TTHE SHAPE OF RANDOM PATTERN-AVOIDING PERMUTATIONS
SAM MINER (cid:63)
AND IGOR PAK (cid:63)
Abstract.
We initiate the study of limit shapes for random permutations avoiding a givenpattern. Specifically, for patterns of length 3, we obtain delicate results on the asymptoticsof distributions of positions of numbers in the permutations. We view the permutations as0-1 matrices to describe the resulting asymptotics geometrically. We then apply our resultsto obtain a number of results on distributions of permutation statistics.
Introduction
The
Catalan numbers is one of the most celebrated integer sequences, so much that it is hard tooverstate their importance and applicability. In the words of Thomas Koshy, “Catalan numbersare [..] fascinating. Like the North Star in the evening sky, they are a beautiful and bright lightin the mathematical heavens.” Richard Stanley called them “the most special” and his “favoritenumber sequence” [Kim]. To quote Martin Gardner, “they have the delightful propensity forpopping up unexpectedly, particularly in combinatorial problems” [Gar]. In fact, Henry Gould’sbibliography [Gou] lists over 450 papers on the subject, with many more in recent years.Just as there are many combinatorial interpretations of Catalan numbers [S1, Exc. 6.19] (seealso [P2, Slo, S2]), there are numerous results on statistics of various such interpretations (seee.g. [B2, S1]), as well as their probabilistic and asymptotic behavior (see [Drm, FS]). The latterresults usually come in two flavors. First, one can study the probability distribution of statistics,such as the expectation, the standard deviation and higher moments. The approach we favoris to define the shape of a large random object, which can be then analyzed by analytic means(see e.g. [A2, Ver, VK]). Such objects then contain information about a number of statistics,under one roof.In this paper we study the set S n ( π ) of permutations σ ∈ S n avoiding a pattern π . Thisstudy was initiated by Percy MacMahon and Don Knuth, who showed that the size of S n ( π ) isthe Catalan number C n , for all permutations π ∈ S [Knu, Mac]. These results opened a wayto a large area of study, with numerous connections to other fields and applications [Kit] (seealso Subsection 9.2).We concentrate on two classical patterns, the - and -avoiding permutations. Naturalsymmetries imply that other patterns in S are equinumerous with these two patterns. Weview permutations as 0-1 matrices, which we average, scale to fit a unit square, and study theasymptotic behavior of the resulting family of distributions. Perhaps surprisingly, behavior ofthese two patterns is similar on a small scale (linear in n ), with random permutations approxi-mating the reverse identity permutation ( n, n − , . . . , n α away from the diagonal), the asymptotics of shapes of random permutationsin S n (123) and S n (132), are substantially different. This explains, perhaps, why there are atleast nine different bijections between two sets, all with different properties, and none truly“ultimate” or “from the book” (see Subsection 9.3).Our results are rather technical and contain detailed information about the random pat-tern avoiding permutations, on both the small and large scale. We exhibit several regimes (cid:63) Department of Mathematics, UCLA, Los Angeles, CA, 90095. Email: { samminer,pak } @math.ucla.edu . a r X i v : . [ m a t h . C O ] N ov SAM MINER AND IGOR PAK (or “phases”), where the asymptotics are unchanged, and painstakingly compute the preciselimits, both inside these regimes and at the phase transitions. Qualitatively, for -avoidingpermutations, our results are somewhat unsurprising, and can be explained by the limit shaperesults on the brownian excursion (see Subsection 9.7); still, our results go far beyond what wasknown. However, for the -avoiding permutations, our results are extremely unusual, andhave yet to be explained even on a qualitative level (see Subsection 9.8).The rest of the paper is structured as follows. In the next section we first present examplesand calculations which then illustrate the “big picture” of our results. In Section 2 we giveformal definitions of our matrix distributions and state basic observations on their behavior.We state the main results in Section 3, in a series of six theorems of increasing complexity, forthe shape of random permutations in S n (123) and S n (132), three for each. Sections 4 and 5contain proofs of the theorems. In the next three sections (sections 6, 7 and 8), we give a longseries of corollaries, deriving the distributions for the positions of 1 and n , the number andlocation of fixed points, and the generalized rank. We conclude with final remarks and openproblems (Section 9). 1. The big picture
In this section we attempt to give a casual description of our results, which basically makesthis the second, technical part of the introduction. The setup.
Let P n ( j, k ) and Q n ( j, k ) be the number of - and -avoiding permu-tations, respectively, of size n , that have j in the k -th position. These are the main quantitieswhich we study in this paper.There are two ways to think of P n ( · , · ) and Q n ( · , · ). First, we can think of these as familiesof probability distributions1 C n P n ( j, · ) , C n P n ( · , k ) , C n Q n ( j, · ) , and 1 C n Q n ( · , k ) . In this setting, we find the asymptotic behavior of these distributions, where they are con-centrated and the tail asymptotics; we also find exactly how they depend on parameters j and k .Alternatively, one can think of P n ( · , · ) and Q n ( · , · ) as single objects, which we can view as abistochastic matrices:P n = 1 C n (cid:88) σ ∈S (123) M ( σ ) , Q n = 1 C n (cid:88) σ ∈S (132) M ( σ ) , where M ( σ ) is a permutation matrix of σ ∈ S n , defined so that M ( σ ) jk := (cid:40) σ ( j ) = k σ ( j ) (cid:54) = k. This approach is equivalent to the first, but more conceptual and visually transparent, sinceboth P n and Q n have nice geometric asymptotic behavior when n → ∞ . See Subsection 9.1 formore on this difference.Let us present the “big picture” of our results. Roughly, we show that both matrices P n and Q n are very small for ( j, k ) sufficiently far away from the anti-diagonal ∆ = { ( j, k ) | j + k = n + 1 } , and from the lower right corner ( n, n ) in the case of Q n . However, already on the next level ofdetail there are large differences: P n is exponentially small away from the anti-diagonal, whileQ n is exponentially small only above ∆, and decreases at a rate Θ( n − / ) on squares below ∆. Here and always when in doubt, we follow Gian-Carlo Rota’s advice on how to write an Introduction [Rota].
HAPE OF PATTERN-AVOIDING PERMUTATIONS 3
At the next level of detail, we look inside the “phase transition”, that is what happens when( j, k ) are near ∆. It turns out, matrix P n maximizes at distance Θ( √ n ) away from ∆, wherethe values scale as Θ( n − / ), i.e. much greater than average 1 /n . On the other hand, on theanti-diagonal ∆, the values of P n scale as Θ( n − / ), i.e. below the average. A similar, but muchmore complicated phenomenon happens for Q n . Here the “phase transition” splits into severalphases, with different asymptotics for rate of decrease, depending on how the distance from( j, k ) to ∆ relates to Θ( √ n ) and Θ( n / ) (see Section 3).At an even greater level of detail, we obtain exact asymptotic constants on the asymptoticbehavior of P n and Q n , not just the rate of decrease. For example, we consider P n at distanceΘ( n / ε ) from ∆, and show that P n is exponentially small for all ε >
0. We also show thatbelow ∆, the constant term implied by the Θ notation in the rate Θ( n − / ) of decrease of Q n ,is itself decreasing until “midpoint” distance n/ P n ( j, k ) and Q n ( j, k ) which we give in lemmas 4.2 and 5.3.These are proved by direct combinatorial arguments. From that point on, the proofs of theasymptotic behavior of P n and Q n are analytic and use no tools beyond Stirling’s formula andthe Analysis of Special Functions.1.2. Numerical examples.
First, in Figures 11 and 12 we compute the graphs of P and Q (see the Appendix). Informally, we name the diagonal mid-section of the graph of P the ca-noe ; this is the section of the graph where the values are the largest. Similarly, we use the wall for the corresponding mid-section of Q minus the corner spike. The close-up views of thecanoe and the wall are given in Figures 13 and 14, respectively. Note that both graphs here arequite smooth, since n = 250 is large enough to see the the limit shape, with C ≈ . × ,and every pixelated value is computed exactly rather than approximated.Observe that the canoe is symmetric across both the main and the anti-diagonal, and containthe high spikes in the corners of the canoe, both of which reach 1 /
4. Similarly, the wall issymmetric with respect to the main diagonal, and has three spikes which reach 1 /
4. Theseresults are straightforward and proved in the next section.To see that the canoe is very thin, we compare graphs of the diagonal sections P n ( k, k ) /C n for n = 62 , ,
250 and 500, as k varies from 90 to 160 (see Figures 7 to 10 in the Appendix).Observe that as n increases, the height of the canoes decreases, and so does the width and“bottom”. As we mentioned earlier, these three scale as Θ( n − / ), Θ( n − / ), and Θ( n − / ),respectively. Note also the sharp transition to a near flat part outside of the canoe; this isexplained by an exponential decrease mentioned earlier. The exact statements of these resultsare given in Section 3.Now, it is perhaps not clear from Figure 13 that the wall bends to the left. To see thisclearly, we overlap two graphs in Figure 1. Note that the peak of P ( k, k ) is roughly in thesame place of Q ( k, k ), i.e. well to the left of the midpoint at 125. The exact computationsshow that the maxima occur at 118 and at 119, respectively. Note also that Q ( k, k ) has asharp phase transition on the left, with an exponential decay, but only a polynomial decay onthe right.1.3. Applications.
We mention only one statistic which was heavily studied in previous years,and which has a nice geometric meaning. Permutation σ is said to have a fixed point at k if σ ( k ) = k . Denote by fp( σ ) the number of fixed points in σ .For random permutations σ ∈ S n , the distribution of fp is a classical problem; for example E [fp] = 1 for all n . In an interesting paper [RSZ], the authors prove that the distribution of fpon S (321) and on S (132) coincide (see also [E2, EP]). Curiously, Elizalde used the generating SAM MINER AND IGOR PAK
Figure 1.
Comparison of P ( k, k ) /C and Q ( k, k ) /C .function technique to prove that E [fp] = 1 in both cases, for all n , see [E1]. He also finds closedform g.f. formulas for the remaining two patterns (up to symmetry).Now, the graphs of P n ( k, k ) /C n and Q n ( k, k ) /C n discussed above, give the expectationsthat k is a fixed point in a pattern avoiding permutation. In other words, fixed points ofrandom permutations in S n (123) and S n (132) are concentrated under the canoe and under thewall, respectively. Indeed, our results immediately imply that w.h.p. they lie near n/ S n (321) and S n (231), the fixed points lie in the ends of thecanoe and near the corners of the wall, respectively. In Section 7, we qualify all these statementsand as a show of force obtain sharp asymptotics for E [fp] in all cases, both known and new.2. Definitions, notations and basic observations
Asymptotics.
Throughout this paper we use f ( n ) ∼ g ( n ) to denotelim n →∞ f ( n ) g ( n ) = 1 . We use f ( n ) = O ( g ( n )) to mean that there exists a constant M and an integer N such that | f ( n ) | ≤ M | g ( n ) | for all n > N. Also, f ( n ) = Θ( g ( n )) denotes that f ( n ) = O ( g ( n )) and g ( n ) = O ( f ( n )). Similarly, f ( n ) = o ( g ( n )) is defined by lim n →∞ f ( n ) g ( n ) = 0 . Recall Stirling’s formula n ! ∼ √ πn (cid:16) ne (cid:17) n . We use C n to denote the n -th Catalan number: C n = 1 n + 1 (cid:18) nn (cid:19) , and C n ∼ n √ πn . Pattern avoidance.
Let n and m be positive integers with m ≤ n , and let σ = ( σ (1) , σ (2) , . . . , σ ( n )) ∈ S n , and pattern τ = ( τ (1) , τ (2) , . . . , τ ( m )) ∈ S m . We say that σ contains τ if there exist in-dices i < i < . . . < i m such that ( σ ( i ) , σ ( i ) , . . . , σ ( i m )) is in the same relative order as( τ (1) , τ (2) , . . . , τ ( m )). If σ does not contain τ then we say σ is τ - avoiding , or avoiding pat-tern τ . In this paper we use only τ ∈ S ; to simplify the notation we use and todenote patterns (1 , ,
3) or (1 , , σ = (2 , , , ,
3) contains , HAPE OF PATTERN-AVOIDING PERMUTATIONS 5 since the subsequence ( σ (1) , σ (2) , σ (5)) = (2 , ,
3) has the same relative order as (1 , , σ = (5 , , , ,
2) is -avoiding.Denote by S n ( π ) the set of π -avoiding permutations in S n . For the case of patterns oflength 3, it is known that regardless of the pattern π ∈ S , we have |S n ( π ) | = C n . Theorem 2.1 (MacMahon, Knuth) . For all π ∈ S , we have | S n ( π ) | = C n . While the equalities |S n (132) | = |S n (231) | = |S n (213) | = |S n (312) | and |S n (123) | = |S n (321) | are straightforward, the fact that |S n (132) | = |S n (123) | is more involved.2.3. Symmetries.
Recall that P n ( j, k ) and Q n ( j, k ) denote the number of permutations in S n (123) and S n (132), respectively, of size n that have j in the k -th position. In this section wediscuss the symmetries of such permutations. Proposition 2.2.
