Permutations avoiding 312 and another pattern, Chebyshev polynomials and longest increasing subsequences
aa r X i v : . [ m a t h . C O ] J a n PERMUTATIONS AVOIDING 312 AND ANOTHER PATTERN,CHEBYSHEV POLYNOMIALS AND LONGEST INCREASINGSUBSEQUENCES
TOUFIK MANSOUR AND G ¨OKHAN YILDIRIM abstract
We study the longest increasing subsequence problem for random permutations avoiding thepattern 312 and another pattern τ under the uniform probability distribution. We determinethe exact and asymptotic formulas for the average length of the longest increasing subse-quences for such permutation classes specifically when the pattern τ is monotone increasingor decreasing, or any pattern of length four.1. Introduction
The study of longest increasing subsequences for uniformly random permutations is a wonder-ful example of a research program which begins with an easy-to-state question whose solutionmakes surprising connections with different branches of mathematics, and culminates withmany astonishing results that have interesting applications in statistics, computer science,physics and biology, see [1, 10–12]. Let σ = σ σ . . . σ n be a permutation of [ n ] := { , . . . , n } .We denote by L n ( σ ) the length of a longest increasing subsequence in σ , that is, L n ( σ ) = max { k ∈ [ n ] : there exist 1 ≤ i < i < · · · < i k ≤ n and σ i < σ i < · · · < σ i k } . Note that, in general, such a subsequence is not unique. Erd¨os-Szekeres theorem [15] statesthat every permutation of length n ≥ ( r − s −
1) + 1 contains either an increasing subse-quence of length r or a decreasing subsequence of length s . After this celebrated result,many researchers worked on the problem of determining the asymptotic behavior of L n on S n , the set of all permutations of length n , under the uniform probability distribu-tion [16, 22, 29, 34, 36]. The problem has been studied by several distinct methods fromprobability theory, random matrix theory, representation theory and statistical mechanics,see [1, 4, 21, 28] and references therein. It is finally known that E ( L n ) ∼ √ n [22, 30, 36] and n − / ( L n − E ( L n )) converges in distribution to the Tracy-Widom distribution as n → ∞ [3, 33]. For a thorough exposition of the subject, see the books [4, 28].We shall study the longest increasing subsequence problem for some pattern-avoiding permu-tation classes. First, we shall recall some definitions. For permutations τ ∈ S k and σ ∈ S n ,we say that τ appears as a pattern in σ if there is a subsequence σ i σ i · · · σ i k of length k in σ which has the same relative order of τ , that is, σ i s < σ i t if and only if τ s < τ t for all1 ≤ s, t ≤ k . For example, the permutation 312 appears as a pattern in 52341 because it Mathematics Subject Classification.
Key words and phrases.
Longest increasing subsequence problem, Pattern-avoiding permutations, Cheby-shev polynomials, Generating functions.
TOUFIK MANSOUR AND G ¨OKHAN YILDIRIM has the subsequences 523 − − , 52 − − or 5 − − . If τ does not appear as a pattern in σ ,then σ is called a τ - avoiding permutation. We denote by S n ( τ ) the set of permutations oflength n that avoid the pattern τ . More generally, for a set T of patterns, we use the notation S n ( T ) = ∩ τ ∈ T S n ( τ ). Pattern-avoiding permutations have been studied from combinatoricsperspectives for many years, for an introduction to the subject, see [8, 35]. Recently prob-abilistic study of random pattern-avoiding permutations has also been initiated, and manyinteresting results have already appeared [5, 17–20, 23, 24, 26].The longest increasing subsequence