Consecutive Up-down Patterns in Up-down Permutations

In this paper, we study the distribution of the number of consecutive pattern matches of the five up-down permutations of length four, 1324, 2314, 2413, 1432, and 3412, in the set of up-down permutations. We show that for any such τ , the generating function for the distribution of the number of consecutive pattern matches of τ in the set of up-down permutations can be expressed in terms of what we call the generalized maximum packing polynomials of τ. We then provide some systematic methods to compute the generalized maximum packing polynomials for such τ .

There have been several papers that have studied the number of up-down permutations σ ∈ A n which avoid a given pattern.For example, Mansour [19] and Deutsch and Reifegerste (see [27, Problem h 7 ] or [14]) showed that for any τ ∈ S 3 , the number of up-down permutations σ ∈ A n which avoid τ is always a Catalan number.For example, the number of up-down permutations σ ∈ A n that avoid 132 (231) is C ⌊n/2⌋ .In [18], it was shown that the number of σ ∈ A 2n that avoids 1234 or 2143 is 2(3n)!n!(n+1)!(n+2)! .There has been somewhat less work on the distribution of τ -matches in up-down permutations.Carlitz [5] found the generating function for the number of rises in the peaks of the updown permutations where a rise in the peaks of an up-down permutation is just 213-match in σ.
In fact, the main ideas of this paper can be extended to study the distributions of τmatches in the set of up-down permutations where τ is a natural analogue of a minimal overlapping permutation as studied by Duane and Remmel [8].This work appears in Duane's thesis [7].We have chosen to focus on the five up-down permutations of length four because the arguments are simpler and the formulas are more tractable than in the general case considered in Duane's thesis.However such an extension will be the subject of a forthcoming paper.Let τ ∈ A 4 .If σ ∈ A n where n 4, then τ -matches can only start at odd positions.If σ ∈ A 2n and τ -mch(σ) = n − 1, then we say that σ is a maximum packing for τ .Thus if σ ∈ A 2n is a maximum packing for τ , then σ has τ -matches starting at positions 1, 3, . . ., 2n − 3. We let MP 2n,τ denote the set of maximum packings for τ in A 2n and we let mp 2n,τ = |MP 2n,τ |.We shall see that it follows from results of Harmse and Remmel [13] that mp 2n,τ (1) = mp 2n,τ Our main theorem will show that for each i ∈ [5], the generating functions A τ (i) (t, x) and B τ (i) (t, x) can be expressed in terms of what we call generalized maximum packings for τ (i) .We say that σ ∈ S 2n is a generalized maximum packing for τ (i) if we can break σ into consecutive blocks σ = B 1 . . .B k such that 1. for all 1 j k, B j is either an increasing sequence of length 2 or red(B j ) is a maximum packing for τ (i) of length 2s for some s 2 and 2. for all 1 j k − 1, the last element of B j is less than the first element of B j+1 .
Our main theorem is the following.
We shall prove Theorem 1 by applying the so-called homomorphism method which has been developed in a series of papers [1,3,4,13,15,21,22,24,26,27].In particular, we shall show that the generating functions in Theorem 1 arise by applying certain ring homomorphisms defined on the ring of symmetric functions Λ in infinitely many variables to simple symmetric function identities.For example, let h n denote the nth homogeneous symmetric function in Λ and e n denote the nth elementary symmetric function in Λ.That is, h n and e n are defined by the generating functions Then we shall show that (6) arises by applying a ring homomorphism θ to the simple symmetric function identity the electronic journal of combinatorics 21(3) (2014), #P3.2 For example, we shall show that for an appropriately chosen ring homomorphism θ.Typically, one proves equations like (9) by interpreting the left-hand side of ( 9) in terms of a signed weighted sum of filled brick tabloids and then applying an appropriate sign-reversing weight-preserving involution to show that the combinatorial interpretation of (2n)!θ(h 2n ) reduces to the desired polynomial.The situation in this paper is a bit different from previous examples of the homomorphism method in that it requires two involutions to show that our combinatorial interpretation of (2n)!