C`(n 4 `) for n 4 + `. For example, when ` =, we have that and, for ` = 3, we have that Q (0,,,) n,3 (x) x n 5 = 7 + 4(n 6) for n 6 Q (0,,3,) n,3 (x) x n 6 = (n 7) for n 7, which agrees with the series we computed. We can also find the formulas for Q (0,,,) n,3 (0) and Q (0,,,) n,3 (x) x. Theorem 4. (i) Q (0,,,) n,3 (0) = (n ) n + for n. (ii) Q (0,,,) n,3 (x) x = (n 9n + 4) n 3 3 n for n 4.

24 INTEGERS: 5 (05) 4 Proof. In this case, the recursion (3) becomes Q (0,,,) n,3 (x) = Q (0,,,) n n,3 (x) + X i= Q (0,,,0) i,3 (x)q(0,0,,) n i,3 (x). It was proved in [0, Theorem 5] that Q (0,,,0) 3 (t, x) = Q (,0,,0) 3 (t, x) so that Q (0,0,,) n,3 (x) = Q (0,,,0) n,3 (x) = Q (,0,,0) n,3 (x) for all n. Thus we have that Q (0,,,) n,3 (x) = Q (0,,,) n n,3 (x) + X i= Q (,0,,0) i,3 (x)q(,0,,0) n i,3 (x). (6) By [0, Theorems 7 and 8], we know that Q (,0,,0) n,3 (0) = n for n and that Q (,0,,0) n,3 (x) x = (n 3) n + for n 3. Our calculations show that our formulas hold for n apple 5. Then for n 6, we have by induction that n n,3 (0) =Q (0,,,) n,3 (0) + Q(,0,,0) n,3 (0) + X Q (,0,,0) i,3 (0)Q(,0,,0) n i,3 (0) Q (0,,,) i= =(n ) n n + i n i i= =(n ) n n + (n ) n 3 =(n ) n +. Similarly, if we separate out the i =,, 3, n, n terms from the sum in recursion (6), we find by induction that Q (0,,,) n,3 (x) x =Q (0,,,) n,3 (x) x + Q (,0,,0) n,3 (x) x + Q (,0,,0) n,3 (x) x + Q (,0,,0) n 3,3 (x) x+ Q (,0,,0) n,3 (x) x + Q (,0,,0) n 3,3 (x) x+ 3 Q (,0,,0) i,3 (x) xq (,0,,0) i=4 n 3 n i,3 (0) + X i=4 Q (,0,,0) i,3 (0)Q(,0,,0) n i,3 (x) x =((n ) 9(n ) + 4) n 4 3 (n ) + ((n 4) n 3 + )+ ((n 5) n 4 + ) + 4((n 6) n 5 + )+ 3 3 ((i 4) i 3 + ) n i + i ((n i 3) n i + ) i=4 i=4 =((n ) 9(n ) + 4) n 4 3 (n ) + (6n 30) n 4 +! n 4 (i 4) + (n i 3) + n i + i=4 i=4 i=4 i=4 i

25 INTEGERS: 5 (05) 5 n 6 = (n n 4 ) 9(n ) n (n ) ( n 4 4) =(n 9n + 4) n 3 3 n. The sequence (Q (0,,,) n,3 (0)) n starts out,, 5, 3, 33, 8, 93, 449,.... This is the sequence A00583 in OEIS and it counts the number of permutations of length n which avoids the patterns 3 and 43. We obtain the interpretation that this sequence also counts the number of permutations of length n which avoid the patterns 3 and 43 (which are precisely the permutations counted by Q (0,,,) n,3 (0), since an occurrence of the pattern MMP(0,,, ) in a 3-avoiding permutation implies an occurrence of the pattern 43, and vice versa). Problem. Find a bijection between permutations of length n avoiding the patterns 3 and 43, and permutations of length n avoiding the patterns 3 and 43. The sequence (Q (0,,,) n,3 (x) x ) n 4 starts out, 8, 39, 50, 50, is sequence A05558 which counts the number of directed column-convex polyominoes of area n + 5 having along the lower contour exactly reentrant corners. Problem 3. Find a bijective correspondence between the number of permutations in S n (3) which have exactly one occurrence of the pattern MMP(0,,, ) and the polyominoes described in A05558 in the OEIS. We have computed the following: Q (0,,,) 3 (t, x) = + t + t + 5t 3 + 4t 4 + (0 + x)t x + x t x + 5x + x 3 t x + 78x + 33x 3 + x 4 t x + 947x + 95x x 4 + x 5 t x + 49x + 90x x x 5 + x 6 t 0 + ; Q (0,,,) 3 (t, x) = + t + t + 5t 3 + 4t 4 + 4t 5 + 4(3 + x)t x + 4x t x + 48x + 4x 3 t x + 35x + 56x 3 + 4x 4 t x + 06x + 483x x 4 + 4x 5 t 0 + ;