problem for the pattern-avoiding permutations was firststudied for the patterns of length tree, that is, for τ ∈ S on S n ( τ ) with uniform probabilitydistribution in [13]. The case S n ( τ , τ ) with τ , τ ∈ S is studied for all possible casesin [24]. One of the corollaries of our main result, Theorem 2.2, covers the case S n (312 , τ )with either τ ∈ S (312) or τ ∈ S (312) and hence add some new results to this researchprogram.Note that for any σ ∈ S n , we have(1.1) L n ( σ ) = L n ( σ rc ) = L n ( σ − )where the reverse, complement and inverse of σ are defined as σ ri = σ n +1 − i , σ ci = n + 1 − σ i and σ − i = j if and only if σ j = i , respectively. These symmetries significantly reduce thenumber of cases needed to be studied.In a different direction of research, the longest increasing subsequence problem has also beenstudied on S n under some non-uniform probability distributions such as Mallow distribution[6, 7, 27] and in some other context such as colored permutations [9], independent-identicallydistributed sequences, and random walks [2, 14].The paper is organized as follows. We present our main result, Theorem 2.2, in Section 2 andas a first case apply it to S n (312 , τ ) with τ ∈ S (312) which gives an alternative proof forsome cases considered in [24] through generating functions. In Section 3, we consider threespecific longer patterns where τ is the monotone increasing/decreasing pattern or the pattern( m − m ( m − m − · · · S n (312 , τ )with τ ∈ S (312).For the rest of the paper, we only deal with random variables defined on sets S n (312 , τ )under the uniform probability distribution. That is, for any subset A ⊂ S n (312 , τ ), P τ ( A ) = | A || S n (312 ,τ ) | . The notation E τ ( X ) denotes the expected value of a random variable X on S n (312 , τ ) under P τ . We denote the coefficient of x n in a generating function G ( x ) by [ x n ] G .For two sequences { a n } n ≥ and { b n } n ≥ , we write a n ∼ b n if lim n →∞ a n b n = 1.2. General Results
Note that if τ / ∈ S k (312), then S n (312 , τ ) = S n (312) for all n ≥
1. For any τ ∈ S k (312) with k ≥
2, we define the generating function(2.1) F τ ( x, q ) = X n ≥ X σ ∈ S n (312 ,τ ) x n q L n ( σ ) with F ( x, q ) ≡ E τ ( L n ) = 1 s n [ x n ] ∂∂q F τ ( x, q ) (cid:12)(cid:12)(cid:12) q =1 and E τ ( L n ) = 1 s n [ x n ] (cid:18) ∂ ∂q F τ ( x, q ) (cid:12)(cid:12)(cid:12) q =1 + ∂∂q F τ ( x, q ) (cid:12)(cid:12)(cid:12) q =1 (cid:19) ONGEST INCREASING SUBSEQUENCES 3 where s n = | S n (312 , τ ) | . Note also that s n = [ x n ] F τ ( x, w = w w · · · w m of m -distinct integers, we define the corresponding reducedform to be the unique permutation v = v v · · · v m where v i = ℓ if the w i is the ℓ -th smallestterm in w . For example, the reduced form of 253 is 132. For any sequence w , we define F w ( x, q ) to be F v ( x, q ) where v is the reduced form of w .In order to determine F τ ( x, q ) explicitly, we shall introduce some notations. Let w , w betwo sequences of integers, we write w < w or w > w if w i < w j for all possible i, j . Recallthat for any permutation τ = τ · · · τ k , τ i is called a right-to-left minimum if τ i < τ j for all j > i . Note that, by definition, the last entry τ k is a right-to-left minimum. Let τ ∈ S k (312)and let m = 1 < m < . . . < m r be the right-to-left minima of τ written from left to right.Then τ can be represented as τ = τ (0) m τ (1) m · · · τ ( r ) m r , where m < τ (0) < m < τ (1) < · · · < m r < τ ( r ) , and τ ( j ) (may possibly be empty) avoids312 for each j = 0 , , . . . , r . In this case we call this representation the normal form of τ . Forinstance, if τ = 214365, then the normal form of τ is τ (0) τ (1) τ (2) τ (0) = 2, τ (1) = 4and τ (2) = 6.Assume that τ ∈ S k (312) is given in its normal form, that is, τ = τ (0) m τ (1) m · · · τ ( r ) m r .We use Θ ( j ) to denote τ (0) m τ (1) m · · · τ ( j ) m j , which we call the j th prefix of τ . We use Θ
