aa r X i v : . [ m a t h . C O ] O c t Pattern avoidance in ascent sequences
Paul Duncan
25 Vega CourtIrvine, CA 92617, USA [email protected]
Einar Steingr´ımsson ∗ Department of Computer and Information SciencesUniversity of Strathclyde, Glasgow G1 1XH, UK [email protected]
November 12, 2018
Mathematics Subject Classification: 05A05, 05A15, 05A18, 05A19
Abstract
Ascent sequences are sequences of nonnegative integers with restric-tions on the size of each letter, depending on the number of ascents pre-ceding it in the sequence. Ascent sequences have recently been related to(2 + 2)-free posets and various other combinatorial structures. We studypattern avoidance in ascent sequences, giving several results for patternsof lengths up to 4, for Wilf equivalence and for growth rates. We estab-lish bijective connections between pattern avoiding ascent sequences andvarious other combinatorial objects, in particular with set partitions. Wealso make a number of conjectures related to all of these aspects. An ascent sequence is a sequence x x . . . x n of nonnegative integers satisfying x = 0and, for all i with 1 < i ≤ n , x i ≤ asc( x x . . . x i − ) + 1 , where asc( x x . . . x k ) is the number of ascents in the sequence x x . . . x k , that is,the number of places j ≥ x j < x j +1 . An example of such a sequence is0101312052, whereas 0012143 is not, because the 4 is greater than asc(00121) + 1 = 3.Replacing x j < x j +1 in the definition of asc by x j > x j +1 gives the number of descents .Ascent sequences became prominent after they were related to the (2 + 2)-free posets,by Bousquet-M´elou, Claesson, Dukes and Kitaev [1], who also managed to find the ∗ Steingr´ımsson was supported by grant no. 090038013 from the Icelandic Research Fund. enerating function counting these, which was quite a feat. Ascent sequences havesince been studied in a series of papers by various authors, connecting them to manyother combinatorial structures. These connections, and generalizations of them, haveexposed what seem to be deep structural correspondences in these apparently dis-parate combinatorial objects, including certain integer matrices, set partitions andpermutations, in addition to the (2 + 2)-free posets. A good source of references andfurther information is [6, Section 3.2.2]; see also [3, 4, 5, 7].In this paper we study ascent sequences avoiding certain patterns. Our patterns areanalogues of permutation patterns, but they seem to provide a greater variety of count-ing sequences, which may perhaps be explained by the fact that ascent sequences lackthe so called trivial symmetries on permutations—such as reversing a permutation toobtain another one—that imply numerical equivalence between sets of permutationsavoiding different patterns.A pattern is a word on nonnegative integers, where repetitions are allowed. An occur-rence of a pattern p in an ascent sequence x = x x . . . x n is a subsequence x i x i . . . x i k in x , where k equals the length of p , whose letters appear in the same relative orderof size as those in p . For example, the ascent sequence 0123123 has three occurrencesof the pattern 001, namely in the subsequences 112, 113 and 223. Note that in anoccurrence of a pattern p , letters corresponding to two equal letters in p must be equalin the occurrence, such as the 22 in 223, which correspond to the two 0’s in 001. Anascent sequence x avoids the pattern p if x has no occurrences of p . As an example,012321 avoids 001.Since we write our patterns with nonnegative integers, whereas permutation patternshave traditionally been written with positive integers, it is important to note that thetraditional permutation patterns have different names here. For example, 123 becomes012, and 231 becomes 120. We use the latter notation, with nonnegative rather thanpositive integers, since ascent sequences are traditionally defined in such a way as tocontain zeros.To the best of our knowledge, pattern avoidance has not been studied for ascent se-quences so far. Given the success of such studies for other combinatorial structures,such as permutations and set partitions, and the strong connections of ascent se-quences to other combinatorial objects mentioned above, it is reasonable to hope forresults of similar significance in the case of ascent sequences. The initial results andconjectures presented here, together with the great variety in the integer sequencescounting ascent sequences avoiding various patterns, seem to indicate that this isfertile ground for interesting research.We exhibit a connection between ascent sequences and set partitions, which does notseem to have been studied before, although such a connection, of a very differentnature, is considered in [5]. A standard way to represent set partitions of the set { , , . . . n } , or any other ordered set, is to write the elements of each block in increas-ing order, and the blocks in order of increasing minima, such as in 124-36-5, whichrepresents the partition of { , , , , , } into blocks { , , } , { , } and { } . Thisstandard representation of a partition of { , , . . . n } can be encoded by the string a a . . . a n , where a i = k if i belongs to block number k , counting from left to right, ith the leftmost block numbered 0. In our previous example of 124 36 5 this stringwould be 001021. It is easy to see that a string x of nonnegative integers encodes aset partition if and only if the first occurrence of each letter k > x is precededby some occurrence of k −
1. Such a string is called a restricted growth function , orRGF. A non-crossing partition is a partition that does not have letters a < b < c < d with a, c in one block and b, d in another.If p is a pattern, we let S p ( n ) be the set of ascent sequences of length n that avoid p and A p ( n ) the number of such sequences. Also, S p is the union of S p ( n ) for all n ≥ left-to-right maxima in a sequence of numbers a a . . . a n is the set of a i such that a i > a j for all j < i . The set of right-to-left minima is the set of a i suchthat a i < a j for all j > i . Left-to-right minima and right-to-left maxima are definedanalogously. Also, we define LR max( x ) to be the number of left-to-right maxima ina sequence x , and LR min, RL max, RL min analogously.Following is an overview of our main results, which appear in Section 2. See alsoTable 1 at the end of the paper, where we list the counting sequences connected toour results and conjectures. • If p is any one of the patterns 10 , , , ,