For all n, j, k positive integers such that ≤ j, k ≤ n , we have P n ( j, k ) = P n ( k, j ) and Q n ( j, k ) = Q n ( k, j ) . Also, P n ( j, k ) = P n ( n + 1 − k, n + 1 − j ) , for all k, j as above. The proposition implies that we can interpret P n ( j, k ) as either |{ σ ∈ S n (123) s.t. σ ( j ) = k }| or |{ σ ∈ S n (123) s.t. σ ( k ) = j }| , and we use both formulas throughout paper. The analogous statement holds with Q n ( j, k ) aswell. Note, however, that Q n ( j, k ) is not necessarily equal to Q n ( n + 1 − k, n + 1 − j ); forexample, Q (1 ,
2) = 2 but Q (2 ,
3) = 1. In other words, there is no natural analogue of thesecond part of the proposition for Q n ( j, k ), even asymptotically, as our results will show in thenext section. Proof.
The best way to see this is to consider permutation matrices. Observe that P n ( j, k )counts the number of permutation matrices A = ( a rs ) which have a kj = 1, but which have norow indices i < i < i nor column indices j < j < j such that a i j = a i j = a i j = 1.If A is such a matrix, then B = A T is also a matrix which satisfies the conditions for P n ( k, j ),since b jk = 1 and B has no indices which lead to the pattern . This transpose map is clearlya bijection, so we have P n ( j, k ) = P n ( k, j ).Similarly, since any pattern in a permutation matrix A will be preserved in B = A T ,we have Q n ( j, k ) = Q n ( k, j ). Finally, observe that P n ( j, k ) = P n ( n + 1 − k, n + 1 − j ), since σ is -avoiding if and only if ρ = ( n + 1 − σ ( n ) , n + 1 − σ ( n − , . . . , n + 1 − σ (1)) is -avoiding. (cid:3) Maxima and minima.
Here we find all maxima and minima of matrices P n ( · , · ) and Q n ( · , · ). We separate the results into two propositions. Proposition 2.3.
For all n ≥ , the value of P n ( j, k ) is minimized when ( j, k ) = (1 , or ( n, n ) . Similarly, P n ( j, k ) is maximized when ( j, k ) = (1 , n ) , (1 , n − , (2 , n ) or ( n − , , ( n, , ( n, . Proof.
For any n , the only σ ∈ S n (123) with σ (1) = 1 is σ = (1 , n, n − , . . . , , P n (1 ,
1) = 1. Similarly, P n ( n, n ) = 1, since the only such permutation is ( n − , n − , . . . , , , n ).For every j, k ≤ n , the maximum possible value of P n ( j, k ) is C n − , since the numbers from1 to n excluding j must be -avoiding themselves. Let us show that P n (1 , n ) = P n (2 , n ) = P n (1 , n −
1) = C n − , proving that this maximum is in fact achieved by the above values of j and k .If σ ∈ S n (123) has σ (1) = n , then n cannot be part of a -pattern, since it is the highestnumber but must be the smallest number in the pattern. Therefore, any σ ∈ S n − (123) can beextended to a permutation τ ∈ S n (123) in the following way: let τ (1) = n , and let τ ( i ) = σ ( i − ≤ i ≤ n . Since |S n − (123) | = C n − , we have P n (1 , n ) = C n − . Similarly, if σ (2) = n , then SAM MINER AND IGOR PAK n cannot be part of a -pattern, so P n (2 , n ) = C n − . The same is true if σ (1) = ( n − P n (1 , n −
1) = C n − . By Proposition 2 .
2, we also have P n ( n − ,
1) = P n ( n,
1) = P n ( n,
2) = C n − , as desired. (cid:3) Proposition 2.4.
For all n ≥ , the value of Q n ( j, k ) is minimized when ( j, k ) = (1 , .Similarly, Q n ( j, k ) is maximized when ( j, k ) = (1 , n ) , (1 , n − , ( n − , , ( n, , or ( n, n ) . Proof.
For ( j, k ) = (1 , -avoiding permutation is σ = (1 , , , . . . , n − , n ).Therefore, Q n (1 ,
1) = 1 for all n .For the second part, we use the same reasoning as in Proposition 2.3, except for ( j, k ) = ( n, n ).For ( j, k ) = ( n, n ), we have Q n ( n, n ) = C n − as well, since n in the final position cannot bepart of a 132-pattern. Observe that unlike P n (2 , n ), Q n (2 , n ) < C n − , since σ (2) = n requires σ (1) = n −
1, in order to avoid a -pattern. (cid:3) Main results
In this section we present the main results of the paper.3.1.
Shape of 123-avoiding permutations.
Let 0 ≤ a, b ≤ , ≤ α <
1, and c ∈ R s.t. c (cid:54) = 0for α (cid:54) = 0 be fixed constants. Recall that P n ( j, k ) is the number of permutations σ ∈ S n (123)with σ ( j ) = k . Define F ( a, b, c, α ) = sup (cid:26) d ∈ R + (cid:12)(cid:12)(cid:12) lim n →∞ n d P n ( an − cn α , bn − cn α ) C n < ∞ (cid:27) for α (cid:54) = 0 or a + b (cid:54) = 1 , and F ( a, b, c, α ) = sup (cid:26) d ∈ R + (cid:12)(cid:12)(cid:12) lim n →∞ n d P n ( an − cn α + 1 , bn − cn α ) C n < ∞ (cid:27) for α = 0 and a + b = 1 . Similarly, let L ( a, b, c, α ) = lim n →∞ n F ( a,b,c,α ) P n ( an − cn α , bn − cn α ) C n for α (cid:54) = 0 or a + b (cid:54) = 1 , and L ( a, b, c, α ) = lim n →∞ n F ( a,b,c,α ) P n ( an − cn α + 1 , bn − cn α ) C n for α = 0 and a + b = 1 , defined for all a, b, c, α as above, for which F ( a, b, c, α ) < ∞ ; let L be undefined otherwise. Theorem 3.1.
For all ≤ a, b ≤ , c ∈ R and ≤ α < , we have F ( a, b, c, α ) = ∞ if a + b (cid:54) = 1 , ∞ if a + b = 1 , c (cid:54) = 0 , α > , if a + b = 1 , c = 0 , − α if a + b = 1 , c (cid:54) = 0 , α ≤ . Here F ( a, b, c, α ) = ∞ means that P n ( an − cn α , bn − cn α ) = o ( C n /n d ), for all d >
0. Thefollowing result proves the exponential decay of these probabilities.
HAPE OF PATTERN-AVOIDING PERMUTATIONS 7
Theorem 3.2.
Let ≤ a, b ≤ s.t. a + b (cid:54) = 1 , c ∈ R , and < α < . Then, for n large enough,we have P n ( an − cn α , bn − cn α ) C n < ε n , where ε = ε ( a, b, c, α ) is independent of n , and < ε < . Similarly, let ≤ a ≤ , c (cid:54) = 0 , and < α < . Then, for n large enough, we have P n ( an − cn α , (1 − a ) n − cn α ) C n < ε n α − , where ε = ε ( a, c, α ) is independent of n and < ε < . These theorems compare the growth of P n ( an − cn α , bn − cn α ) to the growth of C n . Clearly, n (cid:88) j =1 P n ( j, k ) = |S n (123) | = C n for all 1 ≤ k ≤ n . Therefore, if P n ( j, k ) were constant across all values of j, k between 1 and n , we would have P n ( j, k ) = C n /n for all 1 ≤ j, k ≤ n . Theorem 3.1 states that for 0 ≤ a, b ≤ a + b (cid:54) = 1, wehave P n ( an, bn ) = o ( C n /n d ), for every d ∈ R . For a + b = 1, we have P n ( an, bn ) = Θ( C n /n ).Theorem 3.1 is in fact stating slightly more. When we consider P n ( an − cn α , bn − cn α ) insteadof P n ( an, bn ), we have P n ( an − cn α , bn − cn α ) = Θ (cid:16) C n /n − α (cid:17) , for all α ≤ . This relationship can be seen in Figures 2 and 3.01 1 ab γ γ : { a + b = 1 − c √ n } γ (cid:48) : { a + b = 1 + c √ n } γ (cid:48) Figure 2.
Region where P n ( an, bn ) ∼ C n /n d for some d .In Figure 3, on γ , we have P n (cid:0) an − c √ n, (1 − a ) n − c √ n (cid:1) = Θ (cid:18) C n √ n (cid:19) . On γ , where a + b = 1 − cn − α , for some 0 ≤ α ≤ , we have P n ( an − cn α , (1 − a ) n − cn α ) = Θ (cid:18) C n n − α (cid:19) . On γ , we have P n ( an, (1 − a ) n ) = Θ( C n /n ). Behavior is symmetric about the line a + b = 1.The following result is a strengthening of Theorem 3 . a, b, c ,and α as above, s.t. F ( a, b, c, α ) < ∞ , we calculate the value of L ( a, b, c, α ). SAM MINER AND IGOR PAK
01 1 γ : { a + b = 1 − c √ n } γ γ : { a + b = 1 − cn − α for some 0 ≤ α ≤ } γ γ γ : { a + b = 1 } ab Figure 3.
Region where P n ( an, bn ) ∼ C n /n d for some d . Theorem 3.3.
For all ≤ a ≤ , c ∈ R , and ≤ α ≤ / , we have L ( a, − a, c, α ) = ξ ( a, c ) if c = 0 or α = 0 ,η ( a, c ) if c (cid:54) = 0 and < α < ,η ( a, c ) κ ( a, c ) if c (cid:54) = 0 and α = , where ξ ( a, c ) = (2 c + 1) √ π ( a (1 − a )) , η ( a, c ) = c √ π ( a (1 − a )) and κ ( a, c ) = exp (cid:20) − c a (1 − a ) (cid:21) . Let us note that for α = 0 or c = 0 as in theorem, we actually evaluate P n ( an − cn α , (1 − a ) n − cn α + 1) rather than P n ( an − cn α , (1 − a ) n − cn α ) . We do this in order to ensure that we truly measure the distance away from the anti-diagonalwhere j + k = n + 1. This change only affects the asymptotic behavior of P n ( · , · ) when α = 0or c = 0.3.2. Shape of 132-avoiding permutations.
Recall that Q n ( j, k ) is the number of permuta-tions σ ∈ S n (132) with σ ( j ) = k . Let a, b, c and α be defined as in Theorem 3.1. Define G ( a, b, c, α ) = sup (cid:26) d ∈ R + (cid:12)(cid:12)(cid:12) lim n →∞ n d Q n ( an − cn α , bn − cn α ) C n < ∞ (cid:27) . Let M ( a, b, c, α ) = lim n →∞ n G ( a,b,c,α ) Q n ( an − cn α , bn − cn α ) C n , defined for all a, b, c, α as above for which G ( a, b, c, α ) < ∞ ; M is undefined otherwise. Theorem 3.4.
For ≤ a, b ≤ , c ∈ Z and α ≥ , we have G ( a, b, c, α ) == ∞ if ≤ a + b < , if < a + b < , α if a = b = 1 , < α < , c (cid:54) = 0 , if a = b = 1 , α = 0 , if a + b = 1 , c = 0 , ∞ if a + b = 1 , < α < , c > , if a + b = 1 , ≤ α ≤ , c (cid:54) = 0 , if a + b = 1 , ≤ α ≤ , c < , α if a + b = 1 , < α < , c < , − α if a + b = 1 , ≤ α ≤ , c > . As in Theorem 3.1, here G ( a, b, c, α ) = ∞ means that Q n ( an − cn α , bn − cn α ) = o ( C n /n d ), forall d >
0. The following result proves exponential decay of these probabilities (cf. Theorem 3.2.)
HAPE OF PATTERN-AVOIDING PERMUTATIONS 9
Theorem 3.5.
Let ≤ a, b < such that a + b < , c (cid:54) = 0 , and < α < . Then, for n largeenough, we have Q n ( an − cn α , bn − cn α ) C n < ε n , where ε = ε ( a, b, c, α ) is independent of n , and < ε < . Similarly, let ≤ a ≤ , < c , and < α < . Then, for n large enough, we have Q n ( an − cn α , (1 − a ) n − cn α ) C n < ε n α − , where ε = ε ( a, c, α ) is independent of n , and < ε < . The above theorems compare the relative growth rates of Q n ( i, j ) and C n , as n → ∞ .Theorem 3.4 states that for a + b < , Q n ( an, bn ) = o ( C n /n d ) for all d >
0. For 1 < a + b < Q n ( an, bn ) = Θ (cid:18) C n n (cid:19) .Q n ( an, bn ) is the largest when a = b = 1 or when a + b = 1. The true behavior of Q n ( i, j )described in Theorems 3 . . i and j into account. In factwe have that Q n ( n − cn α , n − cn α ) = Θ (cid:18) C n n α (cid:19) when α ≤ . For a + b = 1, the asymptotic behavior of Q n ( an − cn α , bn − cn α ) varies through severalregimes as α varies between 0 and 1, and c varies between positive and negative numbers. Thisrelationship is illustrated in Figures 4 and 5.01 1 ab γ γ : { a + b = 1 − c √ n } Figure 4.