Assume that τ ∈ S k (312) is given in its normal form, τ = τ (0) m τ (1) m · · · τ ( r ) m r .Then σ = σ ′ σ ′′ avoids both and τ if and only if there exists i , ≤ i ≤ r , such that σ ′ avoids Θ ( i ) and contains Θ ( i − , while σ ′′ avoids Θ .Proof. We denote the set of all permutations, including the empty permutation, that avoidboth 312 and τ by T τ , and that avoid both 312 and τ and contain τ ′ by T τ ; τ ′ . Let σ be anynonempty permutation in T τ . We can write σ as σ = σ ′ σ ′′ . Note that σ avoids 312 if andonly if σ ′ < σ ′′ and both σ ′ and σ ′′ avoid 312. Note also that Θ ( j ) is a prefix of Θ ( j +1) andΘ
1, and Θ ( r ) = Θ < > = τ . We have T τ = T Θ (0) ∪ T τ ;Θ (0) ,T τ ;Θ ( s ) = T Θ ( s +1) ;Θ ( s ) ∪ T τ ;Θ ( s +1) , s = 0 , , , . . . , r − , with T τ ;Θ ( r ) = T τ ; τ = ∅ . Therefore, T τ = T Θ (0) ∪ T Θ (1) ;Θ (0) ∪ T Θ (2) ;Θ (1) ∪ · · · ∪ T Θ ( r ) ;Θ ( r − . Thus, since σ ∈ T τ we have that σ ′ ∈ T τ , so there exists i , 0 ≤ i ≤ r , such that σ ′ ∈ T Θ ( i ) ;Θ ( i − . But then we must have σ ′′ ∈ T Θ . Hence, σ ∈ T τ implies that( ∗ ) there exists i , 0 ≤ i ≤ r , such that σ ′ ∈ T Θ ( i ) ;Θ ( i − and σ ′′ ∈ T Θ .Note that in the case i = 0, σ ′ (0) , while σ ′′ avoids Θ < > , and we defined Θ ( − = ∅ .Then clearly σ ′ ( − .On the other hand, let both σ ′ σ ′′ avoid 312 and satisfy the condition ( ∗ ), that is, thereexists i , 0 ≤ i ≤ r , such that σ ′ ∈ T Θ ( i ) ;Θ ( i − and σ ′′ ∈ T Θ . Thus, both σ ′ σ ′′ TOUFIK MANSOUR AND G ¨OKHAN YILDIRIM avoid τ . If σ = σ ′ σ ′′ contains τ , then σ ′ ( j − and σ ′′ contains Θ
Let τ ∈ S k (312) be given in its normal form τ (0) m τ (1) m · · · τ ( r ) m r with k ≥ . • If τ (0) = ∅ , then F τ ( x, q ) = 1 + xq + x ( F τ ( x, q ) −
1) + xq ( F Θ < > ( x, q ) − x r X j =1 ( F Θ ( j ) ( x, q ) − F Θ ( j − ( x, q ))( F Θ
1. If σ = σ ′ σ ′ = ∅ , then we have the contribution of x ( F τ ( x, q ) − σ = 1 σ ′′ with σ ′′ = ∅ , thenwe have the contribution of xq ( F Θ < > ( x, q ) − σ = σ ′ σ ′′ with σ ′ , σ ′′ = ∅ . By Lemma 2.1, the contribution ofthis case is given by x r X j =2 ( F Θ ( j ) ( x, q ) − F Θ ( j − ( x, q ))( F Θ
1) + x ( F Θ (1) ( x, q ) − F Θ < > ( x, q ) − x r X j =1 ( F Θ ( j ) ( x, q ) − F Θ ( j − ( x, q ))( F Θ
Case τ (0) = ∅ : If σ = σ ′ σ ′ = ∅ , then we have the contribution of x ( F τ ( x, q ) − r ≥
1, and x ( F τ (0) ( x, q ) −
1) when r = 0. If σ = 1 σ ′′ with σ ′′ = ∅ , then we have thecontribution of xq ( F τ ( x, q ) − σ = σ ′ σ ′′ with σ ′ , σ ′′ = ∅ . By Lemma 2.1, the contribution of this case is given by x r X j =2 ( F Θ ( j ) ( x, q ) − F Θ ( j − ( x, q ))( F Θ
For any k ≥ and τ ∈ S k (312) , the generating function F τ ( x, q ) is a rationalfunction in x and q . Note that F ( x, q ) = 1 (the only permutation that avoids the pattern 1 is the empty permu-tation). Theorem 2.2 with τ = 21 gives F ( x, q ) = 1 + xq + x ( F ( x, q ) −
1) + xq ( F ( x, q ) −
1) + x ( F ( x, q ) − F ( x, q ) − , where F ( x, q ) = 1. Thus, F ( x, q ) = − xq . Theorem 2.2 with τ = 12 gives F ( x, q ) = 1 + xq + x ( F ( x, q ) −
1) + xq ( F ( x, q ) −
1) + x ( F ( x, q ) − F ( x, q ) − . Thus, F ( x, q ) = xq − x − x = 1 + xq − x . We summarize these results in the following corollaryfor future references. Corollary 2.4.