012 then A p ( n ) = 2 n − . • If p is any one of the patterns 101 , ,
021 then A p ( n ) is the n -th Catalannumber, and the ascent distribution on S p ( n ) is given by the Narayana numbers.Moreover, the distribution of the bistatistic counting ascents and right-to-leftminima is the same on S ( n ) as it is on permutations of length n avoiding 132. • If p is any one of the patterns 102 , , A p ( n ) = (3 n + 1) / • Ascent sequences avoiding 101 (or, equivalently, 0101) are precisely the RGFsof the non-crossing partitions. • The set S p ( n ) consists solely of RGFs if and only if p is a subpattern of 01012.The Catalan numbers, which we will frequently refer to, are given by C n = n +1 (cid:0) nn (cid:1) .The Narayana numbers are given by N ( n, k ) = n (cid:0) nk (cid:1)(cid:0) nk − (cid:1) and they refine the Catalannumbers in that C n = P k N ( n, k ). It is well known that the Narayana numbers recordthe distribution of the number of ascents on permutations avoiding any given one ofthe patterns 132, 213, 231 and 312.In addition to the results mentioned above we conjecture, in Section 3, avoidancesequences for the patterns 210, 0123 and 0021, in terms of entries in the OnlineEncyclopedia of Integer Sequences [10], and also that 0021 and 1012 are
Wilfequivalent , that is, have the same avoidance sequence. In particular, we conjecture thatascent sequences avoiding 210 are equinumerous with partitions avoiding so called 3-crossings (a 2-crossing is simply a crossing in the usual sense).Moreover, we mention what little we know about growth rates of the counting se-quences for pattern avoiding ascent sequences. In all cases we know, or conjecture,the growth is exponential, as has been proved for permutations avoiding any classi-cal pattern [9]. Finally, we conjecture that modified ascent sequences (defined in [1]) voiding 101 are in bijection with set partitions, and that there is a bijection takingthe number of non-ascents in such a modified sequence to the number of blocks in thecorresponding partition. Note that this conjecture would imply that modified ascentsequences avoiding 101 have super-exponential growth, that is, the number of suchsequences of length n is not bounded by C n for any constant C . We first dispose of the easy results concerning patterns of length less than three. Thereis, of course, no (nonempty) ascent sequence avoiding the only pattern of length one.There are three patterns of length two, namely 00, 01 and 10. It is easy to see thatthe only ascent sequences avoiding 00 are the strictly increasing sequences 0123 . . . ,one for each length n . There is also precisely one sequence of each length n avoiding01, namely the all zero sequence 00 . . . n − n , so the number of such ascent sequences is 2 n − and the number of such sequences with exactly k ascents is (cid:0) n − k (cid:1) . The same is trueof three of the sequences of length 3 as we now show. Theorem 2.1 If p is any one of the patterns 10, 001, 010 or 011, then A p ( n ) = 2 n − .Moreover, the number of sequences in S p ( n ) with k ascents is (cid:0) n − k (cid:1) .Proof. We have already proved the claim for the pattern 10. A proof for each of theremaining three cases is easy to construct, given the following characterizations ofsequences avoiding each one of the patterns in question: • Ascent sequences avoiding 001 are precisely those that start with a strictlyincreasing sequence—necessarily 012 . . . k for some k —followed by an arbitraryweakly decreasing sequence of letters smaller than or equal to k . An exampleis 01234444211. • Ascent sequences avoiding 010 are precisely those that are weakly increasingand thus of the form 00 . . . . . . . . . kk . . . k for some k . • Ascent sequences avoiding 011 are precisely those that consist of a strictly in-creasing ascent sequence arbitrarily interspersed with 0’s, for example 000100230400500.In the first two cases we use the fact that the number of weakly increasing (or weaklydecreasing) sequences of nonnegative integers of length a , and not exceeding b invalue, is (cid:0) a + ba (cid:1) . ✷ It turns out that A ( n ) is also 2 n − , but the ascent distribution of sequences avoiding012 is different from the cases in Theorem 2.1. heorem 2.2 We have A ( n ) = 2 n − . The number of sequences in S ( n ) with k ascents is (cid:0) n k (cid:1) .Proof. Ascent sequences avoiding 012 are those that have no increasing subsequenceof length 3. If an ascent sequence avoids 012, then, after the initial 0, there can beno letters other than 0 and 1, since any letter a larger than 1 must be preceded by1, in which case we have a subsequence 01a, forming a 012. Since the initial 0 can befollowed by any sequence of 0’s and 1s, the sequences avoiding 012 are precisely thosethat consist of an arbitrary binary string after the initial 0, of which there are 2 n − .To prove the result about the ascent distribution, suppose we are given a 2 k -elementsubset S = { x < x < · · · < x k } of the set { , , . . . , n } , and we will use this toconstruct a binary string of length n , starting with 0, with k ascents. The string tobe constructed will consist of 0’s in all places preceding and including place x , thenthe letters in places x + 1 up to and including x will be 1s. Generally, the letters inplaces x i + 1 up to and including x i +1 , or after place x k , will be 0’s and those inplaces x i +1 + 1 up to and including x i +2 will be 1s. The sequence thus constructedwill clearly have ascents precisely in the k places x i +1 , where i ranges from 0 to k −
1. It is straightforward to construct the set S of size 2 k from a binary string with k ascents, which shows this is a bijection. ✷ Recall that two patterns p and q are Wilf equivalent if A p ( n ) = A q ( n ) for all n .Theorems 2.1 and 2.2 lead to the following result. Corollary 2.3
The patterns 10, 001, 010, 011 and 012 are Wilf equivalent.