θ(h 2n ) reduces to the right-hand of (9).Equation ( 7) is proved in a similar manner except that we apply θ to a more complicated symmetric function identity.The outline of the paper is as follows.In Section 2, we shall provide the necessary background on symmetric functions that is required for our proofs.In Section 3, we shall prove Theorem 1.In Section 4, we shall show how to compute mp n,τ (i) for i = 1, 2, 3, 4, 5.In Section 5, we shall develop recursions for GMP n,τ (i) (x) for i = 1, 2, 4. The simplest case is the set of the recursions for GMP n,τ (1) (x) described above.In that case, we shall show that GMP 2n,τ (1 n−2 where for any formal power series f (x) = n 0 f n x n , we write f (x)| x n the coefficient of x n in f .Using these facts, we can compute the generating functions for the number of up-down permutations with no τ (1) -matches or with exactly one τ (1) -match.For example, we shall show that where N 2n,τ (1) is the number of σ ∈ A 2n with no τ (1) -matches.Finally, in Section 6, we shall study the distribution of double rise pairs and double descent pairs in up-down permutations.That is, if σ = σ 1 . . .σ n ∈ A n is an up-down permutation, then we say that a pair (2i − 1)(2i) is a double rise pair if both σ 2i−1 < σ 2i+1 and σ 2i < σ 2i+2 .Thus (2i−1)(2i) is a double rise pair in σ if and only if there is a 1324 match starting at position 2i − 1 in σ.We say that a pair (2i − 1)(2i) is a double descent pair if both σ 2i−1 > σ 2i+1 and σ 2i > σ 2i+2 .Thus (2i − 1)(2i) is a double descent pair in σ if and only if there is a D-match starting at position 2i − 1 in σ where D = {2413, 3412}.

Symmetric Functions
In this section we give the necessary background on symmetric functions needed for our proofs of the generating functions (6) and (7).
Let Λ denote the ring of symmetric functions over infinitely many variables x 1 , x 2 , . . .with coefficients in some field F .We let Λ n denote the space of homogeneous symmetric functions of degree n so that Λ = ⊕ n 0 Λ n .
If the sum of these integers is n, we say that λ is a partition of n and write λ ⊢ n.For any partition λ = (λ 1 , . . ., λ ℓ ), let e λ = e λ 1 • • • e λ ℓ and h λ = h λ 1 • • • h λ ℓ .The well-known fundamental theorem of symmetric functions says that {e λ : λ is a partition} is a basis for Λ or, equivalently, that {e 0 , e 1 , . ..} is an algebraically independent set of generators for Λ.Since {e 0 , e 1 , . ..} is an algebraically independent set of generators for Λ, we can specify a ring homomorphism θ on Λ by simply defining θ(e n ) for all n 0.
Let B λ,n denote the set of all λ-brick tabloids of shape (n) and let B λ,n = |B λ,n |.We shall write B = (b 1 , . . ., b k ) if B is a brick tabloid of shape n such that the lengths of the bricks in B are b 1 , . . ., b k as we read from left to right.Egecioglu and Remmel proved in [9] that Next we define a class of symmetric functions p n,ν which have a relationship with e λ that is analogous to the relationship between h n and e λ .These functions were first introduced in [17] and [21].Let ν be a function which maps the set of non-negative integers into the field F .Recursively define p n,ν ∈ Λ n by setting p 0,ν = 1 and By multiplying series, this means that where the last equality follows from the definition of p n,ν .Therefore, the electronic journal of combinatorics 21(3) (2014), #P3.2 or, equivalently, When taking ν(n) = 1 for all n 1, (13) becomes which implies that p n,1 = h n for all n.Other special cases for ν give well-known generating functions.For example, if ν(n) = n for n 1, then p n,ν is the power symmetric function p n = i x n i .For any statement A, we let The coefficient of e λ in p n,ν has a nice combinatorial interpretation similar to that of h n .Suppose T is a brick tabloid of shape (n) and type λ and that the final brick in T has length ℓ.Define the weight of a brick tabloid w ν (T ) to be ν(ℓ) and let It was proved in [17] and [21] that 3 The proof of Theorem 1 In this section, we shall prove Theorem 1. Fix τ ∈ A 4 .We start out by proving (6).Define a ring homomorphism θ from Λ into Q(x) by setting θ(e 0 ) = 1, θ(e 2n+1 ) = 0 for all n 0, and Then we claim that θ(h 2n−1 ) = 0 and x τ -mch(σ) (16) for all n 1.Note that by (10), the electronic journal of combinatorics 21(3) (2014), #P3.2 Clearly if µ is a partition of 2n − 1, then µ must have an odd part so that θ(e µ ) = 0. Thus θ(h 2n−1 ) = 0 for all n 1.Note also that so that there is no loss if we restrict the sum on the right-hand side of (17) to partitions µ where every part of µ is even, i.e., to partitions of the form 2λ where λ is a partition of n and 2λ = (2λ 1 , . . ., 2λ ℓ(λ) ) if λ = (λ 1 , . . ., λ ℓ(λ) ).Thus Next we want to give a combinatorial interpretation to the right-hand side of (18).We start with a brick tabloid T = (2b 1 , . . ., 2b ℓ(λ) ) of type 2λ.Then the binomial coefficient 2n 2b 1 ,...,2b ℓ(λ) allows us to pick a set partition U = (U 1 , . . ., U ℓ(λ) ) of {1, . . ., 2n} where |U i | = 