For τ ∈ S , the generating functions are given by F ( x, q ) = 11 − xq and F ( x, q ) = 1 + xq − x . Next we will consider the application of Theorem 2.2 to the patterns of length three, that is, τ ∈ S (312). In each case we will also use Corollary 2.4 and F ( x, q ) = 1. Pattern τ = . We have Θ (0) = 1, Θ (1) = 12, Θ (2) = 123, and Θ < > = 123, Θ < > = 12,Θ < > = 1. Thus, F ( x, q ) = 1 + xq + x ( F ( x, q ) −
1) + xq ( F ( x, q ) −
1) + x ( F ( x, q ) − F ( x, q ) − , which gives F ( x, q ) = 1 + xq/ (1 − x ) + x q (1 − x ) . Pattern τ = . We have Θ (0) = 1, Θ (1) = 132, and Θ < > = 132, Θ < > = 21. Thus, F ( x, q ) = 1 + xq + x ( F ( x, q ) −
1) + xq ( F ( x, q ) −
1) + x ( F ( x, q ) − F ( x, q ) − , which gives F ( x, q ) = − x − x − xq . Pattern τ = . We have Θ (0) = 21, Θ (1) = 213, and Θ < > = 213, Θ < > = 1. Thus, F ( x, q ) = 1 + xq + x ( F ( x, q ) −
1) + xq ( F ( x, q ) − , which gives F ( x, q ) = − x − x − xq . TOUFIK MANSOUR AND G ¨OKHAN YILDIRIM
Pattern τ = . We have Θ (0) = Θ < > = 231. Thus, F ( x, q ) = 1 + xq + x ( F ( x, q ) −
1) + xq ( F ( x, q ) −
1) + x ( F ( x, q ) − F ( x, q ) − , which gives F ( x, q ) = − x − x − xq . Pattern τ = . We have Θ (0) = Θ < > = 321. Thus, F ( x, q ) = 1 + xq + x ( F ( x, q ) −
1) + xq ( F ( x, q ) −
1) + x ( F ( x, q ) − F ( x, q ) − , which gives F ( x, q ) = − xq (1 − xq ) − x q .Hence, we can state the following result. Corollary 2.5.
For τ ∈ S (312) , the generating functions are given by F ( x, q ) = 1 + xq − x + x q (1 − x ) ,F ( x, q ) = F ( x, q ) = F ( x, q ) = 1 − x − x − xq ,F ( x, q ) = 1 − xq (1 − xq ) − x q . The results in Corollary 2.5 indeed extend the relevant results of Simion and Schmidt [31] forthe permutations avoiding two patterns of length three. Here we find the generating functionsfor the number of permutations σ in S n (312 , τ ) with τ ∈ S (312) according to the length ofthe longest increasing subsequence in σ .Our next result considers a specific type of pattern in which the last entry is 1. Corollary 2.6.
Assume τ = ρ ∈ S k (312) and k ≥ . Then F τ ( x, q ) = − xq − x ( F ρ ( x,q ) − .Moreover, ∂∂q F τ ( x, q ) (cid:12)(cid:12)(cid:12) q =1 = xF τ ( x, (cid:18) ∂∂q F ρ ( x, q ) (cid:12)(cid:12)(cid:12) q =1 (cid:19) . Proof.