The following lemma is interesting for its own sake, as it characterizes those patternswhose avoidance by a sequence x guarantees that x is an RGF, that is, a restrictedgrowth function that encodes a set partition. Its second part also turns out to beuseful in proving some of our other results. Lemma 2.4
Let p be a pattern. The S p ( n ) consists solely of RGF sequences if andonly if p is a subpattern of 01012. In particular, in a sequence x avoiding any of thesepatterns, every occurrence of each letter k ≥ is preceded by some occurrence of eachof the letters , , . . . , k − .Proof. Clearly the sequence 01013, not being RGF, must be excluded from any setof RGF sequences. This is guaranteed by the avoidance of p only if p is a pattern in01013, which is equivalent to being a subpattern of 01012. Thus, no other patternsthan subpatterns of 01012 have avoidance sets consisting solely of RGFs. We showthat the sequences avoiding any one of these patterns are RGFs, thereby establishingthe claim. We prove the contrapositive, showing that if an ascent sequence is not anRGF it must contain 10102. That is sufficient, since a sequence containing 10102 ofcourse contains all its subpatterns.Suppose then that x is an ascent sequence that is not an RGF. Then there must bea leftmost letter in x that violates this. If that letter is k then it is not preceded by −
1, but for some i ≥ , , . . . , k − i appear in x preceding k , and theirfirst appearances are in increasing order. In order for an ascent sequence to contain k but not be preceded by k −
1, it must contain more ascents than those provided bythe first occurrences of each of 1 , , . . . , k − i . Thus, one of the letters 1 , , . . . , k − i appears at least twice as the rightmost letter in an ascent. Suppose this letter is c andthat the second one of these ascents is of the form . . . ac . . . . Thus, the first occurrenceof c must precede this occurrence. But, the first occurrence of c is preceded by thefirst occurrence of a . Thus, we have a subsequence . . . a . . . c . . . ac . . . k . . . , and acack is an occurrence of 01012.The second part of the lemma follows directly from the definition of RGFs. ✷ It turns out, as we will now show, that the ascent sequences avoiding 101 are the sameas those avoiding 0101. Moreover we will show that these ascent sequences, which areRGFs according to Lemma 2.4, are precisely those that encode non-crossing partitions.
Theorem 2.5
The ascent sequences avoiding are the same as those avoiding0101, and A ( n ) = A ( n ) = C n , the n -th Catalan number. Moreover, the distri-bution of the number of ascents on these sequences is given by the Narayana numbers.Proof. Clearly, an ascent sequence containing 0101 contains 101. We show that the con-verse is also true, thereby showing that an ascent sequence avoids 101 if and only if itavoids 0101. So, let . . . b . . . a . . . b . . . be an occurrence of 101, so a < b . By Lemma 2.4,since 101 is a pattern in 01012, the first b in the occurrence . . . b . . . a . . . b . . . mustbe preceded by an occurrence of a . This gives the subsequence abab , which is anoccurrence of 0101.We now exhibit a bijection from 101-avoiding ascent sequences of length n to 312-avoiding permutations of length n . Given a 101-avoiding ascent sequence x , replaceits 0’s from left to right with the numbers k, ( k − , . . . , ,
1, in decreasing order,where k is the number of 0’s in x . Now repeat this for the 1’s in x , with the numbers( ℓ + k ) , ( ℓ + k − , . . . , ( k +1), where ℓ is the number of 1’s in x and so on. For example,the sequence 01023200 is mapped to 45378621. Suppose the resulting permutation π contains an occurrence of 312, say in a subsequence zxy , where x < y < z . Thenthis subsequence in π would correspond to a subsequence in x of the form bab or cab ,where a < b < c , depending on whether z = y + 1 or z > y + 1. In the first case bab is an occurrence of 101 in x . In the second case the c in cab must be preceded by b ,by Lemma 2.4, again giving an occurrence of 101 in x , a contradiction showing thatthis map produces a 312-avoiding permutation.This map is easily seen to be invertible. Namely, given a permutation π = a a . . . a n avoiding 312, note that if a = k , then the letters k, k − , . . . , π , for it to avoid 312. Thus, the places of these letters arefilled with 0’s in the ascent sequence x corresponding to π . Iterating this process wenext find the leftmost letter a i in π that is larger than k and fill the places in x corresponding to the places of the letters a i , a i − , . . . , k + 1 in π by 1’s, and so on.It is straightforward to verify that the sequence x thus constructed avoids 101. t is also easy to verify that this map preserves ascents. Thus, the 101-avoiding ascentsequences are enumerated by the Catalan numbers, and have ascent distribution givenby the Narayana numbers, as is the case for 312-avoiding permutations. ✷ Theorem 2.6
The ascent sequences avoiding 0101 (equivalently, 101) are RGFs, andas such they encode precisely all non-crossing partitions.Proof.