Region where Q n ( an, bn ) ∼ C n /n d for some d .In Figure 5, on the curve γ , we have Q n ( an − c √ n, (1 − a ) n − c √ n ) = Θ (cid:18) C n n (cid:19) . Similarly, on γ , we have Q n ( an − cn α , (1 − a ) n − cn α ) = Θ (cid:18) C n n − α (cid:19) . In the space between γ and γ , we have Q n ( an + k, (1 − a ) n + k ) = Θ (cid:18) C n n (cid:19) ,
01 1 γ : { a + b = 1 − c √ n } γ γ γ : { a + b = 1 − cn − α , where < α < } γ γ : { a + b = 1 − cn } γ γ : { a + b = 1 + c √ n } γ γ : { a + b = 1 + cn − α , where < α < } ab Figure 5.
Region where Q n ( an, bn ) ∼ C n /n d for some d .where − cn ≤ k ≤ cn . Finally, on γ , we have Q n ( an + cn α , (1 − a ) n + cn α ) = Θ (cid:18) C n n α (cid:19) . As in Theorem 3 .
3, the following result strengthens Theorem 3.4 in a different direction. For a, b, c, α where G ( a, b, c, α ) < ∞ , we calculate the value of M ( a, b, c, α ). Theorem 3.6.
For a, b, c, α as above, we have M ( a, b, c, α ) == v ( a, b ) if < a + b < ,w ( c ) if a = b = 1 , < α < ,u ( c ) if a = b = 1 , c ≥ , α = 0 ,w ( c ) if a + b = 1 , c < , < α,x ( a, c ) if a + b = 1 , c < , α = , z ( a ) if a + b = 1 , ≤ α < ,z ( a ) if a + b = 1 , c < , ≤ α < ,z ( a ) + y ( a, c ) if a + b = 1 , c > , α = ,y ( a, c ) if a + b = 1 , c > , < α < ,y ( a, c ) κ ( a, c ) if a + b = 1 , c > , α = , where u ( c ) = c (cid:88) s =0 (cid:18) s + 12 c + 1 − s (cid:19) (cid:18) c + 1 − sc + 1 (cid:19) s − c − , v ( a, b ) = 12 √ π (2 − a − b ) ( a + b − ,w ( c ) = 12 c √ π , x ( a, c ) = 14 πa (1 − a ) (cid:90) ∞ s ( s + 2 c ) exp (cid:20) − s a (1 − a ) (cid:21) ds,y ( a, c ) = 2 c √ πa (1 − a ) , z ( a ) = Γ( )2 πa (1 − a ) , and κ ( a, c ) is defined as in Theorem 3.3. Observe that for c = 0 or α = 0, values Q n ( an − cn α , (1 − a ) n − cn α ) behave the sameasymptotically as Q n ( an − cn α , (1 − a ) n − cn α +1). We explain this in more detail in Lemma 5.8.This is in contrast with the behavior of P n ( · , · ), where we need to adjust when on the anti-diagonal. Note also that for a = b = 1, α = 0 and c = 0, we have u (0) = 14 = lim n →∞ Q n ( n, n ) C n = lim n →∞ C n − C n , which holds since Q n ( n, n ) = C n − given in the proof of Proposition 2.4. We prove the theoremin Section 5. HAPE OF PATTERN-AVOIDING PERMUTATIONS 11 Analysis of -avoiding permutations
Combinatorics of Dyck Paths.
We say a Dyck path of length 2 n is a path from (0 , n,
0) in Z consisting of upsteps (1,1) and downsteps (1,-1) such that the path never goesbelow the x -axis. We denote by D n the set of Dyck paths of length 2 n . We can express a Dyckpath γ ∈ D n as a word of length 2 n , where u represents an upstep and d represents a downstep.Recall that P n ( j, k ) is the number of permutations σ ∈ S n (123) with σ ( j ) = k . Let f ( n, k ) = P n (1 , k ) (or P n ( k, b ( n, k ) be the number of lattice paths consisting of upsteps anddownsteps from (0 ,
0) to ( n + k − , n − k ) which stay above the x -axis. Here b ( n, k ) are the ballot numbers , given by b ( n, k ) = n − k + 1 n + k − (cid:18) n + k − n (cid:19) . Lemma 4.1.
For all ≤ k ≤ n , we have f ( n, k ) = b ( n, k ) .Proof. We have that f ( n, k ) counts the number of permutations σ ∈ S n (123) such that σ (1) = k .By the RSK-correspondence (see e.g. [B2, S1]), we have f ( n, k ) counts the number of Dyck paths γ ∈ D n whose final upstep ends at the point ( n + k − , n + 1 − k ). Remove the last upstepfrom path γ , and all the steps after it. We get a path γ (cid:48) from (0 ,
0) to ( n + k − , n − k ) whichremains above the x -axis. These paths are counted by b ( n, k ), and the map γ → γ (cid:48) is clearlyinvertible, so f ( n, k ) = b ( n, k ), as desired. (cid:3) Lemma 4.2.
For all ≤ j, k ≤ n , we have P n ( j, k ) = b ( n − k + 1 , j ) b ( n − j + 1 , k ) , where j + k ≤ n + 1 . Similarly, we have P n ( j, k ) = b ( j, n − k + 1) b ( k, n − j + 1) , where j + k > n + 1 . Proof.
Let us show that the second case follows from the first case. Suppose j + k > n + 1. Byassuming the first case of the lemma, we have P n ( j, k ) = P n ( n + 1 − j, n + 1 − k )= b ( n − ( n + 1 − k ) + 1 , n + 1 − j ) b ( n − ( n + 1 − j ) + 1 , n + 1 − k )= b ( k, n − j + 1) b ( j, n − k + 1) , by Proposition 2.2. Therefore, it suffices to prove the lemma for j + k ≤ n + 1.Let j + k ≤ n + 1, and let σ be a -avoiding permutation with σ ( j ) = k . We usedecomposition σ = τ kρ , where τ = { σ (1) , . . . , σ ( j − } and ρ = { σ ( j + 1) , . . . , σ ( n ) } . We now show that σ ( i ) > k , for all 1 ≤ i < j . Suppose σ ( i ) < k for some i < j . Then there areat most ( j −
2) numbers x < j with σ ( x ) > k . Since σ ( j ) = k , in total there are ( n − k ) numbers y such that σ ( y ) > k . Since j − < n − k , there must be at least one number z > j with σ ( z ) > k . However, this gives a pattern consisting of ( i, j, z ), a contradiction. Therefore, σ ( i ) > k , for all 1 ≤ i < j .Consider the values of σ within τ . From above, the values within τ are all greater than k .Given a possible τ , the values within ρ which are greater than k must be in decreasing order,to avoid forming a -pattern starting with k . Therefore, to count possible choices for τ , itsuffices to count possible orderings within σ of the numbers x with k ≤ x ≤ n . The number ofsuch orderings is b ( n − k + 1 , j ), since the smallest number is in the j -th position. Therefore,there are b ( n − k + 1 , j ) possible choices for τ .Now consider the values of σ within { k } ∪ ρ . We have ( n − j + 1) numbers to order, and( k −
1) of them are less than k . Our only restriction on ρ is that we have no -patterns.There are b ( n − j +1 , k ) of these orderings, since the k -th smallest number is in the 1-st position.Therefore, we have b ( n − j + 1 , k ) possible choices for ρ . Once we have chosen τ and ρ , this completely determines the permutation σ . Therefore,there are b ( n − j + 1 , k ) b ( n − k + 1 , j ) choices of such σ , as desired. (cid:3) Example 4.3.
Let us compute P (4 , σ ∈ S (123) with σ (4) = 3 . The numbers 1 and 2 must come after 3 in any such permutation, since otherwise a willbe created with 3 in the middle. There are b (7 − ,
4) = 28 (cid:18) (cid:19) = 14ways to order the numbers between 3 and 7, shown here:(47635) (54736) (57436) (57634) (64735) (65437) (65734)(67435) (67534) (74635) (75436) (75634) (76435) (76534) . For each of these, there are b (7 − ,
3) = 26 (cid:18) (cid:19) = 5ways to place the numbers between 1 and 3, shown here:( ∗ ∗ ∗ ∗
2) ( ∗ ∗ ∗ ∗ ) ( ∗ ∗ ∗ ∗
1) ( ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ , where the asterisks represent the positions of 4 , ,
6, and 7. In total, we have P (4 ,
3) = b (5 , b (4 ,
3) = (14)(5) = 70.4.2.
Proof of theorems 3.1, 3.2, and 3.3.
The proof follows from several lemmas: onetechnical lemma and one lemma for each case from Theorem 3.1.Let h : [0 , → R be defined as h ( a, b ) = (1 − a + b ) (1 − a + b ) (1 − b + a ) (1 − b + a ) a a (1 − a ) (1 − a ) b b (1 − b ) (1 − b ) . Lemma 4.4 (Technical lemma) . We have h ( a, b ) ≤ , for all ≤ a, b ≤ . Moreover, h ( a, b ) = 4 if and only if b = 1 − a .Proof. Observe that h ( a, − a ) = 4. Furthermore, we consider the partial derivatives of h withrespect to a and b . We find that h has local maxima at each point where b = 1 − a . In factthese are the only critical points within [0 , . We omit the details. (cid:3) Lemma 4.5 (First case) . Let a, b ∈ [0 , , c (cid:54) = 0 , and ≤ α < , such that a + b (cid:54) = 1 . Then F ( a, b, c, α ) = ∞ . Moreover, for n sufficiently large, we have P n ( an − cn α , bn − cn α ) /C n < ε n , where ε is independent of n and < ε < .Proof. By lemmas 4.1 and 4.2, we have P n ( an − cn α , bn − cn α ) = b ( n − ( bn − cn α ) + 1 , an − cn α ) b ( n − ( an − cn α ) + 1 , bn − cn α )= ( n (1 − a − b ) + 2 cn α + 2) n (1 − b + a )(1 − a + b ) (cid:18) n (1 − b + a ) n − ( bn − cn α ) + 1 (cid:19)(cid:18) n (1 − a + b ) n − ( an − cn α ) + 1 (cid:19) . Applying Stirling’s formula gives P n ( an − cn α , bn − cn α ) ∼ r ( n, a, b ) · h ( a, b ) n , where r ( n, a, b ) == ( n (1 − a − b ) + 2 cn α + 2) ( an − cn α )( bn − cn α ) (cid:112) (1 − a + b )(1 − b + a )2 πn ( n (1 − a ) + cn α + 1)( n (1 − b ) + cn α + 1)(1 − a + b )(1 − b + a ) (cid:112) ab (1 − a )(1 − b ) . The proof follows similar (and even somewhat simplified) steps as the proof of Lemma 5.5.
HAPE OF PATTERN-AVOIDING PERMUTATIONS 13
Using C n ∼ n √ πn , we obtain n d P n ( an − cn α , bn − cn α ) C n ∼ √ π n d + r ( n, a, b ) h ( a, b ) n − n . Clearly, for h ( a, b ) <
4, the r.h.s. → n → ∞ , for all d ∈ R + . By Lemma 4 .
4, we have h ( a, b ) < b = 1 − a . Therefore, since a + b (cid:54) = 1, we have F ( a, b, c, α ) = ∞ . Also, when n is large enough, P n ( an − cn α , bn − cn α ) C n < (cid:18) h ( a, b ) + 48 (cid:19) n , as desired. (cid:3) Lemma 4.6 (Second case) . For all a ∈ (0 , , < c, and < α < , we have F ( a, − a, c, α ) = ∞ . Moreover, for n large enough, we have P n ( an − cn α , (1 − a ) n − cn α ) C n < ε n α − , where ε is independent of n and < ε < .Proof. Let k = cn α . Evaluating P n ( an − k, (1 − a ) n − k ), we have P n ( an − k, (1 − a ) n − k ) = (2 k + 2) (2(1 − a ) n )(2 an ) (cid:18) − a ) n (1 − a ) n + k + 1 (cid:19)(cid:18) anan + k + 1 (cid:19) . Using Stirling’s formula and simplifying this expression gives P n ( an − k, (1 − a ) n − k ) ∼ ( k + 1) π ( a (1 − a )) n n (cid:18) anan + k (cid:19) an + k (cid:18) anan − k (cid:19) an − k × (cid:18) (1 − a ) n (1 − a ) n + k (cid:19) (1 − a ) n + k (cid:18) (1 − a ) n (1 − a ) n − k (cid:19) (1 − a ) n − k . Clearly, ln (cid:34)(cid:18) anan + k (cid:19) an + k (cid:18) anan − k (cid:19) an − k (cid:35) ∼ − k an as n → ∞ . Therefore, n d P n ( an − k, (1 − a ) n − k ) C n ∼ n d ( k + 1) √ π ( a (1 − a ) n ) exp (cid:20) − k a (1 − a ) n (cid:21) . Substituting k ← cn α , gives n d P n ( an − k, (1 − a ) n − k ) C n ∼ c n d √ π ( a (1 − a )) n − α exp (cid:20) − c a (1 − a ) n − α (cid:21) . For α > , this expression → n → ∞ , for all d . This implies that F ( a, − a, c, α ) = ∞ . Infact, we have also proved the second case of Theorem 3.6, as desired. (cid:3) Lemma 4.7 (Third case) . For all a ∈ (0 , , c > , and α ∈ [0 , , we have F ( a, − a, , α ) = F ( a, − a, c,
0) = 32 . Furthermore, we have L ( a, − a, c,
0) = ξ ( a, c ) . Proof.