2b i for i = 1, . . ., ℓ(λ).Next we use the factor ℓ(λ) j=1 GMP 2b j ,τ (x) to choose a sequence of permutations σ = (σ (1) , . . ., σ (ℓ(λ)) ) such that σ (j) ∈ S 2λ j is a generalized maximum packing for τ for j = 1, . . ., ℓ(λ).Then for each j, we let α (j) be the sequence that arises by replacing the rth largest element of σ (j) by the rth largest element of U j and then we place the elements of α (j) in the cells of brick 2b j from left to right.For example, we have illustrated this process in Figure 2 for τ = τ (1) = 1324 where the brick tabloid is T = (2,8,6).We have also indicated the block structure in each brick by underlining those elements in a common block.The weight w(T, U, σ) of such a triple (T, U , σ) is ℓ(λ) j=1 w(σ (j) ).We can interpret w(T, U, σ) as 2n j=1 L(j) where L : {1, . . ., 2n} → Q[x] is a labeling of the cells of T which is defined as follows.First we define a labeling L : {1, . . ., 2n} → Q[x] where L(j) = 1 if cell j does not start a τ -match that is contained in its brick and L(j) = x − 1 if cell j starts a τ -match that is contained in its brick.Then we define L(j) = − L(j) if j is the first cell of its block and that block is not the last block in its brick and L(j) = L(j) otherwise.Thus the RHS of ( 18) can be interpreted as the sum of the weights of all triples (T, α, L) such that 1. T = (d 1 , . . ., d k ) is a brick tabloid of shape (2n) where each brick d j has even length, 2. α is a permutation of S 2n such that in each brick d j , the sequence of elements in brick d j reduces to a permutation in GMP d j ,τ , and is the labeling of the cells of T described above.
For example, in Figure 2, T = (2, 8, 6), α = 1 3 4 5 7 10 9 12 11 13 2 8 6 14 15 16, and L is the labeling where all the cells which do not have an explicit label in them are assumed to have label 1.We let T 2n,τ denote the set of all such triples constructed in this way.It then follows that Next we will define two involutions I and J which will show that right-hand side of ( 19) is equal to the right-hand side of (16).We define I : T 2n,τ → T 2n,τ as follows.Suppose that we are given a triple (T, α, L) where T = (d 1 , . . ., d k ).Then read the bricks from left to right until you find the first brick d j such that either (i) the generalized maximum packing corresponding to the elements in d j consists of more than one block or (ii) the generalized maximum packing corresponding to the elements in d j consists of a single block and the last element of d j is less than the first element of the following brick d j+1 .In case (i), split d j into two bricks d * and d * * where d * contains the cells of the first block in the generalized maximum packing corresponding to the elements in d j and d * * contains the remaining cells of d j .We keep all the labels the same except that we change the label on the first cell of d * from −1 to 1 if the first block of d j is of length 2 and from −(x − 1) to (x−1) if the first block of d j has length 4. In case (ii), we combine bricks d j and d j+1 into a single brick d.Note that since the last element of d j is less than the first element of d j+1 , the elements in the new brick d will still reduce to a generalized maximum packing for τ .We keep all the labels the same except that we change the label on the first cell of d j from 1 to −1 if d j is of length 2 and from (x − 1) to −(x − 1) if d j has length 4. In both cases, we do not change the underlying permutation α.If neither case (i) nor case (ii) applies, then I(T, α, L) = (T, α, L).For example, if (T, α, L) is the element of T 16,τ (1) pictured in Figure 2, then we are in case (ii) since we can combine the first and second bricks so that I(T, α, L) is pictured in Figure 3.It is easy to see that if Thus we must examine the fixed points of I. Clearly, if (T, α, L) is a fixed point of I, then the elements of each brick d in T must reduce to a generalized maximum packing of τ which consists of a single block.Second, we must not be able to combine any two bricks so that if T = (d 1 , . . ., d k ), then the last element of d j is greater than the first element of d j+1 for j = 1, . . ., k − 1.But this means that the underlying permutation α is an up-down permutation.It follows that the fixed points of I consists of triples (T, α, L) such that (I) α is an up-down permutation of length 2n, (II) T = (d 1 , . . ., d k ) where each d j has even length and the elements of d j reduce to a generalized maximum packing of τ which consists of a single block, and (III) the label of L(j) of the jth cell of T is (x − 1) if j is the start of τ -match in α that lies in its brick and is equal to 1 otherwise.
