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ASYMPTOTIC ENUMERATION IN PATTERN AVOIDANCE AND IN THE THEORY OF SET PARTITIONS AND ASYMPTOTIC UNIFORMITY By MICAH SPENCER COLEMAN A DISSERTATION PRESENTED TO THE GRADUATE SCHOOL OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY UNIVERSITY OF FLORIDA 2008 S2008 Micah Spencer Coleman I dedicate this dissertation with love and pride to the memory of Juliana Cole, whose curiosity and lifelong devotion to learning have defined my family and instilled in me what was critical to survive my graduate career. Enjoy the '., _ipes, Grandmother. ACKNOWLEDGMENTS My deepest love and admiration to my wife Hiroko. While studying in a foreign language in a foreign land, she took up two of the hardest imaginable roles, that of military spouse and that of mathematician caretaker. Daisuki! I thank my parents Bob and Bobbi Coleman, my brother Matt, and mi vecino Abby for their patience, humor, and integrity. Great thanks go to Professors Julie Miller, Tina Carter, and Norm Levin, for first introducing me to I ii iii to our Graduate Coordinator Paul Robinson, and to the greatest advisory committee ever assembled, Professors David Drake, Meera Sitharam, Andrew Vince, and Neil White. I am honored and humbled to be associated with each of them. Finally, my deepest respect and appreciation are held for my advisor, B6na Mikl6s. TABLE OF CONTENTS page ACKNOW LEDGMENTS ................................. 4 LIST OF FIGURES .................................... 7 A B ST R A CT . . . . . . . . .. . 8 CHAPTER 1 INTRODUCTION ...................... .......... 10 1.1 Asymptotic Enumeration ............................ 10 1.2 Notation for Asymptotic Growth Rates ......... ........... 10 1.3 Generating Functions .................. ......... .. .. 11 2 PATTERN AVOIDANCE IN PERMUTATIONS AVOIDING A MONOTONE PATTERN . . . . . . . . .. 12 2.1 Permutations and Permutation Patterns ........ ........... 12 2.2 An Open Problem by M. Atkinson .................. ..... 24 2.3 Generating Trees .................. ............. .. 26 2.4 "Hat" Notation .................. .............. .. 31 2.5 Monotone Increasing Patterns q ........ ........ .. .. 33 2.6 The Pattern q = 123 ............. .......... .. 38 3 PATTERN PACKING ........... ..... . ..... .. 53 3.1 General Pattern Packing .......... . . ... 53 3.2 Pattern Packing in 123avoiding Permutations . . ..... 57 3.3 Pattern Packing in qavoiding Permutations ................ .. 59 3.4 Packing Density and Further Directions .............. .. .. 61 4 ASYMPTOTIC NORMALITY AND UNIFORMITY . . ..... 63 4.1 Probability Theory .................. ............ .. 63 4.2 Triangular Arrays .................. ............. .. 64 4.3 Asymptotic Normality .................. .......... .. 65 4.4 Asymptotic Uniformity ..... . . ..... ........... 66 4.5 Generating Polynomials with Real, NonPositive Roots . .... 68 4.6 Asymptotic Normality Implies Asymptotic Uniformity . ... 71 5 ON THE ROOTS OF THE BELL POLYNOMIALS . . ..... 74 5.1 Stirling Numbers of the Second Kind ................ .. .. 74 5.2 Bell Polynomials ............... . . .... 77 5.3 Bounds on the Roots of the Bell Polynomials . . ...... 80 5.4 Asymptotics of the Roots of the Bell Polynomials . . ..... 83 REFERENCES ....................................... 88 BIOGRAPHICAL SKETCH .......... ... ................ 91 LIST OF FIGURES Figure 21 The permutation 3142. ......... 22 The permutation 532614 . . 23 The permutation 865321947 . . 24 A rooted tree. ............ 25 The complete binary tree . . 26 The Fibonacci tree . ..... 27 T(123,132). ............. 28 Tree in W(123,231) rooted at 42153 . 29 The 1 it, i permutation 213654 with 31 The permutation W(6) = 342516 . lv, irs 21, 3, and 654 page 13 25 27 28 29 30 32 40 41 55 Abstract of Dissertation Presented to the Graduate School of the University of Florida in Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy ASYMPTOTIC ENUMERATION IN PATTERN AVOIDANCE AND IN THE THEORY OF SET PARTITIONS AND ASYMPTOTIC UNIFORMITY By Micah Spencer Coleman May 2008 ('C! r: Mikl6s B6na Major: Mathematics We demonstrate ..i! iiill ic properties of some popular combinatorial objects, including a partial answer to an open problem posed by Michael Atkinson and a general result on conditions for the coincidence of ..imptotic normality and uniformity. For a permutation 7r S, written in one line notation as 7r = 7Tr2 Tr, we ,i 7 contains the pattern a E Sm if there exists a subsequence 7i, such that for all 1 < j, k < m it holds that Ti, < 7 if and only if ca < Uk. Denote by s,((, a) the number of npermutations which do not contain either of the patterns a and r. Letting a' denote the pattern (m + 1)a, we construct classes of patterns for which the limit supremum of sT(123 .* r, a) /" agrees with the limit supremum of sT(123 .. r, a')1/" for several classes of patterns a. We also construct classes of permutations which avoid 123 ... r and contain i,,r, '': patterns. iM ,I: combinatorial sequences are of the form (aT,k) where n ranges over the nonnegative integers N and, for each n, there exists m = m(n) such that k ranges from 1 to m. We call such a sequence a combinatorial distribution. M ilv combinatorial distributions, upon rescaling, approach in distribution the normal distribution as n grows to infinity, a phenomenon we call ';,,,/'l/. '.: no I ,,,l.:/, A combinatorial distribution is said to be i~;,,'/,/I J.. all ;, u [.,rm if, for each positive integer q and each residue class modulo q, the sum of coefficients an,k with k r (mod q) approaches 1/q as n grows to infinity. We call this 1' ;,/,/./ : unifoiiii;, We prove that if the generating polynomials for a combinatorial distribution have real, nonnegative zeros, .,i~ii l, I ic normality implies .i!,ill I ic uniformity. We apply this result to several sequences from the literature. Finally, we present original results on the zeros of the Bell polynomials which were first attained in proving the . i:!! I,.1 ic uniformity of the Stirling numbers of the second kind Sn,k. CHAPTER 1 INTRODUCTION 1.1 Asymptotic Enumeration Enumerative combinatorics involves counting discrete objects, determining the cardinalities of sets which are indexed by one or more integers. In many cases exact formulae are known. However, some formulae are so convoluted as to obscure their most telling information. In either of these cases .. mptotics provide powerful methods for understanding the classes under question. With that said, we could define .,iii!,l ,l ic combinatorics as the study of the growth of combinatorial sequences indexed by an integer n as n grows without bound, n  oo. For a survey of these techniques, the reader is reffered to the chapter by Odlyzko [26] in the Handbook of Combinatorics. Definition 1. We introduce some notation to be used throughout. Let [n] denote the set {1, 2,... n}, and [a, b] denotes the set {a, a + 1,..., b}. N denotes the natural numbers n > 0, while P denotes the positive integers n > 1. 1.2 Notation for Asymptotic Growth Rates Let f : P P be a function with some sort of predictable behavior. What does it mean to understand the ..I:ii!1.1 ic behavior or growth of f? If it is anything like most functions we encounter, as n grows arbitrarily large, the growth of f probably follows some pattern, for example a straight line or logarithmic curve. Perhaps f acts erratically on every small interval but has a smooth overall growth which we can mimic with some other function g which is easier to understand. Example 1. D. I,, the function ((n) = n2 + (1). This function is .., to write but not so ..r; to ig'''l, However, as n grows very I7",,, we see that ((n) is very close to the '"* 'lJi'. 'ii !, Jl n 2. Such a situation motivates Big 0 notation. Given two functions f and g defined on the positive integers, we write f(n) = O(g(n)) if there exists a nonzero constant M such that If(n) < IMg(n) for all n > 1. Similarly, f(n) = (g(n)) if there exists a constant M such that If(n)l > IMg(n) for all n > 1. Finally, f (n) ~ (n) if lim f(n) 1 noo g(n) holds. 1.3 Generating Functions A powerful area of combinatorics is Generatingfunctionology. For surveys of the field, the reader is referred to the texts by Wilf [40], Stanley [33], [34], and B6na [8]. Definition 2. Given a sequence (an)n>o, the associated ordinary generating function is the power series f(x) = anxT. n>O Similarly, for a sequence (a,,k) 1, ,.,' l for all n > 0 and 0 < k < m(n) for some function m(n), the associated generating polynomial for each n is the I" l;'l, ,.;;'.:l m(n) ,q(x) Z Y ank. k>O The power is in handling these power series and polynomials to reveal information about the sequence in question. Generating polynomials will p1 iv a fundamental role in the last two chapters. CHAPTER 2 PATTERN AVOIDANCE IN PERMUTATIONS AVOIDING A MONOTONE PATTERN 2.1 Permutations and Permutation Patterns One aspect of enumerative combinatorics, one of the simple beauties that make it such an attractive discipline, is the ease with which it can be explained to the nonspecialist or even nonmathematician / nonscientist. We seek to count the objects in some class of size n. For example, we may estimate the number of objects in some class which are of size n and have statistics 1, k as n grows arbitrarily large. In fact, there are some combinatorial objects which are more easily explained to a nonmathematician than to some of our mathematical colleagues. The case in mind is the permutation. In the field of permutation patterns, one views a permutation in oneline notation; i.e., as an arrangement of the numbers 1 through n for some n. That is it. While we readily admire and appreciate the wisdom and complexity of our friends the algebraists and topologists, there is a certain frustration and loss of momentum when trying to describe the simple properties of permutations that we are dealing with here to such an audience. There is no consideration of cycle structure, what set is acted upon, etc. From such a rough definition we can pose many of the fundamental questions dealt with in this field of research. In the first three chapters of this dissertation we introduce the concept of a permutation pattern, survey some of the 1n i, r results and trends in the field of permutation patterns, develop a foundation for the work contained herein, and pose further questions to be explored. Definition 3 (rough). A permutation of length n or a permutation on [n] or an npermutation is an arr i.,, i,, n,: of the numbers 1 through n. Definition 4 (rigorous). An npermutation r is an injective I'""Ij'.:'" from the set [n] = 1, 2,..., n} onto itself. We denote 7(i) by it and write w as the concatenation 7T172 7' IT l .':/ the numbers 71 ,... ,7r the entries of r. It should be noted that by convention we allow there to be one permutation on the mpl"; set. F.:,.11, for each n > 0 we denote by S, the set of npermutations. We use the convention that an ascent in a permutation 7~ is the index of an entry 7~ with 7i < 7i+,. Likewise a descent in 7~ is the index of an entry 7ri with 7ri > 7i+1. We define an ascendee (resp. descended) to be the index of an entry 7~ with 7~i < T (resp. 7i1 > ii). Example 2. For ~ = 3142, 2 is an ascent; 1 and 3 are descents. 3 is an ascendee; 2 and 4 are descendees. As these are geometric as opposed to ri/, braic concepts, they i,,~ni best be understood ;.',all;i as in Fr.:';,,, 21 4 3 2 1 Figure 21. The permutation 3142. In all contexts considered here, a permutation pattern or simply a pattern is itself a permutation but there are subtle differences which we shall exploit. Given a permutation 7T = 1 TT,, a subsequence of 7T is an ordered subset of the entries of T, ( T~T ,..., ) for some k, which we write in the same order as they appear in r, so il < i2 < < ik. Example 3. Consider the 5permutation r = 13425. If we want to construct a subse quence, we have two choices of what to do with each entry 7i, whether to include 7r in our subsequence or not. Therefore, we have 25 = 32 subsequences of 7. We list them here in an obvious but i, i,'.',.l. l. ordering: Length 0 : 0 Length 1: 1 3 4 25 Length 2 13 14 12 15 34 32 35 42 45 25 Length 3: 134 132 135 142 145 125 342 345 325 425 Length 4 1342 1345 1325 1425 3425 Length 5 : 13425 Roughly speaking, each such subsequence can be associated with some pattern, and from this we generate vast fields of research. With such IJlu: i let us formalize the concept of the permutation pattern. Definition 5. We call a finite sequence X of distinct positive integers reduced if the elements of the sequence form the set [n] for some n, i.e. if X is a permutation of [n]. Let X = X1X2 ... X, be a finite sequence of n distinct positive integers. Then, there exists a unique permutation 7r E S, such that for all pairs of indices i and j with 1 < i Xi < Xj if and only if 7ri < rj. So, X and 7r are in the same "order", or order isomorphic. As we assumed 7r to be a permutation in S,, 7r is reduced, and we call 7r the reduction of X. Definition 6. Let 7 E S,. We .,;/ that 7 contains the pattern a E Sm (m < n) if there exists a subsequence 7 i .. 7 (i < i2 < ... < i,) whose reduction is a. If there is no such subsequence, we .,;, that 7 avoids the pattern a. One more note is in order on the above definition, which implicitly defines patterns. A pattern is itself simply a permutation, but we use the two words to describe distinct sets. We ask questions such as does the permutation 7 contain the pattern a, etc. These meanings will be clear in the context. Please also note that we will use the convention wherever possible of letting m denote the length of a pattern a and letting n denote the length of a permutation 7 for which we want to determine aavoidance, etc. Much of this dissertation will be devoted to questions surrounding the avoidance of a set of two patterns, one of which will be monotone increasing. We will conventionally let q denote the monotone increasing pattern in question and let r denote its length. Example 4. The permutation 34521 contains the pattern 123, as the first three entries of 34521 are 345, which reduces to 123. Example 5. We exhaust all patterns contained in the permutation 13425 from Example 3 by reducing all subsequences, in the same ordering as above: Length 0 : 0 Length 1:1 1 1 1 1 Length 2: 12 12 12 12 12 21 12 21 12 12 Length 3: 123 132 123 132 123 123 231 123 213 213 Length 4: 1342 1234 1324 1324 2314 Length 5 : 13425 Of course, we could continue in this fashion. It is a great exercise for the beginner in this area to exhaust the subsequences of a permutation to determine what patterns the permutation contains or avoids and pose some conjectures. This is how one learns anything in combinatorics, by w 11 ig our hands dirty", doing enough manual labor on our combinatorial objects to get a feel for their growth and other properties. Definition 7. Let a E S, be a pattern. For each n > 0 we denote by S,(cr) the set of all npermutations which avoid a, and write s,(a) = S, (a)l, i.e. the number of npermutations which avoid a. More generill let E be a set of patterns. For each n > 0 we denote by S,(E) the set of all npermutations which avoid all patterns in E, enumerated by s,,(). If Z = {uli 2, 2. Jk we write S,(Z) as S.(ui, 02,. ok) and likewise write s,n() as sn(ul1, 2, *, ok), dropping the brackets. On a historical note, the S, and s, above may have been coined in honor of Simion and Schmidt, whose 1985 paper [31] launched pattern avoidance and contained some results which are still hallmarks of the field. As the name would imply, enumerative combinatorialists most enjoy enumerating sets, that is, determining a precise formula for the cardinality of each set which depends only on the index or indices of that set. Unfortunately, we often find quite interesting discrete objects whose nature is complex enough to elude precise formulae. Alas, we will see that for most patterns a, the sequence s,(a) falls into the latter category. However, all is not lost. As was briefly discussed in the introductory chapter, great information can still be had by the imptotics of a sequence, and many of the current results in the field of permutation patterns involve bounds and limits which are not as strong as precise formulae, but carry power and beauty of their own. Let us first see some examples of patterns for which we can give an exact formula. We will treat all patterns in S, for 0 < m < 3. We will make use of the Kronecker delta 6ij, defined by 1 if i = j i,j 0 if 0 ifi e/. Proposition 6. We have the following exact formulae for a E Sm, m = 0, 1,2: n(0) = 0, Sn(l) = 6,o0, s,(12) 1, s,(21) 1. Proof. As we can take an empty subsequence of any permutation, including the empty permutation itself, no permutation can avoid the pattern 0, and s,(0) = 0. Similarly, every non n/l' ;, permutation has a nonempty subsequence of length 1, and s,(l) = 0 unless n = 0, in which case we have so(l) = 1, counting the sole permutation in So. Note that a permutation 7r S,, avoiding the pattern 12 is equivalent to 7 having no ascents, implying 7 is the unique monotone descending sequence of length n, so s,(12) = 1. The dual statement is that avoiding the pattern 21 is equivalent to avoiding descents, thus s,(21)= 1. E The reader may (should) have been amazed by the fact that s,(12) = s,(21) and their dual p"... '[ In fact, the duality involved was that 12 and 21 are reverses of each other, and for each n the unique permutation in S,(12), the monotone decreasing permutation, is the reverse of the monotone increasing permutation, the unique element of S,(21). Of course, one could also prove the equality with the fact that 12 and 21 are complements of each other. Perhaps these statements also apply to longer, more interesting patterns? Definition 8. Let 7 = 71rT2 ... 