By Theorem 2.2 with τ = ρ r = 0, m = 1 and τ (0) = ρ ), we have F τ ( x, q ) = 1 + xq + x ( F ρ ( x, q ) −
1) + xq ( F τ ( x, q ) −
1) + x ( F ρ ( x, q ) − F τ ( x, q ) − , which implies F τ ( x, q ) = 11 − xq − x ( F ρ ( x, q ) − . In particular, F τ ( x,
1) = − xF ρ ( x, , as shown in [25]. Moreover, by differentiating F τ ( x, q )with respect to q and evaluating at q = 1, we obtain ∂∂q F τ ( x, q ) (cid:12)(cid:12)(cid:12) q =1 = x (cid:18) ∂∂q F ρ ( x, q ) (cid:12)(cid:12)(cid:12) q =1 (cid:19) (1 − xF ρ ( x, = xF τ ( x, (cid:18) ∂∂q F ρ ( x, q ) (cid:12)(cid:12)(cid:12) q =1 (cid:19) , which completes the proof.By Corollary 2.4 and 2.5, we recover the relevant results in [24]. ONGEST INCREASING SUBSEQUENCES 7
Theorem 2.7.
For all n ≥ , we have E ( L n ) = 2( n − n + 1) n − n + 2 , E ( L n ) = 2(2 n − n + 1) n − n + 2 , E ( L n ) = E ( L n ) = E ( L n ) = n + 12 , E ( L n ) = E ( L n ) = E ( L n ) = n ( n + 3)4 , E ( L n ) = 3 n , E ( L n ) = n (9 n + 1)16 . Special cases of longer patterns
The main result of this paper, Theorem 2.2, can be used to obtain general results for severallonger patterns. In this subsection, as an example, we apply it to the following three specificpatterns 12 · · · m , m ( m − · · ·
21 and ( m − m ( m − · · · U j (cos θ ) = sin(( j +1) θ )sin θ .It is well known that these polynomials satisfy U ( t ) = 1 , U ( t ) = 2 t, and U m ( t ) = 2 tU m − ( t ) − U m − ( t ) for all integers m, (3.1)and U n ( t ) = 2 n n Y j =1 (cid:18) t − cos (cid:18) jπn + 1 (cid:19)(cid:19) . (3.2)3.1. Monotone increasing pattern τ = 12 · · · m . In this subsection, we study the pattern τ = 12 · · · m . By Corollaries 2.4 and 2.5, we see that F ( x, q ) = 1, F ( x, q ) = 1 + xq − x and F ( x, q ) = 1 + xq − x + x q (1 − x ) . By Theorem 2.2 with τ = 12 · · · m , we have F ··· m ( x, q ) = 1 + xq + x ( F ··· m ( x, q ) −
1) + xq ( F ··· ( m − ( x, q ) − x m X j =2 ( F ··· j ( x, q ) − F ··· ( j − ( x, q ))( F ··· ( m − j +1) ( x, q ) − . which is equivalent to F ··· m ( x, q ) = 1 + xqF ··· ( m − ( x, q )+ x m X j =2 ( F ··· j ( x, q ) − F ··· ( j − ( x, q )) F ··· ( m − j +1) ( x, q ) . Define G ( x, q, v ) = P m ≥ F ··· m ( x, q ) v m . Then, by multiplying the above recurrence by v m and summing over m , we obtain X m ≥ F ··· m ( x, q ) v m = X m ≥ v m + xqv X m ≥ v m − F ··· ( m − ( x, q )+ x X m ≥ v m m X j =2 ( F ··· j ( x, q ) − F ··· ( j − ( x, q )) F ··· ( m − j +1) ( x, q ) , which implies G ( x, q, v ) = v − v + xqvG ( x, q, v ) + xv ( G ( x, q, v ) − v ) G ( x, q, v ) − x ( G ( x, q, v )) . TOUFIK MANSOUR AND G ¨OKHAN YILDIRIM
Thus, G ( x, q, v ) satisfies v − v + (1 + x − qxv ) G ( x, q, v ) − x (1 − v ) v G ( x, q, v ) = 0 . By solving the above equation for G ( x, q, v ) we obtain G ( x, q, v ) = (1 + x − qxv − p (1 + x − qxv ) − x ) v x (1 − v ) . Then G ( x, q, v (1 − x ) qx ) − v (1 − x )2 qxv (1 − x )2 qx − − x ) v = 1 − v (1 − x ) − p − v (1 + x ) + v (1 − x ) xv (1 − v ) , which, by definition of Narayana numbers (see Sequence A001263 in [32]), leads to G ( x, q, v (1 − x ) qx ) − v (1 − x )2 qxv (1 − x )2 qx − − x ) v = 1 + X n ≥ n X k =1 n (cid:18) nk (cid:19)(cid:18) nk − (cid:19) x k − v n . Therefore, by replacing v by xqv/ (1 − x ) , we have G ( x, q, v ) − vv − xqv − x = 1 + X n ≥ n X k =1 n (cid:18) nk (cid:19)(cid:18) nk − (cid:19) x n + k − (1 − x ) n q n v n . which implies G ( x, q, v ) = v − v qxv − x + X n ≥ n X k =1 n (cid:18) nk (cid:19)(cid:18) nk − (cid:19) q n +1 x n + k v n +1 (1 − x ) n +1 . By finding the coefficient of v m , we obtain the following result. Corollary 3.1.