Since 0101 is a pattern in 01012, the ascent sequences avoiding it are RGFs,by Lemma 2.4.The occurrence of the pattern 0101 in an ascent sequence x causes a crossing, sinceit implies that we have letters abab , occurring in that order, in x . Thus, if the placeswhere these four letters occur are x, y, z, w , so that x < y < z < w , we have that x and z are in the same block, and y and w in the same block, different from the firstone, which constitutes a crossing.Conversely, suppose a set partition has a crossing consisting of letters a < b < c < d ,with a and c in the same block and b and d together in a different block. Then either a < b < c or b < c < d gives an occurrence of 101, which is equivalent to having anoccurrence of 0101. ✷ In order to give a formula for the number of ascent sequences avoiding one of thepatterns in Theorem 2.9 we need the following lemma.
Lemma 2.7
The number of ternary sequences of length n on the letters , , withan even number of 2’s is (3 n + 1) / .Proof. It’s easy to see that the number of ternary sequences of length n with exactly k (cid:0) nk (cid:1) n − k . Thus, the difference between the number of such sequences with aneven number of 2’s and those with an odd number is given by( − n X k (cid:18) nk (cid:19) ( − n − k = ( − n (1 − n = 1 . Since the total number of ternary sequences of length n is 3 n , the claim follows. ✷ Theorem 2.8
We have A ( n ) = (3 n − + 1) / .Proof. We exhibit a bijection between sequences in S ( n ) and ternary strings oflength n − lifted binary strings , namely, strings composed from two integers differing n size by one. More precisely, 102-avoiding ascent sequences are those that can bewritten as x x . . . x k b b . . . b i b b . . . b i . . . b m b m . . . b mi m , where x x . . . x k is a weakly increasing ascent sequence, the b j form a strictly de-creasing sequence with x k > b > b > · · · > b m ≥ b qp equals either b q or b q + 1. An example of such a sequence is0012234566 56656 4454 2 01001 , where we have separated the respective lifted binary strings by spaces, for clarity.It is easy to see that a sequence of this type avoids 102. We now explain why any102-avoiding sequence has this structure.Any ascent sequence x starts with a weakly increasing sequence x x . . . x k of length atleast 1. In what follows, we choose k such that this initial weakly increasing sequenceis of maximal length. By Lemma 2.4, since 102 is a pattern in 01012, all the integersbetween 0 and the maximum letter in x must appear in x .If there are no letters following this initial weakly increasing subsequence of x then x clearly has the form described, with an empty sequence of lifted binary stringsafter the initial weakly increasing sequence. Otherwise there is a descent immediatelyfollowing the initial sequence x x . . . x k , that is, x k is followed by b where x k > b .If any letter x after b satisfied x > b + 1, then, for some i , the subsequence x i , b , x would form a 102. This is because the letter ( b + 1) ≤ x k must appear in the initialsequence x x . . . x k .Thus, if x k is not the last letter of x , then x k is followed by a sequence b b . . . b i , eachof whose letters is either b or ( b + 1), and where b < x k . An analogous argumentnow shows that b i must be followed by a sequence b b . . . b i of letters b and b + 1,where b < b , and so on.We now encode a 102-avoiding ascent sequence with a ternary string t = t t . . . t n ,of length n − t starts with 2. Recall that k is the index of the lastcopy of the largest letter in x , that is, the last letter of the initial weakly increasingsequence. In what follows we refer to the prefix x x . . . x k of x as its first part, andthe remainder as its second part, and likewise for the corresponding parts of t . We leteach of the letters t i in the first part of t be a 0 if x i = x i − , and 1 otherwise, that is,if x i = x i − + 1. Some of the 1s in this first part of t will later be changed to 2’s. Werefer the reader to the example after this proof for clarification of this and the rest ofthe proof.The rest of t is constructed from what remains of x , that is, from the tail x k +1 . . . x n .This is the part of x that consists of lifted binary strings with entries that decreasein size between two consecutive such strings. Here, we let t i be a 2 if x i is the firstletter ℓ in one of these lifted binary strings, a 0 if x i = ℓ but is not the first in thestring, or a 1 if x i = ℓ + 1 in the string.We now change some of the 1s in the first part of t to 2’s as follows: Suppose theleftmost 2 in (the second part of) t is t i . Then the integer x i + 1 occurs in the first art of x , and we change t j from 1 to 2, where x j is the leftmost occurrence of x i + 1in x .The inverse of this map is described as follows, where we construct a 102-avoidingascent sequence of length n from a ternary sequence of length n −