Next we want to modify our interpretation of the right-hand side of (20) to consist of all triples (T ′ , α, L ′ ) such that (I ′ ) α is an up-down permutation of length 2n, (II ′ ) T = (d 1 , . . ., d k ) where each d j has even length and the elements of d j reduce to a generalized maximum packing of τ which consists of a single block, and (III ′ ) the label of L(j) of the jth cell of T is either x or −1 if j is the start of τ -match in α that lies in its brick and is equal to 1 otherwise.
We let F I 2n,τ denote the set of triples (T ′ , α, L ′ ) satisfying (I ′ )-(III ′ ).Then for any (T ′ , α, L ′ ) ∈ F I 2n,τ , we define the weight w(T ′ , α, L ′ ) of (T ′ , α, L ′ ) to be 2n j=1 L ′ (j).For example, Figure 4 pictures an element of F I 16,τ (1) whose weight is x, where again the cells which do not have labels are assumed to have label 1.It then follows that Next we define an involution J : F I 2n,τ → F I 2n,τ .Given an element (T, α, L) ∈ F I 2n,τ , scan the cells of T = (d 1 , . . ., d k ) from left to right looking for the first cell c such that either (A) the label of c is −1 or (B) c is the second to last element of a brick d j such that the elements of bricks d j and d j+1 reduce to a generalized maximum packing of τ which consists of a single block.Note that in case (B), c must have label 1 since it does not start match of τ in α that lies in its brick.In case (A), if c is in brick d j , then break d j into two bricks d * and d * * where d * contain the cells of d j up to and including cell c + 1 and d * * contains the rest of the cells of d j .We then replace the −1 label on cell c by 1.In case (B), we replace the bricks d j and d j+1 by a single brick d and replace the label of 1 on c by −1.In either case, we do not change the underlying permutation α.If neither case (A) nor case (B) applies, then we let J(T, α, L) = (T, α, L).For example, if we consider the triple (T, α, L) pictured in Figure 4, we cannot combine bricks d 1 and d 2 because α does not have a τ (1) -match starting cell 1 and we cannot combine bricks d 2 and d 3 because α does not have a τ (1) -match starting cell 3. Thus we are in case (B) where d j = d 3 and c = 7.Thus we split d 3 at cells 8 and 9 so that J(T, α, L) = (T ′ , α, L ′ ) is the filling pictured in Figure 5.Note that it will automatically be the case that the first action that we can take for (T ′ , α, L ′ ) is to combine the two bricks that made up the d 3 in (T, α, L).It is easy to see that if the electronic journal of combinatorics 21(3) (2014), #P3.2 Thus we must examine the fixed points of J.If J(T, α, L) = (T, α, L), then clearly (T, α, L) can have no cells which have a −1 label.Thus in a brick d of T of length 4, the start of every τ -match contained in d is labeled with an x.Moreover, we claim that there cannot be a τ -match that involves cells in two different bricks d j and d j+1 .That is, the only way a τ -match could span cells in both d j and d j+1 is if that τ match started in cell c which is the second to last cell of d j .But this would imply that the elements of d j and d j+1 would reduce to a generalized maximum packing for τ with a single block and, hence, case (B) of our involution would apply to c. Hence if (T, α, L) is a fixed point of J, then w(T, α, L) = x τ -mch(α) .