26 INTEGERS: 5 (05) 6 Q (0,,3,) 3 (t, x) = + t + t + 5t 3 + 4t 4 + 4t 5 + 3t 6 + (49 + 0x)t x + 0x t x + 3x + 0x 3 t x + 900x + 43x 3 + 0x 4 t x + 5x + 96x x 4 + 0x 5 t + ; Q (0,,,) 3 (t, x) = + t + t + 5t 3 + 4t 4 + 4t 5 + 4t 6 (3 + x) x + 4x t x + 57x + 4x 3 t x + 453x + 73x 3 + 4x 4 t x + 650x + 77x x 4 + 4x 5 t x + 749x x 3 + 4x 4 + 7x 5 + 4x 6 t + ; Q (0,,,) 3 (t, x) = + t + t + 5t 3 + 4t 4 + 4t 5 + 3t 6 + (4 + 8x)t x + 4x t x + 0x + 8x 3 t x + 885x + 6x 3 + 8x 4 t x x + 77x 3 + 4x 4 + 8x 5 t x + 887x + 845x x x 5 + 8x 6 t + ; Q (0,,3,) 3 (t, x) = + t + t + 5t 3 + 4t 4 + 4t 5 + 3t t 7 + (40 + 0x)t x + 0x t x + 8x + 0x 3 t x + 68x + 3x 3 + 0x 4 t x + 444x + 930x x 4 + 0x 5 t +. We can explain the coe enough n. cient of the highest power of x in Q (0,k,`,m) n,3 (x) for large Theorem 5. For n ` + k + m +, the highest power of x that appears in Q (0,k,`,m) n,3 (x) is x n k ` m, which occurs with a coe cient of C k C`C m. Proof. It is easy to see the maximum number of matches of MMP (0, k, `, m) for n > k+`+m that are possible in a 3-avoiding permutation occur in a permutation of the form where is a permutation of n k +,..., n such that red( ) S k (3), S m (3), and is some permutation of k +, k +,..., n m, which has the maximum number of matches of MMP (0, 0, `, 0). Such a must be of the form (`+a)(`+a+)(`+a+)... (n m) where is a pemutation of k+,..., k+` such that red( ) S`(3). Thus the highest power of x that occurs in Q (0,k,`,m) n (x) for n > n k ` m is x n k ` m which occurs with a coe cient of C k C`C m. For example, for n > n ` 3, the highest power of x in Q (0,,`,) n,3 (x) is x n ` 3 which

27 INTEGERS: 5 (05) 7 has a coe cient of C` and for n > n ` 4, the highest power of x in Q (0,,`,) n,3 (x) is x n ` 4 which has a coe cient of 4C`. This agrees with our computations. For su ciently large n, we can also explain the coe cient of the second highest term in Q (0,,`,) n,3 (x). Theorem 6. (i) For n 6, Q (0,,,) n,3 (x) x n 5 = 3 + n 3 (ii) For all ` and n 5+`, Q (0,,`,) n,3 (x) x n 4 ` = C`+ +5C`+4C`(n 5 `). Proof. For (i) and (ii), we note that the recursion for Q (0,,`,) n,3 (x) is Q (0,,`,). n n,3 (x) = Q (0,,`,) n,3 (x) + Q(0,,`,) n,3 (x) + X i= Q (0,,`,0) i (x)q(0,0,`,),3 n i,3 (x). By [0, Theorems 6 and 5], the highest power of x that occurs in Q (0,,`,0) n,3 (x) is x n `, and by [0, Theorem 7], the highest power of x that occurs in Q (0,0,`,) n,3 (x) is n `. It follows that the highest power of x that occurs in Q (0,,`,0) i,3 n i,3 (x) is less than x n 4 ` for i = 3,..., n 3. Thus, we have to consider five cases when computing Q (0,,`,) n,3 (x) x n 4 `. Case. Q (0,,`,) n,3 (x) xn 4 `. By part Theorem 5, Q (0,,`,) n,3 (x) x n 4 ` = C` for n ` + 5. Case. Q (0,,`,) n,3 (x) xn 4 `. We have shown in Theorem 3 that Q (0,,`,) n,3 (x) x n 4 ` = C` for n ` + 5. Case 3. i = n. In this case, Q (0,,`,0) i,3 n i,3 (x) equals Q(0,,`,0) n 3,3 (x). We have shown in [0, Theorems 6 and 5] that Q (0,,`,0) n 3,3 (x) x n 4 = ` C` for n ` + 5 so that we get a contribution of C` in this case. Case 4. i =. In this case, Q (0,,`,0) i shown in [0, Theorem 7] that (x)q(0,0,`,),3 n i,3 Q (0,0,`,) n,3 (x) x n 4 ` = C` for n ` + 5. Case 5. i =. In this case, Q (0,,`,0) i n shown in [0, Theorem 7] that for n ` + 5, (x)q(0,0,`,),3 i,3 (x) equals Q(0,0,`,) (x). We have n,3 (x) equals Q(0,0,`,) (x). We have n,3 Q (0,0,`,) n,3 (x) x n 4 ` = ( 6 + n 3 if ` =, C`+ + 8C` + 4C`(n 5 `) if `.