1, with 2 k copiesof 2’s: Let t e be the ( k + 1)-st 2 in t . Then, for 2 ≤ i < e , x i = x i − if t i = 0, and x i = x i − + 1 if t i is 1 or 2. We then let x e be x j − x j is the k -th 2 in t , thatis, the last 2 in the first part of t . The binary string of 0’s and 1s strictly between x e and x d , where x d is the next 2 in t after t e , now determines the letters x e +1 through x d − , where t i = 0 means x i = x j − t i = 1 means x i = x j . This is now repeatedfor the remainder of t : x d is set to x m −
1, where t m is the k − t , and thebinary string between t d and the next 2 in t determines the corresponding letters in x in a way analogous to the previous case. This is then repeated for the rest of t . ✷ Here is an example of how the bijection in the proof of Theorem 2.8 works: t = 2 1 0 0 1 2 1 0 2 2 2 1 0 0 2 0 1 2 2 x = 0 1 2 2 2 3 4 5 5 6 7 6 7 6 6 5 5 6 3 0For example, the letter below the penultimate 2 in t is 3 = 4 −
1, because the second 2in t is in a place where x has a 4. Since the third 2 in t has a 6 below it in x , theletter below the third last 2 in t is 6 − x mentioned here are 56, corresponding to the 01 in the corresponding part of t .We next show that the patterns 102, 0102 and 0112 are Wilf equivalent. Theorem 2.9
The patterns 102, 0102 and 0112 are Wilf equivalent. In particular,we have S ( n ) = S ( n ) .Proof. Of course, an occurrence of 0102 implies an occurrence of 102. It suffices toshow that the converse is also true. Let b be a letter involved in an occurrence of 102in an ascent sequence x , and let a and c be letters such that bac is such an occurrence,so a < b < c . Now, 102 is a subpattern of 01012, so, by Lemma 2.4, b must be precededby a , and we therefore have abac as a subsequence in x , constituting an occurrenceof 0102.We next show that the number of 0112-avoiding ascent sequences of length n is givenby (3 n − + 1) /
2, which, together with Theorem 2.8, completes the proof.We claim that the ascent sequences avoiding 0112 are precisely those that consist ofa strictly increasing sequence 012 . . . k followed by a weakly decreasing sequence, theentire sequence arbitrarily interspersed with 0’s. It is clear that a strictly increasingsequence followed by a weakly decreasing sequence cannot contain a 0112, and alsothat interspersing such a sequence with 0’s cannot create a 0112. We thus only needto show that an occurrence of 0112 in an ascent sequence prevents it from havingthis prescribed form. That is straightforward, since the letter corresponding to thesecond 1 in such an occurrence could only belong to the weakly decreasing sequence,but that prevents it from being followed by a larger letter constituting the 2. o count the sequences described in the previous paragraph, observe that such asequence is either the all zero sequence or else it consists, apart from the interspersedzeros, of a strictly increasing sequence 1 , , . . . i for some i ≥
1, followed by a weaklydecreasing sequence of letters a j such that 1 ≤ a j ≤ i . If there are n − − k zeros afterthe initial 0 in such a sequence x , then we can pick their positions in (cid:0) n − n − − k (cid:1) = (cid:0) n − k (cid:1) ways and there are k places left for the increasing and weakly decreasing sequence.Thus, the length of the increasing sequence can range from 1 to k . If the length of theincreasing sequence is i then the weakly decreasing sequence has length k − i and itsletters are positive and not exceeding i . As pointed out in the proof of Theorem 2.1, thenumber of weakly decreasing sequences with such parameters is (cid:0) ( k − i )+( i − k − i (cid:1) = (cid:0) k − k − i (cid:1) .Thus, the total number of ascent sequences avoiding 0112 is1 + n − X k =1 (cid:18) n − k (cid:19) k X i =1 (cid:18) k − k − i (cid:19) , which easily simplifies to (3 n − + 1) / ✷ We now show that ascent sequences avoiding 021 are counted by the Catalan numbers,but first a definition of a set of sequences that turn out to be equinumerous, for each n , with S ( n ). Definition 2.10 A restricted ascent sequence is an ascent sequence x x . . . x n with x i ≥ m i − for all i > , where m i is the maximum value among x , x , . . . , x i − .We denote the set of such sequences of length n by R n . Theorem 2.11
There is a bijection preserving ascents between R n and S ( n ) , forall n . Consequently, A ( n ) = C n , the n -th Catalan number.Proof. In a restricted ascent sequence, the only letters that can appear between twosuccessive LR-maxima x i and x k or after its rightmost LR-maximum x i are x i − x i . Moreover, any ascent sequence satisfying this condition is a restricted ascentsequence.On the other hand, an ascent sequence x is easily seen to avoid 021 if and only itsnonzero entries are weakly increasing. Thus, in particular, x is such a sequence if andonly if it has just two kinds of letters, namely x i and 0, between successive LR-maxima x i and x k , and after its rightmost LR-maximum x i .Interchanging x i − |R n | = C n , which completes the proof. ✷ Theorems 2.5 and 2.11 now imply the following.