For all m ≥ , F ··· m ( x, q ) = 1 + qx − x + m − X j =2 q j x j (1 − x ) j − j − X k =1 j − (cid:18) j − k (cid:19)(cid:18) j − k − (cid:19) x k − ! . By Corollary 3.1, we see that the generating function F ··· m ( x,
1) has a pole at x = 1 oforder 2 m −
3. Thus,[ x n ] F ··· m ( x, ∼ n m − (2 m − m − m − X k =1 (cid:18) m − k (cid:19)(cid:18) m − k − (cid:19) , which, by definition of Narayana numbers, implies that[ x n ] F ··· m ( x, ∼ n m − (2 m − c m − , where c n = n +1 (cid:0) nn (cid:1) is the n th Catalan number.
ONGEST INCREASING SUBSEQUENCES 9
Also, by Corollary 3.1, we see that the generating function ∂∂q F ··· m ( x, q ) | q =1 has a pole at x = 1 of order 2 m −
3. Thus,[ x n ] ∂∂q F ··· m ( x, q ) | q =1 ∼ ( m − n m − (2 m − m − m − X k =1 (cid:18) m − k (cid:19)(cid:18) m − k − (cid:19) = ( m − n m − (2 m − c m − . Hence, we can state the following result.
Theorem 3.2.
Let m ≥ . When n → ∞ , we have E ··· m ( L n ) ∼ m − . Monotone decreasing pattern τ = m ( m − · · · . In this subsection, we study thepattern m = m ( m − · · · Corollary 3.3.
Let m = m ( m − · · · . Then F m ( x, q ) = U m − ( t ) − √ xU m − ( t ) √ x ( U m − ( t ) − √ xU m − ( t )) , where t = x − xq √ x .Proof. The proof is given by induction on m . Clearly, F ( x, q ) = 1 and F ( x, q ) = − xq , sothe claim holds for m = 1 ,
2. Assume that the claim holds for 1 , , · · · , m and let’s us prove itfor m + 1. Since m + = ( m + 1)1, then Corollary 2.6 gives F m + ( x, q ) = − xq − x ( F m ( x,q ) − .Thus by induction assumption, we obtain F m + ( x, q ) = 11 − xq − x ( F m ( x, q ) − √ x ( U m − ( t ) − √ xU m − ( t )) x (2 tU m − ( t ) − U m − ( t )) − x √ x (2 tU m − ( t ) − U m − ( t ))= √ x ( U m − ( t ) − √ xU m − ( t )) x ( U m ( t ) − √ xU m − ( t ))= U m − ( t ) − √ xU m − ( t ) √ x ( U m ( t ) − √ xU m − ( t ))where we used the fact (3.1) and 2 t √ x = 1 + x − xq .By Corollary 3.3 with q = 1 and (3.1), we have F m ( x,
1) = U m − ( √ x ) √ xU m ( √ x ) , as shown in [25].Moreover, by Corollary 2.6, we have ∂∂q F m ( x, q ) (cid:12)(cid:12)(cid:12) q =1 = xF m ( x, (cid:18) ∂∂q F m − ( x, q ) (cid:12)(cid:12)(cid:12) q =1 (cid:19) with ∂∂q F ( x, q ) (cid:12)(cid:12)(cid:12) q =1 = 0. Thus, by induction on m , we can state the following result. Corollary 3.4.