In this case, to ensure that cn α measures the distance from the anti-diagonal, we needto analyze P n ( an − cn α , (1 − a ) n − cn α + 1). Evaluating as in Lemma 4.6, gives n d P n ( an − cn α , (1 − a ) n − cn α + 1) C n ∼ n d − (2 k + 1) √ π ( a (1 − a )) exp (cid:20) − k a (1 − a ) n (cid:21) . For c = 0, we get F ( a, − a, , α ) = 3 / L ( a, − a, , α ) = ξ ( a, α = 0, we get F ( a, − a, c,
0) = 3 / L ( a, − a, c,
0) = ξ ( a, c ), as desired. (cid:3) Lemma 4.8 (Fourth case) . For all a ∈ (0 , , c > and < α ≤ , we have F ( a, − a, c, α ) = 32 − α. Furthermore, for < α < , we have L ( a, − a, c, α ) = η ( a, c ) and L (cid:18) a, − a, c, (cid:19) = η ( a, c ) κ ( a, c ) , where η ( a, c ) and κ ( a, c ) are defined as in Theorem 3.3.Proof. As in Lemma 4.6, we have n d P n ( an − k, (1 − a ) n − k ) C n ∼ c n d √ π ( a (1 − a )) n − α exp (cid:20) − c a (1 − a ) n − α (cid:21) . We can rewrite this expression as c n d √ π ( a (1 − a )) n − α exp (cid:20) − c a (1 − a ) n − α (cid:21) ∼ η ( a, c ) n d − ( − α ) exp (cid:20) − c a (1 − a ) n − α (cid:21) . For α < , we clearly have exp (cid:20) − c a (1 − a ) n − α (cid:21) → n → ∞ , so F ( a, − a, c, α ) = − α and L ( a, − a, c, α ) = η ( a, c ). For α = , by the definition of κ ( a, c ), we have exp (cid:20) − c a (1 − a ) n − α (cid:21) → κ ( a, c ) , as n → ∞ , so F ( a, − a, c, /
2) = 1 /
2, and L ( a, − a, c, /
2) = η ( a, c ) κ ( a, c ), as desired. (cid:3) Let us emphasize that the results of the previous two lemmas hold for c < c > P n ( j, k ) displayed in Lemma 2.2, and since c only appears inthe formulas for η ( a, c ) and κ ( a, c ) as c . Therefore, we have proven all cases of Theorems 3.1and 3.3. 5. Analysis of -avoiding permutations
Combinatorics of Dyck paths.
We recall a bijection ϕ between S n (132) and D n , whichwe then use to derive the exact formulas for Q n ( j, k ). This bijection is equivalent to that in [EP],itself a variation on a bijection in [Kra] (see also [B2, Kit] for other bijections between thesecombinatorial classes).Given γ ∈ D n , for each downstep starting at point ( x, y ) record y , the level of ( x, y ). Thisdefines y γ = ( y , y , . . . , y n ).We create the -avoiding permutation by starting with a string { n, n − , . . . , , } andremoving elements from the string one at a time each from the y i -th spot in the string, creatinga permutation ϕ ( γ ). Suppose this permutation contains a -pattern, consisting of elements a, b , and c with a < b < c . After a has been removed from the string, the level in the string must HAPE OF PATTERN-AVOIDING PERMUTATIONS 15 be beyond b and c . Since we can only decrease levels one at a time, we must remove b beforeremoving c , a contradiction. Therefore, the map ϕ is well-defined, and clearly one-to-one. ByTheorem 2.1, this proves that ϕ is the desired bijection. Example 5.1.
Take the Dyck path γ = ( uuduuddudd ). Then z γ = (2 , , , , -avoiding permutation ϕ ( γ ) by taking the string { , , , , } and removing elements one at a time. First we remove the 2-nd element (4), then we remove the3-rd element from the remaining list { , , , } , which is 2, then the 2-nd from the remaininglist { , , } , which is 3, then the 2-nd from { , } , which is 1, then the last element (5), andwe obtain ϕ ( γ ) = (4 , , , , Figure 6.
Dyck Path γ with downsteps at y γ = (2 , , , , g ( n, k ) be the number of permutations σ ∈ S n (132) with σ (1) = k , so g ( n, k ) = Q n (1 , k ).Recall that since Q n ( j, k ) = Q n ( k, j ), we can also think of g ( n, k ) as the number of permutations σ ∈ S n (132) with σ ( k ) = 1. Let b ( n, k ) denote the ballot numbers as in Lemma 4.1. Lemma 5.2.
For all ≤ k ≤ n , we have g ( n, k ) = b ( n, k ) . Proof.
Let σ ∈ S n (132) with σ (1) = k . Using bijection ϕ , we find that ϕ − ( σ ) is a Dyck pathwith its final upstep from ( n + k − , n − k ) to ( n + k − , n + 1 − k ). The result now followsfrom the same logic as in the proof of Lemma 4.1. (cid:3) Lemma 5.3.
For all ≤ j, k ≤ n , Q n ( j, k ) = (cid:88) r b ( n − j + 1 , k − r ) b ( n − k + 1 , j − r ) C r , where the summation is over values of r such that max { , j + k − n − } ≤ r ≤ min { j, k } − . Proof.
Since our formula is symmetric in j and k except for the upper limit of summation,proving the lemma when j ≤ k will suffice. When j ≤ k the upper limit is j −
1, rather than k − j > k . Q n ( j, k ) represents the number of permutations σ ∈ S n (132) with σ ( j ) = k .Let q n ( j, k, r ) be the number of -avoiding permutations σ counted by Q n ( j, k ) such thatthere are exactly r values x with x < j such that σ ( x ) < k . Below we show that q n ( j, k, r ) = b ( n − j + 1 , k − r ) b ( n − k + 1 , j − r ) C r for all 0 ≤ r ≤ j − , which implies the result.Let σ ∈ S n (132) such that σ ( j ) = k and there are exactly r numbers x i with x i < j and σ ( x i ) < σ ( j ) = k . We use decomposition σ = τ πkφ , where τ = { σ (1) , . . . , σ ( j − r − } , π = { σ ( j − r ) , . . . , σ ( j − } , and φ = { σ ( j + 1) , . . . , σ ( n ) } . Observe that either all elements of π are smaller than k , or there is some element of π greaterthan k , and some element of τ smaller than k . Suppose the second case is true, with a anelement of τ smaller than k , and b an element of π larger than k . Then a, b , and k form a -pattern, a contradiction. Therefore, all elements of π must be smaller than k .Suppose some element x of π is smaller than ( k − r ). Then some number y with k − r ≤ y < k is an element of φ . Then we have a -pattern, formed by x, k , and y , which is a contradiction, so π consists of { k − r, k − r + 1 , . . . , k − , k − } . There are C r possible choices for π , since π must avoid the -pattern.Now consider the values of σ within τ . Observe that regardless of τ , the numbers s in φ with s > k must be in decreasing order in φ , in order to avoid a -pattern that starts with k .Therefore, the number of possible choices for τ is equal to the number of possible orderings ofthe numbers between k and n that avoid , with k in the ( j − r )-th position (since π onlyconsists of numbers smaller than k ). There are exactly b ( n − k + 1 , j − r ) such possible choices.Finally, consider values of σ within { k } ∪ φ . Here we need to order n − j + 1 numbers sothat they avoid the -pattern, with the first number being the ( k − r )-th smallest. Thereare exactly b ( n − j + 1 , k − r ) ways to do this. Choosing π , τ , and φ completely determines σ .Therefore, there are b ( n − j + 1 , k − r ) b ( n − k + 1 , j − r ) C r possible choices for σ , as desired. (cid:3) Example 5.4.
Let us compute Q (4 , σ ∈ S n (132) with σ (4) =3. We count the permutations separately depending on how many numbers smaller than 3come ahead of 3. First suppose r = 0, so there are no numbers smaller than 3 ahead of 3 inthe permutation. This means that 3 is in the 4-th position among those numbers greater thanequal to 3. There are b (7 − , −
0) = 5 − − (cid:18) − (cid:19) = 28 (cid:18) (cid:19) = 14ways to order the numbers between 3 and 7, displayed here:(45637) (54637) (56437) (56734) (64537) (65437) (65734)(67435) (67534) (74536) (75436) (75634) (76435) (76534) . For each of these there are b (7 − , −
0) = 26 (cid:18) (cid:19) = 5ways to place the numbers between 1 and 3, shown here:( ∗ ∗ ∗ ∗ ) ( ∗ ∗ ∗ ∗ ) ( ∗ ∗ ∗ ∗
1) ( ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ , where the stars represent the positions of 4,5,6 and 7. In total we find that q (4 , ,
0) = b (5 , b (4 ,
3) = (14)(5) = 70.Similarly, for r = 1 there is 1 number smaller than 3 ahead of 3 in the permutation. Thisnumber has to be 2, since otherwise a pattern would be formed with 1,2, and 3. Alsothis number must be directly in front of 3 in the permutation, since otherwise a would beformed with 2 as the 1 and 3 as the 2. This means that 3 is now in the 3-rd position amongthose numbers greater than equal to 3. There are b (7 − , −
1) = 37 (cid:18) (cid:19) = 9ways to order the numbers between 3 and 7, displayed here:(45367) (54367) (56347) (64357) (65347) (67345) (74356) (75346) (76345) . For each of those orderings there are b (7 − , − C = 35 (cid:18) (cid:19) = 3ways to place the numbers between 1 and 3, displayed here:( ∗ ∗ ∗ ∗ ) ( ∗ ∗ ∗ ∗ ) ( ∗ ∗ ∗ ∗ . Therefore, overall we have q (4 , ,
1) = b (5 , b (4 , C = (9)(3) = 27. In other words, there are27 distinct -avoiding permutations of length 7 with 3 in the 4-th position, and one numbersmaller than 3 and ahead of 3. HAPE OF PATTERN-AVOIDING PERMUTATIONS 17
The final case is when r = 2, in which case both 1 and 2 come ahead of 3 in the permutation.This means that 3 is in the 2-nd position among those numbers between 3 and 7. Thereforethere are b (7 − , −
2) = 46 (cid:18) (cid:19) = 4ways to order the numbers between 3 and 7, displayed here:(43567) (53467) (63457) (73456) . For each of these we have b (7 − , − C = (cid:18) (cid:19) (2) = 2ways to place the numbers between 1 and 3, displayed here:( ∗ ∗ ∗∗ ) ( ∗ ∗ ∗∗ ) . Together we have that q (4 , ,
2) = b (5 , b (4 , C = (4)(1)(2) = 8, so there are 8 different -avoiding permutations of length 7 with 3 in the 4-th position, and two numbers smallerthan 3 and ahead of 3. We have shown that Q (4 ,
3) = q (4 , ,
0) + q (4 , ,
1) + q (4 , ,
2) = 70 + 27 + 8 = 105 . Proof of theorems 3.4, 3.5, and 3.6.
The proof again involves one technical lemmaand several cases corresponding to the statements of Theorems 3.4 and 3.6.Let h : [0 , → R be defined so that h ( a, s, t ) = 4 ast (1 − at + a − ast ) − at + a − ast (1 − a + at − ast ) − at + a − ast (1 − at ) (1 − at ) ( a − ast ) ( a − ast ) (1 − a ) (1 − a ) ( at − ast ) ( at − ast ) . Lemma 5.5.
For all ( a, s, t ) ∈ [0 , , we have h ( a, s, t ) ≤ . Moreover, h ( a, s, t ) = 4 if and only if s = at + a − at . Proof.
Take the logarithmic derivative of h to obtain d (ln h ) ds = at ln 4 + ( − at (1 + ln (1 − at + a − ast ))) + ( − at (1 + ln (1 − a + at − ast ))) − [ − at (1 + ln ( a − ast )) − at (1 + ln ( at − ast ))]= at ln [4( a − ast )( at − ast )] − at ln [(1 − at + a − ast )(1 − a + at − ast )] . Set this derivative equal to 0 to get4( a − ast )( at − ast ) − (1 − at + a − ast )(1 − a + at − ast ) = 0 , or ( ast − ( at + a − ast − ( at + a + 1)) = 0 , giving s = at + a − at or s = at + a + 13 at . Since at + a + 13 at = 13 + 13 t + 13 at ≥
13 + 13 + 13 = 1 , this value of s is greater than 1, and is only equal to 1 if a = s = t = 1. Similarly, the ratio( at + a − /at is between 0 and 1 if at + a >
1. It is easy to see that the second derivative d (ln h )( ds ) < s = at + a − at , which implies that this value of s does indeed maximize h ( a, s, t ). We can also verify that h (cid:16) a, at + a − at , t (cid:17) = 4 ( at + a − (2 − at ) (2 − at ) (2 − a ) (2 − a ) (1 − at ) (1 − at ) (1 − at ) (1 − at ) (1 − a ) (1 − a ) (1 − a ) (1 − a ) = 4 . Observe that h ( a, s, t ) < , except for where a = s = t = 1, completingthe proof. (cid:3) Lemma 5.6.