Corollary 2.12
The patterns 101, 0101 and 021 are Wilf equivalent. n example of the bijection mentioned in the last paragraph of the proof of Theo-rem 2.11 is given by the following two sequences, where the top one is a restrictedascent sequence, and the bottom one a sequence avoiding 021, and where the onlydifferences between the two sequences are in the places of the zeros in the bottomsequence (apart from the initial zero in each):0 1 2 3 2 3 4 4 3 4 6 50 1 2 3 0 3 4 4 0 4 6 0Our goal is to give a bijection from the set of 021-avoiding ascent sequences to 132-avoiding permutations, and it should be noted that 021 and 132 are equivalent as pat-terns. Since there are very transparent bijections between restricted and 021-avoidingascent sequences, as shown above, and between 132-avoiding and 231-avoiding per-mutations, by reversing each permutation, we choose to exhibit a bijection betweenrestricted ascent sequences and 231-avoiding permutations, which is more convenient.First a definition and a lemma. Definition 2.13
A letter x i in an ascent sequence x x . . . x n is maximal if x i is aslarge as it can be for its place, that is, if x i = asc( x x . . . x i − i ) + 1 . In particular,the initial zero in an ascent sequence is defined to be maximal. If a maximal letter x i satisfies x i = x i +1 = x i +2 = · · · = x i + k = x i + k +1 , or if i + k = n , for some k ≥ ,then we say that x i is a repeated maximal letter , and that x i + k is its last repetition .A maximal letter that is not repeated is its own last repetition. For example, the maximal letters in the following ascent sequence are in bold and thelast repetitions of repeated maximal letters are overlined: Lemma 2.14
In a restricted ascent sequence x = x x . . . x n , let x i be the last rep-etition of the rightmost maximal letter in x , and let x ′ = x i +1 x i +2 . . . x n . Then, if x ′ is nonempty, we must have x i +1 = x i − and the sequence obtained by subtracting x i +1 from each letter in x ′ is a restricted ascent sequence.Proof. Since x i is the last repetition of the rightmost maximal letter, x i +1 must besmaller than x i . Since x is a restricted ascent sequence, x i +1 cannot be smaller than x i − k > i + 1 then x k < x i + 1 + a , where a is the number of ascents in x ′ that strictly precede x k , for otherwise x k would be a maximal letter, contrary to theassumption about x i . Equivalently, x k ≤ x i +1 + 1 + a . Also, x k ≥ x i − x i +1 ,since x is a restricted ascent sequence. Thus, subtracting x i +1 from each letter in x ′ produces a sequence of nonnegative integers that satisfies the condition defining ascentsequences. The resulting sequence is a restricted ascent sequence because subtractingthe same number from each letter doesn’t affect the condition for restricted ascentsequences. ✷ As an example, the last repetition of the rightmost maximal letter in the restricted as-cent sequence x = 00101332232434665 is the second 3, so x ′ = 2232434665. Reducingeach letter in x ′ by 2 we obtain 0010212443, which is a restricted ascent sequence. e now define a bijective map φ : R n → P n (231), where P n (231) is the set of 231-avoiding permutations of length n . Because we construct π = φ ( x ) recursively weneed to define an auxiliary map ω , which takes two arguments, a restricted ascentsequence y and an interval of integers whose length equals that of y . We then set φ ( x ) = ω ( x , [1 , n ]), where n is the length of x .The map ω is called recursively on three segments into which x is partitioned, eachof which is fed to ω together with a part of the original interval. We let ǫ stand forthe empty string and the empty sequence, and set ω ( ǫ, ∅ ) = ǫ . Given a nonemptyrestricted ascent sequence x , let m be its rightmost maximal letter. Then x can beexpressed uniquely as the concatenation LmR , where L is the part of x precedingthe last repetition of m , and R the part following that m . We let R be the restrictedascent sequence obtained by subtracting the first letter of R from each of its letters.Suppose the last repetition of the rightmost maximal letter m is in place p in x .Intuitively, φ , via ω , writes, in the first place in π , the number 1 if m is a repeatedmaximal letter, but p otherwise. That first letter in π is then followed by ω ( L, I ), andthen by ω ( R, J ), where I and J are the appropriate intervals. Formally, we define ω as follows when x = LmR , [ a, b ] is an interval of integers whose length equals thelength of x , ℓ is the length of L and ⊕ is concatenation: ω ( LmR, [ a, b ]) = a ⊕ ω ( L, [ a + 1 , a + ℓ ]) ⊕ ω ( R, [ a + ℓ + 1 , b ]) , if m is repeated , ( a + ℓ ) ⊕ ω ( L, [ a, a + ℓ − ⊕ ω ( R, [ a + ℓ + 1 , b ]) , otherwise . Here is an example of how φ works, via ω . Note that the maximal letters in 011213232are the leftmost occurrences of 0, 1, 2 and 3, with the second 1 the only repeatedmaximal letter: φ (011213232) = ω (011213232 , [1 , ⊕ ω (01121 , [1 , ⊕ ω (232 , [7 , ⊕ ⊕ ω (011 , [1 , ⊕ ω (1 , [5 , ⊕ ω (010 , [7 , ⊕ ⊕ ω (011 , [1 , ⊕ ω (0 , [5 , ⊕ ⊕ ω (0 , [7 , ⊕ ω (0 , [9 , ⊕ ⊕ ⊕ ω (01 , [2 , ⊕ ω ( ∅ , ∅ ) ⊕ ω (0 , [5 , ⊕ ⊕ ω (0 , [7 , ⊕ ω (0 , [9 , ⊕ ⊕ ⊕ ⊕ ω (0 , [2 , ⊕ ω ( ∅ , ∅ ) ⊕ ω ( ∅ , ∅ ) ⊕ ω (0 , [5 , ⊕ ⊕ ω (0 , [7 , ⊕ ω (0 , [9 , ⊕ ⊕ ⊕ ⊕ ⊕ ∅ ⊕ ∅ ⊕ ⊕ ⊕ ⊕ . Theorem 2.15