Let m = m ( m − · · · . Then ∂∂q F m ( x, q ) (cid:12)(cid:12)(cid:12) q =1 = 1 U m ( √ x ) m − X j =1 U j ( 12 √ x ) . Since the smallest pole of 1 /U n ( x ) is cos (cid:16) πn +1 (cid:17) , it follows, by Corollary 3.4, that the coefficientof x n in the generating function ∂∂q F m ( x, q ) (cid:12)(cid:12)(cid:12) q =1 is given by,[ x n ] ∂∂q F m ( x, q ) (cid:12)(cid:12)(cid:12) q =1 ∼ α m n (cid:18) (cid:18) πm + 1 (cid:19)(cid:19) n as n → ∞ . Let v =
14 cos ( πm +1 ) . The constant α m can be computed explicitly as α m = lim x → v (cid:16) − x cos (cid:16) πm +1 (cid:17)(cid:17) U m ( √ x ) m − X j =1 U j ( 12 √ x )= P m − j =1 U j (cid:16) cos (cid:16) πm +1 (cid:17)(cid:17)(cid:16) (cid:16) πm +1 (cid:17)(cid:17) m Q m − j =2 (cid:18) − cos ( jπm +1 ) cos ( πm +1 ) (cid:19) = P m − j =1 U j (cid:16) cos (cid:16) πm +1 (cid:17)(cid:17) m cos (cid:16) πm +1 (cid:17) Q m − j =2 (cid:16) cos (cid:16) πm +1 (cid:17) − cos (cid:16) jπm +1 (cid:17)(cid:17) . Moreover, the coefficient of x n in the generating function F m ( x,
1) = U m − ( √ x ) √ xU m ( √ x ) is given by[ x n ] F m ( x, ∼ ˜ α m (cid:18) (cid:18) πm + 1 (cid:19)(cid:19) n as n → ∞ , (3.3)where ˜ α m = lim x → v (cid:16) − x cos (cid:16) πm +1 (cid:17)(cid:17) U m − ( √ x ) √ xU m ( √ x )= U m − (cos( πm +1 ))2 m − cos m − ( πm +1 ) Q m − j =2 (cid:18) − cos ( jπm +1 ) cos ( πm +1 ) (cid:19) = U m − (cos( πm +1 ))2 m − cos (cid:16) πm +1 (cid:17) Q m − j =2 (cid:16) cos (cid:16) πm +1 (cid:17) − cos (cid:16) jπm +1 (cid:17)(cid:17) . Thus we have E m ( L n ) ∼ α m ˜ α m n . By substituting expressions of α m and ˜ α m , it leads to thefollowing result. Theorem 3.5.
Let m ≥ . When n → ∞ , we have E m ( L n ) ∼ P m − j =1 U j (cid:16) cos (cid:16) πm +1 (cid:17)(cid:17) m +1 cos (cid:16) πm +1 (cid:17) U m − (cos( πm +1 )) Q m − j =2 (cid:16) cos (cid:16) πm +1 (cid:17) − cos (cid:16) jπm +1 (cid:17)(cid:17) n. For example, Theorem 3.5 for m = 3 gives that E ( L n ) ∼ n as shown in [24], and for m = 4, we have E ( L n ) ∼ (cid:16) − √ (cid:17) n . ONGEST INCREASING SUBSEQUENCES 11
Pattern τ = ( m − m ( m − m − · · · . In this subsection, we study the patternˆ m = ( m − m ( m − m − · · · Corollary 3.6.
Let ˆ m = ( m − m ( m − m − · · · with m ≥ . Then F ˆ m ( x, q ) = (1 − x ) U m − ( t ) − √ x (1 − x + xq ) U m − ( t ) √ x ((1 − x ) U m − ( t ) − √ x (1 − x + xq ) U m − ( t )) , where t = x − xq √ x .Proof. We proceed the proof by induction on m . By Example τ = 231 in Section 2, we have F ˆ ( x, q ) = − x − x − xq , so the result holds for m = 3. Assume that the result holds for m − m . By Corollary 2.6, we have F ˆ m ( x, q ) = 11 + x − xq − xF ˆ m − ( x, q ) . Thus by induction hypothesis, we have F ˆ m ( x, q ) = 12 √ xt − √ x (1 − x ) U m − ( t ) −√ x (1 − x + xq ) U m − ( t )(1 − x ) U m − ( t ) −√ x (1 − x + xq ) U m − ( t ) = (1 − x ) U m − ( t ) − √ x (1 − x + xq ) U m − ( t ) √ x ((1 − x )(2 tU m − ( t ) − U m − ( t )) − √ x (1 − x + xq )(2 tU m − ( t ) − U m − ( t ))) , which, by (3.1), implies F ˆ m ( x, q ) = (1 − x ) U m − ( t ) − √ x (1 − x + xq ) U m − ( t ) √ x ((1 − x ) U m − ( t ) − √ x (1 − x + xq ) U m − ( t )) , which completes the proof.By Corollary 3.6 with q = 1 and (3.1), we have F ˆ m ( x,
1) = U m − ( √ x ) √ xU m ( √ x ) . By induction on m , we can state the following result. Corollary 3.7.