Let a, b ∈ [0 , , c (cid:54) = 0 and ≤ α < , such that a + b < . Then G ( a, b, c, α ) = ∞ .Moreover, for n sufficiently large, we have Q n ( an − cn α , bn − cn α ) C n < ε n , where ε is independent of n , and < ε < .Proof. By Lemma 5.3, we have Q n ( atn, an ) = (cid:88) r b ( n − atn + 1 , an − r ) b ( n − an + 1 , atn − r ) C r = (cid:88) r (cid:20) n − atn + 1 − ( an − r ) + 1 n − atn + 1 + ( an − r ) − (cid:18) n − atn + an − rn − atn + 1 (cid:19)(cid:21) × (cid:20) n − an + 1 − atn + r + 1 n − an + 1 + atn − r − (cid:18) n − an + 1 + atn − r − n − an + 1 (cid:19)(cid:21) × (cid:20) r + 1 (cid:18) rr (cid:19)(cid:21) , where the summation is over values of r such thatmax { , j + k − n − } ≤ r ≤ min { j, k } − . Let r = atsn , so s varies from 0 to (cid:0) − atn (cid:1) by increments of atn . Applying Stirling’s formula,we get Q n ( atn, an ) ∼ (cid:88) r χ ( n, atn, an, astn ) h ( a, s, t ) n , where χ ( n, atn, an, astn ) = (cid:115) (1 − at + a − ats )(1 − a + at − ats )(1 − at )( a − ats )(1 − a )( at − ats )( ats ) × a t (1 − s )(1 − st )( n (1 − a − at + ats ) + 2) πn ) / ( n (1 − at ) + 1)( n (1 − a ) + 1)(1 + a − at − ats )(1 − a + at − ats )( atsn + 1) . We now have n d Q n ( an − cn α , bn − cn α ) C n ∼ √ πn d + an − cn α − (cid:88) r =0 ν r ( n ) , where ν r ( n ) = χ ( n, an − cn α , bn − cn α , r ) h ( b, r/an, a/b ) n − n . From Lemma 5.5, we have that h ( b, r/an, a/b ) < r (cid:54) = ( a + b − n . For values of r where h ( b, r/an, a/b ) < ν r ( n ) decreases exponentially as n → ∞ for fixed d . Therefore, for thesevalues of r , √ πn d + ν r ( n ) → n → ∞ for all d. The only values of r which could potentially have lim n →∞ ν r ( n ) (cid:54) = 0, are when r ∼ ( a + b − n ,as n → ∞ . Observe that since a + b − <
0, there are no such possible values of r . In this case,lim n →∞ n d Q n ( an − cn α , bn − cn α ) C n = 0 for all d > . HAPE OF PATTERN-AVOIDING PERMUTATIONS 19
This implies G ( a, b, c, α ) = ∞ when a + b <
1. Also, for n large enough, we have Q n ( an − cn α , bn − cn α ) C n < (cid:18) h ( b, , a/b )2 (cid:19) n , as desired. (cid:3) Lemma 5.7.
Let a ∈ [0 , , c > , and < α < . Then G ( a, − a, c, α ) = ∞ . Moreover, for n large enough, we have Q n ( an − cn α , (1 − a ) n − cn α ) C n < ε n α − , where ε = ε ( a, c, α ) is independent of n , and < ε < .Proof. Let k = cn α . Then Q n ( an − k, (1 − a ) n − k ) = q n ( an − k, (1 − a ) n − k, S a,n,k , where S a,n,k = (cid:32) an − k − (cid:88) r =1 g r ( n ) (cid:33) , and g r ( n ) = (cid:18) q n ( an − k, (1 − a ) n − k, r ) q n ( an − k, (1 − a ) n − k, (cid:19) . Observe that q n ( an − k, (1 − a ) n − k,
0) = P n ( an − k, (1 − a ) n − k ) . Applying Stirling’s formula and using the Taylor expansion for ln(1 + x ) gives g r ( n ) ∼ (2 k + r + 2) C r (2 k + 2) r exp (cid:20) − (4 kr + r )4 a (1 − a ) n (cid:21) . Therefore, n d Q n ( an − cn α , bn − cn α ) C n ∼ (cid:18) n d P n ( an − cn α , bn − cn α ) C n (cid:19) × (cid:32) an − k − (cid:88) r =1 (2 k + r + 2) C r (2 k + 2) r exp (cid:20) − (4 kr + r )4 a (1 − a ) n (cid:21)(cid:33) . For α > , by Theorem 3.1, we havelim n →∞ n d P n ( an − cn α , (1 − a ) n − cn α ) C n = 0 for all d. Therefore, G ( a, − a, c, α ) = ∞ for α > . Also, we have proven the second case of Theorem 3.5,as desired. (cid:3) For the next three cases, we denote r = hn p , and let h and p be fixed as n → ∞ . For p > ,we have n d g r ( n ) C n → n → ∞ , for all d, since g r ( n ) decreases exponentially for fixed d .For α < p < , we have g r ∼ r √ πk ∼ (cid:32) √ h √ π (cid:33) n p − α . For p ≤ α < or p < α = , we have g r = Θ (cid:16) r − (cid:17) = Θ (cid:16) n − p (cid:17) . Similarly, for p = , we obtain g r = Θ (cid:16) n − α (cid:17) . Lemma 5.8.
Let a ∈ [0 , , c ∈ R and ≤ α < . Then G ( a, − a, c, α ) = 34 and M ( a, − a, c, α ) = z ( a ) , where z ( a ) = Γ( )2 π [ a (1 − a )] as in Theorem 3.6 . Proof.
As in Lemma 5.7, we write Q n ( an − cn α , (1 − a ) n − cn α ) = P n ( an − cn α , (1 − a ) n − cn α ) S a,n,k , where S a,n,k = 1 + an − k − (cid:88) r =0 g r ( n ) . Again, as n → ∞ , we have g r ( n ) ∼ (2 k + r + 2) C r (2 k + 2) r exp (cid:20) − (4 kr + r )4 a (1 − a ) n (cid:21) . Fix s >
0, and observe that for any 0 ≤ α < , we have g n δ ( n ) = o ( g s √ n ( n )) , for every δ (cid:54) = 12 . Therefore, as n → ∞ , t →
0, and u → ∞ , we have S a,n,k ∼ u √ n (cid:88) r = t √ n g r ( n ) . Interpreting this sum as a Riemann sum, we have S a,n,k ∼ √ n (cid:90) ut g v √ n ( n ) dv ∼ √ n (cid:90) ut (2 k + v √ n + 2) (2 k + 2) √ π ( v √ n ) (cid:18) exp (cid:20) − (4 kv √ n + ( v √ n ) )4 a (1 − a ) n (cid:21)(cid:19) dv. Therefore, we have S a,n,k ∼ √ n (cid:90) ut v n k √ π (cid:18) exp (cid:20) − v a (1 − a ) (cid:21)(cid:19) dv. A direct calculation gives S a,n,k = n − α c z ( a ) (cid:16) √ π [ a (1 − a )] (cid:17) . Now we see that n d Q n ( an − cn α , (1 − a ) n − cn α ) C n ∼ n d P n ( an − cn α , (1 − a ) n − cn α ) C n S a,n,k ∼ z ( a ) n d − , by the proof of Theorem 3.1 and the analysis of S a,n,k . Therefore, G ( a, − a, c, α ) = . For α < this also gives M ( a, − a, c, α ) = z ( a ), as desired. (cid:3) This case displays why we do not need to adjust our analysis to be on the anti-diagonal.Since the behavior of Q depends on values of r on the order of √ n , adding 1 to the secondcoordinate is a lower-order term and does not affect G or M at all. In fact, the whole value of cn α has no effect on G or M for this case. HAPE OF PATTERN-AVOIDING PERMUTATIONS 21
Lemma 5.9.
Let a ∈ [0 , , c > , and < α ≤ . Then G ( a, − a, c, α ) = 32 − α . Moreover, M ( a, − a, c, α ) = (cid:40) y ( a, c ) if < α < ,y ( a, c ) κ ( a, c ) if α = , where y ( a, c ) and κ ( a, c ) are defined as in Theorem 3.3.Proof. As in the previous lemma, we have Q n ( an − cn α , (1 − a ) n − cn α ) = P n ( an − cn α , (1 − a ) n − cn α ) S a,n,k . We analyze S a,n,k to see which values of r contribute the most. In this case,for p = , we have g r = Θ( n − α ), which is on the order of n d with d strictly less than − .Therefore, even if we sum over all values of r where p = , we will end up with an expressionon the order of n d + which is lower order than a constant. For p ≤ α < or p < α = , since g r = Θ( n − p ), the terms with the highest order will come when p = 0. Therefore, the values of r which contribute the most to S a,n,k will be constants in this case. From this, we have S a,n,k ∼ s (cid:88) r =1 g r ∼ s (cid:88) r =1 C r r ∼ , as n → ∞ . Therefore, n d Q n ( an − k, (1 − a ) n − k ) C n ∼ n d P n ( an − k, (1 − a ) n − k ) C n . Referring back to Theorems 3.1 and 3.3 gives us the desired results. (cid:3)
Lemma 5.10.
Let a ∈ [0 , and c > . Then G ( a, − a, c, /
8) = and M ( a, − a, c, /
8) = z ( a ) + y ( a, c ) .Proof. Here we are essentially on the intersection of the last two cases, which provides someintuition to the reason that M ( a, − a, c, /
8) = z ( a ) + y ( a, c ). Again we let r = hn p . When α = , values of g r which contribute the highest order to S a,n,k are when p = 0 and when p = . We get z ( a ) from the terms where p = , and y ( a, c ) when p = 0. S a,n,k ∼ z ( a ) c (cid:16) √ π [ a (1 − a )] (cid:17) + 2 , so n d Q n ( an − k, (1 − a ) n − k ) C n ∼ c √ π ( a (1 − a )) n d − + z ( a ) n d − , so G ( a, − a, c, α ) = and M ( a, − a, c, α ) = z ( a ) + y ( a, c ), as desired. (cid:3) Lemma 5.11.
Let c > and ≤ α < . Then G (1 , , c, α ) = 32 α. Moreover, for α > , we have M (1 , , c, α ) = w ( c ) , and M (1 , , c,
0) = u ( c ) , where w ( c ) and u ( c ) are defined as in Theorem 3.6.Proof. We first consider Q n ( n − k, n − k ) with k = cn α , and α >
0. We have Q n ( n − k, n − k ) = n − k − (cid:88) i = n − k − b ( k + 1 , n − k − i ) C i , which is equivalent to Q n ( n − k, n − k ) = C k C k C n − k − (cid:32) k (cid:88) r =1 h r ( n ) (cid:33) , where h r ( n ) = ( r + 1) (cid:0) k − rk (cid:1) (cid:0) kk (cid:1) (cid:18) C n − k − r C n − k − (cid:19) . Observe that h r ( n ) ∼ r exp (cid:104) − r k (cid:105) , so h k δ ( n ) = o ( h k ( n )) for all δ (cid:54) = . From here we have n d Q n ( n − k, n − k ) C n ∼ n d πk (cid:90) ∞ n α s exp (cid:20) − s c (cid:21) ds ∼ n d − α πc (cid:90) ∞ s exp (cid:20) − s c (cid:21) ds. Therefore G (1 , , c, α ) = α and M (1 , , c, α ) = 12 π c = w ( c ) , as desired.For α = 0, Q n ( n − k, n − k ) = Q n ( n − c, n − c ) = n − c − (cid:88) i = n − c − b ( c + 1 , n − c − i ) C i , = c (cid:88) s =0 (cid:18) s + 12 c + 1 − s (cid:18) c + 1 − sc + 1 (cid:19)(cid:19) C n − c − s . Therefore, we have Q n ( n − k, n − k ) C n ∼ c +1 c (cid:88) s =0 (cid:18) s + 12 c + 1 − s (cid:18) c + 1 − sc + 1 (cid:19)(cid:19) s as n → ∞ , completing the proof. (cid:3) Lemma 5.12.