7 be a permutation. We /. I;,: the reverse, complement, and algebraic inverse, resp., of 7 as R 7T R= 7nn1 7T1, 7T C (n 71l + l)(n 7"2 + 1) (n 7{n + 1), 1 1 ... n, where, for all 1 < i < n, r, = i. Example 7. Let r = 24315. Then, we have the eight related permutations r = 24315, R = 51342, c = 42351, 1 41325, 7rR( 7CR) 15324, (7R)1 25341, (rC)1 52314, (7RC) 1 14352. Of course, the algebraic inverse r1 is just what the name would imply, the inverse of r considered as an element of the p'. '"' S,. Such a definition as the one above may seem crude, but we had promised to only consider these permutations as linear orders, not group elements. In our example of r = 24315, we made easy use of the obvious fact that R and C commute as operators. We leave it as an exercise that they each commute with the algebraic inverse as well. The definition of reverse should be clear. We note that we can write each entry of rR as r R = 7 i+1. We can think of complement as "flippiin the permutation in a vertical sense. Now that we have these definitions, we return to the question of equivalences among patterns. Suppose a permutation 7 = 71r r, contains the pattern a = a1 ... r,. Then, we have a set of indices 1 < pl < pm < n such that the reduction of the subsequence pU, 7, is precisely a, i.e. for all i < j, ., < 7rp, if and only if ai < Kj. By the remarks following Definition 8, this is equivalent to the statement that, for each pair < j, R R R R R, +1 > I,,+1 if and only if ani+1 > a7_i,. As we run through all pairs i < j, we see that this collection of statements is equivalent to R > R if and only if ae > a1. Altogether, 7 containing a is equivalent to rR containing aR. Likewise, r containing a is equivalent to rC containing aC and 71 containing a1. Of course, we can replace the word containingg by the word ,vpli..[Ii; in each statement. Critically, as the three maps R, .c and 1 are bijections S,  Sn, it follows that the number of npermutations which avoid a is the same as the number of npermutations avoiding aR, aC, a . We have achieved the following result. Lemma 1. Let a be a pattern. Then, for all n > 0, s,(ao) = s,(aR) = s,(oaC) = s,(1~). In fact, we can ? that each a E Sm belongs to an equivalence class of mpatterns whose sequences sT are equal. Definition 9. We .r;, the patterns a and r are Wilfequivalent if s,(ac) = Sn,() holds for all n > 0. This leads nature /ll/; to the 1/ fl ,.:/.:,In of the Wilfequivalence class of a pattern a as the set of all patterns which are Wilfequivalent to a. For our first nontrivial results, we move to m = 3. S3 consists of the six patterns 123, 132, 213, 231, 312, and 321. One quickly recognizes that there are at most two Wilf equivalence classes in S3 by noting 123 = 321, 132R 231, 132 312, 132Rc 213. In fact, there is only one Wilf equivalence class in S3 as is seen in the following lemma. Lemma 2. (Simion and Schmidt [31]) For all n > 0, ,(123) s(132). Proof. We call a permutation entry which is less than all entries preceding it a lefttoright minimum. The remaining entries we call remaining entries. Let 7 be a 132avoiding npermutation. Form a new npermutation 7' by fixing the lefttoright minima and transposing (swapping) pairs of remaining entries until they are in decreasing order. They will still all be remaining entries, because at each transposition the left, smaller entry is preceded by some lefttoright minimum which will still precede the larger remaining entry after the transposition, and the smaller remaining entry moves to the right, still not a lefttoright minimum. Note that the lefttoright minima are in decreasing order, as 7r < rj with i < j would imply that 7rj is not a lefttoright minimum. Therefore, v' is composed of two decreasing sequences, so by the PigeonHole Principle there cannot be an increasing subsequence of three entries (three pigeons cannot fit into two pigeon holes). So, we have mapped our 132avoiding 7 to the 123avoiding 7' and can apply this process to all r S,,(132). To see that this process is reversible, we fix the lefttoright minima of r', leaving blanks at the other indices and placing the remaining entries in a sack. Moving left to right, we fill each blank with the least entry still in our sack which is larger than the rightmost lefttoright minimum to the left of said blank. Now we have a permutation kr, which we claim to be 132avoiding. Indeed, if there were a copy of 132, then there would be a copy of 132 beginning with a lefttoright minimum, however this is a contradiction as the entries following and larger than each lefttoright minimum are increasing. Thus, we have a bijection. O So, in fact we see that for all a, rT E S3 and n > 0, s,(a) = s,(7). One may be tempted to suspect such a statement holds for patterns of every length m. However, with a computer check or a few pages of scribbling, one obtains S6(1342) = 512 / s6(1234) = 513. We now turn to .,i.,il'uitics. In 1980, Richard Stanley and Herb Wilf independently conjectured that for each pattern a there exists a constant c, such that, for all n > 0, we have sn(a7) < c. In [2], Arratia proved that this was equivalent to the following, long known as the StanleyWilf Conjecture. Theorem 8 ( r! ircusTardos). Let a be a pattern. Then, the limit lim 8,(a)1 n*oo exists (and is finite). The validity of the StanleyWilf Conjecture was established by the 2003 proof by Marcus and Tardos [23] of the FiirediHajnal Conjecture on permutation matrices. That the FiirediHajnal Conjecture implies the StanleyWilf conjecture was proven by Klazar [21]. For a clear and concise treatment of all, see section 4.5 of [5]. The reader is also encouraged to see Doron Zeilberger's alternative rendition [41]. There is still hope for a tighter proof of the StanleyWilf Conjecture, as the FiirediHajnal Conjecture only proves there exists a constant, but the constants which we get from the proof are astronomically larger than the observed constants. It does still give structure to our work to know that for any pattern a, s,(a) is at most exponential, and additionally the limit lim s,(O)l/ n*oo exists. We denote this limit by L(a) as it is critical to the sequel. Now, for a finite set of at least two patterns, we do not have such a strong general result. For any such set of patterns E, it is readily seen that s,(E) < s,(c) for each pattern a E Z, so we have the following corollary to Theorem 8. Corollary 1. Let E be a no", mnil;i finite set of patterns. Then, there exists a constant cy such that for all n > 0, Sn(Z) < CY. So, for any such set of patterns E, we see that the sequence s,(E) is at most subexponential, i.e. bounded above by an exponential function, but we did not mention the existence of a limit of s,(Z)1/'. For many classes of such sets we cannot prove the existence of such a limit, an open problem with much interest, discussed in [37]. Our work in this chapter is restricted to sets of the form {q, a}, where q is a monotone increasing pattern, i.e. q = 12 ... r for some r > 3. We have the following result on a class of such sets, a generalization of Arratia's proof [2] that the exponential bound (i.e. the MarcusTardos theorem) implies the existence of the limit for the case of a single pattern. This generalization was ir. 1 1 by M. Klazar [22]. Definition 10. We .';, a permutation 7r E ST is decomposable if there exists 1 < k < n such that for all 1 < i < k < j, ri > Tj. Lemma 3. Let q = 12 r and a E Sm for some r > 3 and m > 3. Then, if o is not decomposable, the limit [ L(q, a) lim s,(q, a)1" noo exists. Proof. Let m, n > 1. For each pair of permutations a E Sm(q, a) and 3 E S,(q, a), construct the (m + n)permutation F (al + ((a + ) ( n) (.m.( + n) 1 0 3 2 /3.. .n By our assumptions on a and 3, 7r avoids q and a in its first m entries and in its last n entries. Furthermore, as q and a are not decomposable, there is no <" *i of q or a which contains entries from these two sets. Thus, 7r is (q, o)avoiding and uniquely determined by a and 3, proving the following inequality. Sm+n(q, o) > Sm(q, )Sn(q, ). We have the following lemma by Fekete [17] for superadditive sequences. The analog of this lemma for subadditive sequences was used in Arratia's proof for the case of a single pattern. Lemma 4 (Fekete). For every sequence {a,}fn N '.:.''' ..'1 an+m > anam for all m, n > 0, the limit lim noo 7 exists. Applying Fekete's lemma to the sequence log s(q, oa), we thus have that the sequence s,(q, )1/" has a limit. As this sequence is bounded above, this limit is finite. D With these facts in mind, we define L on sets of two patterns as follows. Definition 11. Let a and r be patterns. Then, /. fI;,. the following ,I;il.:/l; L(q, ) =lim sups,(,r)1 "' n*oo Of course, for pairs of patterns a and r for which the limit exists, such as those in Lemma 3, the limit and lim sup agree, and L is as we want it to be. 2.2 An Open Problem by M. Atkinson This work was originated in response to a question posed by Michael Atkinson at the fifth International Conference on Pattern Avoiding Permutations. We answer the question in the affirmative for all increasing patterns q on some classes of patterns a and specifically for q = 123 for larger classes of patterns a. For a pattern a of length m, define a' to be the pattern (m + 1)a. For example, for a = 2431, we have a' = 52431. Given a monotone ascending pattern q and a pattern a, does it hold that L(q,a') = L(q, a)? In any permutation the entries preceding the first ascendee form a decreasing sequence. For a permutation 7r, if the first ascendee of 7 is the index i, we call Tr the threshold of 7 and call the decreasing sequence preceding the threshold the front end. The set of entries following the threshold we call the back end. Example 9. For 7 = 532614, our first ascendee is the index 4, so the threshold of 7 is the entry 7F4 = 6, the front end is 532, and the back end is 14, as shown in Figure 22. Note that in our previous example each entry of the back end is less than the threshold. Creating our own good luck, we chose 532614 for our example specifically because it avoids 123. In fact, every 123avoiding permutation shares this property, a simple structure of which we shall take great advantage in our handling of these permutations. To see this property, suppose our threshold is rt and there is an entry rj in the back end (equivalently j > t) with Tr > ft. By definition of ascendee, it1 < 7Tt, and the subsequence 7rt_1 t Tj forms a 123. We borrow the next few definitions from the recent paper by Vatter [37]. * Threshold Front End Back End Figure 22. The permutation 532614. Definition 12. An interval in a permutation 7r is a set of consecutive indices {i, i + 1,..., i +r} such that the set of values {1i, iTii, .. i+r} is a set of consecutive integers. A decreasing interval is then a set of consecutive indices whose values appear in decreasing order. F.:,,,ll, a maximal decreasing interval is a decreasing interval {i, i + 1,..., i + r} such that {i 1,..., i + r} and {i,..., i + r + 1} are not decreasing intervals. Increasing intervals and maximal increasing intervals are 1, 17,.,, ,..irl,l;, I.;,1; We denote an interval by its set of indices, written with set notation, or by its entries, written in oneline notation. Example 10. Let r = 3756124. Then, 7r contains the interval {2, 3, 4}, i.e. 756. Indeed, {2,3, 4} is a set of consecutive indices, and the set of values {12, 3, } 4= {7, 5, 6} is a set of consecutive integers. Note that 7r also contains the maximal increasing interval {5, 6}, i.e. 12. Definition 13. Given a permutation w E S, with interval I = {i, i + 1,..., i + r}, we /. I;,.' the deflation of 7 at I to be the reduction ofrl ri_1 i 7ri+r+l 7"r,. Similarly, the inflation of 7 at the index i by the permutation a E Sm is the permutation obtained from 7 by increasing by m 1 each entry greater than ri and replacing the entry Tr with the interval whose reduction is a. Example 11. The /. fl.,[/:.n of the permutation 264513 at the interval 645 is the reduction is 2413. The :,fl/rl,.,n of the permutation 3124 at index 3 by the permutation 321 is 514326. It was remarked above that the front end of a permutation is monotone decreasing. Thus, the front end has a unique factorization as the concatenation of one or more maximal decreasing intervals. Example 12. Let 7 = .. :21947. Then, the front end of T is 865321, the concatenation of maximal decreasing intervals 8, 65, and 321. 2.3 Generating Trees A tree is simply a connected, 1 i i': simple graph, or equivalently a connected simple graph on n vertices with n 1 edges. A forest is a collection of trees. A rooted, labeled tree is a tree with an assignment of labels (typically nonnegative integers) to the nodes and one node designated the root, giving an orientation to the entire graph. We  that a node y is a child of the node x if the final edge in the unique path from the root to y is the edge {x, y}. Likewise, we call x the parent of y and define the depth of y to be the number of children. The descendants of y is the set of nodes x such that the unique path from the root to x passes through y. Finally, we call the set of all nodes whose unique path to the root contains k edges the kth generation of the tree. In [39] Julian West defines a generating tree as a rooted, labeled tree having the property that the labels of the children of each node x can be determined from the label of x itself. This leads to the characterization of a generating tree by the label of its root and 40 3 2 Figure 23. The permutation s.. :21947. a set of succession rules which determine the number of children and labels of children for each node of a given length and label. The classic task for a combinatorial enumerologist is to determine the number of some combinatorial objects of size n, perhaps further indexed with respect to some property or some statistic k. Typically, one is presented with an initial object of some small size and a recursion rule which z,i how many objects of each successive generation (objects of size n + 1) can be created inductively from those of the previous generation (objects of size n). Define the gth levelnumber of a tree to be the number of nodes in the gth generation. Thus the generating tree is easily seen as a tool which lends itself quite readily to combinatorial enumeration. We consider the nodes of our tree to be the combinatorial objects themselves. There are many situations when the number of (n + 1)objects which root 0 *' children Third generation of tree Figure 24. A rooted tree. can be generated from any nobject is all we need to know, so we might as well label each node with its depth. In [39] West begins with a trivial example, the complete binary tree. We begin with a root with label (2). Our succession rule is that each node with label (2) has two children also labeled (2). Example 13 ([39], Example 1). The complete binary tree is determined by the set of rules Root: (2) Rule: (2) (2)(2) A long celebrated integer sequence is that of the Fibonacci numbers (F,)n>_o 0, 1, 1,2, 3, 5,..., where we have the initial assumptions F0 = 0 and F1 = F2 = 1, and each additional number F, is the sum of the preceding two numbers of the sequence. They will 1p1 i a role in our studies of pattern avoidance, so we show their generating tree as a less trivial example of generating trees and a slow introduction to this sequence. (2) (2) (2) (2 (2) (2 (2) Figure 25. The complete binary tree. Example 14 ([39], Example 3). The Fibonacci tree is determined by the set of rules Root : (1) Rules: (1) (2) (2) (1)(2) The observant reader undoub', l./; noticed that our succession rules are not the same as the recursion we gave to 1. F,'.: our sequence. We verify that these rules are in fact equivalent to the statement F~i+l F= + F_1. For each n > 1 we have a set of G$) objects labeled (2) and a set of G$) objects labeled (1). Each object in either set produces an offspring of size n + 1 with label (2). So, we have (2) G(1) + G(2) = . Gn+l n n Furthermore, each object in G,1) produces an offspring of size n + 1 with label (1), so G(1) (2) n_ 1. where the second ..,; ,'l/:/l follows from our previous statement. We have thus accounted for all Fn+i objects, and we have our recurrence. (1) (2) (2) (1 (2) Figure 26. The Fibonacci tree. For a detailed exposition on the use of generating trees in the study of pattern avoidance, see [39], [12], [36], [10] and [24]. Here we define the generating trees which will be used throughout. These definitions depend on the patterns a and q which are being avoided, so we assume the patterns to be given. This will be clear from context. First we explain the motivations. Recall our / notation. For a pattern a E S,, the pattern a' E S,+1 is obtained by prepending a with the entry (m + 1). The fundamental question here is whether various limits (or limit supreme) for the number of permutations which avoid some pattern a are the same as those which avoid a' (assuming for now that the limits exist). It was noted above that aavoidance implies a'avoidance, but there are a' avoiders which contain a. So, our question boils down to just how many of these there are, in particular what are the .i,iii1ll 1.ics of these permutations with respect to the set of aavoiders. We would like to make use of generating trees to study the set of a'avoiders which contain a, i.e. {1 : w avoids a'} \ {w : avoids a}. For each n > 0, let T, = T,(q, a) be the set of all (q, a)avoiding permutations of length n, enumerated by t,. We construct the generating tree T = T(q, a) whose nt level is T,. A permutation 7 e T,,+ is a child of 7 e T, if and only if 7 can be obtained by inserting n + 1 at one of the n + 1 open sites in 7. Similarly, for each n > 0, let U" be the set of all (q, a')avoiding permutations of length n, enumerated by u, and construct the generating tree U = U(q, a) with levels U, and succession defined as for T. Atkinson's question is thus answered in the affirmative for some pattern by showing that u, does not grow ..i ,i 11..I i. ly faster than t,. We will focus our attention on the sets W, of all (q, u')avoiding permutations of length n which contain at least one copy of a, i.e. W, = Un\T,, enumerated by 1',,. This motivates the forest W = U\T. Note that each tree in W is rooted at a qavoiding permutation which avoids a' and contains a but whose parent in U avoids a. For q = 123 and a = 132, we have the tree shown in Figure 27. An active site in a permutation is a valid insertion point, that is, a site where we can insert n + 1 and obtain a child which is still in the current generating tree, so for our purposes an active site is such that the insertion will not cause an occurence of any pattern which we seek to avoid. The depth of a permutation 7 is the number of active sites in 7, equivalent to the notion of depth defined above on generating trees. We note that the depth depends on both the permutation itself and on the tree, specifically the pattern being avoided which determines the tree. 2.4 "Hat" Notation Given a permutation 7 and a pattern a, we denote by a any copy of a in T. For each index p, we use the notation ap to refer to an entry which serves the role of ap in some a. As there may be more than one <". , of the pattern a in some permutation, ap generally does not refer to any specific entry. For example, with a = 132, the permutation r = 24135 contains one a, namely 243. In this case a1 refers to the entry 7t = 2, a2 refers 1 (2) 21 (3) 12 (2) 321 (4) 231 (1) 213 (1) 312 (2) 4321 (5) 3421 (1) 3241 (1) 3214 (1) 4231 (2) 4213 (2) 4312 (3) 3412 (1) Figure 27. T(123,132). to 72 = 4, and 63 refers to T4 = 3. On the other hand, the permutation 1432 contains three 's, namely 143, 142, and 132. In this case we can refer to the a1, the entry 1, but we have several a2's and several a3's. It should also be noted that one entry could be both a i and a aj for some i / j. Example 15. With a = 132, in the permutation 7 = 25431, r3 = 4 is the 2 = of the subsequence 254 as well as the 3 2 of the subsequence 243. Often we will determine the depth of a permutation 7 by the index of the rightmost a1 in 7, by which we mean the entry with the greatest index of those entries rk such that there exists a a which begins at rk. Example 16. For a = 231 and 7 = 43521, the entries 71 and 72 are both I1 's, and we call 72 the rightmost i1. Any confusion over hat notation should fade upon seeing its motivation in the following proofs. As long as we use the ^ notation carefully, the meaning should alvi be clear. 