The map φ described above is a bijection from R n to P n (231) and asc( x ) = des( φ ( x )) for any x ∈ R n . Consequently, the distribution of the number ofascents on S ( n ) (or, equivalently, R n ) is given by the Narayana numbers.Proof. We first prove the claim asc( x ) = des( φ ( x )), then show that φ produces a231-avoiding permutation, and finally show that φ is bijective. By Theorem 2.11 theclaim about ascents applies equally to sequences in R n and S ( n ). t follows by induction from the definition of φ , via the definition of ω , that φ yieldsa permutation of the letters 1 through n , where n is the length of x . When ω writesout a letter f , that letter is followed by two words, constructed from L and R . In thecase when the rightmost maximal letter m of x is repeated, and so is not the secondletter in an ascent in x , f will be smaller than all the letters eventually following itin π , thus not causing a descent in π . If m is not repeated, and is not the first letterin x , then m is the second letter in an ascent in x , and the letter f = a + ℓ writtenout by ω is greater than all the letters arising from applying ω to L , and will thus bethe first letter in a descent in π . Thus φ , via ω , translates ascents in x into descentsin π , and non-ascents into non-descents.Analyzing the two cases in the previous paragraph shows that a letter f written outby ω at any stage can not be the 2 in an occurrence of the pattern 231 in π , since f is either smaller than everything following it, or else followed by letters smaller than f that then are followed by letters larger than f . Since no letter in π can be the 2 inan occurrence of 231, there can be no occurrence of 231 in π , so φ is indeed a map to P n (231).We now show that φ is injective. This is trivially true for n = 1. Suppose x = LmR and x ′ = L ′ m ′ R ′ are restricted ascent sequences of length at least 2, where m and m ′ arethe respective last repetitions of their rightmost maximal letters. If π = φ ( x ) = φ ( x ′ )then either both m and m ′ are repeated maximal letters, in which case π (1) = 1, orneither is. If neither is repeated then π (1) = p >
1, where m and m ′ are in place p in x and x ′ , respectively. Thus, we must have that ω ( L, I ) = ω ( L ′ , I ), where I is theinterval passed with L and L ′ to ω , and likewise for R, R ′ and the interval J passedwith them to ω . By induction, since L, L ′ , R, R ′ are all strictly shorter than x , we mayassume that this implies that L = L ′ and R = R ′ . That, in turn, implies that x = x ′ .If m and m ′ are repetitions of the rightmost maximal letter in x and x ′ , respectively,then π (1) = 1. In fact, π then starts with 12 . . . k where k is the number of repetitionsof the rightmost maximal letters m and m ′ , which must therefore be repeated thesame number of times. Once ω has been applied successively to these repetitions weare back in the case where m and m ′ are not repeated and, by an argument identicalto the one applied to that case above, we can infer that x = x ′ .Since φ : R n → P n (231) and φ is injective, and we know that R n and P n (213) areequinumerous, the map φ is a bijection. ✷ There is an easy, and well known, bijection from 231-avoiding permutations to non-crossing partitions, that is, set partitions of { , , . . . , n } that do not contain a
2) + a ( n −
3) given for sequence
A080937 in OEIS [10].Moreover, Conjecture 3.5 implies a growth rate of 5 for 0021 (and 1012), given therecursive formula for sequence
A007317 in [10].It is a tempting conjecture that the Marcus-Tardos Theorem mentioned above alsoholds for ascent sequences, but this requires more work to be justified.By reversing a 231-avoiding permutation we obtain a 132-avoiding permutation, andturn each descent into an ascent, and conversely. This, together with Theorem 2.15,shows that the number of ascents has the same distribution on 021-avoiding (a.k.a.132-avoiding) ascent sequences and 132-avoiding permutations. We conjecture thatthis result can be strengthened as follows.
Conjecture 3.1
The bistatistic (asc , RL min) has the same distribution on S ( n ) and on 132-avoiding permutations of length n . After a preliminary version of this paper was posted on the internet Bruce Saganproved Conjecture 3.1 (personal communication). et fwd be the length of the maximal final weakly decreasing sequence in an ascentsequence. For example, fwd(01123035523220) = 4, since 3220 has length 4. are therightmost occurrences of 0 , , x ) is the number of 0’s in an ascent sequence x . Conjecture 3.2
We have A ( n ) = C n , the n -th Catalan number. Moreover, thebistatistic (asc , fwd) on S ( n ) has the same distribution as (asc , RL max) does onpermutations avoiding the pattern 132. In particular, this implies that the number ofascents has the Narayana distribution on S ( n ) . Also, the bistatistics (asc , fwd) and (asc , zeros) have the same distribution on 0012-avoiding ascent sequences. We should point out that the bistatistics (asc , RL min) on 132-avoiding permutationsand (asc , RL max) on 231-avoiding permutations mentioned in Conjectures 3.1 and3.2 have the same distribution as double upsteps and the number of returns to the x -axis on Dyck paths. We have found more apparent equidistributions of bistatisticsof a similar kind to those in Conjectures 3.1 and 3.2, and it seems certain that manymore could be found easily. Conjecture 3.3
The number A ( n ) equals the number of non-3-crossing set par-titions of { , , . . . , n } . See sequence A108304 in [10], where non-3-crossing is alsodefined. Conjecture 3.4
The number A ( n ) equals the number of Dyck paths of semilength n and height at most 5. See sequence A080937 in [10]. Conjecture 3.5