For any α ∈ A 2n , there is a unique fixed point (T, α, L) of J whose underlying permutation is α.That is, we must define the bricks d 1 , d 2 , . . .inductively as follows.We let d 1 be of length 2 if there is no τ -match in α starting at 1 and d 1 be of length 2s if there are τ -matches starting at positions 1, 3, . . ., 2s − 3 but not at 2s − 1 in α.Then having defined bricks d 1 , . . ., d r where d r ends at cell c = 2k < 2n, we let d r+1 be of length 2 if there is no τ -match in α starting at 2k + 1 and d r+1 be of length 2s if there are τ -matches starting at positions 2k + 1, 2k + 3, . . ., 2k + 2s − 3 but not at 2k + 2s − 1 in α.Hence x τ -mch(α) .

It then follows that
which is what we wanted to prove.
To prove (7), we will use the same ring homomorphism θ with weight function ν : We have designed ν so that ν(2n)θ(e 2n ) = (−1) 2n−1 (2n−1)!GMP 2n−1,τ (x).Then we claim that for all n 0, θ(p 2n+1,ν ) = 0 and Note that by (14), the electronic journal of combinatorics 21(3) (2014), #P3.2 Clearly if µ is a partition of 2n + 1, then µ must have an odd part so that θ(e µ ) = 0. Thus θ(p 2n+1,ν ) = 0 for all n 0. It also follows that when we want to compute θ(p 2n+2,ν ), we can restrict ourselves to considering partitions of the form 2λ where λ is a partition of n + 1.Thus As before, we want to give a combinatorial interpretation to the right-hand side of (24).We start with a brick tabloid T = (2b 1 , . . ., 2b ℓ(λ) ) of length 2n + 2 and type 2λ.Then the binomial coefficient GMP 2b j ,τ (x) to choose a sequence of permutations σ = (σ (1) , . . ., σ (ℓ(λ)) ) such that σ (j) ∈ S 2b j is a generalized maximum packing for τ for j = 1, . . ., ℓ(λ) − 1 and σ (ℓ(λ)) ∈ S 2b ℓ(λ) −1 is a generalized maximum packing for τ .Then for each j, we let α (j) be the sequence that arises by replacing the rth largest element of σ (j) by the rth largest element of U j and then we place the elements of α (j) in the cells of brick 2b j from left to right.This means that for the last brick 2b ℓ(λ) , we will fill in all but the last cell which we leave blank.For example, we have illustrated this process in Figure 6 for τ (1) = 1324 where the underlying brick tableau T = (2, 8, 6).We have also indicated the block structure in each brick by underlying those elements in a common block.The weight w(T, U, σ) of such a triple (T, U , σ) is ℓ(λ) j=1 w(σ (j) ).Again we can interpret w(T, U, σ) to be 2n+1 j=1 L(j) where L : {1, . . ., 2n} → Q[x] is a labeling of the cells of T .To define L, we label the blank cell with 1 and then we label the remaining cells exactly as we did before.Thus the RHS of (24) can be interpreted as the sum of the weights of all triples (T, α, L) such that Note that the only difference between the fillings of even length in the proof of (16) and our current fillings is that, in our current fillings, the last brick ends in a blank cell and the block structure of the reduction of the sequence of elements in the last brick must end in a block of size 1.This means that we can define the two involutions I and J exactly as before since the fact that the last block of the final brick is length 1 does not change things.Using the same reasoning as in our proof of (16), it is easy to check that our involutions I and J show that It then follows that θ( Dividing the second and the last elements in the string of equalities in ( 27) by t gives (7) which is what we wanted to prove.
4 Computing mp n,τ (i) In this section, we shall consider the problem of computing mp τ (i) ,n since we will need such computations to compute GMP τ (i) ,n .
The problem of computing mp τ (i) ,2n has been studied by Harmse and Remmel [13] in a different context.Harmse and Remmel studied maximum packings in column strict arrays.That is, let F n,k denote the set of all fillings of a k × n rectangular array with the integers 1, . . ., kn such that the elements increase from bottom to top in each column.We let (i, j) denote the cell in the ith row from the bottom and the jth column from the left of the k × n rectangle and we let F (i, j) denote the element in cell (i, j) of F ∈ F n,k .