2.5 Monotone Increasing Patterns q For any pattern a we have by construction that a' contains a, so it follows immediately from our introductory remarks on pattern avoidance that sn(() < s,,(t'). for all n, and L() < L('). Of course, these bounds also extend to L(a, n) < L(7', n). for any set of patterns II, etc. Let q be the increasing pattern 123 ... r. We will show that the statement L(q, a') = L(q, a) holds if a pattern a begins with its greatest entry. Our proof builds on B6na's proof for the case of single patterns, found in [4]. Definition 14. A lefttoright maximum is an entry in a permutation which is greater than each entry to its left. The remaining entries of a permutation are those which are not lefttoright maxima A weak class (weak nclass) is a set of permutations (resp. npermutations) whose lefttoright maxima are the same and are in the same respective positions. Example 17. The permutations 32415 and 31425 both have lefttoright maxima 3, 4, and 5, at the first, third, and fifth entries, so we C,1 that 32415 and 31425 are in the same weak class or weak 5class. Example 18. The permutations 3412 and 2413 both have lefttoright maxima in the first two positions, but as the maxima themselves are not the same, 3412 and 2413 are not in the same weak 4class. Lemma 5. For each r > 1, there exists a py. '1';. ;,,;. fr(x) such that for all n > 1 the number of weak nclasses of permutations with ,.. /i;, r lefttoright maxima is less than frn). Proof. Fix n. We can easily count the weak nclasses with one or two lefttoright maxima. An npermutation with only one lefttoright maximum must begin with n, so there is only one such weak nclass. Next we claim that there are (2) weak nclasses with exactly two lefttoright maxima. Indeed, pick two numbers a, b E [n] with a < b. Place the entry a in the first position, and place the entry n in the (n + 1 b)th position. Given such constraints, we can ahvli place the remaining entries to find at least one permutation in each weak nclass. For r > 2 we take the same approach but allow overcounting. There are at most (/) 2 v; to choose the entries and positions (numerical and geographical values) for the r lefttoright maxima, as this includes all possible sequences of lefttoright maxima, as well as some sequences which cannot possibly be the set of lefttoright maxima of a permutation. As (7)2 is a polynomial in n for any fixed r, the statement holds. O Corollary 2. For each r > 1, the number of weak nclasses with less than r lefttoright maxima is bounded by a ./, ;;.,;;,,.l in n. Proof. For each 1 < k < r, we have a polynomial upper bound fk(n) on the number of weak nclasses with exactly k lefttoright maxima. The sum over all such k is clearly an upper bound for the number of weak nclasses with fewer than r lefttoright maxima. As a sum of (a fixed number of) polynomials is a polynomial, we are finished. O Proposition 19. Let q be the ascending pattern 1 2 ... r, and let a be a permutation which begins with its greatest entry. Then, L(q, a') L(q, a). Proof. We first note that a permutation which avoids q has fewer than r lefttoright maxima. Now, if a permutation avoids a', then its remaining entries avoid a. Indeed, by definition each remaining entry is preceded by a lefttoright maximum. If the remaining entries of a permutation contain a, then we may prepend this a with any lefttoright maximum which is to the left of and greater than a1 to obtain a a', as ai is itself greater than all other entries of a by the fact that a begins with its largest entry. Therefore, we can overcount npermutations which avoid a' by multiplying the number of weak nclasses which have fewer than r lefttoright maxima by the number of possible (q, a)avoiding permutations of the remaining entries. By Corollary 2 the number of such weak nclasses is at most a polynomial function f(n). Therefore our overcount of npermutations is f(n)s,_i(q, a). We are now in position to take our limits. L(q,a') = lim sup s,(q, a')1// n*oo < lim sup (f (n) s,(q, ))1/T lim sup f(n)1/"s,_I(q, a) 1/n = lim sup 1 s_l(q, a)1/n7 = L(q, a). Combined with the knowledge that L(q, a') > L(q, o), we are finished. O The following lemma from [4] and [5] provides an upper bound on the number of permutations of length n which avoid the increasing pattern of length r. Lemma 6. s,(1 2 .. r) < (r 1)2n for all r, n > 2. Proof. We define a rank function on the entries of each permutation 7t E S,(12 .. r) by setting the rank of an entry 7r to be the length of any maximal increasing subsequence 7i,17,2 ... 7i. Note that this generalizes the concept of the lefttorightmaxima, which are precisely those entries with rank 1. If an entry 7i has rank t, then there exists some increasing subsequence 7rjTrj2 I rj1~Ti of length t, so for any entry trk > 7Ti with k > i, the rank of trk is at leas t + 1, as we have the increasing subsequence 7rj'Trj2, itrj k. Therefore, for each 1 < t < n, we see that the entries of rank t form a decreasing sequence. There are at most n such decreasing sequences, and as sets of indices they form a set partition of [n]. As 7r is assumed to be 12 .. ravoiding, 7r has no entry of rank r or greater, and there are at most r 1 blocks of our set partition. Assigning each entry of 7r to one of r 1 blocks, we may overcount and see that there are at most (r 1)" possible assignments of the indices to the blocks and at most (r 1)" possible assignments of the values to the blocks. O With this lemma in hand, we subtly alter another proof of B6na to achieve: Proposition 20. Let q be the ascending pattern 1 2 .. r, let a be a pattern, and let c be a constant such that for all n > 1, s,(q, a) < c". Then, for all n > 1, Sn(q, U') < (c + (r 2)2)1. Proof. In order for a permutation to avoid a', it is necessary that it avoid a in the region to the right of n. So, we may overcount the number of (q, a')avoiding permutations by choosing where to place n, which entries to place to the left of n such that they are 1 2 ... (r 1)avoiding (because any 1 2 ... (r 1) among them could be postpended with n to create a q), and how to arrange the entries to the right of n such that they are (q, a)avoiding. We let k be the position of n, so there are (" ) possibilities for the entries preceding n. By Lemma 6 there are at most (r 1)2(k1) possible permutations of these entries. Finally, by our original hypothesis on S,(q, a), there are at most c"k possible permutations of the n k entries which follow n. Altogether, these work out to the binomial expansion s,(qo)') < (r 2)2(k1)Ck' kl k= 1 (c+ (r 2)2n1 concluding the proof. O In particular, for q = 123, we have (r2)2 = 1, so with the assumption s,(123, a) < c" for all n, we find s,(123, a') < (c + 1)"1 for all n. The StanleyWilf Conjecture (il rcusTardos Theorem) tells us that for any pattern a or set of patterns E there is such a constant c as in the above hypothesis. In the case of avoiding a single pattern a, Arratia showed in [2] that the sequence s(o)1/"' is increasing. However, there are sets of patterns E for which the sequence s,(E)1/" is not increasing. Thus, taking c to be the least constant such that s,(E) < c" for all n < N for some N, it may be that there is a constant d < c such that s,(E) < d" for all n > N. In particular, the constant c may be significantly greater than L(E), so the new constant d is closer to our limit and thus a better indicator of the ..imptotic behavior of our sequence s,(E). Such a situation motivates a strengthening of the previous proposition. Proposition 21. Let q be the ascending pattern 1 2 ... r, let a be a pattern, and let d be a constant such that for some N and all n > N, s,(q, a) < d". Then, there exists a constant D such that, for all n > N, s,(q, a') < D(d + (r 2)1. Proof. The set {s,(q, a) : 1 < n < N} is finite, so it is bounded above, and we can choose a constant D such that s,(q, a) < Dd" for all 1 < n < N. We retain all machinery from the proof of the previous proposition, except instead of counting at most cnk possible permutations of the n k entries which follow n, we count them by Ddnk. Then our expansion becomes n t (r sn(q, a') < nj [ r 2)2 k1)Ddnk k 1 k 1 D t) (r 2)2(k1)dnk D(d + (r 2)2). 2.6 The Pattern q 123 In this section we restrict our attention to the 123avoiding environment. Some statements will be generalized to longer ascending patterns q in the following section. Consider the active sites of a permutation 7r W. As any child of 7 is 123avoiding, there can be no active site to the right of the first ascendee. As any child of 7 is o'avoiding, there can be no active site to the left of a &1. However, the consecutive sites satisfying these two criteria are all active. So, our understanding of depth reduces to understanding where these two bounds lie. Furthermore, if 7 has depth d, inserting n + 1 into one of the d active sites will not increase the depth. Indeed, n + 1 is inserted to the left of the first ascendee of 7 and itself becomes the first ascendee in the child. The rightmost a1 of 7 remains in place in the child, so the child will have a rightmost a1 which is at the same position or to the right of that of T. This demonstrates the following lemma. Recall our definition of the threshold of a permutation as the leftmost ascendee of the permutation. Lemma 7. Let 7r W for some pattern a. Then, the depth of 7 is the distance from the ,i.:1, i,,i , 1 to the threshold, i.e. the difference of the index of the threshold and the index of the rightmost a1. We proceed with a lemma which will be used extensively as it provides a polynomial bound for the levelnumbers of each tree in the forest W. Lemma 8. Let a be a pattern of length m which does not begin with m. Let 7r be a (123, a')avoiding permutation with depth d which contains a but whose parent avoids a. Then, the number of descendants of 7 at the jth generation, i.e. the jth levelnumber of the tree in W rooted at 7, is bounded above by (d + which is a i ;, .. ;;,.:' in j of degree d. Proof. (We are in fact bounded by the lesser polynomial (d+ 1), however we dropped the l's for neatness.) Let n be the length of Tr. First we note that 7r does not begin with n. Indeed, as the parent of 7r has no a but 7r has a a, n is the rh in any a in Tr. 7r contains a copy of a, which does not begin with its largest entry, so n appearing to the left of any a would complete a a', contradicting the assumption that 7r is u'avoiding. In fact, by this argument we see that n is the threshold and a must contain entries in the front end of Tr. We are trying to show that the number of descendants at each generation is bounded by our polynomial, so we might as well assume the worst case scenario, that inserting at the kth active site alvii produces a child of depth k, i.e. that the rightmost 1 of a child is alv,v at the same position as the rightmost a1 of the parent. Then, for all 1 < k < d, we in fact have the succession rules Root (d) Rule (k) (1)(2)...(k) An example of a tree in such a forest is shown in Figure 28. 42153 (3) 462153 (1) 426153 (2) 421653 (3) 4762153 4726153 4276153 4721653 4271653 4217653 Figure 28. Tree in W(123,231) rooted at 42153 Denoting by aj,k the number of permutations at the jth level with depth k, 0 < j and 1 < k < d, we have the recursive system ao,d = 1 ao,k = 0 Vk / d d aj,k = Y aj1,t Vj > 1, 1 < k < d. tk From this recursive system we see that aj,k (dk+l+j) For each d and j we may sum over all k to attain the levelnumber ( d+_l). However, a combinatorial proof is preferable. We know that ( dj1) counts jmultisets of [d], and such a multiset written in nonincreasing order spells out the order of active sites chosen in the lineage from 7r to a permutation of length n + j. E Definition 15. ([5] D. ii7.l' 5.33) A permutation 7 is called 1 i, ,. if it can be written as the concatenation qlq2 .. qk where each qi is a decreasing sequence of consecutive integers and the leading entry of qi is smaller than the leading entry of qi+l for 1 < i < k1. Example 22. An example of a 7.';,. ,, I permutation is 213654, which has three 7.';, ,, as shown in Fi:,,,n 29. 6 5 4 3 2 Figure 29. The 1 i, ,I permutation 213654 with lIv. r 21, 3, and 654 AT ii: results are known concerning pattern avoidance and pattern packing for 1 iv. I permutations. One can see Section 5.2.2 of [5], [27], [6], and [20]. Our next result is on 1 ,i,, ,.1 patterns with just two I1V, rs, equivalently nonmonotone 1 ,i ri patterns which avoid 123. Proposition 23. Let a be a pattern of the form d (d 1)...1 m (m 1)...(d + 1) for some d > 1. Then, L(123, a') L(123, a). Proof. Note that any permutation which contains a has depth at most d. In fact, we will use a stronger property of the roots of the trees in W in which the depth is exactly d. Indeed, let the Npermutation p be such a root. We know that p contains at least one copy of a, while the parent of p in U avoids a. As the only change between the parent of p and p is the insertion of N, N must act as m in any a in p. This implies that there is a set of d entries preceding N which serve as the d, (d 1),..., 1 in a a in p. In fact, as the front ends of both a and p are decreasing (because a and p are 123avoiding), and as every front end entry of a is less than every back end entry of a, the d entries .: I,, .:Vl,. /;/ preceding N serve these roles. Therefore, the rightmost o cannot be more than d positions to the left of N. As N is the only im in p, the entry which is d positions to the left of N is clearly the rightmost o1. That there is a fixed depth which applies to every root in W makes our job easy. For each N, the total number of trees in W whose roots have length N is less than NuN1, as each such root in W has a parent in U on the (N l)t level of U, and an (N 1)permutation can have depth at most N. Each such tree has levelnumbers which are bounded above by the polynomial (d+tNl1) So, the ^th levelnumber of WI, i is bounded by I',, NUNI N= v ,a N=m (fd+n < d u, times a fixed polynomial f(n), a polynomial that s,(123, a) < s, (123, ') for all n, we have n N 1 d1 +nN1d  1 t n 1 which only depends on a. With the fact The nth roots of this fixed polynomial approach 1, so we may apply the Squeeze Theorem to achieve lim sup '1/ = lim sup u1/'. n*oo n*oo We just saw that, if the front end is composed of entries which are all lower than any entry in the back end, then there is a polynomial function which depends only on the pattern and dictates the number of descendants for any permutation. Now consider how the situation changes if there is more than one maximal decreasing interval in the front end. Our shortest and prototypical example is 3142. The critical difference here is that our depth is no longer bounded as it was for the patterns considered in Proposition 23. For example, for any d > 2, we have the (123, 3142)avoiding permutation (d + 1) (d  1) (d 2) ... 