Let a ∈ [0 , , c < and < α < . Then G ( a, − a, c, α ) = (cid:40) if < α ≤ , α if < α < , and M ( a, − a, c, α ) = z ( a ) if < α < ,x ( a, c ) if α = ,w ( c ) if < α < , where z ( a ) , x ( a, c ) , and w ( c ) are defined as in Theorem . .Proof. We now analyze Q n ( an + k, (1 − a ) n + k ), where k = cn α . We have Q n ( an + k, (1 − a ) n + k ) = an + k − (cid:88) r =2 k − b ((1 − a ) n − k + 1 , (1 − a ) n + k − r ) b ( an − k + 1 , an + k − r ) C r . We can rewrite this as Q n ( an + k, an + k ) = an − k (cid:88) d =0 b ((1 − a ) n − k +1 , (1 − a ) n − k − d +1) b ( an − k +1 , an − k − d +1) C d +2 k − . Denote by g d the d -th term of this summation. Then we can express Q n ( an + k, an + k ) as Q n ( an + k, an + k ) = g (cid:32) an − k (cid:88) d =1 g d g (cid:33) . HAPE OF PATTERN-AVOIDING PERMUTATIONS 23
Denote h d as h d = g d g . We can express h d as h d = ( d + 1) (cid:0) an − k ) − dan − k (cid:1)(cid:0) − a ) n − k ) − d (1 − a ) n − k (cid:1)(cid:0) an − k ) an − k (cid:1)(cid:0) − a ) n − k )(1 − a ) n − k (cid:1) (cid:18) C d +2 k − C k − (cid:19) . We now have h d ∼ d exp (cid:20) − d an (cid:21) exp (cid:20) − d − a ) n (cid:21) (cid:18) C d +2 k − d C k − (cid:19) . Fix s > ≤ β <
1. We set d = sn β , and consider the behavior of h d as n → ∞ . Observethat for any α , h n δ = o ( h sn )when δ (cid:54) = . For = β < α , we have h d ∼ d exp (cid:20) − s a (1 − a ) (cid:21) . For = β = α , we have h d ∼ d exp (cid:20) − s a (1 − a ) (cid:21) (cid:18) cs + 2 c (cid:19) . Similarly, for α < β = , h d ∼ d (2 k ) exp (cid:20) − s a (1 − a ) (cid:21) . Therefore, as n → ∞ , t →
0, and u → ∞ , Q n ( an + k, an + k ) ∼ g un (cid:88) d = tn h d . Recall that g ∼ n − (2 πa (1 − a ) k ) n . We are now ready to analyze G ( a, − a, c, α ). For α < we have n d Q n ( an + cn α , (1 − a ) n + cn α ) C n ∼ n d π (2 a (1 − a ) c ) n (1+ α ) (cid:90) ∞ n (2 k ) s exp (cid:20) − s a (1 − a ) (cid:21) ds ∼ n d − π ( a (1 − a )) (cid:32) Γ( )(4 a (1 − a )) (cid:33) , so G ( a, − a, c, α ) = and M ( a, − a, c, α ) = z ( a ), as desired. For α > we have n d Q n ( an + cn α , (1 − a ) n + cn α ) C n ∼ n d π (2 a (1 − a ) c ) n (1+ α ) (cid:90) ∞ n s exp (cid:20) − s a (1 − a ) (cid:21) ds ∼ n d − α π (2 a (1 − a ) c ) (cid:32) π (4 a (1 − a )) (cid:33) , so G ( a, − a, c, α ) = α and M ( a, − a, c, α ) = w ( c ). For α = we have n d Q n ( an + cn α , (1 − a ) n + cn α ) C n ∼ n d π (2 a (1 − a ) c ) n (1+ α ) (cid:90) ∞ n s exp (cid:20) − s a (1 − a ) (cid:21) (cid:18) cs + 2 c (cid:19) ds ∼ n d − π ( a (1 − a )) (cid:90) ∞ s ( s + 2 c ) exp (cid:20) − s a (1 − a ) (cid:21) ds ∼ n d − x ( a, c ) , proving that G ( a, − a, c, ) = and M ( a, − a, c, ) = x ( a, c ). This completes the proof. (cid:3) Our final case is where 1 < a + b < Lemma 5.13.
Let a, b ∈ [0 , , such that < a + b < , and let c ∈ R , < α < . Then G ( a, b, c, α ) = 32 and M ( a, b, c, α ) = v ( a, b ) , where v ( a, b ) is defined as in Theorem 3.6.Proof. Let k = cn α . We analyze Q n ( an + cn α , bn + cn α ) in the same manner as the proof ofLemma 5.12, by looking at Q n ( an + k, bn + k ). We obtain: Q n ( an + k, bn + k ) = C (1 − a ) n + k C (1 − b ) n + k C ( a + b − n − k − (1 − b ) n (cid:88) d =1 g d , where g d = ( d + 1) (cid:0) − a ) n + k ) − d (1 − a ) n + k (cid:1)(cid:0) − b ) n + k ) − d (1 − b ) n + k (cid:1)(cid:0) − a ) n + k )(1 − a ) n + k (cid:1)(cid:0) − b ) n + k )(1 − b ) n + k (cid:1) (cid:18) C ( a + b − n − d C ( a + b − n − (cid:19) . We find that g d ∼ d exp (cid:20) − d (2 − a − b )4(1 − a )(1 − b ) n (cid:21) , so g n β = o ( g n ) if β (cid:54) = . Therefore, n d Q n ( an + cn α , bn + cn α ) C n ∼ n d − π ((1 − a )(1 − b )( a + b − (cid:90) ∞ n s exp (cid:20) − s (2 − a − b )4(1 − a )(1 − b ) (cid:21) ds . We can now see that G ( a, b, c, α ) = and M ( a, b, c, α ) = v ( a, b ), as desired. We have nowproved all cases of Theorems 3.4 and 3.6. (cid:3) Expectation of basic permutation statistics
Results.
The following result describes the behavior of the first and last elements of σ and τ . Theorem 6.1.
Let σ ∈ S n (123) and τ ∈ S n (132) be permutations chosen uniformly at randomfrom the corresponding sets. Then (1) E [ σ (1)] = E [ σ − (1)] = E [ τ (1)] = E [ τ − (1)] → n − as n → ∞ , (2) E [ σ ( n )] = E [ σ − ( n )] → , as n → ∞ , and (3) E [ τ ( n )] = E [ τ − ( n )] = ( n + 1)2 for all n. We remark here that the above theorem can be proved by using the exact formulas for P n ( j, k ) and Q n ( j, k ) shown in Lemmas 4.2 and 5.3. We will instead prove the theorem byanalyzing bijections between S n (123) and S n (132). Before proving Theorem 6.1 we need somenotation and definitions. HAPE OF PATTERN-AVOIDING PERMUTATIONS 25
Definitions.
Let
A, B be two finite sets. We say A and B are equinumerous if | A | = | B | .Let α : A → Z and β : B → Z . We say α and β are statistics on A and B , respectively. Wesay that two statistics α and β on equinumerous sets are equidistributed if | α − ( k ) | = | β − ( k ) | for all k ∈ Z ; in this case we write α ∼ β . Note that being equidistributed is an equivalencerelation.For example, let A = S n (132), let α : A → Z , such that α ( σ ) = σ (1), and let β : A → Z suchthat β ( σ ) = σ − (1). Then α ∼ β , by Proposition 2.2. Given π ∈ S n , letrmax( π ) = { i s.t. π ( i ) > π ( j ) for all j > i, where 1 ≤ i ≤ n } . We call rmax( π ) the number of right-to-left maxima in π . Letldr( π ) = max { i s.t. π (1) > π (2) > . . . > π ( i ) } . We call ldr( π ) the leftmost decreasing run of π .6.3. Position of first and last elements.
To prove Theorem 6.1, we first need two proposi-tions relating statistics on different sets. Let T n be the set of rooted plane trees on n vertices.For all T ∈ T n , denote by δ r ( T ) the degree of the root vertex in T . Recall that D n is the setof Dyck paths of length 2 n . For all γ ∈ D n , denote by α ( γ ) the number of points on the line y = x in γ . Recall that |D n | = |S n (123) | = |S n (132) | = C n . Recall that |T n +1 | = C n (see e.g. [B2, S1]), so {T n , D n , S n (123) , S n (132) } are all equinumerous. Proposition 6.2.
Define the following statistics: δ n : T n +1 → Z such that δ n ( T ) = δ r ( T ) ,α n : D n → Z such that α n ( γ ) = α ( γ ) , rmax n : S n (132) → Z such that rmax n ( σ ) = rmax( σ ) , ldr n : S n (123) → Z such that ldr n ( σ ) = ldr( σ ) , ldr (cid:48) n : S n (132) → Z such that ldr (cid:48) n ( σ ) = ldr( σ ) , and first n : S n (123) → Z such that first n ( σ ) = σ (1) . Then, we have δ n ∼ α n ∼ rmax n ∼ ldr n ∼ ldr (cid:48) n ∼ ( n + 1 − first n ) for all n. Proof of Proposition 6.2.
Throughout the proof we refer to specific bijections described andanalyzed in [CK]; more details and explanation of equidistribution are available there. We nowprove equidistribution of the statistics one at a time. • δ n ∼ α n Recall the standard bijection φ : T n +1 → D n . Observe that φ : δ n → α n . Therefore, δ n ∼ α n , as desired. • α n ∼ rmax n Let ψ : D n → S n (132) be the bijection ϕ presented in Section 5.1. Observethat ψ : α n → rmax n . Therefore, α n ∼ rmax n , as desired. • rmax n ∼ ldr n Let Φ be the West bijection between S n (132) and S n (123). Observe thatΦ : rmax n → ldr n . Therefore, rmax n ∼ ldr n , as desired. • ldr n ∼ ldr (cid:48) n Let Ψ be the Knuth-Richards bijection between S n (123) and S n (132). Observethat if Ψ( σ ) = τ for some σ ∈ S n (123), then ldr n ( σ ) = ldr (cid:48) n ( τ − ). Since S n (132) is closed underinverses, we get ldr n ( S n (123)) ∼ ldr (cid:48) n ( S n (132)), as desired. • ldr (cid:48) n ∼ ( n + 1 − first n ) Let ∆ be the Knuth-Rotem bijection between S n (132) and S n (321),and let Γ : S n (321) → S n (123) such that Γ( σ ) = ( σ ( n ) , . . . , σ (1)). Observe that if Γ ◦ ∆( σ − ) = τ for some σ ∈ S n (132), then ldr (cid:48) n ( σ − ) = n + 1 − first n ( τ ). Since σ − ranges over all S n (132),we get ldr (cid:48) n ∼ first n , as desired. (cid:3) The second proposition explains the behavior of τ ( n ), for τ ∈ S n (132). Proposition 6.3.
Let τ ∈ S n (132) be chosen uniformly at random. Let last n : S n (132) → Z be defined as last n ( τ ) = τ ( n ) . Then we have last n ∼ n + 1 − last n . Proof.
By applying Lemma 5.3 for k = n , we obtain Q n ( j, n ) = C j − C n − j , implying that Q n ( j, n ) = Q n ( n + 1 − j, n ) for all integers j and n where j ≤ n. Since | last − n ( j ) | = Q n ( j, n ) and | ( n + 1 − last n ) − ( j ) | = Q n ( n + 1 − j, n ) for all 1 ≤ j ≤ n, we have last n ∼ ( n + 1 − last n ), by the definition of equidistribution. (cid:3) Proof of Theorem 6.1.
Let σ ∈ S n (123) and τ ∈ S n (132) be chosen uniformly atrandom. It is known that E [ δ n ] → n → ∞ , where δ n is defined as in Proposition 6.2 (see e.g. [FS, Example III.8]). Since δ n and ( n + 1 − first n ) are equidistributed by Proposition 6.2, we have E [ δ n ] = n + 1 − E [ σ (1)] . Therefore, we have E [ σ (1)] → ( n −
2) as n → ∞ , as desired.Due to the symmetries explained in the proof of Proposition 2 .
2, we have E [ σ − (1)] = E [ σ (1)] → n − E [ σ ( n )] = E [ σ − ( n )] = n + 1 − E [ σ (1)] → n → ∞ , completing the proof of (1).Since P n (1 , k ) = b ( n, k ) = Q n (1 , k ) for all n and k by lemmas 4.1 and 5.2, we have E [ τ (1)] = E [ σ (1)] → n − n → ∞ . Also, by Proposition 2.2, we have E [ τ − (1)] = E [ τ (1)] → n − n → ∞ , completing the proof of (2).To complete the proof of Theorem 6 .
1, we need to prove (3). By Proposition 6 .
3, we havelast n ∼ ( n + 1 − last n ), so E [last n ] = E [ n + 1 − last n ] . By the linearity of expectation, we have E [last n ] = ( n + 1) /
2, so E [ τ ( n )] = n + 12 , as desired. By the symmetry in Proposition 2.2, we have E [ τ − ( n )] = E [ τ ( n )] = n + 12 , completing the proof. HAPE OF PATTERN-AVOIDING PERMUTATIONS 27 Fixed points in random permutations
Results.
Let σ ∈ S n . The number of fixed points of σ is defined asfp n ( σ ) = { i s.t. σ ( i ) = i, ≤ i ≤ n } . In [E1, § . S n (321) , S n (132), and S n (123). In the following threetheorems, the first part is due to Elizalde, while the second parts are new results.We use I n,ε ( a ) = [( a − ε ) n, ( a + ε ) n ] to denote the intervals of elements in { , . . . , n } . Theorem 7.1.