If F is any filling of a k × n-rectangle with distinct positive integers such that elements in each column increase, reading from bottom to top, then we let red(F ) denote the element of F n,k which results from F by replacing the ith smallest element of F by i.For example, Figure 7 demonstrates a filling, F , with its corresponding reduced filling, red(F ).If F ∈ F n,k and 1 c 1 < • • • < c j n, then we let F [c 1 , . . ., c j ] be the filling of the k × j rectangle where the elements in column a of F [c 1 , . . ., c j ] equal the elements in column c a in F for a = 1, . . ., j.If P ∈ F j,k and F ∈ F n,k where j n, then we say there is a P -match in F starting at position i if red(F [i, i + 1, . . ., i + j − 1]) = P .We let P -mch(F ) denote the number of P -matches in F .For example, if we consider the fillings P ∈ F 3,3 and F, G ∈ F 6,3 shown in Figure 8, then it is easy to see that there are no P -matches in F and there are 2 P -matches in G starting at positions 1 and 2 so P -mch(F ) = 0 and P -mch(G) = 2.
If P ∈ F 2,k , then we define MP P n to be the set of F ∈ F n,k with P -mch(F ) = n − 1, i.e. the set of F ∈ F n,k with the property that there are P -matches in F starting at positions 1, 2, . . ., n − 1.We let mp P n = |MP P n | and, by convention, we define mp P 1 = 1.Given an F ∈ F n,2 , we let σ(F ) be the permutation We then let P (i) denote the element of F 2,2 such that σ(P (i) ) = τ (i) .For example, P (1) , . . ., P (5) are pictured in Figure 9.It is then easy to see that for any maximum packing F ∈ F n,2 of P (i) , σ(F ) is an up-down permutation in A 2n which is a maximum packing for τ (i) .Vice versa, if σ = σ 1 . . .σ 2n is a maximum packing of τ (i) , then the 2 × n array F σ where F σ (1, i) = σ 2i−1 and F σ (2, i) = σ 2i is a maximum packing for P (i) .An example of this correspondence is pictured at the top of Figure 9 for τ (1) = 1324 and It follows that for i = 1, . . ., 5, mp 2n,τ (i) = mp P (i) n .Now Harmse and Remmel [13] proved that for n 2, mp P Our next goal is to prove the following.
Proof.To compute mp 2n+1,τ (i) , we must exploit some of the techniques used by Harmse and Remmel [13] to compute mp P n for P ∈ F 2,k .To help us visualize the order relationships within P (i) , we form a directed graph G P (i) on the cells of the 2 × 2 rectangle by drawing a directed edge from the position of the number j to the position of the number j + 1 in P for j = 1, 2, 3.For example, in Figure 10, the graph G P (1) is pictured immediately to the right of P (1) .Then G P (i) determines the order relationships between all the cells in P (i) since P (i) (r, s) < P (i) (u, v) if there is a directed path from cell (r, s) to cell (u, v) in G P (i) .Now suppose that F ∈ MP P (i) n where n 3.Because there is a P (i) -match starting in column j for each 1 j < n, we can superimpose G P (i) on the cells in columns j and j + 1 to determine the order relations between the elements in those two columns.If we do this for every pair of columns, j and j + 1 for j = 1, . . ., n − 1, we end up with a directed graph on the cells of the 2 × n rectangle which we will call G n,P (i) .For example, in Figure 10, G 6,P (1) is pictured in the second row.It is then easy to see that if F ∈ MP P (i) n and there is a directed path from cell (r, s) to cell (u, v) in G n,P (i) , then it must be the case that F (r, s) < F (u, v).Note that G n,P (i) will always be a directed acyclic graph with no multiple edges.Harmse and Remmel [13] proved that the problem of computing mp P (i) n for any P (i) ∈ F 2,2 of shape 2 2 can be reduced to finding the number of linear extensions of a certain poset associated with P (i) .That is, the graph G n,P (i) induces a poset W G n,P (i) = ({(i, j) : 1 i 2 & 1 j n}, < W ) on the cells of the 2 × n rectangle by defining (i, j) < W (s, t) if and only if there is a directed path from (i, j) to (s, t) in G n,P (i) .Harmse and Remmel proved that there is a 1:1 correspondence between the elements of MP n and the linear extensions of W n,P (i) .That is, if n , then it is easy to see that (a 1 , b 1 ), . . ., (a 2n , b 2n ) where extension of W G n,P (i) , then one can define F so that F (a i , b i ) = i and it will automatically be the case that F ∈ MP P (i) n .We can define a similar poset for maximum packings of τ (i) of length 2n + 1.Note that in a maximum packing F ∈ MP P (i)  n , the element a in the top right-hand corner of F corresponds to the last element of σ(F ) so that, to account for the last element in a permutation α = α 1 . . .α 2n+1 ∈ A 2n+1 which has τ (i) -matches starting at positions 1, 3, . . .2n − 3, we must add an extra element b to graph G n,P (i) with a directed arrow from b to a since we know that α 2n > α 2n+1 .We let G + n,P (i) denote this extended graph.For example, the graph G + 6,P (1) is pictured in the third line of Figure 10.It follows that mp 2n+1,τ (i) equals the number of linear extensions of W G + n,P (i) .