1 (d + 2) d, which has depth d. This means we can are not guaranteed a fixed bound to the degrees of our generating polynomials of the levelnumbers of our trees. All is not lost, however, as we can show that for any of a class of patterns a and any k there is an upper bound to the number of trees in W whose root has depth which is k less than its length. For the pattern 3142 it follows easily from the fact that the depth of a permutation in W is completely determined by the length of the second maximal decreasing interval in the front end. First, we calculate L(123, 3142). As was mentioned in our introduction to generating trees, a famous combinatorial sequence is the Fibonacci numbers F, defined for all n > 0 by the recursive system Fo 0 F, 1 F,~+ = F, + F_ 1 V n > 1. Multiplying both sides of the recursive equation F,+ = F, + F,_1 by x", summing over all n > 1, and solving for the ordinary generating function F(x) = Zn>o Fnx", we find F(x) (()  from which we see that F, is .i mptotically approximated by the nth power of the dominant term, the golden ratio, F (1+ Furthermore, as the golden ratio enjoys the property of being a solution to the equation 2 = + so that ( )= 1 3+, we immediately achieve an approximation for every other Fibonacci number 2F n By iteratively applying the recursive equation for F,, we achieve a recursion on F2,1: F2n1 F2n2 + F2n3 = F2n3 + F2n3 + F2n4 = F2n3 + F2n3 + F2n5 + F2n6 SF2n3 + F2k1 k=0 S2F2n3 + F2n5 + F2n7 + + F1. Proposition 24. For all n > 1, s,(123, 3142) = F2n1. where F, is the the nth Fibonacci number. Furthermore, 3+5 L(123, 3142) 3 2 Proof. As the number of (123, 3142)avoiding permutations of lengths 1, 2, and 3 are 1,2, and 5, respectively, we see that the boundary conditions are satisfied. Writing sr for s,(123, 3142), it suffices to show that our sequence satisfies the appropriate recursion, ni Sn = Sn, + Sn 2s,_1 + Sn2 + + S1. k=I Given a (123, 3142)avoiding permutation of length n 1, we can ahv, prepend n, as n can never be the first entry of a 123 pattern nor a 3142 pattern, so the new permutation will remain (123, 3142)avoiding. This is clearly an bijection between S,_1(123, 3142) and permutations in S,(123, 3142) which begin with n, accounting for the first term, s_l. For a (123, 3142)avoiding permutation 7r of length n which does not begin with n, let j be the index of n, so 7j = n, (and j ranges from 2 to n as 7r ranges over all such permutations). As 7r is 123avoiding, n is in fact the threshold. Therefore, the front end of 7 consists of one maximal decreasing interval. Indeed, if the front end contains more than one maximal decreasing interval, then there exist entries in the front end ri and i1i+l, with the entry (~i+1 1) in the back end, and the reduction of r ri+1 n ( 1) is 3142, contradicting our assumption on r. As the front end of 7 is an interval, the deflation of 7 at the front end is a (123, 3142)avoiding permutation r of length n j + 2. The largest entry of is of course k2 = (n j + 2). Removing this entry, we obtain a (123, 3142)avoiding permutation of length n j + 1 (which may or may not begin with its largest entry). We claim that this is a bijection. Indeed, given a permutation 7 E S,_j+1(123, 3142), the inflation of 7 at the first index by the permutation (j 1) (j 2) ... 1 is a (123, 3142)avoiding permutation 7 of length n 1. Inserting n at index j regains a permutation in S,(123, 3142), and this is an injection. Such a construction over all j accounts for the terms s,_l + + si in our recursion. The limit L(123, 3142) = 3+ follows immediately from the above discussion of the .,i',i i. i cs of the Fibonacci numbers. Proposition 25. Let a be the pattern 3142. Then, the limit lim s,(123, a)1" noo exists and L(123, a') L(123,a), i.e. L(123, 53142) L(123, 3142) 2 Proof. Suppose p is the root of a tree in W and has depth d and length n. Recall that being a root implies p contains one copy of a, but the parent of p in U is aavoiding. We will map p to a permutation of length nd+1 by removing certain superfluous entries and reducing. By superfluous we mean that there is a subset of the entries of p which do not add any information to the identity of p in the sense that they are completely determined by the other entries and the fact that p has length n and depth d. Consider the structure of the front end of p. We claim that its factorization consists of exactly two maximal decreasing intervals. Indeed, n is both a 4 and the threshold of p, so there exist 3 and 1 in the front end and a 2 in the back end, so the front end itself cannot be a decreasing interval and thus contains at least two maximal decreasing intervals. On the other hand, suppose the front end of p contains three maximal decreasing intervals II, I2, and 13. Then, we can choose indices i1 E II, i2 E 12, and i3 E 13 and indices j < k such that pil > pj > Pi2 and pi, > k > Pi3, and these five entries form a a', contradicting the assumption that p is a'avoiding. We will proceed by removing the superfluous entries, reducing accordingly at each step. We know n contained no information, because its location is determined by the depth d of p and the location of the rightmost a. Remove n, ridding the permutation of any a. We retain the first (leftmost) maximal decreasing interval, but we do not need all the entries from the second maximal decreasing interval, as we can detect their presence (or absence) from d. So, deflate this interval. We now have a permutation of size n d + 1 which we call the 1' *'I. /i /,: of p. Definition 16. Given a root p of W, the permutation obtained by /. fl.[i':.,i the second maximal decreasing interval of p and removing its l',,l, . entry is called its prototype. By this process, we map our roots of length n and depth d into the set of aavoiding permutations of length n d + 1 which contain at least two maximal decreasing intervals in the front end and contain an ascent (we alvx leave a 1 and a 2). To see that the mapping is injective, let 7 be such an (n d + 1)permutation. We will construct the permutation of length n and depth d of which 7 is the prototype. Let p be the lowest index in the second maximal decreasing interval in the front end of 7. Inflate 7 at p by the decreasing permutation (d 1) (d 2) ... 1, obtaining a permutation of length n 1. Finally, insert n before the (p + d l)st entry. As this reverse mapping holds for any aavoiding permutation which contains an ascent and at least two maximal decreasing intervals in its front end, we can overcount them as follows. We know that Sk(123, 3142) is F2k1, F2k1 < 3k for all k > 1, and F2 1 03, where 3 (3 + 5)/2. So, for any N, the number of roots in W of length N and depth d is less than F2(Nd+1)1 and, letting k = N d + 1 be the length of the prototype of each root, they each have levelnumber polynomial d+jl1 N k + + j 1 d1 Nk+11 (Nk+) (N k+(n N) Nk (n k). Nk We can now overcount all permutations in W by the roots of their trees and the prototypes of the roots: nl n ( k\ <, < F2k1 Y k k=3 N=k+1 nl < F2k12 nk k 3 n1 < 3k2 k k= 3 n1 < k 3 S 3).in In the end, we see that indeed the nth root approaches 3 = L(123, 3142). D As was mentioned prior to this proposition, the pattern 3142 makes our work easier because the front end of the pattern itself is the concatenation two maximal decreasing intervals. The above proof provided a warmup for the main result of this section. The following lemma allows us to work with longer front ends. Lemma 9. Let a E Sm be a pattern with a decreasing back end. Let p E S, be a root of W. Then, the front end of p has at most m maximal decreasing intervals. Proof. We first note that if {i, i + 1,..., i + r} is a maximal decreasing interval in the front end of p, with i > 2, then by maximality and the fact that the front end is decreasing, pi + 1 is in the back end of p. Assume the front end of p has at least m + 1 maximal decreasing intervals. We show that this implies the parent of p in U contains a, a contradiction. Label the rightmost m + 1 maximal decreasing intervals in the front end of p by II, 2,... Im in order of their greatest entries. So, the front end of p is the concatenation pl I.m+1mlm1 _l I1 or m1+11, II if p C ,m+1. For each 1 < i < m, if the entry i is in the front end of a, we choose an entry in Ii to be our i. If the entry i is in the back end of a, we choose an entry in the back end of p which is greater than every entry of Ii and less than every entry of Iji+ (if i < m) to be our i. Doing so, we achieve our 6. O Example 26. If a = 231, and p = 97642531, then our four labeled maximal decreasing intervals are II = 2, 12 = 4, = 76, and 14 = 9. Thus the front end of p is 14131211. We construct our 6. As the entry 3 is greater than every entry of 11 and less than every entry of 12, we choose 3 as our 1. Next we choose 4 as our 2. F.:,iull as 5 is greater than every entry of 13 and less than every of 14, we choose 5 as our 3, and 453 indeed reduces to a= 231. We now answer M. Atkinson's question in the affirmative for patterns a which avoid 123 and contain the entry 1 in the front end, noting that this restriction implies the back end is decreasing. Any such pattern is not decomposable, so we do know that the limit L exists and the sequence s,(123, o)1/"' is nondecreasing. Theorem 27. Let a E Sm be a pattern with the entry 1 in its front end. Then, L(123, a') = L(123, a). Proof. We generalize the proof of Proposition 25. Let do denote the depth of a. Note that any root in W contains in its front end a (< li' of the front end of a and therefore has depth which is at least do. Let d > do. We begin with a pl, l. ;, l'.' 7, a permutation in Sk(123, a) for some k, and insert d do entries in 7 to construct a root in W with depth d. Let 7 e Sk(123, a) have a child p(0) in U which is a root in W and has the same depth as a, do. So, there exists an index t such that inserting k + 1 at t in 7 gives us p(o), and 7 contains a copy of \ m. If d = do, we take p(0) to be our root in W. Otherwise, we build p(ddo) inductively. Let r be the index of the rightmost 1 of p(o), and note that by construction t is the index of the threshold of p(o), and 7172 7 t1 pl P2 pt1 Besides the threshold, we will insert each entry using one of two methods. One method is to insert at a maximal decreasing interval, i.e. choose an index in a maximal decreasing interval in the front end of p() and inflate the permutation at that index by the permutation 21. Example 28. Let p() = 6431752. We i,,~r insert an entry by :ifl,i,; p(0) at index 2 to obtain 75431862, where the '1,..1fl.,.. 5 is the inserted entry. The second method may be applied for any pair of entries in the back end of a which (0) have consecutive indices and consecutive values. Suppose pp) and pn+ correspond to such a pair in the (".li, of a in p(0) and are the least such pair. By being the least such pair we mean that if p = q, and the greatest entry in the front end of a which is less than aq is ai, then the maximal decreasing interval containing ai contains the greatest entry in the front end of p(0) which is less than dq. As the depth of p(O) is the depth of u, there is no entry in the front end of p(O) whose value is between those of p() and p1, so we have p) be the greatest entry in the front end of p(o) which is less than p +1. Then, our second method of insertion will be to insert the entry p(0) at the index i and increase by one every entry which is at least as large as p(). Example 29. Let a = 41532 and p() = 521643. We i,,mii insert an entry in the front end of po) to obtain 6421753, where the boldfaced 4 is the inserted entry. Of course, in p(O) the pair pp and (0) have consecutive values, and this may not be the case in p(j) for greater values of j, as we may have already inserted at an index q using the second method and corresponding to the pair pp(o) and p 1. In this case, we inflate p() at the index q by the permutation 21. By Lemma 9, we know that the front end of p(O) has at most m maximal decreasing intervals. In fact, there are exactly M maximal decreasing intervals in the subsequence yr+1 .. 7t1 for some M < m. Suppose there are P pairs of entries for which we may apply our second method above, so altogether we have M + P choices for each insertion, and we order them as C1, C2,..., CM+P. So, let {xi,... Xddo} be a multiset of [M + P] with xi > x 2> > Xddo. For each 1 < j < d do, inductively define p(J) to be the insertion into p(1) determined by the choice Cx,. So, we obtain p(ddo) by inserting entries into some of the maximal decreasing intervals of p(o). As these entries were inserted between the rightmost o1 and the threshold, we have that p(J) has depth do + j for each j, and each one is a root in W. Altogether, we obtain our root of depth d, p = pdd) Now, we have our root and we can overcount the (123, c')avoiding permutations of length n. We are in the same situation as in the proof of Proposition 25 except that we've made (M+P+ddo1) choices of how to insert the entries into our prototype to obtain our root. Of course, M + P < m and d do < n, so this is bounded above by (mn ). Note that since a contains the pattern 132, L(123, u) > L(123, 132) = 2. Altogether, we have <' < m < (mn + m < (m + m )nl n Sk(123, a) N k 3 N= k+ nl r) sk(123, a)2k k 3 n1 k 3 n (n 3)s (123, e). Taking the limit, lim W < nnoo n*oo nlim m +n ( 3)1l/s,(123, a)1/ noo rm lim sT(123, ))1" n oo L(123, 7). Experimental results support the following conjecture of Atkinson. Conjecture 30. Let a be in,; pattern. Then, L(123, 7') = L(123, 7). CHAPTER 3 PATTERN PACKING 3.1 General Pattern Packing In the previous section we posed anew some classic questions in the field of pattern avoidance with the restriction that all permutations considered are given to avoid some increasing pattern q. Along the same line, we can consider the question of pattern packing on the set of permutations which avoid q. We define the function pat on all permutations by letting pat(r) denote the number of distinct patterns contained in the permutation 7. (Note that we i distinct to avoid confusion with the number of unique patterns, an entirely distinct (and unique) area of research.) Example 31. We have pat(1432) = 7, as 1432 contains the patterns 0, 1, 12, 21,132, 321, 1432 and no others. An easy result on the function pat follows. Proposition 32. Let q be the monotone increasing permutation 12... r. Then, pat(q)= r +1. Proof. As q is i i i ii:. so is every subsequence, so the only patterns we find in q are the increasing patterns of length 0, 1,..., r. E In the previous chapter we discussed equivalence classes for pattern avoidance. If the permutation 7 contains the pattern a, then 71 contains a1. Similarly for rR and 7'. From this we immediately achieve for all permutations 7 pat(Tr) = pat(uT) = pat( R') =pat( T). So, 7, 7r, 7 and 7r are in the same equivalence class with respect to the function pat. Next, we define the function maxpat over all nonnegative integers by maxpat(n) = max {pat(}) 1 E S, }. An obvious upper bound to maxpat(n) is the total number of subsequences of a permutation, 2". We are therefore interested in the i~mptotics of maxpat(n), or the growth rate of maxpat(n)1/" as n i oo. This question was originally posed by Herb Wilf at the Conference on Permutation Patterns, Dunedin, Otago, New Zealand in 2003. He presented the following class of permutations which give a lower bound for maxpat(n). To construct Wilf's class, begin with the empty permutation, which we label W(o), and the permutation 1, which we label W(1). Inductively assume that we have constructed W("1). If n is even, we postpend W("1) with n, i.e. W(t) = W(l)n. If n is odd, we increase each entry by one and postpend W("1) with 1. The first few permutations of our class are evidently W(1)= 1, W(2) 12, W(3) 231, W(4) 2314, W(5) = 34251, W(6) 342516. See Figure 31 for an example. The claim is that for all n > 1, pat(W(")) > F,, where F, is the 'th Fibonacci number, as defined in the previous section. Proposition 33. (Wilf) For all n > 0, we have the lower bound maxpat(n) > F,. Furthermore, we have the bounds 1+ / S < liminfmaxpat(n)1 / < limsupmaxpat(n)1 < 2. 2 noo 0o 3W 2 Figure 31. The permutation W(6) = 342516 Proof. We have pat(1) = 1 = Fi and pat(2) > 1 = F2, so the first statement follows immediately for n = 1 and n = 2. Now, for each n > 3, the number of distinct patterns we obtain from subsequences of length at least 2 which end with W,) is equal to the number of distinct patterns in W("1), as any such pair of subsequences are of the form aWli ) and 31Wp ,) where a and 3 are necessarily distinct subsequences of W("1). Therefore, the number of distinct patterns which we find in subsequences ending with the last entry of W(") is pat(W("l)), which is at least F,_1 by induction. Likewise, the number of distinct patterns of length at least 2 in W(") which end with W,1$ is at least pat(W(2)) > Fn2 by similar arguments. It remains to show that these two sets are disjoint. We note that, for n even (odd), each subsequence which ends at W,) ends at its greatest (resp. least) entry, whereas each subsequence which ends at W ) ends at its least (resp. greatest) entry. These are mutually exclusive conditions by our assumption that each pattern has length at least 2. We see that indeed we have a lower bound which satisfies the Fibonacci recurrence. From our previous work, we know that F, = v .5 The upper bound on the limit supremum is trivial for maxpat, which is necessarily bounded above by the number of subsequences of an npermutation, 2". O Based on empirical evidence, (see sequence A088532 in the OnLine Encyclopedia of Integer Sequences [32]), it seemed this number may 1Ilio .... !i" the trivial upper bound of 2" (the number of all subsets of an npermutation). In [15] this author constructed a class of permutations over which it was shown that lim maxpat(n)1" = 2. n*oo However, while confirming our suspicion that the nth root approaches 2, this result still left open the possibility that m axpn) 0 as n  oo, i.e. the possibility that maxpat(n) = o(2"). Recently Miller [25] and Albert et al. [14] independently proved with a refinement of the original class from [15] and more delicate counting techniques that indeed lim =axpat() 1, i.e. maxpat(n) ~ 2 . noo 2n In particular, in [25], Miller showed the wonderfully exact bounds 2" (r22 ) < maxpat(n) < 2 (n2" . Definition 17. We generalize the function maxpat by 1. /I,.':.'