Let ε > , and let σ ∈ S n (321) be chosen uniformly at random. Then E [fp n ( σ )] = 1 for all n. Moreover, for a ∈ ( ε, − ε ) , P ( σ ( i ) = i for some i ∈ I n,ε ( a )) → as n → ∞ . Theorem 7.2.
Let ε > , and let σ ∈ S n (132) be chosen uniformly at random. Then E [fp n ( σ )] = 1 , as n → ∞ . Moreover, for a ∈ (0 , / − ε ) ∪ (1 / ε, − ε ) , P ( σ ( i ) = i for some i ∈ I n,ε ( a )) → as n → ∞ . Theorem 7.3.
Let ε > , and let σ ∈ S n (123) be chosen uniformly at random. Then E [fp n ( σ )] → , as n → ∞ . Moreover, for a ∈ (0 , / − ε ) ∪ (1 / ε, , P ( σ ( i ) = i for some i ∈ I n,ε ( a )) → as n → ∞ . In fact, Elizalde obtains exact formulas for E [fp n ( σ )] in the last case as well [E1, Prop. 5.3].We use an asymptotic approach to give independent proofs of all three theorems.The final case to consider, of fixed points in -avoiding permutations, is more involved.In the language of Section 1, the expectation is equal to the sum of entries of Q n along anti-diagonal ∆, parallel to the wall. It is larger than in the case of the canoe since the decay fromthe wall towards ∆ is not as sharp as in the case of the canoe.In [E1], Elizalde calculated the (algebraic) generating function for the expected number offixed points in S n (231), but only concluded that E [fp n ( σ )] > n ≥
3. Our methods allowus to calculate the asymptotic behavior of this expectation, but not the location of fixed points.
Theorem 7.4.
Let σ ∈ S n (231) be chosen uniformly at random. Then E [fp n ( σ )] ∼ ) √ π n , as n → ∞ . Recall that if σ ∈ S n is chosen uniformly at random, then E [fp n ( σ )] = 1, so the numberof fixed points statistic does not distinguish between random permutation in S n (132), S n (321)and random permutations in S n . On the other hand, permutations in S n (123) are less likelyto have fixed points, in part because they can have at most 2 of them, and permutations in S n (231) are much more likely to have fixed points than the typical permutation in S n . Proof of Theorem 7.1.
Let σ ∈ S n (321) be chosen uniformly at random, and σ (cid:48) =( σ ( n ) , . . . , σ (1)). Since σ ∈ S n (321), we have σ (cid:48) ∈ S n (123). For τ ∈ S n , define afp( τ ) = { i s.t. τ ( i ) = n + 1 − i, ≤ i ≤ n } . Let afp n : S n (123) → Z such that afp n ( τ ) = afp( τ ). Letfp n : S n (321) → Z such that fp n ( σ ) = fp( σ ). By definition, we have fp n ∼ afp n .Clearly, E [afp n ] = 1 C n n (cid:88) k =1 P n ( k, n + 1 − k ) . Observe that P n ( k, n + 1 − k ) = C k − C n − k by Lemma 4.2. By the recurrence relation for the Catalan numbers, we have n (cid:88) k =1 P n ( k, n + 1 − k ) = C n . Therefore, E [afp n ] = 1 , so E [fp n ] = 1 . For the proof of the second part, let ε > a ∈ ( ε, − ε ), and define δ = min { ε, a − ε } .Let P = P ( σ ( i ) = i for some i ∈ [ δn, (1 − δ ) n + 1]) . By the definition of P , we have P ( σ ( i ) = i for some i ∈ I n,ε ( a )) ≤ P for all n, so it suffices to show that P → n → ∞ . Observe that for all n , we have P = 1 C n (1 − δ ) n +1 (cid:88) k = δn P n ( k, n + 1 − k ) . By Lemma 4.2, we have P n ( δn, (1 − δ ) n + 1) C n = C δn − C (1 − δ ) n C n ∼ √ π ( δ (1 − δ ) n ) , as n → ∞ . Observe that for d ∈ ( δ, − δ ) and for n sufficiently large, we have P n ( dn, (1 − d ) n + 1)
We omit the proofs of the probabilities tending to0 since they are very similar to the proof of Theorem 7 .
1, and instead prove the followingproposition.
Proposition 7.5.
Let σ ∈ S n (123) and τ ∈ S n (132) be chosen uniformly at random. Then E [fp n ( τ )] = 1 for all n, and E [fp n ( σ )] → , as n → ∞ . Proof.
Let σ ∈ S n (123) and τ ∈ S n (132) be chosen uniformly at random. By a bijection between -avoiding and -avoiding permutations in [EP] (see also [CK, Rob, RSZ]), fixed pointsare equidistributed between S n (132) and S n (321). Therefore, by the proof of Theorem 7 .
1, wehave E [fp n ( τ )] = 1.Now we prove that E [fp n ( σ )] → , as n → ∞ . HAPE OF PATTERN-AVOIDING PERMUTATIONS 29
Observe that for all n , we have E [fp n ( σ )] = n (cid:88) k =1 P n ( k, k ) C n . Let C and C be constants such that 0 < C < C . By Theorems 3 . .
3, we have E [fp n ( σ )] ∼ n/ − C √ n (cid:88) k = n/ − C √ n P n ( k, k ) C n + n/ C √ n (cid:88) k = n/ C √ n P n ( k, k ) C n + e ( C , C , n ) , where the error term e ( C , C , n ) → C → , C → ∞ , and n → ∞ . By the symmetry inProposition 2 .
2, and by Lemma 4 .
8, we have E [fp n ( σ )] ∼ n/ − C √ n (cid:88) k = n/ − C √ n P n ( k, k ) C n + e ( C , C , n ) ∼ C √ n (cid:88) k = C √ n k n − √ π e − k n + e ( C , C , n ) . As C → C → ∞ , we have E [fp n ( σ )] ∼ √ n (cid:90) ∞ c √ π √ n e − c dc = 16 √ π (cid:90) ∞ c e − c . Since (cid:90) ∞ c e − c dc = √ π , we get E [fp n ( σ )] → /
2, as desired. (cid:3)
Proof of Theorem 7.4.
Proof.
Let σ ∈ S n (231) be chosen uniformly at random. Observe that τ = ( σ ( n ) , σ ( n − , . . . , σ (1)) ∈ S n (132) and is distributed uniformly at random. Therefore, E [fp n ( σ )] = E [afp n ( τ )] = n (cid:88) k =1 Q n ( k, n + 1 − k ) C n . Let A k = Q n ( k, n − k + 1) /C n , so E [fp n ( σ )] = (cid:80) nk =1 A k . By Lemma 5.3 and Stirling’s formula,we have E [fp n ( σ )] ∼ n/ (cid:88) k =1 n √ πk ( n − k ) (cid:32) k − (cid:88) r =1 √ r √ π exp (cid:20) − r n k ( n − k ) (cid:21)(cid:33) . Let B K = (cid:80) K − k =1 A k , C K = (cid:80) nk = K A k , and let K = √ n . We show that B K = o ( C K ). For k < K , we have (cid:32) k − (cid:88) r =1 √ r √ π exp (cid:20) − r n k ( n − k ) (cid:21)(cid:33) = O ( k ) , since C √ k (cid:88) r =1 √ r √ π ∼ Ck √ π . Therefore, A k = O ( k − ), and B K = O ( √ n ( √ n ) − ) = O ( n ) . For k → ∞ as n → ∞ , we have (cid:32) k − (cid:88) r =1 √ r √ π exp (cid:20) − r n k ( n − k ) (cid:21)(cid:33) ∼ k √ π (cid:90) ∞ √ x exp (cid:20) − x n k ( n − k ) (cid:21) dx = √ (cid:0) (cid:1) √ π (cid:18) k ( n − k ) n (cid:19) . Therefore, for these values of k , we have A k ∼ √ (cid:0) (cid:1) π · n k ( n − k ) . Consequently, we have C K = n (cid:88) k = √ n A k ∼ √ (cid:0) (cid:1) π n (cid:90) ( x − x ) − dx ∼ √ (cid:0) (cid:1) π n (cid:32) √ π Γ (cid:0) (cid:1) Γ (cid:0) (cid:1) (cid:33) = 2 Γ (cid:0) (cid:1) √ π n . As a result, we have E [fp n ( σ )] = B K + C K ∼ (cid:0) (cid:1) √ π n , as desired. (cid:3) Generalized rank and the longest increasing subsequence
Results.
For λ >
0, define the rank λ of a permutation σ ∈ S n as the largest integer r such that σ ( i ) > λr for all 1 ≤ i ≤ r . Observe that for λ = 1, we have rank λ = rank as definedin [EP] (see also [CK, Kit]). Theorem 8.1.
Let c , λ > and < ε < . Also, let σ ∈ S n (123) and τ ∈ S n (132) be chosenuniformly at random. Let R λ,n = rank λ ( σ ) and S λ,n = rank λ ( τ ) . Then E [ R ,n ] = E [ S ,n ] forall n . Furthermore, there exists a constant c > such that for n sufficiently large, we have nλ + 1 − c n + ε ≤ E [ R λ,n ] ≤ nλ + 1 − c √ n. The following corollary rephrases the result in a different form.
Corollary 8.2.
Let λ and R n be as in Theorem 8.1. Then we have lim n →∞ log (cid:16) nλ +1 − E [ R n ] (cid:17) log n = 12 . For λ = 1, the corollary is known and follows from the following theorem of Deutsch,Hildebrand and Wilf. Define lis( σ ), the length of the longest increasing subsequence in σ ,to be the largest integer k such that there exist indices i < i < . . . < i k which satisfy σ ( i ) < σ ( i ) < . . . < σ ( i k ). Let lis n : S n (321) → Z such that lis n ( σ ) = lis( σ ). Theorem 8.3 ([DHW]) . Let σ ∈ S n (321) . Define X n ( σ ) = lis n ( σ ) − n √ n . Then we have lim n →∞ P ( X n ( σ ) ≤ θ ) = Γ( , θ )Γ( ) , where Γ( x, y ) is the incomplete Gamma functionΓ( x, y ) = (cid:90) y u x − e − u du. Theorem 8.3 states that E [lis n σ ] → n + c √ n as n → ∞ , for some constant c >
0. By [EP],we have S ,n ∼ ( n − lis n ). By Knuth-Richards’ bijection between S n (123) and S n (132), we have S ,n ∼ R ,n , so E [ R ,n ] = E [ n − lis n ]. Therefore, E [ R ,n ] → n − ( n c √ n ) = n − c √ n, as n → ∞ . HAPE OF PATTERN-AVOIDING PERMUTATIONS 31
Another technical lemma.
By Knuth-Richards’ bijection (also Simion-Schmidt’s bijec-tion) between S n (123) and S n (132), the rank statistic is equidistributed in these two classes ofpermutations, so E [ R ,n ] = E [ S ,n ] for all n (see [CK, Kit]). Therefore, it suffices to prove theinequalities for R λ,n .We prove the lower bound first, followed by the upper bound. To prove the lower bound, wefirst need a lemma regarding this sum. Lemma 8.4.
Let ε > , c > , λ > , n a positive integer, and i, j be integers such that ≤ i ≤ r, ≤ j ≤ λr, where r = (cid:22) nλ + 1 − c n + ε (cid:23) . Then the function P n ( i, j ) is maximized for ( i, j ) = ( r, λr ) , as n → ∞ .Proof of Lemma 8.4. By reasoning similar to the proof of Lemma 4 . P n ( r, λr ) C n ∼ ( λ + 1) c n ε − λ √ π exp (cid:20) − ( λ + 1) c n ε λ (cid:21) , as n → ∞ . Let 1 ≤ i ≤ r and 1 ≤ j ≤ λr such that i + j ∼ sn for some 0 ≤ s <
1. Then by Lemma 4 . < δ < P n ( i, j ) C n < δ n for n sufficiently large . For n sufficiently large, we have δ n < ( λ + 1) c n ε − λ √ π exp (cid:20) − ( λ + 1) c n ε λ (cid:21) , so P n ( i, j ) < P n ( r, λr ) as n → ∞ .It remains to consider 1 ≤ i ≤ r, ≤ j ≤ λr such that i + j ∼ n . Since r ∼ n/ ( λ + 1), weneed i ∼ nλ + 1 and j ∼ λnλ + 1as well. Let i = n/ ( λ + 1) − c and j = λn/ ( λ + 1) − d , where c = an + ε + α and d = bn + ε + β , ≤ α, β < / − ε , and if α = 0 (or β = 0), then a ≥ c (or b ≥ λc , respectively).We have P n ( i, j ) C n ∼ ( λ + 1) ( c + d + 2) λ √ πn exp (cid:20) − ( λ + 1) ( c + d ) λn (cid:21) , by similar logic to that used in the proof of Lemma 4.6. Plugging in for c and d in the exponentgives P n ( i, j ) C n ∼ ( λ + 1) ( c + d + 2) λ √ πn exp (cid:20) − ( λ + 1) ( an α + bn β ) n ε λ (cid:21) . Clearly if α > β > P n ( i, j ) < P n ( r, λr ) as n → ∞ . Similarly, if α = β = 0 but a + b > ( λ + 1) c , then we again have P n ( i, j ) < P n ( r, λr ) as n → ∞ . Therefore, the function P n ( i, j ) is indeed maximized at ( i, j ) = ( r, λr ), as desired. (cid:3) Proof of the lower bound in Theorem 8.1.