Let ˆ m = ( m − m ( m − m − · · · with m ≥ . Then ∂∂q F ˆ m ( x, q ) (cid:12)(cid:12)(cid:12) q =1 = 1 U m ( √ x ) U ( 12 √ x ) + m − X j =2 U j ( 12 √ x ) . By similar arguments as in the proof of Theorem 3.5, we obtain the following result.
Theorem 3.8.
Let m ≥ . As n → ∞ , we have E ˆ m ( L n ) ∼ U (cid:16) cos (cid:16) πm +1 (cid:17)(cid:17) + P m − j =1 U j (cid:16) cos (cid:16) πm +1 (cid:17)(cid:17) m +1 cos (cid:16) πm +1 (cid:17) U m − (cos( πm +1 )) Q m − j =2 (cid:16) cos (cid:16) πm +1 (cid:17) − cos (cid:16) jπm +1 (cid:17)(cid:17) n. τ F τ ( x, q ) E τ ( L n ) Reference1234 1 + xq − x + x q (1 − x ) + x (1+ x ) q (1 − x ) n − n +9 n − n +4) n − n +11 n − n +12 → xq ( qx (2 x − − x ) )(1 − x − qx ) (1 − x ) 2 n − ( n − n +4)( n − n − +1 ∼ n Example 4.22134 Theorem 2.22314 1 + xq (1 − x )(1 − x ) − qx ∼ n √ Theorem 2.213422143,3214 − x − qx (1 − qx ) − x ∼ n √ Theorem 2.22431,32413421,14322341,4321 (1 − x ) (1 − x ) − xq (1 − x ) − x q ∼ ( − a +22 a − n Theorem 2.2 a ≈ . · · · , a − a + 5 a − Table 1.
A summary of the results for S n (312 , τ ) with τ ∈ S (312).4. The case S n (312 , τ ) where τ ∈ S (312)In this section, we present the results for random permutations from S n (312 , τ ) where τ ∈ S (312). A summary of the results for all τ ∈ S (312) is given in Table 1. We present thedetails only for the two patterns, τ = 1234 and τ = 1243. Since the computations for othercases are very similar, the details are omitted. Example 4.1.
By Theorem 2.2 with τ = 1234 , we have F ( x, q ) = 1 + xqF ( x, q ) + x ( F ( x, q ) − F ( x, q )) F ( x, q )+ x ( F ( x, q ) − F ( x, q )) F ( x, q ) + x ( F ( x, q ) − F ( x, q )) F ( x, q ) . By Corollaries 2.4 and 2.5, we have F ( x, q ) = 1 + xq − x + x q (1 − x ) + x (1 + x ) q (1 − x ) , which agrees with Theorem 3.1 with m = 4 . Therefore, we have E ( L n ) = 3( n − n + 9 n − n + 4) n − n + 11 n − n + 12 and E ( L n ) = 3(3 n − n + 23 n − n + 4) n − n + 11 n − n + 12 . Example 4.2.
By Theorem 2.2 with τ = 1243 , we have F ( x, q ) = 1 + xqF ( x, q ) + x ( F ( x, q ) − x ( F ( x, q ) − F ( x, q ) −
1) + x ( F ( x, q ) − F ( x, q ))( F ( x, q ) − , which, by Corollaries 2.4 and 2.5, leads to F ( x, q ) = 1 + xq ( qx (2 x −
1) + (1 − x ) )(1 − x − qx ) (1 − x ) . ONGEST INCREASING SUBSEQUENCES 13
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