/ for ,:1,; pattern a, the function 1,,,., ,,,I(n) = max {pat(r) : E ST,(a)}. Our task is to construct a class of permutations which maximize as much as possible the function ,,,., .,/, (n) for increasing patterns q. First, consider what sort of upper bounds we can find on i,,,,1 i i I (n) for a few simple patterns a. Proposition 34. For q = 123, we have the upper bound 11'.uI(n) < t(n) in (Cm, ( ,)) 0n o where cm is the mth Catalan number and counts 123avoiding permutations of length m. Furthermore, we have the limit lim t(n)/" = 2. n*oo Proof. For each 0 < m < n, the number of mpatterns in any npermutation 7 is at most the number of subsequences of 7 of length m, i.e. the number of msubsets of [n], ('). Similarly, there are at most Cm possible 123avoiding mpatterns. As 7 itself is 123avoiding, we know that any pattern contained in 7 is also 123avoiding. So, for each m, the number of mpatterns satisfies both these bounds, and the total number of patterns is at most the sum of the minima over all m. The significance of the above limit is that one may hope to pack just as many patterns into 123avoiding permutations as in the general case. Furthermore, for any increasing pattern q = 12 ... r with r > 3, the same upper bound holds, albeit trivial. 3.2 Pattern Packing in 123avoiding Permutations Proposition 34 gives us hope that, even with the 123avoiding restriction, we may pack as many patterns as we would like in a permutation of sufficient length. Experimentation supports such a conjecture. Here we give constructions for qavoiding permutations with in ,iy" patterns for increasing patterns q. Our constructions are modeled on those of [15],[16], [14], and [25] and meet or surpass the original lower bound given for the general case by Wilf, (1 ). We note that Wilf's original construction is 132avoiding, so we already have it established that maxpatl32(n) > ( I )n. We begin with a family of 123avoiding permutations which show that the limit infimum of maxpat123(n)1/" is at least 5 a n  Oc. We define our permutations P(") inductively. Let P() = 1 and p(2) 12. For each odd n > 3, let P(t) nP(" ), i.e. p("1) prepended with n. For each even n > 4, let P(n) be p("1) with n inserted immediately after 1. So, we have P(1) 1, p(2) 12, p(3) 312, p(4) 3142, p(5) 53142, p(6) 531642. So, each P(") consists of two decreasing subsequences, and thus avoids 123. The next proposition gives a lower bound on pat(n) with a proof similar to the proof given for Proposition 33. Proposition 35. For all n > 1, pat(P(')) > F,. Furthermore, we have the bounds 1 + < liminfmaxpat123(n)1/' < limsupmaxpati23(n)1/' < 2. 2 n__oo 0__o Proof. Our induction hypothesis will be the stronger statement that P(") contains at least F,1 patterns cor;lr.',,.:'. the entry 1. We have that P(1) contains the pattern 1, and p(2) contains the pattern 12, so the statement of the proposition holds for n = 1 and n = 2. Assume the statement holds up to and including n > 2. Then, P('+l) contains a set A of at least pat(P(")) > F,_ patterns which contain the entry (n + 1) and the entry 1. p(n+1) also contains a set B of at least pat(P("1) > F_2 patterns which do not contain the entry (n + 1) but do contain the entries n and 1. We claim that the sets A and B are disjoint. Indeed, if n + 1 is even (odd), then for each pattern in A the largest entry occurs after (resp. before) the entry 1, while for each pattern in B the largest entry occurs before (resp. after) the entry 1. Therefore, we have p(++1) > F,_1 + Fn2 = F.. The bounds follow immediately. O 3.3 Pattern Packing in qavoiding Permutations Next we extend our construction to longer increasing patterns q. Suppose q = 12 ... (r + 1) is the increasing pattern of length r + 1. We define the permutations Q(") inductively. Set Q () 1, Q(2) = 12, ..., Q(r) = 12.. r. For n > r, our permutation Q(") consists of r maximal decreasing subsequences. If n + 1 = p (mod r) with 1 < p < r, we construct Q((+I) by inserting n + 1 at the beginning of the pth maximal decreasing subsequence. Example 36. If q = 12345, i.e. r = 4, we have Q(5) = 51234, Q(6) = 516234, Q(7) = 5162734, Q() = 51627384, Q(9) = 951627384. Definition 18. For k > 2, the Fibonacci kstep numbers, denoted FT ) for n > 0, are , 1 [ .1 by Fk)= 0 V n < 0, F(k) .. = (k) = I k 1 and, for all n > k + 1, F (k) (k)+ F(k) + ... F(k) ,n nk nk+1 n1 The classical Fibonacci numbers are the special case k = 2. It is well known that the limit lim (Fk)) 1 n OC00 is the unique real root greater than 1 of the equation Xk x x2 + x 1. This root, which we denote by ak, is called the kanacci constant. We have that ak increases with k and lim k = 2. koo Proposition 37. Let q be the increasing pattern of length r + 1. Then, pat(Q(")) > F r) nr+2" Furthermore, we have the bounds or < liminfn,,,l .,i, (n)1/ < limsup i,., i.,l (n)1/" < 2. nfoo l*l00 Proof. For each 1 < n < r, we only count the pattern Q(") =1 ... n itself and have pat(Q()) > 1 > F(r) nr+2" For n > r + 1, we count the patterns of subsequences which contain the entries 1, 2,..., r, i.e. subsequences which contain the lowest entry of each maximal decreasing subsequence in Q("). Fix n > r + 1 and assume by induction that, for all N < n, the number of such patterns is at least F(r) r2 For 0 < i < r 1, let Ai be the set of subsequences of Q(n) which contain the entries 1, 2,..., r and whose greatest entry is n i, and let Bi be the set of patterns of subsequences in Ai. We claim that the Bi's are pairwise dli ~iil Indeed, for i / j, suppose ai E Bi and aj E Bj, where n i p (mod r) and n j q (mod r), 0 < p, q < r 1. Then, the greatest entry of ai occurs after p ascents, and the greatest entry of aj occurs after q ascents. Also, for each i there is a bijection between the patterns in Bi and the patterns counted for Q("')l. So, if we count all patterns in the union of the Bi's, we have pat(Q )) > pat(Q ")) + +pat(Q(nr)). As we have the recursion for the Fibonacci rstep numbers, the induction follows. O It may be noted that these classes of permutations resemble the classes used in [15], [16], [14], and [25], except that whereas in each of these papers the construction consisted of k rows of k entries each, or a stripped down version of that, the permutations we are using here have restricted row lengths (number of maximal decreasing subsequences) due to the qavoiding restriction. It was noted that ak  oo. From this fact we see that as we let r oo, our constructions for increasing patterns of length r + 1 ipl ... !i" the constructions for the general case and the lim infs approach 2. 3.4 Packing Density and Further Directions Pattern packing is in a weak sense a dual concept to that of pattern avoidance. We i the duality is only weak as pattern avoidance, when compared to the total number of permutations, can be understood as the p' '.1,1, l.:1.:1, that a permutation will have the property that it avoids the given pattern, whereas the question of pattern packing is extremal, asking what is the maximum of a certain statistic, the number of patterns contained in a permutation, over all permutations. Perhaps the proper dual concept to pattern avoidance question is a question also posed by Herb Wilf, at the 1992 SIAM Conference on Discrete Mathematics. For a given pattern a, how many copies of a can a permutation contain? The answer depends on the packing 1. ,, .:/1 of the pattern a. Definition 19. The packing density p(a) of a pattern a E S, is 1. I;,' ., by g ((a, n) p(a) lim g n noo (7) where g(a, n) is the maximum number of copies of a in ,:n;, npermutation. For more on the work in this area, please see the Ph.D. dissertations of Dan Warren [38] and Alkes Price [27]. Herb Wilf has recently ,. 1i ,1 that the notion of packing density be examined in the qavoiding environment. Why do we raise the issue of these dualities? We saw in the section of pattern avoidance that in the restricted environments of 123avoiding permutations, or more generally qavoiding permutations, it is more likely that a randomly chosen permutation avoids some pattern T, or, critically, the chance of avoiding a approaches or equals the chance of avoiding a', a statement which is usually false in the general case. So, restricting to qavoiding permutations makes our life easier in that work. However, pattern packing becomes more difficult. Indeed, if we consider all approaches used to prove pattern packing or superpatterns, they take advantage of a lattice or checkerboard structure in the class of constructions to bound maxpat. If we restrict the length of the increasing sequences allowed, we lose this structure. So, the metaquestion is whether we lose our .,1ill ic limits or simply need more delicate techniques to see them. CHAPTER 4 ASYMPTOTIC NORMALITY AND UNIFORMITY 4.1 Probability Theory In this paper we only deal with discrete probabilities and their limits. For the sake of simplicity we avoid the measure theoretic foundations of probability theory and define a smaller class of random variables, giving associated properties which we will need in this chapter. Definition 20. A random variable X is a function from the unit interval [0, 1] to the real line R. For each r E R, we denote by P(X = r) the Lebesgue measure of the set {w E [0,1] : X(u) = r} and call this the probability that X = r. The set of values {X(w) : E [0, 1]} we call the range of X. A random variable X is called discrete if its ri,.,,. is countable or if X has the weaker condition that there is a countable subset B of the reals with P(X e B) = 1. F.:,,,ill; a 01 random variable or indicator random variable is a random variable whose ru,,I., is the set {0, 1}. Definition 21. Let X be a random variable with finite r,.ig R. The mean or expected value of X is I,. ./ by E(X) = r. r 'ER The variance of X is E((X E(X))2), which, by the i. ,,,, i of expectation, is also given by Var X = E(X2) (E(X))2. The square root of the variance is the standard deviation. We note that for a 01 random variable I, 12 = I, a fact that eases computation of mean and variance for such variables. Indeed, given a 01 random variable I with P(I = 1) = p, we have E(I) p and Var I = p p2. As we will only be dealing (at least before taking limits) with random variables with finite range, we will not develop all the measure theory behind these definitions. For this, the reader is referred to the texts by Halmos [18], Taylor [35], and Chiing [13]. For a treatment of discrete probabilities, the reader is referred to the texts by Alon and Spencer [1] and CI', 1 .,, i1,. [11]. 4.2 Triangular Arrays Definition 22. Given a random variable Y whose rr,,,'. R is a finite set of integers, 1, i,'.: its probability generating polynomial to be py(x) = P(Y r)xr. rER By a 1, .:.,,i,,,l,;r i ,,,r, of nonnegative real numbers (a,,k) we mean a sequence of numbers which are indexed by n = 0,1, 2,..., and k = 0,1,... m m(n) for some function m defined on N, so for each fixed n there is a finite number of terms an,k. The term triangular array comes to us from the probabilists. When the sequence consists of nonnegative integers, we use the term combinatorial distribution found in [19]. In combinatorial applications an,k counts objects of size n with some statistic k. For example, in the next chapter we will study the numbers S,,k, counting set partitions of an n element set into k blocks. Given such a sequence an,k, we set Sn = an,i +an,2 + .. +an,m for each n and construct a new sequence bn,k = a interpreting bn,k as the 1 /d, I.:./.; that a randomly selected nobject has statistic k. Example 38. A fundamental combinatorial distribution is that of the binomial coefficients (~). It is well known that for each n we have the sum ( ) + () + + () = 2'. The generating i .l;,,'... ;,.:'l (x + 1) T () k k0 can be interpreted as the 1"' al.,,d.:.I/;, generating / .l;,,,;...;,.: for a random variable X, which is the sum of n coin flips, i.e. X, is the sum of n independent 01 random variables Y,,k, k 1, 2,... ,n, with 1 P(Y.,k 1) 2 for all n and k. The term triangular array can be seen in the wellknown Pascal's triangle listing the binomial coefficients: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 We will only deal here with triangular arrays whose generating polynomials have real, nonpositive roots only. The motivation for this will become clear when it is seen what strong statements can be made concerning such sequences and the clever, powerful techniques used to deal with them. 4.3 Asymptotic Normality For a sequence of random variables X, such that, for each n, X, is the sum of m = m(n) independent, identical random variables, with m  oo, the associated distribution approaches the normal distribution, most easily understood as the limit (in distribution) of the binomial coefficients. M ii;: significant results in probability theory are of the form of Central Limit Theorems proving sufficient conditions for convergence to the normal distribution. Definition 23. For a random variable X, let X denote the normalized random variable XE(X) For a sequence of random variables X,, we write X,  N(O, 1) to mean that X, \Var(X) converges in distribution to the standard normal variable. F.:,ill; we .'; that a sequence of random variables X,, is .,i','1iii 1 ically normal if X,  N(0, 1). We have the following theorem, appearing as Theorem 6.7.11 of [35]. Theorem 39 (Central Limit Theorem for Triangular Arrays). Given a t, :.,i!,sl., r r ,,1; of random variables X,i~,..., Xn,m with respective variances on,,..., X,,m, let a,2 o + + oT, and Xn X,, + + X,,m. Then, the distribution converges ;i, l./; to the unit normal distribution if the Lindeberg condition is .,i/.:f7 , Ve > 0 z x2Qk (dx) 0 as n oo, k=1 Xn >,r7n where, for each k, Qn,k is the cumulative distribution function of Xn,k. 4.4 Asymptotic Uniformity Another ..ii !l l ic property of some combinatorial distributions which provides us with both qualitative and quantitative understanding of their ..imptotic nature is the concept of . mptotic uniformity. We build up to this with B6na's definition of balance ([7], [9]). Definition 24. For each positive integer q, we .r;, that the combinatorial distribution an,k is qbalanced if for each 0 < r < q, we have the limit lim Zkr an,k 1 noo k an,k q where all congruences are modulo q. So, if aT,k counts nobjects with statistic k, qbalance implies that there is ;. ,.,lli equal probability of a randomly selected nobject to have kstatistic in any one of the residue classes of q. Example 40. We give a quick proof that the binomial coefficients are 2balanced. Evalu ating their generating r. I;,,/'; ... ,. (x + 1)" at x = 1 gives us the difference of binomial coefficients for even and odd values of k: (1+1)" + (C)kkC k even k odd As (1 + 1)" = 0" = 0, we have that for all n > 1 the set of evens and the set of odds each account for ir. I.I. half of the total. Of course, most cases are not so clean as this and we must examine the limits. However, we will see that evaluating the generating polynomial at the qth root of unity as above is integral to our work. Definition 25. We .,r; the combinatorial distribution a,,k is imptotically balanced if it is qbalanced for all q E P. The following lemma and proof are based on those of B6na in [7] and [9] Lemma 10. Let q E P. Suppose the combinatorial distribution an,k has generating j,, ,J,, ,' !,,.,1d pn(x) /, ,'. ,/ by Pn(x) Z an,,x. k0 Then the sequence na,k is qbalanced if for all 0 < t < q lim P. = 0, (41) noo p,(1) where ( = exp a primitive qth root of ii.l Proof. Let 0 < r < q 1. For now we assume q divides m. Consider the sum q1 s,,,(() := Yp((t)"r (42) t=0 S( ak) t (43) t=0 k=1 m q1 Ya. ^ Y (44) k=1 t=0 m q1 Z ,, (Y) (45) = 1an,k 5 Enclosed in parentheses in (45) is a geometric sum, and we take advantage of the property that the (k r)roots of unity add up to 1 to see that 1 O if (k, . t=0if kr Therefore, we have annihilated all terms in (45) for which q does not divide k r, and (42) reduces to m/q S,r(() q an,jqr, or (46) j=1 s,,(C) Zj1 an,jq+r qp ) P (47) We recognize the right hand side of (47) as the function from the definition of qbalance. So, qbalance is equivalent to this expression converging to for all r, and assuming (41) we can drop our assumption that q divides m, and we have the limit sin zr ( 1~)lit 1 ) tr nzo qp. (t) q n^00 ^ P(1 q = n0oo p(l) q1 1 tr lir aP ) q toP = (1 + 0 + + 0) q 1 q 4.5 Generating Polynomials with Real, NonPositive Roots Throughout this section we will assume that a,n,k is a triangular array of nonnegative reals whose generating polynomials have real, nonpositive roots and set s, = a,,o + a,1i + .. + a,m,. Also, X,z will refer to the random variable defined by P(X, = k)= a1k V k > 0, Sn with probability generating polynomial Pn(X) = ank k=0 Sn The notion of . I i llic normality or normal convergence is that the normalized sequence _ converges in distribution to the normal distribution. For each n we will find a set of 01 random variables whose sum is X, and use these to prove ..i!,i,;,il'' uniformity. We develop here the method attributed to Harper [19] and also used in [30]. By our assumption on the roots of our generating polynomials, we can factor the probability generating polynomial pn(x) as Pn (X) a H (x + Ank) an,,,m x + A,k Sl with 0 < A1 < A2 < .. < Am c R. For each n, define the mutually independent 01 random variables I,ji,..., In,m, by P(In,k 0) = A,k and 1 + An,fk 1 P(In,k 1) t +,,k 1 + An,k From this the mean and variance of each ,n,k follow immediately. 