28 INTEGERS: 5 (05) 8 Thus, for ` =, we get Q (0,,,) n 3 n,3 (x) x n 5 = 3 + and, for `, for n 6 Q (0,,`,) n,3 (x) x n 4 ` = C`+ + 5C` + 4C`(n 5 `) for n 5 + `. For example, when ` =, we get and, for ` = 3, we get Q (0,,,) n,3 (x) x n 6 = (n 7) for n 7 Q (0,,3,) n,3 (x) x n 7 = (n 8) for n 8, which agrees with the series that we computed. 4. Q (`,k,0,m) n,3 (x) where k, `, m Suppose that k, `, m and n k + m. It is clear that n cannot match MMP(`, k, 0, m) for k, `, m in any S n (3). If =... n S n (3) and i = n, then we have three cases, depending on the value of i. Case. i < k. It is easy to see that as we sum over all the permutations in S n (i) (3), our choices for the structure for A i ( ) will contribute a factor of C i to Q (`,k,0,m) n,3 (x) since the elements in A i ( ) do not have enough elements to the left to match MMP(`, k, 0, m) in. Similarly, our choices for the structure for B i ( ) will contribute a factor of Q (`,k i,0,m) n i,3 (x) to Q (`,k,0,m) n,3 (x) since... i will automatically be in the second quadrant relative to the coordinate system with the origin at (s, s) for any s > i. Thus, the permutations in Case will contribute to Q (`,k,0,m) n,3 (x). kx (`,k i,0,m) C i Q n i,3 (x) i= Case. k apple i apple n m. It is easy to see that as we sum over all the permutations in S n (i) (3), our choices for the structure for A i ( ) will contribute a factor of Q (`,k,0,0) i,3 (x) to Q (`,k,0,m) n,3 (x) since the elements in B i ( ) will all be in the fourth

29 INTEGERS: 5 (05) 9 quadrant and i = n is in the first quadrant relative to a coordinate system centered at (r, r) for r < i in this case. Similarly, our choices for the structure for B i ( ) will contribute a factor of Q (`,0,0,m) n i,3 (x) to Q(`,k,0,m) n,3 (x) since... i will automatically be in the second quadrant relative to the coordinate system with the origin at (s, s) for any s > i. Thus, the permutations in Case will contribute to Q (`,k,0,m) n,3 (x). X n m Q (`,k,0,0) i,3 (x)q (`,0,0,m) n i,3 (x) i=k Case 3. i n m +. It is easy to see that as we sum over all the permutations in S n (i) (3), our choices for the structure for A i ( ) will contribute a factor of (`,k,0,m (n i)) Q i,3 (x) to Q (`,k,0,m) n,3 (x) since i = n will be in the first quadrant and the elements in B i ( ) will all be in the fourth quadrant relative to a coordinate system centered at (r, r) for r < i in this case. Similarly, our choices for the structure for B i ( ) will contribute a factor of C n i to Q (`,k,0,m) n,3 (x) since j where j > i does not have enough elements to its right to match MMP(`, k, 0, m) in. Thus, the permutations in Case 3 will contribute i=n m+ (`,k,0,m (n i)) Q i,3 (x)c n i to Q (`,k,0,m) n,3 (x). Thus, we have the following. For n k + m, kx (`,k i,0,m) n,3 (x) = C i Q n i,3 (x) + Q (`,k,0,m) i= i=n m+ X n m Q (`,k,0,0) i,3 (x)q (`,0,0,m) n i,3 (x)+ i=k (`,k,0,m (n i)) Q i,3 (x)c n i. (7) Multiplying (7) by t n and summing over n will yield the following theorem. Theorem 7. For all `, k, m, k+m X Q (`,k,0,m) 3 (t, x) = C p t p + p=0 X k t i=0 C i t i (`,k i,0,m) Q 3 (t, x) (`,k,0,0) t Q 3 (t, x)! kx C a t a a=0 k i+m X r=0 C r t r! + Q (`,0,0,m) 3 (t, x) mx b=0 C b t b! +