Let ε > , σ ∈ S n (123), and let 0 < c .Consider P ( R λ,n ≤ r ) , where r = nλ + 1 − c n + ε . By the union bound, P ( R λ,n ≤ r ) ≤ r (cid:88) i =1 λr (cid:88) j =1 P n ( i, j ) C n . By Lemma 8.4, we have P ( R λ,n ≤ r ) ≤ λr P n ( r, λr ) C n , and by Lemma 4.6, there exists δ > n sufficiently large, we have λr P n ( r, λr ) C n < λr δ n ε → . Therefore, P (cid:18) R λ,n ≤ nλ + 1 − c n + ε (cid:19) → n → ∞ , and E [ R λ,n ] ≥ nλ + 1 − c n + ε , as desired.8.4. Proof of the upper bound in Theorem 8.1.
We can express E [ R λ,n ] as E [ R λ,n ] = n/ ( λ +1) (cid:88) k =0 k P ( R λ,n = k ) = nλ + 1 − n/ ( λ +1) (cid:88) k (cid:48) =0 k (cid:48) P (cid:18) R λ,n = nλ + 1 − k (cid:48) (cid:19) , if we let k (cid:48) = n/ ( λ + 1) − k . From here, for every 0 < a < b , we have E [ R λ,n ] ≤ nλ + 1 − b √ n (cid:88) k (cid:48) = a √ n k (cid:48) P (cid:18) R λ,n = nλ + 1 − k (cid:48) (cid:19) ≤ nλ + 1 − a √ n b √ n (cid:88) k (cid:48) = a √ n P (cid:18) R λ,n = nλ + 1 − k (cid:48) (cid:19) . Therefore, it suffices to show that for some choice of 0 < a < b , we have P (cid:18) nλ + 1 − b √ n ≤ R λ,n ≤ nλ + 1 + a √ n (cid:19) = A > , for some constant A = A ( a, b, λ ).Let F = (cid:22) λ − λ + 1 n (cid:23) . Let σ ∈ S n (123), and suppose we have i, j < nλ + 1 , σ ( i ) = i + F, and σ ( j ) = j + F. Then for any r > j , we have σ ( r ) < j + F , since otherwise a -pattern would exist with( i, j, r ). However, this is a contradiction, since σ : Z ∩ [ j + 1 , n ] → Z ∩ [1 , j + F −
1] must beinjective, but n − j > j + F −
1. Consequently, σ can have at most one value of i < n/ ( λ + 1)with σ ( i ) = i + F . HAPE OF PATTERN-AVOIDING PERMUTATIONS 33
Let k (cid:48) = n/ ( λ + 1) − d √ n for some constant d . Then we have P ( R λ,n ≤ k (cid:48) ) ≥ k (cid:48) (cid:88) i =1 P n ( i, i + F ) C n , since (cid:80) k (cid:48) i =1 P n ( i, i + F ) counts the number of values i ≤ k (cid:48) such that σ ( i ) = i + F for some σ ∈ S n (123), and each σ is counted by at most one i . In the notation of Theorem 3 .
3, we obtain P ( R λ,n ≤ k (cid:48) ) ≥ (cid:90) ∞ d η (cid:18) λ + 1 , t (cid:19) κ (cid:18) λ + 1 , t (cid:19) dt as n → ∞ . For any d >
0, this integral is a positive constant which is maximized at d = a for d ∈ [ a, b ].Therefore, P (cid:18) nλ + 1 − b √ n ≤ R λ,n ≤ nλ + 1 + a √ n (cid:19) → (cid:90) ba η (cid:18) λ + 1 , t (cid:19) κ (cid:18) λ + 1 , t (cid:19) dt as n → ∞ . Denote A ( a, b, λ ) = (cid:90) ba η (cid:18) λ + 1 , t (cid:19) κ (cid:18) λ + 1 , t (cid:19) dt. For any λ we can choose 0 < a ( λ ) < b ( λ ) so that A ( a, b, λ ) is bounded away from 0. Pluggingback into our upper bound gives E [ R λ,n ] ≤ nλ + 1 − a ( λ ) A ( a, b, λ ) √ n, completing the proof of the upper bound and of Theorem 8.1.9. Final remarks and open problems C n +1 /C n → n → ∞ , see [Eul]. In the second half of the 20th century, the studyof various statistics on Catalan objects, became of interest first in Combinatorics and then inAnalysis of Algorithms. Notably, binary and plane trees proved to be especially fertile groundfor both analysis and applications, and the number of early results concentrate on these. Werefer to [A3, BPS, Dev, DFHNS, DG, FO, GW, GP, Ort, Tak] for an assortment of both recentand classical results on the distributions of various statistics on Catalan objects, and to [FS]for a compendium of information on asymptotic methods in combinatorics.The approach of looking for a limiting object whose properties can be analyzed, is standardin the context of probability theory. We refer to [A1, A2] for the case of limit shapes ofrandom trees (see also [Drm]), and to [Ver, VK] for the early results on limit shapes of randompartitions and random Young tableaux. Curiously, one of the oldest bijective approach topattern avoidance involves infinite “generating trees” [West].9.2. The study of pattern avoiding permutations is very rich, and the results we obtain herecan be extended in a number of directions. First, most naturally, one can ask what happensto patterns of size 4, especially to classes of equinumerous permutations not mapped into eachother by natural symmetries (see [B1]). Of course, multiple patterns with nice combinatorialinterpretations, and other generalizations are also of interest (see e.g. [B2, Kit]). Perhaps, onlya few of these will lead to interesting limit shapes; we plan to return to this problem in thefuture.Second, there are a number of combinatorial statistics on S n (123) and S n (123), which havebeen studied in the literature, and which can be used to create a bias in the distribution. In other words, for every such statistic α : S n ( π ) → Z one can study the limit shapes of theweighted average of matrices (cid:88) σ ∈S n ( π ) q α ( σ ) M ( σ ) , where q ≥ α which counts the number of times pattern ω occurs in a permutation σ . When π is empty, that is when the summation above is over thewhole S n , these averages interpolate between S n for q = 1, and S n ( ω ) for q →
0. We referto [B3, Hom] for closely related results (see also [MV1, MV2])Finally, there are natural extension of pattern avoidance to 0-1 matrices, see [KMV, Spi],which, by the virtue of their construction, seem destined to be studied probabilistically. Weplan to make experiments with the simple patterns, to see if they have interesting limit shapes.9.3. There are at least nine different bijections between S n (123) and S n (132), not countingsymmetries which have been classified in the literature [CK] (see also [Kit, § x ( a, c ) in Theorem 3.5 is not easilyevaluated by elementary methods. After a substitution, it is equivalent to (cid:90) ∞ z ( z + c ) e − z dz, which can be then computed in terms of hypergeometric and Bessel functions ; we refer to [AS]for definitions. Similarly, it would be nice to find an asymptotic formula for u ( c ) in Theorem 3.6.9.5. Let us mention that the results in Section 3 imply few other observations which are notimmediately transparent from the figures. First, as we mentioned in Subsection 1.1, our resultsimply that the curve Q n ( k, k ) is symmetric for (1+ ε ) n/ < k < (1 − ε ) n , reaching the minimumat k = 3 n/
4, for large n . Second, our results imply that the ratio Q n ( n/ − √ n, n/ − √ n ) P n ( n/ − √ n, n/ − √ n ) → n → ∞ , which is larger than the apparent ratios of peak heights visible in Figure 1. Along the maindiagonal, the location of the local maxima of P n ( k, k ) and Q n ( k, k ) seem to roughly coincideand have a constant ratio, as n → ∞ . Our results are not strong enough to imply this, as extramultiplicative terms can appear. It would be interesting to see if this is indeed the case.9.6. The generalized rank statistics rank λ we introduce in Section 8 seem to be new. Ournumerical experiments suggest that for all λ > n , rank λ is equidistributed between S n (132) and S n (123). This is known for λ = 1 (see § σ ) ≤ n/ σ ∈ S n , since otherwise M ( σ ) is singular. Using the samereasoning, we obtain rank λ ( σ ) ≤ n/ (1 + λ ) for λ ≤
1. It would be interesting to see if thereis any connection of generalized ranks with the longest increasing subsequences, and if theseinequalities make sense from the point of view of the Erd˝os-Szekeres inequality [ES]. We referto [AD, BDJ] for more on the distribution of the length of the longest increasing subsequencesin random permutations. For more discussion of this integral, see http://tinyurl.com/akpu5tk
HAPE OF PATTERN-AVOIDING PERMUTATIONS 35 P n ( j, k ) for j + k ≤ n + 1, coincided with theprobability that a random Dyck path of length 2 n passes through point ( n − j + k − , n + j − k − √ πn (1 + o (1))(see [Chu, DI]). Similarly, the exponential decay of P n ( k − t, n − k − t ) for t = n / ε , followsfrom the setting, and seems to correspond to tail estimates for the expected maximal distance.However, because of the emphasis on the maxima and occupation time of Brownian excursions,it seems there are no known probabilistic analogues for results such as our Theorem 3.3 despitea similarities of some formulas. For example, it is curious that for c (cid:54) = 0 and α = 1 /
2, theexpression η ( a, c ) κ ( a, c ) = c √ π a (1 − a ) exp (cid:20) − c a (1 − a ) (cid:21) is exactly the density function of a Maxwell-distributed random variable, which appears in thecontour process of the Brownian excursion (cf. [GP]).9.8. Unfortunately, there seem to be no obvious way to interpret Lemma 5.3 probabilistically.One can of course, use bijections to random binary trees, but the corresponding statistics arenot very natural. It would be interesting to find a good probabilistic model with the same limitshape as S n (132). χ -squared statistic on S n : χ ( σ ) := n (cid:88) i =1 r ( i ) , where r ( i ) = min (cid:8) ( n + 1 − σ ( i ) − i ) , (2 n − σ ( i ) − i ) (cid:9) . This statistic measures how far the permutation σ is from the reverse identity permutation(in cyclic order). Curiously, in contrast with the number of fixed points, this statistic candistinguish our sets of pattern avoiding permutations. Theorem 9.1.
For χ defined as above and n → ∞ , we have: E [ χ ( σ )] = Θ( n ) , where σ ∈ S n (123) uniform , E [ χ ( σ )] = Θ( n . ) , where σ ∈ S n (132) uniform , E [ χ ( σ )] = Θ( n ) , where σ ∈ S n uniform . The remaining four patterns of length 3 have also these asymptotics by the symmetries. Theproof of Theorem 9.1 will appear in a forthcoming thesis [Min] by the first author.9.10. After this paper was written and posted on the arXiv , we learned of two closely relatedpapers. In [ML], the authors set up a related random pattern avoiding permutation modeland make a number of Monte Carlo simulations and conjectures, including suggesting an em-piric “canoe style” shape. Rather curiously, the authors prove the exponential decay of theprobability P (cid:0) τ (1) > . n (cid:1) , for random τ ∈ S n (4231).In [AM], the authors prove similar “small scale” results for patterns of size 3, i.e. exponentialdecay above anti-diagonal and polynomial decay below anti-diagonal for random σ ∈ S n (132).They also study a statistic similar but not equal to rank. The first author surveys these resultsand explores the connections in [Min]. Acknowledgments:
The authors are grateful to Ton´ci Antunovi´c, Drew Armstrong, MarekBiskup, Stephen DeSalvo, Sergi Elizalde, Sergey Kitaev, Jim Pitman and Richard Stanley, for Most recently, such model was found by Christopher Hoffman, Erik Slivken and Doug Rizzolo, using the tunnel concept from [E1] (in preparation). useful remarks and help with the references. Special thanks to Neal Madras for telling us abouthis ongoing work with Lerna Pehlivan, and hosting the first author during his Summer 2013visit. We are indebted to an anonymous referee for careful reading of the paper, comments andhelpful suggestions. The second author was partially supported by the BSF and the NSF.
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Appendix: Numerical calculationsFigure 7.
Values of P ( k, k ). Figure 8.
Values of P ( k, k ). Figure 9.
Values of P ( k, k ). Figure 10.
Values of P ( k, k ). Figure 11.
Surface of P ( j, k ). Figure 12.
Surface of Q ( j, k ). Figure 13.
A closer look at P ( j, k ), 201 ≤ j + k ≤ Figure 14.
A closer look at Q ( j, k ), 201 ≤≤
A closer look at Q ( j, k ), 201 ≤≤ j + k ≤≤