1 E(In,k) 1 + An,k Var Ink E(Ik) (E(In,k))2 1 1 1 + An,k (1 + An,k)2 Arn,k (1 + An,k)2 We have the respective probability generating polynomials gn,i(x),..., gn,m(x) given by gn,k(x) P(I,k = 0) + P(In,k = )x n ,k+ X. 1 + An,k 1+ An,k Now we can rewrite each probability generating polynomial pn(x) in terms of the probability generating polynomials gn,k(x). ,() = an,m H x + A n,k 1 an,mn H gn,k(x) l We have thus expressed X, as the sum of the 01 random variables In,1... ,, n,m. We will be interested in the variance of X,, which we denote o2. We are now in position to state the following theorem of Bender [3] which follows from Theorem 39: Theorem 41 (Bender, 73). Let the combinatorial distribution aT,k be as above. Then, the sequence of random variables X, is r ;,/*i1.I..'.ll//; normal if and only if an 00c as n oo. We have the following computation, where 1 < j, k < m for all j, k. 2 = E(XX) (E(X))2 E((Z I. k) E(Z (I,k))2 k k SE(Ink) + 2 E(IjIk) 2 2E(I )) EI I k j Z= E(I,k) Z(E(,))2 k k S Var ,k k n,k (1 + Af,k)2* With Theorem 41 and keeping the above notation, we have shown the following. Lemma 11. With real, nonpositive zeros, r/,,,, l...:'.' non,,'l.Ui is equivalent to the divergence im = c0. noo (1 + A,,k)2 k 0 4.6 Asymptotic Normality Implies Asymptotic Uniformity With the results of the above two sections, in particular Lemmas 10 and 11, we can state the main result of this chapter. Theorem 42. Let the combinatorial distribution a,,k have generating j'. *'i;,iil,.:.l* with real, nonpositive roots only. Then, i /,, i/,/':.: noi,,n' l.:,i implies ' ;,, /,/l/:. unifo ,iiil.' Proof. We retain the notation of Section 4.5. Assuming .ii~!,i1 normality, Lemma 11 tells us S(1+ oE)2 as n oo. (48) Y 1(1 + A,k )2 By Lemma 10, it suffices to show that this divergence implies ) 0 (4 9) as n  oo for all q and all 0 < r < q, where ( =exp If (r = 1 is a root of p,(x), then q p,((r) 0, so we can assume here that this is not the case in order to avoid pathologies. Taking logarithms, (49) is equivalent to the condition S(log (1 + A,k) log + n,k) 00. (410) k=1 By Lemma 12 below, there exists a constant a > 0 which depends only on (r such that for all n and k we have og(+ ) log (1 + ,),k > a (Ank 2' (1+ An,k) Thus, 5 (log (1 + A,k) log 1r + A.) > a ,' k)k)> (1+A,k)2' k=1 k 1 and (48) implies (410) which is itself equivalent to (49). E Lemma 12. Let z = e0 for some 0 < 0 < 27r. Then, there exists a constant a = a(z) such that for all x > 0, f(x) log(1 + x) loglz + x > a 2 (x + 1)2 Proof. Write f(x) as f(x) +1 dt fx+z dt +1 dt t J t JIx+z t The length of the domain of integration is the function d(x) = x + 1 Ix + z and we have d(x) f(x) > x+l So, we are interested in a constant a such that d(x) > a x+l for all x > 0. We first show that there is a constant a1 such that d(x) > a1 for all x > 1. We have the formula Ix + z = (x2 + 2x cos 0 + 1)1/2. Therefore, for all x > 1, d'(x) ( x + 1 (x2 + 2xcos0 + 1)1/2) dx 2x + 2 cos 0 =1 2(2 + 2x cos 0 + 1)1/2 x X2 + 2X Ccos 0 + cos2 0)1/2 1 (x2 + 2xcos 0 + 1 > 0. Therefore, d(x) is monotonically increasing for positive x, and for all x > 1 it is greater than d(1), which we denote by al. Next we show there exist constants a2 and 6 such that d(x) > a2 for all 0 < x < S. We have d'(0) = 1 cos 0, so there exists S > 0 such that for all 0 < x < 6 1 cos 0 d(x) > x 2 Moreover, as 1 < 1, for all 0 < x < 6, X+1 1 cos 0 x d(x) > 2 x+l' so we let a2 = 1Co. Now we have constants for x E [0, j) U [1, oc). As [6, 1] is a compact set, there exists a constant a3 such that d(x) > a3  for all x C [6, 1]. Finally, set = min {ai, a2, a3}. Example 43. The signless Stirling numbers of the first kind c,,k count npermutations with k 1. For each n, we have the generating pr ,;''."".:,/ n Cn,() = nkXk (. + 1) ... + X n 1) k= 1 So, for each n, C,(x) has the roots 0,, 2,. n + We have the following limit. lim j noo (1 +j)2 + j1 j>_ j=1j 1 1 2 j j>l1 S00 Therefore, by Lemma 11 and Theorem 42, the signless Stirling numbers of the first kind are i,i,;;;,/I. // ll'/ normal and thus ;,,.'.rm. CHAPTER 5 ON THE ROOTS OF THE BELL POLYNOMIALS 5.1 Stirling Numbers of the Second Kind This chapter is concerned with the generating polynomials for a classic combinatorial distribution, the Stirling numbers of the second kind. Recalling the notation [n] {1, 2,..., n}, we begin with the definition of a fundamental combinatorial object. Definition 26. For n E P, a set partition of [n], or an nset partition, is a set of disjoint, nor., ijplI subsets of [n] called blocks whose union is [n]. Example 44. A set partition of [6] into three blocks is {{1, 4, }, {2}, {3, 5, 6}}. The natural question posed by an enumerologist is how many such objects are there. In particular, given nonnegative integers n and k, how many set partitions are there of [n] with k blocks? The answer is the Stirling number of the second kind S,n,k. By convention we set So,o = 1 and Sn,k = 0 for all k > n > 0. The results of this chapter rest primarily on the following recurrence relation. Lemma 13. For all n > 0 and 1 < k < n + 1, Sn+1,k = Sn,k1 + kS,,k. (5 1) Proof. Let r be a set partition counted by Sn+1,k. If {n + 1} is a block in r, we may remove it, leaving a set partition of [n] with k 1 blocks, an object counted by Sn,k1. On the other hand, if there are other elements in that block of r which contains n + 1, we simply remove n + 1 from that block and still have k blocks of n elements, an object counted by Sn,k. This is clearly an injection and thus shows that the left hand side above is less than or equal to the right hand side. Now, let 7 be a set partition counted by the right hand side above. If 7 has k 1 blocks, we append a block containing only n+ 1. Else, we choose one of the k blocks in 7 into which we insert n + 1. Again we have an injection, and the equality holds. O Lemma 14. For all n > 1 and 1 < k < n, we have the exact formula Sk i) (k (52) i=0 Furthermore, we have the i;;l,,/i.i/'.: formulae Sn,k k 0((k 1)") (53) k" and Sn,k k. (54) Proof. A set partition of an nset is constructed by placing the elements into k disjoint sets. Consider instead the case where there are k labeled boxes and n labeled balls which we place in the boxes such that no box is empty. Call this an (n, k)placement and let an,k be the number of (n, k)placements. Then, an,k = k! Sn,k, as there are Sn,k Av, ~ to place the balls into unlabeled boxes, and k! vi, to label these boxes. So, it suffices to show that the summation in (52) is an,k. We apply the method of InclusionExclusion. There are k" vi to distribute the balls to the boxes, as we have k choices of what to do with each of n balls. However, we may have overcounted distributions in which a box was left empty, so we subtract the (k) (k 1)" VI to choose a box to be empty and distribute the balls to the remaining k 1 boxes. Now, we have overcounted the distributions in which there are two empty boxes. Indeed, for some 1 < p < q < k, consider the set of distributions which leave boxes p and q empty. We counted such a distribution once in the first term, but subtracted twice in the second term, once for leaving box p empty and once for leaving box q empty. So, we need to add 1 for each such pair of boxes and each way to leave that pair empty and distribute the balls to the other boxes, () (k 2)". This process continues until the number i of empty boxes reaches zero, and we have the desired sum. For the .iiii!ld lI ics, we write (52) as k k(k t) k(k 1)(k 2) Sk k! k! + 2k! As n + oc, the dominant term is . Furthermore, the dominant term of S,,k is k a constant multiple of (k 1)", demonstrating (5 3). Finally, as n o0, S. k O((k 1)") k_ k1 '. k! k! Lemma 15. For all n > 1 and 1 < k < n, we have the bounds kk < S,k < (55) and k [k/2] 1 k k ( 2i 1)" < S,k < . (56) k! 2i1 k! i 0 Proof. The upper bound, Sn,k < L, follows trivially from (52). By the recurrence (51), for all n > k > 1, we have Sn,k > kSnl,k > k2S_2,k > kkSk,k = kk as Sk,k = 1 for all k, completing (55). The lower bound of (56) is the difference of the dominant term and all oddindexed, hence negative, terms of (52). D 5.2 Bell Polynomials For all n > 1, let B,(x) be the generating polynomial for the Stirling numbers of the second kind: B, (x) Z= S,kXk. k= 1 These are known as the Bell p y.i/;/n'; ,,i,.:.l Example 45. The first few Bell i .1/;/,;,. iii,,.:, are as follows. Bo(xr)= 1, Bl(x) = x, B2(x) =x + 2, B3(x) = x + 32 + x3, B4(x) x+7x2+6 3 +x4. Lemma 16. The following recurrence relation on the Bell './;/ ,:'i "./l/ holds for all n > 0. Bn+(x) = xB,(x) + xBj(x). (57) Proof. Applying (51) to each term of B,+l(x), we have Brz+l () Sn+l,kXk = ((Sn,k1 + kSf,k )Xk S= .Sr,kk + kSfl,kXk = X Sn,k1xkI 1 + x kSn,kXk1 xB,(x) + xB,(x). We have the following wellknown result on the roots of the Bell polynomials. Lemma 17. For all n > 1, B(x) has n distinct, real, nonpositive roots, including zero. Proof. (Wilf, [40]) The statement holds for Bi(x). Assume by induction the statement holds for B,(x). Multiplying each term in (57) by ex, exB, 1(x) = x(exB,(x))'. By Rolle's theorem, B,(x) has n 1 roots, one between each consecutive pair of roots of B,(x). Multiplication by x guarantees a root at zero. Since exB,(x) approaches zero as x i oo, its derivative has one more root to the left of the leftmost root of B,(x), accounting for all n + 1 roots of B,+i(x). O Thus we can factor each Bell polynomial as B,(x) = x(x + A,,2) (x + n,n), (58) where 0 = An,1 < An,2 < < An,, and n > 1. In this chapter we explore the roots An,k, finding bounds and limits for each fixed k. First, we have a theorem on the Stirling numbers which builds on the results of the previous chapter but was held aside for the foregoing definitions and discussion. Theorem 46. The Stirling numbers of the second kind Sn,k are ir;,,,,' I.:/ ,ill/// u:./f.rm. Proof. Harper [19] proved that the Stirling numbers of the second kind are ,mptotically normal. The above discussion shows that their generating polynomials have real, nonpositive roots only, so by Theorem 42 in the previous chapter, we achieve the desired result. O Now, let us consider the roots more closely. Example 47. We solve for the roots of the first few Bell pi..1 ;,,;.. i,,.l BI(x) = x, B2(x) x(x + ), 3+ 3 B3(x) x(x + )(X + 2 2 2 For higher n, the il'. 1.,,, becomes 'n,,. ". ;, as can be ,' l./7:.; seen in the other Bell 1 1;1,,'.i',,. :,1l given in Example 45. As our work often deals with sums of products of reciprocals of these roots, for notational convenience we will write 6n,k = for k = 2, 3,..., n, or simply 6k = if n is clear from context. Lemma 18. For all n > 1 and 2 < k < n 1, we have Sn,k I An,i ... n,i, k where the sum is taken over all (nk)tuples {il, i2, ... ink} with 2 < il < ... < ink < n. Proof. Expanding the factorization in (58), an xk term is achieved by choosing x from k of the terms and A,,i from the remaining n k terms. The sum of all such products is S,,k, the coefficient of xk in B.(x). E We note that, for all n > 1, we have S,, = 1. Lemma 18 could be rewritten to reflect this by allowing the sum over all (n n)tuples to be 1. Lemma 19. For all n > 2 and all no. m1i1i;I subsets M C {2,..., n}, __ ATn,i 6n,i, iEM iEMC where Mc = {2 < i < n i f M} is the complement of M. Proof. We first note that the product of the n 1 nonzero An,k's is S,i = 1, so >2 >2 k>2 k>2 Thus, for any M C {2,..., n}, Sn,i ,J i = 1, SO iEMC iEM ( )1 iEM iEMC j i A. j 6n,j. iEM iEMc D Lemma 20. For all n > 1 and 2 < k < n 1, we have Sn,k >1 n,i .. ,i1, (59) where the sum is taken over all (k 1)tuples {ii, i2, ..., ik} with 2 < il < ... < ik < n. Proof. Apply Lemma 19 to each (n k)tuple in Lemma 18. D For more on Stirling numbers of the second kind and Bell polynomials, particularly their properties and myriad applications, the reader is referred to the classic by Riordan [28] as well as the excellent texts by Roman [29] and Wilf [40]. 5.3 Bounds on the Roots of the Bell Polynomials Lemma 21. For all n > 2, 2"1  < 6n,2 < 2n1. n Proof. By Lemmas 14 and 20 and the nonnegativity of the 6,ji's, n K6,2< ,k Sn,2 < 2" k=2 As 6n,2 is the largest term of the sum, it is at least as large as the mean, so 6n,2 > 2 Finally, this last expression is at least 2 for all n > 2. no l n>2 Lemma 22. For all n > 3, 3n1 4 2T < 6n,3 < 2" 9n2 3n1 2 n Proof. We begin with the upper bound. By Lemmas 14 and 20, 2 25i 1  Sn,3 23n 2 3n1 0(2) < 2 2 (510) In particular, an1 6n,2n,3 < 3` 2 Applying the inequality 6n,2 > T from Lemma 21, 3n1 6n,3 <  26n,2 2" For the lower bound, inequality (55) of Lemma 15 provides C q3 Sn,3 > 3 The next inequality follows from Lemma 20, the fact that the largest of the (2 1) terms in (510) is 6n,2n,3, and our upper bound 6n,2 < 21. Finally, dividing by 2"1 3n3 2n 6n,3 > 3 3n3 6n,3 > 3"1 2" 2 3 3 n(n 1) n(n 1) 4 9n2" Theorem 48. For each k > 2, there exist j.. I;;./.',,..l,.: fk(x) and Fk(x) of degrees 2k2 and 2k2 ,, t1, / :/; such that for all n > k ( k >" 1 < (k "Fk k1 fk(u) k1 Proof. We induct on k, generalizing the proof of Lemma 22. We proved the statement for 6n,2 and 6n,3 in Lemmas 21 and 22, respectively. Let k > 4 and assume by induction that the statement holds for all 6T,j with 2 < j < k. By Lemmas 14 and 20, for all n > k, l .i 6,i41 = Sn,k < k! (511) where the sum is taken over all (k1)tuples {il,,..., ***k with 2 < ii < ... < ik1_ < n. By induction we have the following lower bound for each 2 < j < k. S j 1 f ,i2 (i )" fj(> which holds for all n > k and where fj(x) is a polynomial of degree 2j2 for each j (independent of n). Define the degree 2j1 1 polynomial F(x) = f2(x)f3(x) fk x). Then, k" n,26n,3 n,k k k" 6n,k < k!,n,2 2 6n,k1 k (k 2)T (k 3)T 2T 1 (k k 2)" 12 F (n) k (k 1)" For the lower bound, inequality (55) of Lemma 15 provides Sn,k > knk Sn,k > kn The largest of the (Q ) terms in (511) is n,26n,3 ... 6n,k. By our induction hypothesis, 6,,j obeys the following bound for each 2 < j < k. 6n,< F(n)1 for polynomials Fj(x). Altogether, ]knk 6n,26n,3 6n,k > n1 k) k nk (k 2)n (k 3)n 2n 1 1 (k 1)n (k 2)n t fk(n) where fk(x) (D)F2(x)F3(x) ... k (x). 5.4 Asymptotics of the Roots of the Bell Polynomials The next two lemmas use our bounds on 6,,2 and 6,,3 to obtain the .,imptotic equivalences 6n,2 ~ 2"1 and 6n,3 2n Lemma 23. We have the following limit. lim = 1. noo 2n1 Proof. The inequality (n,2 < 2'1 from Lemma 21 implies n, < 1 for all n > 2, and it suffices to show that the terms 6 are bounded below by a sequence converging to 1. By Lemmas 14 and 20 we know that S,,2 = 2n1 1 is the sum of I,2, 6Tn,3, and n 3 other terms, each of which is less than 6J,3, so for all n > 3 we have the following inequalities (applying the bound n ()" > 6n,3) 6n,2 + n,3 > 2n1 6n,2 + (2 1 > 2"1, (3\n 6n,2 > 2T 2n2 ( 2 6n,2 2 ( 2n >1 4 The final limit, in essence 2n2 (I)T  0 as n  oo, holds as 2n2 is a polynomial and (Q)" is a geometric progression with positive ratio less than one. Lemma 24. We have the following limit. 2" lim n,3 1 1 n>oo ,n Proof. Lemma 20 expresses Sn,3 as the sum of all products 6i6j. We can split this into two sums, one sum containing those pairs which include 6n,2 and one sum containing all other pairs, 3n1 6n,2 6n,i + 6n,i6n,j = S,3 2 (2"). (512) i>3 3 We proceed by showing that the second summation above is negligible with respect to the dominant term 3"1. Lemma 22 gives us the upper bound 6,,3 < n , and by our ordering on the 6's we have 6,,j < 6n,3 for all j > 3, implying 6n,i6n,j < (.n" 2 for each of the (",2) pairs i,j with 3 < i < j < n. Thus, n 2 n3"_1 2 9 n4 S iK <( 2 2 ( j2 (513) 3 and 3 0 i Combining (512) and (513), we have 3n1 " 2 4 Dividing by our .i~iil i..c approximation of 6 ,2 67n,2 ~ 2"1 ,i 2 O Q\ n4 i>3 By Theorem 48, (I)F 4(n) > 6,4 > n,5 > ... for the fixed polynomial of degree 3 F4(x), and so n1 6n,331 0((4) "l) Finally, we divide by 3 and take the limit as n oo. 2n t 0 8) n T\t ; (( n4) 6r,3 31 O 4 liM rn,3 2 lim 1 (( n4 lim 3 n4 n*oo 3fl 1 = 1. The following theorem generalizes these results to all k, giving the ..i~''..ill ic approximation 6n,k (k1 for each fixed k. Theorem 49. For each fixed k, we have the following limit. (k 1)" lim 6n,k ( 1.. noo k 1 Proof. We generalize the proof of the previous lemma to all k > 4. We induct on k, the cases k = 2 and k = 3 given in the previous two lemmas. The induction hypothesis gives us .i',iii ll Iic bounds which improve upon those given in the previous section. Indeed, for all e > 0 there exists N > 0 such that n > N implies that for all 2 < j < k, we have ( j 1) ,; ( 1 (1 ,1 + ). We can rewrite equation (59) of Lemma 20 as k" 6n,26n,3 J 6n,k1 > Jn,j + ^ 6n,j_, = Sn,k = k! O((k 1)"), (514) j>k where the second sum is taken over all (k 1)tuples 2 < ji < .. < jk1 < n containing at least two elements which are each at least k. We first show that this summation is negligible with respect to k". Let 2 < ji < .. < jk1 < n be such a (k 1)tuple. Then, the product of the greatest k 3 6,,j's is bounded above by the product 6,n ,2 6n,k2, and the product of the remaining two J6,j's is bounded above by 52,k* n,ji n,jk < 6n,26n,3 ... n,k26n,k6n,k 3n (k 2)T k" k"( < 2" F3 (n) Fk2(n) Fk(n)Fk(n) 2T (k 3)n (k 1) (k 1) 2 (k 2) ( (k (k1) )2 where G(x) = F3x) Fk2 x)Fk2(x). Set a = k and note that k 1 < a < k. The summation was over a subset of all (k 1)tuples, so we overcount the terms by setting H(x) = G(x) () and we can therefore rewrite equation (514) as k" 6 ,26n,3" 6n,k, 6 ,j= O(a H(n)). (515) j>k Now, by induction we have n,3" 1 (k 1)n1 2n (k 2) (k 1)"1 (k 2)! While the above was an .i!,ii1ll ic approximation, we have the useful information n,26n,3 ". 