The patterns 0021 and 1012 are Wilf equivalent, and A ( n ) = A ( n ) is given by the binomial transform of Catalan numbers, which is sequenceA007317 in [10]. It is easy to show that the Wilf equivalences proved or conjectured here are the onlyones possible for the set of all patterns of length at most four. This is because otherpairs of sequences are seen, by computer testing, to diverge.Thus, the following are all the Wilf equivalences for patterns of length at most four,where the cases of 0012 and the pair (0021 , ∼ , ∼ ∼ ∼ ∼ , ∼ ∼ , ∼ ∼ ∼ , ∼ . hen computing the numbers of ascent sequences avoiding various pairs of patternsof lengths 3 and 4 we found several sequences of numbers recognized by the OEIS [10], such as Motzkin numbers, various transformations of the Catalan and Fibonaccinumbers, and sequences counting permutations avoiding pairs of patterns. We alsofound some apparent Wilf equivalences that might be interesting. We do not list anyof these here, but we will gladly share our data with anybody who might be interested.Finally, we consider the modified ascent sequences defined in [1, Section 4.1]. Givenan ascent sequence x , we successively create sequences x , x , . . . , x k , where k is thenumber of ascents in x . The sequence x i +1 is constructed from x i by increasing by1 each letter in x i that precedes the i -th ascent in x and is larger than or equal tothe larger letter in that ascent. The sequence x k is then defined to be the modifiedascent sequence associated to x . Observe that in constructing each x i we preserve theascents in x . It is easy to see that this is an invertible process so that modified ascentsequences are in bijection with ascent sequences. As an example, the modified ascentsequence of 010221212 successively becomes 010331212 and 010441312, with the firsttwo ascents causing no changes. Conjecture 3.6
On modified ascent sequences the patterns , , , , , are all Wilf equivalent, and the number of ascent sequences avoiding any one ofthese patterns is the n -th Bell number. These sequences are thus equinumerous withpartitions of an n -element set. Moreover, the distribution of the number of ascents onsuch sequences is the reverse of the distribution of the number of blocks on set par-titions. That is, the number of modified ascent sequences of length n with k ascents,and avoiding any single one of these patterns, equals the number of set partitions ofan n -element set with n − k blocks. Bruce Sagan (personal communication) has proved the above conjecture in the caseof the pattern 101 and k = 1.Note that if Conjecture 3.6 is true then the number of modified ascent sequencesavoiding 101 would not have exponential growth, since that is known to be false forthe Bell numbers; they grow faster than C n for any constant C . We are grateful to Anders Claesson, Mark Dukes, Sergey Kitaev and Jeff Remmelfor useful discussions and suggestions. We are also deeply indebted to a referee whocorrected a few errors and made many useful suggestions, which led to a substantialimprovement of the presentation. In particular, that referee pointed out the bijectionnow used in the proof of Theorem 2.5. attern p Number of sequences avoiding p OEIS
Formula Reference001010011012 1 , , , , , , , , , , . . . A000079 2 n − Thms. 2.1, 2.210201020112 1 , , , , , , , , , , . . . A007051 (3 n + 1) / , , , , , , , , , , . . . A000108 1 n + 1 (cid:18) nn (cid:19) Thms. 2.5, 2.11000 1 , , , , , , , , , , , , , , . . .
100 1 , , , , , , , , , , , , , , . . .
110 1 , , , , , , , , , , , , , , . . .
120 1 , , , , , , , , , , , , , , . . .
201 1 , , , , , , , , , , , , , , . . .
210 1 , , , , , , , , , , , , , , . . . A108304 Conj. 3.30123 1 , , , , , , , , , , , , , , . . . A080937 Conj. 3.400211012 1 , , , , , , , , , , , , , . . . A007317 Conj. 3.5
Table 1: Number sequences for pattern avoidance by ascent sequences. All pat-terns of length 3 are listed, a few of length 4.
OEIS refers to entry in [10].17 eferences [1] M. Bousquet-M´elou, A. Claesson, M. Dukes and S. Kitaev: (2+2)-free posets,ascent sequences and pattern avoiding permutations, J. Comb. Theory Ser. A (2010), 884–909.[2] M. Bousquet-M´elou and G. Xin: On partitions avoiding 3-crossings, S´em. Lothar.Combin. (2006), Article B54e, 21 pp.[3] A. Claesson and S. Linusson: n! matchings, n! posets, Proc. Am. Math. Soc. (2011), 435-449.[4] M. Dukes, J. Remmel, S. Kitaev and E. Steingrimsson: Enumerating (2+2)-freeposets by indistinguishable elements, J. Comb. (2011), 139–163.[5] M. Dukes and R. Parviainen: Ascent sequences and upper triangular matricescontaining non-negative integers. Electronic J. Combin. (2010), Monographs in Theoretical Com-puter Science (with a foreword by Jeffrey B. Remmel), Springer-Verlag, ISBN978-3-642-17332-5, 2011.[7] S. Kitaev and J. Remmel: Enumerating (2 + 2)-free posets by the number ofminimal elements and other statistics, Discrete Applied Mathematics, to appear.[8] M. Klazar: Some general results in combinatorial enumeration (English sum-mary), Permutation patterns, 3-40, London Math. Soc. Lecture Note Ser. ,Cambridge Univ. Press, Cambridge, 2010.[9] A. Marcus and G. Tardos: Excluded permutation matrices and the Stanley-Wilfconjecture, J. Combin. Theory Ser. A (2004), 153–160.[10] OEIS Foundation Inc. (2011), The On-Line Encyclopedia of Integer Sequences, http://oeis.org .[11] V. Vatter: Small permutation classes, Proc. London Math. Soc., to appear..[11] V. Vatter: Small permutation classes, Proc. London Math. Soc., to appear.