30 INTEGERS: 5 (05) 30 X m t j=0 C j t j (`,k,0,m j) Q 3 (t, x) k+mx j s=0 C s t s!. (8) Note that we can compute Q (`,k,0,0) 3 (t, x) = Q (`,0,0,k) 3 (t, x) by Theorem 4 so that (8) allows us to compute Q (`,k,0,m) 3 (t, x) for any k, `, m Explicit Formulas for Q (`,k,0,m) n,3 (x) x r In general, the formulas for Q (`,k,0,m) 3 (t, x) are quite complicated. For example, in the simplest case, where Q (,,0,) 3 (t, x) = R(t, x) + S(t, x)p 4xt ( t)( t + p 4xt) 3 R(t, x) =4( 3t + 4t t 3 t 4 3xt + 6xt 3xt 3 + xt 4 ) and S(t, x) =4( 3t + 4t t 4 xt xt 3 + xt 4 ). It follows from Theorem 7 that (`,,0,0) 3 (t, x) = + t + tq 3 (t, x)(q (`,0,0,) 3 (t, x) )+ Q (`,,0,) (`,,0,) t(q 3 (t, x) ), (9) 3 (t, x) = + t + t (`,,0,0) + tq 3 (t, x)(q (`,0,0,) 3 (t, x) ( + t))+ Q (`,,0,) (`,,0,) t(q (`,,0,) 3 (t, x) ( + t) + t(q 3 (t, x) )), and Q (`,,0,) 3 (t, x) = + t + t + 5t 3 + t(q (`,,0,) 3 (t, x) ( + t + t ))+ (`,,0,0) t(q 3 (t, x) )(Q (`,0,0,) 3 (t, x) ( + t))+ (`,,0,) t(q (`,,0,) 3 (t, x) ( + t + t ) + t(q 3 (t, x) ( + t))). One can use these formulas to compute the following. Q (,,0,) 3 (t, x) = + t + t + 5t 3 + (5 + x)t 4 + (7 + 7x + 8x )t 5 + (6 + 44x + 4x + 0x 3 )t 6 + ( x + 9x + 7x x 4 )t 7 + ( x + 305x + 397x x x 5 )t 8 + ( x + 65x + 09x x x x 6 )t 9 + (8 + 39x + 3x + 59x x x x x 7 )t 0 +. It is easy to explain the highest coe cient of x in Q (,k,0,) n,3 (x).

31 INTEGERS: 5 (05) 3 Theorem 8. For n 3 + k, the highest power of x that occurs in Q (,k,0,) n,3 (x) is x n k which occurs with a coe cient of C k+ C n k. Proof. It is easy to see that the maximum number of MMP(, k, 0, ) matches occurs in S n (3) when is of the form... k k+ or... k k+ where =... k+ is a rearrangement of {n k, n k +,..., n} such that red( ) S k+ (3) and is a 3-avoiding permutation on,..., n k. Thus, the highest power of x in Q (,k,0,) n,3 (x) for n k + 3 is x n k which occurs with a coe cient of C k+ C n k. We can also explain the second highest coe cient in Q (,,0,) n,3 (x). Theorem 9. For n 5, Q (,,0,) n,3 (x) x n 4 = 8C n 3 + C n 4. Proof. In this case, the recursion for Q (,,0,) n,3 (x) is Q (,,0,) n,3 (x) = Q (0,,0,) n n,3 (x) + X i= Q (0,,0,0) i,3 (x)q(,0,0,) n i,3 (x). It was proved in [9, Theorem 5.] that for n, the highest power of x that occurs in Q (0,,0,0) n,3 (x) is x n which occurs with a coe cient of C n. It was proved in [0, Theorem ] that for n 3, the highest power of x that occurs in Q (,0,0,) n,3 (x) is x n which occurs with a coe cient of C n. It follows that Q (,,0,) n,3 (x) x n 4 =Q (0,,0,) n,3 (x) x n 4 + Q(,0,0,) n,3 (x) x n 4 + Q(0,,0,0) n,3 (x) x n 4+ Q (0,,0,0) i,3 (x) x i Q(,0,0,) n i,3 (x) x n i. i= The following three equations follow from [0, Theorem 9], [0, Theorem ] and [9, Theorem 5.], respectively. Q (0,,0,) n,3 (x) x n 4 =C n 3 + C n 4 for n 5, Q (,0,0,) n,3 (x) x n 4 =3C n 3 for n 5, and Q (0,,0,0) n,3 (x) x n 4 =C n 3 for n 5.