6n,k 1 o ((k l)"). As a > k 1, this implies 6n,26n,3 6.n,k1 O(c"). Dividing equation (515) by (k1), (k2)! j>k k"1 (k 1)" o (p' H( n)), (516) where k= (k 2) Now it suffices to annihilate all terms on the left hand side except 6,,k. By Theorem 48, (k Fk+(n) > 6n,k+l > . \ k I k"1 kn 1 k (k 1) k"1 (k 1)" i +l i>k+1 0 (P"(H(n) + nFk+1(n))). Taking the limit as n  oo, we achieve our desired result. O(pH((n)) REFERENCES [1] N. Alon and J. Spencer. The Probabilistic Method. WileyInterscience, New York, NY, 2000. [2] R. Arratia. On the StanleyWilf conjecture for the number of permutations avoiding a given pattern. Electronic Journal of Combinatorics, 6(1):N1, 1999. [3] E. A. Bender. Central and local limit theorems applied to .ivmptotic enumeration. Journal of Combinatorial Th(..,;, Ser. A, 15:91111, 1973. [4] M. B6na. Permutations avoiding certain patterns; the case of length 4 and generalizations. Discrete Mathematics, 175:55 67, 1997. [5] M. B6na. Combinatorics of Permutations. C'!h 11i, i1, & Hall, Boca Raton, FL, 2004. [6] M. B6na. The limit of a StanleyWilf sequence is not ahv, rational and 1 ,li 1, patterns beat monotone patterns. Journal of Combinatorial Th(..,It 110:223235, 2005. [7] M. B6na. On a balanced property of derangements. Electronic Journal of Combina torics, 13, 2006. [8] M. B6na. A Walk Th,,..;,i, Combinatorics. World Scientific, River Edge, NJ, 2006. [9] M. B6na. On a balanced property of compositions. Online Journal of A,.,rl;,.:. Combinatorics, 2, 2007. [10] M. BousquetMilou. Four classes of patternavoiding permutations under one roof: Generating trees with two labels. Electronic Journal of Combinatorics, 9(2):R19, 2003. [11] C. A. C'!ii 1 I oilides. Combinatorial Methods in Discrete Distributions. WileyInterscience, Hoboken, NJ, 2005. [12] F.R.K. Chuiin, R.L. Graham, V.E. Hoggatt Jr, and M. Kleiman. The number of Baxter permutations. Journal of Combinatorial Theory (Series A), 24:382394, 1978. [13] K. L. Chuing. A Course In Pi,..,,:7Ii,. Th(..<,; Elsevier, San Diego, CA, 2001. [14] M. Coleman, M. Albert, I. Leader, and R. Flynn. Permutations containing many patterns. Annals of Combinatorics, to appear. [15] Micah Coleman. An answer to a question by Wilf on packing distinct patterns in a permutation. Electron. J. Combin., ll(1):Note 8, 4 pp. (electronic), 2004. [16] H. Eriksson, K. Eriksson, S. Linusson, and J. Wastlund. Dense packing of patterns in a permutation, Proceedings of the 15th Conference on Formal Power Series and Algebraic Combinatorics (\1. !l>ourne, Australia). http://www.i3s.unice.fr/fpsac/ FPSAC02/articles.html, 2002. [17] M. Fekete. Uber die Verteilung der Wurzeln bei gewissen algebraischen gleichungen mit ganzzahligen koeffizienten. Mathematische Zeitschrift, 17:228 249, 1923. [18] P. Halmos. Measure The.. ;. D. Van N..iI i.1I Co., Berlin, 1956. [19] L. H. Harper. Stirling behavior is .,i~ !',ltically normal. Ann. Math. Statist., 38:410414, 1967. [20] P. A. Hasto. The packing density of other 11,i,. I permutations. Electronic Journal of Combinatorics, 9(2):R1, 2002. [21] M. Klazar. The FilrediHajnal conjecture implies the StanleyWilf conjecture. In Formal Power Series and Al/, /,i.:.. Combinatorics, pages 250255, Berlin, Germany, 2000. Springer Verlag. [22] M. Klazar. Personal communication, 2008. [23] A. Marcus and G. Tardos. Excluded permutation matrices and the StanleyWilf conjecture. Journal of Combinatorial Theory Series A, 107:153160, July 2004. [24] D. Marinov and R. Radoicid. Counting 1324avoiding permutations. Electronic Journal of Combinatorics, 9(2):R13, 2002. [25] Alison Miller. Asymptotic bounds for permutations containing many different patterns, 2006. preprint. [26] A. M. Odlyzko. Handbook of Combinatorics, volume 2, chapter Asymptotic Enumeration Methods, pages 10631229. Elsevier, Cambridge, MA, 1995. [27] A. Price. Packing Densities of L'n r Patterns. PhD thesis, University of Pennsylvania, 1997. [28] J. Riordan. An Introduction to Combinatorial A,.l,i!. Wiley, New York, NY, 1980. [29] S. Roman. The Umbral Calculus. Dover, Mineola, NY, 1984. [30] A. Rucifiski. Random Graphs, chapter 2, Proving Normality in Combinatorics, pages 215231. Wiley Interscience, Cambridge, UK, 1992. [31] R. Simion and F. W. Schmidt. Restricted permutations. European Journal of Combinatorics, 6:383 406, 1985. [32] N.J.A. Sloane. The online encyclopedia of integer sequences. http://www. research. att.com/njas/sequences/A088532, 2003. [33] R. Stanley. Enumerative Combinatorics, Volume 1. Cambridge University Press, Cambridge, UK, 1997. [34] R. Stanley. Enumerative Combinatorics, Volume 2. Cambridge University Press, Cambridge, UK, 1999. [35] J. Taylor. An Introduction to Measure and Pi, ..',,7.~li Springer, New York, NY, 1997. [36] V. Vatter. Permutations avoiding two patterns of length three. Electronic Journal of Combinatorics, 9(2):R6, 2003. [37] V. Vatter. Small permutation classes. http://arxiv.org/abs/0712.4006, 2007. [38] D. Warren. O'l.:I, ..:,,,j the Packing Behavior of L.,. ., Permutation Patterns. PhD thesis, University of Florida, 2005. [39] J. West. Generating trees and forbidden subsequences. Discrete Mathematics, 157(1  3):363 374, 1996. [40] H. Wilf. Generril.ilf, iI.'...A...,i~ ; A K Peters, Wellesley, MA, 2006. [41] D. Zeilberger. A loving rendition of the MarcusTardos amazing proof of the FiirediHajnal conjecture. http://www. math. rutgers. edu/~zeilberg/mamarim/ mamarimPDF/martar .pdf, 2003. BIOGRAPHICAL SKETCH A native Florida Cracker, I graduated from Seabreeze High School in 1994 and entered college as a music 1 i i' r. Realizing my lack of the skill and devotion necessary for a professional musician, I enlisted in the United States Navy, stationed for four years in Yokosuka, Japan, where I met my wife Hiroko Shinohara. After my active duty period, I completed an Associate's degree at Daytona Beach Community College and joined my brother and his wife here in Gainesville, entering the world of higher math for the first time at the University of Florida. As a junior I first met my advisor and fell in love with Combinatorics. I graduated summa cum laude in 2004 and accepted a fellowship at UF. I have thoroughly enjoi, d my graduate career, even my eightmonth sabbatical to Baghdad, Iraq last year, fully funded by the N i I1 Reserve. I look forward to a long career of mathematics research and hope to make some small contribution to the Conversation. PAGE 1 1 PAGE 2 2 PAGE 3 3 PAGE 4 MydeepestloveandadmirationtomywifeHiroko.Whilestudyinginaforeignlanguageinaforeignland,shetookuptwoofthehardestimaginableroles,thatofmilitaryspouseandthatofmathematiciancaretaker.Daisuki!IthankmyparentsBobandBobbiColeman,mybrotherMatt,andmivecinoAbbyfortheirpatience,humor,andintegrity.GreatthanksgotoProfessorsJulieMiller,TinaCarter,andNormLevin,forrstintroducingmeto\realmath",toourGraduateCoordinatorPaulRobinson,andtothegreatestadvisorycommitteeeverassembled,ProfessorsDavidDrake,MeeraSitharam,AndrewVince,andNeilWhite.Iamhonoredandhumbledtobeassociatedwitheachofthem.Finally,mydeepestrespectandappreciationareheldformyadvisor,BonaMiklos. 4 PAGE 5 page ACKNOWLEDGMENTS ................................. 4 LISTOFFIGURES .................................... 7 ABSTRACT ........................................ 8 CHAPTER 1INTRODUCTION .................................. 10 1.1AsymptoticEnumeration ............................ 10 1.2NotationforAsymptoticGrowthRates .................... 10 1.3GeneratingFunctions .............................. 11 2PATTERNAVOIDANCEINPERMUTATIONSAVOIDINGAMONOTONEPATTERN ...................................... 12 2.1PermutationsandPermutationPatterns ................... 12 2.2AnOpenProblembyM.Atkinson ...................... 24 2.3GeneratingTrees ................................ 26 2.4\Hat"Notation ................................. 31 2.5MonotoneIncreasingPatternsq 33 2.6ThePatternq=123 .............................. 38 3PATTERNPACKING ................................ 53 3.1GeneralPatternPacking ............................ 53 3.2PatternPackingin123avoidingPermutations ................ 57 3.3PatternPackinginqavoidingPermutations ................. 59 3.4PackingDensityandFurtherDirections .................... 61 4ASYMPTOTICNORMALITYANDUNIFORMITY ............... 63 4.1ProbabilityTheory ............................... 63 4.2TriangularArrays ................................ 64 4.3AsymptoticNormality ............................. 65 4.4AsymptoticUniformity ............................. 66 4.5GeneratingPolynomialswithReal,NonPositiveRoots ........... 68 4.6AsymptoticNormalityImpliesAsymptoticUniformity ........... 71 5ONTHEROOTSOFTHEBELLPOLYNOMIALS ................ 74 5.1StirlingNumbersoftheSecondKind ..................... 74 5.2BellPolynomials ................................ 77 5.3BoundsontheRootsoftheBellPolynomials ................. 80 5.4AsymptoticsoftheRootsoftheBellPolynomials .............. 83 5 PAGE 6 ....................................... 88 BIOGRAPHICALSKETCH ................................ 91 6 PAGE 7 Figure page 21Thepermutation3142. ................................ 13 22Thepermutation532614. ............................... 25 23Thepermutation865321947. ............................. 27 24Arootedtree. ..................................... 28 25Thecompletebinarytree. .............................. 29 26TheFibonaccitree. .................................. 30 27T(123,132). ...................................... 32 28TreeinW(123,231)rootedat42153 ......................... 40 29Thelayeredpermutation213654withlayers21,3,and654 ............ 41 31ThepermutationW(6)=342516 ........................... 55 7 PAGE 8 8 PAGE 9 9 PAGE 10 26 ]intheHandbookofCombinatorics. PAGE 11 40 ],Stanley[ 33 ],[ 34 ],andBona[ 8 ]. 11 PAGE 13 21 Thepermutation3142. Inallcontextsconsideredhere,apermutationpatternorsimplyapatternisitselfapermutationbuttherearesubtledierenceswhichweshallexploit.Givenapermutation=1n,asubsequenceofisanorderedsubsetoftheentriesof,(i1;:::;ik)forsomek,whichwewriteinthesameorderastheyappearin,soi1 PAGE 15 3 byreducingallsubsequences,inthesameorderingasabove:Length0:;Length1:11111Length2:12121212122112211212Length3:123132123132123123231123213213Length4:13421234132413242314Length5:13425Ofcourse,wecouldcontinueinthisfashion.Itisagreatexerciseforthebeginnerinthisareatoexhaustthesubsequencesofapermutationtodeterminewhatpatternsthepermutationcontainsoravoidsandposesomeconjectures.Thisishowonelearnsanythingincombinatorics,by\gettingourhandsdirty",doingenoughmanuallaboronourcombinatorialobjectstogetafeelfortheirgrowthandotherproperties. 15 PAGE 16 31 ]launchedpatternavoidanceandcontainedsomeresultswhicharestillhallmarksoftheeld.Asthenamewouldimply,enumerativecombinatorialistsmostenjoyenumeratingsets,thatis,determiningapreciseformulaforthecardinalityofeachsetwhichdependsonlyontheindexorindicesofthatset.Unfortunately,weoftenndquiteinterestingdiscreteobjectswhosenatureiscomplexenoughtoeludepreciseformulae.Alas,wewillseethatformostpatterns,thesequencesn()fallsintothelattercategory.However,allisnotlost.Aswasbrieydiscussedintheintroductorychapter,greatinformationcanstillbehadbytheasymptoticsofasequence,andmanyofthecurrentresultsintheeldofpermutationpatternsinvolveboundsandlimitswhicharenotasstrongaspreciseformulae,butcarrypowerandbeautyoftheirown.Letusrstseesomeexamplesofpatternsforwhichwecangiveanexactformula.WewilltreatallpatternsinSmfor0m3.WewillmakeuseoftheKroneckerdeltai;j,denedbyi;j=8>><>>:1ifi=j;0ifi6=j: PAGE 17 Thereadermay(should)havebeenamazedbythefactthatsn(12)=sn(21)andtheirdualproofs.Infact,thedualityinvolvedwasthat12and21arereversesofeachother,andforeachntheuniquepermutationinSn(12),themonotonedecreasingpermutation,isthereverseofthemonotoneincreasingpermutation,theuniqueelementofSn(21).Ofcourse,onecouldalsoprovetheequalitywiththefactthat12and21arecomplementsofeachother.Perhapsthesestatementsalsoapplytolonger,moreinterestingpatterns? 17 PAGE 18 18 PAGE 19 8 ,thisisequivalenttothestatementthat,foreachpairi PAGE 20 31 ])Foralln0,sn(123)=sn(132): 20 PAGE 21 So,infactweseethatforall;2S3andn0;sn()=sn():Onemaybetemptedtosuspectsuchastatementholdsforpatternsofeverylengthm.However,withacomputercheckorafewpagesofscribbling,oneobtainss6(1342)=5126=s6(1234)=513:Wenowturntoasymptotics.In1980,RichardStanleyandHerbWilfindependentlyconjecturedthatforeachpatternthereexistsaconstantcsuchthat,foralln0,wehavesn()cn:In[ 2 ],Arratiaprovedthatthiswasequivalenttothefollowing,longknownastheStanleyWilfConjecture. 23 ]oftheFurediHajnalConjectureonpermutationmatrices.ThattheFurediHajnalConjectureimpliestheStanleyWilfconjecturewasprovenbyKlazar 21 PAGE 22 21 ].Foraclearandconcisetreatmentofall,seesection4.5of[ 5 ].ThereaderisalsoencouragedtoseeDoronZeilberger'salternativerendition[ 41 ].ThereisstillhopeforatighterproofoftheStanleyWilfConjecture,astheFurediHajnalConjectureonlyprovesthereexistsaconstant,buttheconstantswhichwegetfromtheproofareastronomicallylargerthantheobservedconstants.Itdoesstillgivestructuretoourworktoknowthatforanypattern,sn()isatmostexponential,andadditionallythelimitlimn!1sn()1=nexists.WedenotethislimitbyL()asitiscriticaltothesequel.Now,foranitesetofatleasttwopatterns,wedonothavesuchastronggeneralresult.Foranysuchsetofpatterns,itisreadilyseenthatsn() PAGE 23 Proof. 17 ]forsuperadditivesequences.TheanalogofthislemmaforsubadditivesequenceswasusedinArratia'sproofforthecaseofasinglepattern. Withthesefactsinmind,wedeneLonsetsoftwopatternsasfollows. 23 PAGE 24 3 ,thelimitandlimsupagree,andLisaswewantittobe. 22 .Notethatinourpreviousexampleeachentryofthebackendislessthanthethreshold.Creatingourowngoodluck,wechose532614forourexamplespecicallybecauseitavoids123.Infact,every123avoidingpermutationsharesthisproperty,asimplestructureofwhichweshalltakegreatadvantageinourhandlingofthesepermutations.Toseethisproperty,supposeourthresholdistandthereisanentryjinthebackend(equivalentlyj>t)withj>t.Bydenitionofascendee,t1 PAGE 25 Thepermutation532614. PAGE 26 39 ]JulianWestdenesageneratingtreeasarooted,labeledtreehavingthepropertythatthelabelsofthechildrenofeachnodexcanbedeterminedfromthelabelofxitself.Thisleadstothecharacterizationofageneratingtreebythelabelofitsrootand 26 PAGE 27 Thepermutation865321947. asetofsuccessionruleswhichdeterminethenumberofchildrenandlabelsofchildrenforeachnodeofagivenlengthandlabel.Theclassictaskforacombinatorialenumerologististodeterminethenumberofsomecombinatorialobjectsofsizen,perhapsfurtherindexedwithrespecttosomepropertyorsomestatistick.Typically,oneispresentedwithaninitialobjectofsomesmallsizeandarecursionrulewhichsayshowmanyobjectsofeachsuccessivegeneration(objectsofsizen+1)canbecreatedinductivelyfromthoseofthepreviousgeneration(objectsofsizen).Denethegthlevelnumberofatreetobethenumberofnodesinthegthgeneration.Thusthegeneratingtreeiseasilyseenasatoolwhichlendsitselfquitereadilytocombinatorialenumeration.Weconsiderthenodesofourtreetobethecombinatorialobjectsthemselves.Therearemanysituationswhenthenumberof(n+1)objectswhich 27 PAGE 28 QQQQQQQQQ Arootedtree. canbegeneratedfromanynobjectisallweneedtoknow,sowemightaswelllabeleachnodewithitsdepth.In[ 39 ]Westbeginswithatrivialexample,thecompletebinarytree.Webeginwitharootwithlabel(2).Oursuccessionruleisthateachnodewithlabel(2)hastwochildrenalsolabeled(2). 39 ],Example1). 28 PAGE 29 Thecompletebinarytree. 39 ],Example3). PAGE 30 AAAAAAAAqqqqqqqqqFigure26. TheFibonaccitree. Foradetailedexpositionontheuseofgeneratingtreesinthestudyofpatternavoidance,see[ 39 ],[ 12 ],[ 36 ],[ 10 ]and[ 24 ].Herewedenethegeneratingtreeswhichwillbeusedthroughout.Thesedenitionsdependonthepatternsandqwhicharebeingavoided,soweassumethepatternstobegiven.Thiswillbeclearfromcontext.Firstweexplainthemotivations.Recallour0notation.Forapattern2Sm,thepattern02Sm+1isobtainedbyprependingwiththeentry(m+1).Thefundamentalquestionhereiswhethervariouslimits(orlimitsuprema)forthenumberofpermutationswhichavoidsomepatternarethesameasthosewhichavoid0(assumingfornowthatthelimitsexist).Itwasnotedabovethatavoidanceimplies0avoidance,butthereare0avoiderswhichcontain.So,ourquestionboilsdowntojusthowmanyofthesethereare,inparticularwhataretheasymptoticsofthesepermutationswithrespecttothesetofavoiders.Wewould 30 PAGE 31 27 .Anactivesiteinapermutationisavalidinsertionpoint,thatis,asitewherewecaninsertn+1andobtainachildwhichisstillinthecurrentgeneratingtree,soforourpurposesanactivesiteissuchthattheinsertionwillnotcauseanoccurenceofanypatternwhichweseektoavoid.Thedepthofapermutationisthenumberofactivesitesin,equivalenttothenotionofdepthdenedaboveongeneratingtrees.Wenotethatthedepthdependsonboththepermutationitselfandonthetree,specicallythepatternbeingavoidedwhichdeterminesthetree. 31 PAGE 32 4231(2)@@@@@@@@213(1) 4213(2)eeeeeeee12(2)BBBBBBBB312(2)4312(3)BBBBBBBB3412(1)qqqqqqqqqqqqqqqqqqqqqqqqFigure27. T(123,132). to2=4,and^3refersto4=3.Ontheotherhand,thepermutation1432containsthree^'s,namely143,142,and132.Inthiscasewecanrefertothe^1,theentry1,butwehaveseveral^2'sandseveral^3's.Itshouldalsobenotedthatoneentrycouldbebotha^ianda^jforsomei6=j. PAGE 33 4 ]. PAGE 34 Proof. Proof. 34 PAGE 35 2 thenumberofsuchweaknclassesisatmostapolynomialfunctionf(n).Thereforeourovercountofnpermutationsisf(n)sn1(q;).Wearenowinpositiontotakeourlimits.L(q;0)=limsupn!1sn(q;0)1=nlimsupn!1(f(n)sn1(q;))1=n=limsupn!1f(n)1=nsn1(q;)1=n=limsupn!11sn1(q;)1=n=L(q;):CombinedwiththeknowledgethatL(q;0)L(q;),wearenished. Thefollowinglemmafrom[ 4 ]and[ 5 ]providesanupperboundonthenumberofpermutationsoflengthnwhichavoidtheincreasingpatternoflengthr. 35 PAGE 36 Proof. Withthislemmainhand,wesubtlyalteranotherproofofBonatoachieve: 6 thereareatmost(r1)2(k1)possiblepermutations 36 PAGE 37 Inparticular,forq=123,wehave(r2)2=1,sowiththeassumptionsn(123;)cnforalln,wendsn(123;0)(c+1)n1foralln.TheStanleyWilfConjecture(MarcusTardosTheorem)tellsusthatforanypatternorsetofpatternsthereissuchaconstantcasintheabovehypothesis.Inthecaseofavoidingasinglepattern,Arratiashowedin[ 2 ]thatthesequencesn()1=nisincreasing.However,therearesetsofpatternsforwhichthesequencesn()1=nisnotincreasing.Thus,takingctobetheleastconstantsuchthatsn()cnforalln PAGE 39 Proof. 28 39 PAGE 40 4762153qqq TreeinW(123,231)rootedat42153 Denotingbyaj;kthenumberofpermutationsatthejthlevelwithdepthk,0jand1kd,wehavetherecursivesystema0;d=1a0;k=08k6=daj;k=dXt=kaj1;t8j1;1kd:Fromthisrecursivesystemweseethataj;k=dk+1+jdk+1.Foreachdandjwemaysumoverallktoattainthelevelnumberd+j1d1.However,acombinatorialproofispreferable.Weknowthatd+j1d1countsjmultisetsof[d],andsuchamultisetwritteninnonincreasingorderspellsouttheorderofactivesiteschoseninthelineagefromtoapermutationoflengthn+j. 40 PAGE 41 5 ]Denition5.33)Apermutationiscalledlayeredifitcanbewrittenastheconcatenationq1q2qkwhereeachqiisadecreasingsequenceofconsecutiveintegersandtheleadingentryofqiissmallerthantheleadingentryofqi+1for1ik1. 29 Thelayeredpermutation213654withlayers21,3,and654 Manyresultsareknownconcerningpatternavoidanceandpatternpackingforlayeredpermutations.OnecanseeSection5.2.2of[ 5 ],[ 27 ],[ 6 ],and[ 20 ].Ournextresultisonlayeredpatternswithjusttwolayers,equivalentlynonmonotonelayeredpatternswhichavoid123. 41 PAGE 42 42 PAGE 43 23 .Forexample,foranyd>2,wehavethe(123,3142)avoidingpermutation(d+1)(d1)(d2)1(d+2)d,whichhasdepthd.Thismeanswecanarenotguaranteedaxedboundtothedegreesofourgeneratingpolynomialsofthelevelnumbersofourtrees.Allisnotlost,however,aswecanshowthatforanyofaclassofpatternsandanykthereisanupperboundtothenumberoftreesinWwhoseroothasdepthwhichisk PAGE 44 2!n1p 2!n!;fromwhichweseethatFnisasymptoticallyapproximatedbythenthpowerofthedominantterm,thegoldenratio,Fn1+p 2!n:Furthermore,asthegoldenratioenjoysthepropertyofbeingasolutiontotheequationx2=x+1,sothat1+p 22=1+p 2+1=3+p 2,weimmediatelyachieveanapproximationforeveryotherFibonaccinumberF2n+13+p 2!n: PAGE 45 2: 45 PAGE 46 2followsimmediatelyfromtheabovediscussionoftheasymptoticsoftheFibonaccinumbers. 2: 46 PAGE 47 47 PAGE 48 48 PAGE 49 Proof. 49 PAGE 50 25 .Letd0denotethedepthof.NotethatanyrootinWcontainsinitsfrontendacopyofthefrontendofandthereforehasdepthwhichisatleastd0.Letdd0.Webeginwithaprototype,apermutationinSk(123;)forsomek,andinsertdd0entriesintoconstructarootinWwithdepthd.Let2Sk(123;)haveachild(0)inUwhichisarootinWandhasthesamedepthas,d0.So,thereexistsanindextsuchthatinsertingk+1attingivesus(0),andcontainsacopyofnm.Ifd=d0,wetake(0)tobeourrootinW.Otherwise,webuild(dd0)inductively.Letrbetheindexoftherightmost^1of(0),andnotethatbyconstructiontistheindexofthethresholdof(0),and12t1=(0)1(0)2(0)t1.Besidesthethreshold,wewillinserteachentryusingoneoftwomethods.Onemethodistoinsertatamaximaldecreasinginterval,i.e.chooseanindexinamaximaldecreasingintervalinthefrontendof(j)andinatethepermutationatthatindexbythepermutation21. 50 PAGE 51 9 ,weknowthatthefrontendof(0)hasatmostmmaximaldecreasingintervals.Infact,thereareexactlyMmaximaldecreasingintervalsinthesubsequencer+1t1forsomeM PAGE 53 Inthepreviouschapterwediscussedequivalenceclassesforpatternavoidance.Ifthepermutationcontainsthepattern,then1contains1.SimilarlyforRandC.Fromthisweimmediatelyachieveforallpermutationspat()=pat(1)=pat(R)=pat(C):So,;1;R;andCareinthesameequivalenceclasswithrespecttothefunctionpat.Next,wedenethefunctionmaxpatoverallnonnegativeintegersbymaxpat(n)=maxfpat()j2Sng: PAGE 54 31 foranexample.Theclaimisthatforalln1,pat(W(n))Fn,whereFnisthenthFibonaccinumber,asdenedintheprevioussection. 2liminfn!1maxpat(n)1=nlimsupn!1maxpat(n)1=n2: PAGE 55 ThepermutationW(6)=342516 PAGE 56 2!n!:Theupperboundonthelimitsupremumistrivialformaxpat,whichisnecessarilyboundedabovebythenumberofsubsequencesofannpermutation,2n. Basedonempiricalevidence,(seesequenceA088532intheOnLineEncyclopediaofIntegerSequences[ 32 ]),itseemedthisnumbermay\approach"thetrivialupperboundof2n(thenumberofallsubsetsofannpermutation).In[ 15 ]thisauthorconstructedaclassofpermutationsoverwhichitwasshownthatlimn!1maxpat(n)1=n=2:However,whileconrmingoursuspicionthatthenthrootapproaches2,thisresultstillleftopenthepossibilitythatmaxpat(n) 2n!0asn!1,i.e.thepossibilitythatmaxpat(n)=o(2n).RecentlyMiller[ 25 ]andAlbertetal.[ 14 ]independentlyprovedwitharenementoftheoriginalclassfrom[ 15 ]andmoredelicatecountingtechniquesthatindeedlimn!1maxpat(n) 2n=1;i:e:maxpat(n)2n:Inparticular,in[ 25 ],Millershowedthewonderfullyexactbounds2nOn22np PAGE 57 Thesignicanceoftheabovelimitisthatonemayhopetopackjustasmanypatternsinto123avoidingpermutationsasinthegeneralcase.Furthermore,foranyincreasingpatternq=12rwithr>3,thesameupperboundholds,albeittrivial. 34 givesushopethat,evenwiththe123avoidingrestriction,wemaypackasmanypatternsaswewouldlikeinapermutationofsucientlength.Experimentationsupportssuchaconjecture.Herewegiveconstructionsforqavoidingpermutationswith\many"patternsforincreasingpatternsq.Ourconstructionsaremodeledonthoseof[ 15 ],[ 16 ],[ 14 ],and[ 25 ]andmeetorsurpasstheoriginallowerbound 57 PAGE 58 2n.WenotethatWilf'soriginalconstructionis132avoiding,sowealreadyhaveitestablishedthatmaxpat132(n)>1+p 2n.Webeginwithafamilyof123avoidingpermutationswhichshowthatthelimitinmumofmaxpat123(n)1=nisatleast1+p 2asn!1.WedeneourpermutationsP(n)inductively.LetP(1)=1andP(2)=12.Foreachoddn3,letP(n)=nP(n1),i.e.P(n1)prependedwithn.Foreachevenn4,letP(n)beP(n1)withninsertedimmediatelyafter1.So,wehaveP(1)=1;P(2)=12;P(3)=312;P(4)=3142;P(5)=53142;P(6)=531642:So,eachP(n)consistsoftwodecreasingsubsequences,andthusavoids123.Thenextpropositiongivesalowerboundonpat(n)withaproofsimilartotheproofgivenforProposition 33 2liminfn!1maxpat123(n)1=nlimsupn!1maxpat123(n)1=n2: PAGE 60 60 PAGE 61 Itmaybenotedthattheseclassesofpermutationsresembletheclassesusedin[ 15 ],[ 16 ],[ 14 ],and[ 25 ],exceptthatwhereasineachofthesepaperstheconstructionconsistedofkrowsofkentrieseach,orastrippeddownversionofthat,thepermutationsweareusingherehaverestrictedrowlengths(numberofmaximaldecreasingsubsequences)duetotheqavoidingrestriction.Itwasnotedthatk!1.Fromthisfactweseethatasweletr!1,ourconstructionsforincreasingpatternsoflengthr+1\approach"theconstructionsforthegeneralcaseandtheliminfsapproach2. 61 PAGE 62 38 ]andAlkesPrice[ 27 ].HerbWilfhasrecentlysuggestedthatthenotionofpackingdensitybeexaminedintheqavoidingenvironment.Whydoweraisetheissueofthesedualities?Wesawinthesectionofpatternavoidancethatintherestrictedenvironmentsof123avoidingpermutations,ormoregenerallyqavoidingpermutations,itismorelikelythatarandomlychosenpermutationavoidssomepattern,or,critically,thechanceofavoidingapproachesorequalsthechanceofavoiding0,astatementwhichisusuallyfalseinthegeneralcase.So,restrictingtoqavoidingpermutationsmakesourlifeeasierinthatwork.However,patternpackingbecomesmoredicult.Indeed,ifweconsiderallapproachesusedtoprovepatternpackingorsuperpatterns,theytakeadvantageofalatticeorcheckerboardstructureintheclassofconstructionstoboundmaxpat.Ifwerestrictthelengthoftheincreasingsequencesallowed,welosethisstructure.So,themetaquestioniswhetherweloseourasymptoticlimitsorsimplyneedmoredelicatetechniquestoseethem. 62 PAGE 63 18 ],Taylor[ 35 ],andChung[ 13 ].Fora 63 PAGE 64 1 ]andCharalambides[ 11 ]. Denition22. 19 ].Incombinatorialapplicationsan;kcountsobjectsofsizenwithsomestatistick.Forexample,inthenextchapterwewillstudythenumbersSn;k,countingsetpartitionsofannelementsetintokblocks.Givensuchasequencean;k,wesetsn=an;1+an;2++an;mforeachnandconstructanewsequencebn;k=an;k 2 64 PAGE 65 35 ]. 65 PAGE 66 7 ],[ 9 ]). PAGE 67 7 ]and[ 9 ] q,aprimitiveqthrootofunity. Proof. 4{5 )isageometricsum,andwetakeadvantageofthepropertythatthe(kr)rootsofunityaddupto1toseethatq1Xt=0(kr)t=8>><>>:0ifkr6=1:qifkr=1: PAGE 68 4{5 )forwhichqdoesnotdividekr,and( 4{2 )reducestoSn;r()=qm=qXj=1an;jq+r;or (4{6)Sn;r() 4{7 )asthefunctionfromthedenitionofqbalance.So,qbalanceisequivalenttothisexpressionconvergingto1 4{1 )wecandropourassumptionthatqdividesm,andwehavethelimitlimn!1Sn;r() PAGE 69 19 ]andalsousedin[ 30 ].Byourassumptionontherootsofourgeneratingpolynomials,wecanfactortheprobabilitygeneratingpolynomialpn(x)aspn(x)=an;m 1+n;k:FromthisthemeanandvarianceofeachIn;kfollowimmediately.E(In;k)=1 1+n;k:VarIn;k=E(I2n;k)(E(In;k))2=1 1+n;k1 (1+n;k)2=n;k 1+n;kx: PAGE 70 3 ]whichfollowsfromTheorem 39 : 41 andkeepingtheabovenotation,wehaveshownthefollowing. PAGE 71 10 and 11 ,wecanstatethemainresultofthischapter. Proof. 4.5 .Assumingasymptoticnormality,Lemma 11 tellsus 10 ,itsucestoshowthatthisdivergenceimplies q.Ifr=1isarootofpn(x),thenpn(r)=0,sowecanassumeherethatthisisnotthecaseinordertoavoidpathologies.Takinglogarithms,( 4{9 )isequivalenttothecondition 12 below,thereexistsaconstant>0whichdependsonlyonrsuchthatforallnandkwehavelog(1+n;k)logjr+n;kj>n;k 4{8 )implies( 4{10 )whichisitselfequivalentto( 4{9 ). 71 PAGE 72 tZjx+zj1dt t=Zx+1jx+zjdt t:Thelengthofthedomainofintegrationisthefunctiond(x)=x+1jx+zj,andwehavef(x)>d(x) x+1forallx0.Werstshowthatthereisaconstant1suchthatd(x)>1forallx>1.Wehavetheformulajx+zj=(x2+2xcos+1)1=2.Therefore,forallx>1,d0(x)=d dx(x+1(x2+2xcos+1)1=2)=12x+2cos x2+2xcos+11=2>0:Therefore,d(x)ismonotonicallyincreasingforpositivex,andforallx>1itisgreaterthand(1),whichwedenoteby1.Nextweshowthereexistconstants2andsuchthatd(x)>2forall0 PAGE 73 x+1;sowelet2=1cos x+1forallx2[;1].Finally,set=minf1;2;3g. 21 11 andTheorem 42 ,thesignlessStirlingnumbersoftherstkindareasymptoticallynormalandthusuniform. PAGE 74 74 PAGE 75 5{2 )isan;k.WeapplythemethodofInclusionExclusion.Thereareknwaystodistributetheballstotheboxes,aswehavekchoicesofwhattodowitheachofnballs.However,wemayhaveovercounteddistributionsinwhichaboxwasleftempty,sowesubtractthek1(k1)nwaystochooseaboxtobeemptyanddistributetheballstotheremainingk1boxes.Now,wehaveovercountedthedistributionsinwhichtherearetwoemptyboxes.Indeed,forsome1p PAGE 76 5{2 )asSn;k=kn 5{3 ).Finally,asn!1,Sn;k kn 5{2 ).Bytherecurrence( 5{1 ),foralln>k1,wehaveSn;kkSn1;kk2Sn2;kknkSk;k=knk;asSk;k=1forallk,completing( 5{5 ).Thelowerboundof( 5{6 )isthedierenceofthedominanttermandalloddindexed,hencenegative,termsof( 5{2 ). 76 PAGE 77 5{1 )toeachtermofBn+1(x),wehaveBn+1(x)=XSn+1;kxk=X(Sn;k1+kSn;k)xk=XSn;k1xk+XkSn;kxk=xXSn;k1xk1+xXkSn;kxk1=xBn(x)+xB0n(x): 77 PAGE 78 Proof. 40 ])ThestatementholdsforB1(x).AssumebyinductionthestatementholdsforBn(x).Multiplyingeachtermin( 5{7 )byex,exBn+1(x)=x(exBn(x))0:ByRolle'stheorem,B0n(x)hasn1roots,onebetweeneachconsecutivepairofrootsofBn(x).Multiplicationbyxguaranteesarootatzero.SinceexBn(x)approacheszeroasx!,itsderivativehasonemoreroottotheleftoftheleftmostrootofBn(x),accountingforalln+1rootsofBn+1(x). ThuswecanfactoreachBellpolynomialas Proof. 19 ]provedthattheStirlingnumbersofthesecondkindareasymptoticallynormal.Theabovediscussionshowsthattheirgeneratingpolynomialshavereal,nonpositiverootsonly,sobyTheorem 42 inthepreviouschapter,weachievethedesiredresult. Now,letusconsidertherootsmoreclosely. 78 PAGE 79 2)(x+3p 2):Forhighern,thealgebrabecomesunwieldy,ascanbereadilyseenintheotherBellpolynomialsgiveninExample 45 .Asourworkoftendealswithsumsofproductsofreciprocalsoftheseroots,fornotationalconveniencewewillwriten;k=1 Proof. 5{8 ),anxktermisachievedbychoosingxfromkofthetermsandn;ifromtheremainingnkterms.ThesumofallsuchproductsisSn;k,thecoecientofxkinBn(x). Wenotethat,foralln1,wehaveSn;n=1.Lemma 18 couldberewrittentoreectthisbyallowingthesumoverall(nn)tuplestobe1. Proof. PAGE 80 Proof. 19 toeach(nk)tupleinLemma 18 FormoreonStirlingnumbersofthesecondkindandBellpolynomials,particularlytheirpropertiesandmyriadapplications,thereaderisreferredtotheclassicbyRiordan[ 28 ]aswellastheexcellenttextsbyRoman[ 29 ]andWilf[ 40 ]. Lemma21. 14 and 20 andthenonnegativityofthen;i's,n;2nXk=2n;k=Sn;2<2n1:Asn;2isthelargesttermofthesum,itisatleastaslargeasthemean,son;2>2n11 80 PAGE 81 9n2n;3<3n1 14 and 20 23n1O(2n)<3n1 21 ,n;3<3n1 5{5 )ofLemma 15 providesSn;3>3n3:ThenextinequalityfollowsfromLemma 20 ,thefactthatthelargestofthen12termsin( 5{10 )isn;2n;3,andourupperboundn;2<2n1.2n1n;3>3n3 9n2: PAGE 82 k1n1 k1nFk(n): 22 .Weprovedthestatementforn;2andn;3inLemmas 21 and 22 ,respectively.Letk4andassumebyinductionthatthestatementholdsforalln;jwith2j PAGE 83 5{11 )isn;2n;3n;k.Byourinductionhypothesis,n;jobeysthefollowingboundforeach2j PAGE 84 2n>n;3)n;2+nn;3>2n1;n;2+n23 2n>2n1;n;2>2n12n23 2n;n;2 4n!1:Thenallimit,inessence2n23 4n!0asn!1,holdsas2n2isapolynomialand3 4nisageometricprogressionwithpositiveratiolessthanone. 20 expressesSn;3asthesumofallproductsij.Wecansplitthisintotwosums,onesumcontainingthosepairswhichincluden;2andonesumcontainingallotherpairs, 22 givesustheupperboundn;3 PAGE 85 5{12 )and( 5{13 ),wehaven;2Xi3n;i=Sn;3O9 4nn4=3n1 4nn4:Dividingbyourasymptoticapproximationofn;2,n;22n1,Xi3n;i=3n1 8nn4:ByTheorem 48 ,4 3nF4(n)>n;4>n;5>forthexedpolynomialofdegree3F4(x),and9 8nn4+(n3)4 3nF4(n)=O4 3nn4;son;3=3n1 3nn4:Finally,wedivideby3n1 9nn4:limn!1n;32n 9nn4=1: 85 PAGE 86 5{9 )ofLemma 20 as (k1)2nG(n);whereG(x)=F3(x)Fk2(x)F2k(x).Set=k2(k2) (k1)2andnotethatk1< PAGE 87 5{15 )by(k1)n1 k1=k2(k2) (k1)3.Nowitsucestoannihilatealltermsonthelefthandsideexceptn;k.ByTheorem 48 ,k+1 87 PAGE 88 [1] N.AlonandJ.Spencer.TheProbabilisticMethod.WileyInterscience,NewYork,NY,2000. 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226ef9845442d7973c3b52c5084bd836 806b3edd319c2ab2fb506169cb5c6412e467bd3d 63460 F20110331_AADESD coleman_m_Page_05.pro ec6fd3acfe71c1171e2a7a731b55e7fe 9ac69291e8bbdd618ebc2e67996cf4dec2aaefdd 1263 F20110331_AADERQ coleman_m_Page_77.txt 588869665d651f0ddcad276a703ef936 88e4ae21ed12565b077276eec9f675b240d44a34 13088 F20110331_AADESE coleman_m_Page_07.pro 6dc55a7970ae47ed4bf7729a8f8e25fe e7731e799720122d9a4d01ac7ca7ced08741d0ae 1644 F20110331_AADERR coleman_m_Page_78.txt 83c9090610887cfb416b285d80aad6e3 0ef016e515dc2d22807b41bb1c02f8265025b5a1 10170 F20110331_AADESF coleman_m_Page_09.pro 21683fcf68719b981c21879072bdfa4c 6e01612df135112363059cbfc9628f7fa927cb74 1769 F20110331_AADERS coleman_m_Page_79.txt f1514acf3f5b477a2a1d1f97e716766b 81a7c3cb111d6815dabde3e1e6c72c821ccf7b84 44891 F20110331_AADESG coleman_m_Page_10.pro a824f7da3fb92f0f7292f63f7444bca0 44335f0aa915852ff97053d3376472fc62fdd0ef 1656 F20110331_AADERT coleman_m_Page_80.txt 1ae0c07ab55f8043041d5552553fd865 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F20110331_AADERX coleman_m_Page_88.txt 1a2ede87bf8ef3b3cb735396089e6a0c 1f238e32871ea53aab6ea2b55c8b3a066704ccc7 38901 F20110331_AADETA coleman_m_Page_35.pro d6daa527d953e0eb1b253f952d8d3950 6c41a549d7de02d71be4ad8333dda63b5940369f 29138 F20110331_AADESM coleman_m_Page_18.pro d96167428b7500348ffe0cf641bf02ec 6140d7a44e2e031f7cb1d7c6bed9a8c37d18a526 2216 F20110331_AADERY coleman_m_Page_89.txt 27b134dd54cd2b80acb56cf98737f640 cac31ae236463717d7e800579ab45341ca793adc 46904 F20110331_AADETB coleman_m_Page_37.pro 830353be1bf022204d665a87cf4390c1 66614582f25f3b6700562bed739b3a67b7429fe1 42140 F20110331_AADESN coleman_m_Page_19.pro 8a82475129732d160b4cd4db6838fdc4 f1cdb3fc2bf12c3ae5e6f2b2365236009767edf9 860 F20110331_AADERZ coleman_m_Page_90.txt d42794a9584bd6827fcebb12193be8fe 6381b0380c25eafe5dc0c96fb1bf559a0343893e 40557 F20110331_AADETC coleman_m_Page_38.pro 2e26d4c11430d9eadfb13d9d632988df 216af00c3ebcbac436ea44263b01c72ee2f45081 40184 F20110331_AADESO coleman_m_Page_20.pro 8454309cb8b6c08b7ae49b002b606ed4 88328c12c0bb9eb7d86bef6bfde2029da72e5a23 20578 F20110331_AADETD coleman_m_Page_40.pro ec86ff23b9b1b0afd3613b5192d184c1 54feae99a41db96b057a89121ff29d1a9a3145b2 43063 F20110331_AADESP coleman_m_Page_21.pro 56fa6fce3a5d75d6af50b20561f7576b 4d788fb725c34ed5996eae51a0bf92237eb1badf 22564 F20110331_AADETE coleman_m_Page_41.pro d5635c967eda848f433ad61e6e9715bf db9d0a904fb1d911c2106e0e0f24038afad9052a 32436 F20110331_AADESQ coleman_m_Page_23.pro 173052e5596558a23786f9abc726c74a 55e46b0ec0010dae8c4036c4fa49704da0393100 37200 F20110331_AADETF coleman_m_Page_42.pro 553bcdb1fda96e7bd286c7b194779022 71e7b99ea43b9a40e610b1032e2c0fe3d57bc0cf 51973 F20110331_AADESR coleman_m_Page_24.pro 0bc4d51ce48ee9ebf8aad5ddc8e678aa 6e0533e23519ad20cc6544664d34cac00fa170cb 25978 F20110331_AADETG coleman_m_Page_44.pro 5180e596c5ab1bf625eaaba0a8c4b87a 5bdcc7c05f49ab37ce7625248e48aa0bff55bcf7 27520 F20110331_AADESS coleman_m_Page_25.pro 12f26ddff36876e687780a088cb7b49c 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9541176c6cb468f192fee85c5e6701d0 201c532e2ab84051cc9c7cc5931f5b81203a98cd 49992 F20110331_AADESZ coleman_m_Page_34.pro d4629877655d04cb3649d92574f87c74 aa90b601b3b2414cee3310f0768bf87a5a7d0f08 34905 F20110331_AADEUD coleman_m_Page_79.pro 852f5821b0af811a48eed0aa5de2886e e596b79623f7742f4c532511638401b3d105e3b0 34208 F20110331_AADETP coleman_m_Page_60.pro fc356ca8bcbb05883291c074798f6a80 60dd3e89b946d7b48ef37a4ca9a76df5dcf904b6 17846 F20110331_AADEUE coleman_m_Page_81.pro dfb29a332143725403311c4c9b9c2901 388f9f54de227df497d6cd4deb84db79eb67a43b 35798 F20110331_AADETQ coleman_m_Page_62.pro 5e9f5ddbe479c9199cbb4404299e54bb 798ed36e52b116713fd51dbb2913d372aaef09a2 28058 F20110331_AADEUF coleman_m_Page_82.pro e0f5bea2cb3d7c958e0e89c2eae13c90 30ff75226535c58f25e5cd82e3d4198743274fb1 46153 F20110331_AADETR coleman_m_Page_63.pro 2f005d8af111474478479fe20101bf7a 1fc3aa29c37a94ed03f15b68eac3dce86ca5433a 1051966 F20110331_AADFAA coleman_m_Page_36.jp2 f28683c0ecfd08fa1935c92aee28fc7d 3020afc532c0fe1af9336e10a42d6dc2b637fb81 23700 F20110331_AADEUG coleman_m_Page_85.pro 430b2c24aea6e64969334c3954868146 b8a00d7047e8b35c470f2f8dea76430da8907500 43035 F20110331_AADETS coleman_m_Page_64.pro b27f4920310c79f4b021627d80858ff5 76fb554615982f805de6625eb79d768b39010049 84152 F20110331_AADFAB coleman_m_Page_38.jp2 ba3a6c18069f5a801a9594d019f0a573 c20f5ec0450694d0617931bc151c16b2210b4af2 35423 F20110331_AADEUH coleman_m_Page_86.pro 198bc828d9156574509a3c3aa64c5ad3 42bce7b5d30159d7ba859dbb6f0488600ed2f067 37913 F20110331_AADETT coleman_m_Page_65.pro 2d0c1bd6b47fa80a23b1c2b9f59d07e0 20846abdc69d4c915bf338e993f8acac4283590f 888425 F20110331_AADFAC coleman_m_Page_39.jp2 e799e1b9645fcfe685021695a4eff9f4 a5197b7a5bfaff9c662fe211641d0aa97cb7b26c 59115 F20110331_AADEUI coleman_m_Page_88.pro 2ef434f2f0499d31cca556ceab4139fd e15c37add655672a8bbcd229f370ac6ae6735948 42389 F20110331_AADETU coleman_m_Page_66.pro b7719ba7d65af9942ee42c3b8be14ffb 478eaebf026bee5c0bd731607795af7ccc0d572f 49410 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