First consider the problem of computing mp P (1)  2n+1 for n 2. In this case, let a be the rightmost element in the top row of G n,P (1) .Since there is a directed path in G n,P (1) from every element other than a to a, it must be the case that a is the last element in any linear extension of W G n,P (1) and, hence, in any F ∈ MP P (1)  n , F (a) = 2n.Note that the same thing happens in G + n,P (1) .That is, there is a directed path in G + n,P (1) from every element other than a to a. Thus it must be the case that a is the last element in any linear extension of W G + n,P (1) so that a would be assigned the label 2n + 1 in any linear extension.
But then it is easy to see that b can be assigned any element in {1, . . ., 2n}.Thus once we pick the value assigned to b, then the number of linear extensions of G + n,P In Figure 11, we have pictured the graphs of G 6,P (2) and G + 6,P (2) in the second line.In this case, it is easy to see that there is a unique linear extension of W G n,P (2) and the rightmost top element a must be the largest element 2n since there is a directed path in G n,P (2) from every element other than a to a.The same thing happens in G + n,P (2) , namely 2n + 1 must be assigned to a since there is a directed path in G + n,P (2) from every element other than a to a.But then it is easy to see that b can be assigned to any element in {1, . . ., 2n}.Thus once we pick a value that is assigned to b, then the number of linear extensions of G + n,P (2) just reduces to the number of linear extensions of G n,P (2) which is just 1.Thus mp 2n+1,τ (2) = 2n.
In Figure 12, we have pictured the graphs of G 6,P (3) and G + 6,P (3) in the second line.Now consider the element b in G + n,P (3) .If we assign b the value 1, then there is no restriction on the linear extensions of the remaining elements so that we get a total of C n−1 linear extensions in that case since mp n,P (3) = C n−1 .However, if b > 1, then it is easy to see that the rightmost bottom element must be the first element in any linear extension since there is a directed path from that element to any other element which is not equal to b.Thus the rightmost bottom element must be assigned to 1.It then follows that we can extend the graph G + n,P (3) to a graph G ++ n,P (3) by adding a new element 0 and adding new directed edges connecting 0 to 1 and 1 to b.This process is pictured on line 4 of Figure 12.It is easy to see that the number of linear extensions of G + n,P (3) where b > 1 is just the number of linear extensions of G ++ n,P (3) which is the same as the number of linear extensions of G n+1,P (3) .Since the number of linear extensions of G n+1,P In Figure 13, we have pictured the graphs of G 6,P (4) and G + 6,P (4) in the second line.In this case, it is easy to see that there is a unique linear extension of W G n,P (4) and the rightmost top element a of G n,P (4) must be the (n + 1)st element in the linear extension of W G 6,P (4) since there are n elements x for which there is a directed path in G n,P (4) from x to a and there are n − 1 elements y such that there is a directed path from a to y in G n,P (4) .Similarly, a must be the (n + 2)nd element in any linear extension of W G + 6,P (4) since there are n + 1 elements x for which there is a directed path in G + n,P (4) from x to a and there are n − 1 elements y such that there is a directed path from a to y in G + n,P (4) .Hence we can assign b to be any element from 1, . . ., n + 1.Once we pick a value for b, then the number of linear extensions of G + n,P (4) just reduces to the number of linear extensions of G n,P (4) which is just 1.Thus mp 2n+1,τ (4) = n + 1.
In Figure 14, we have pictured the graphs of G 6,P (5) and G + 6,P (5) in the second line.In the case of G + n,P (5) , it is easy to see that the rightmost top element a must be the third element in any linear extension of W G + 6,P (5) .Thus we have two linear extensions depending upon how we order the two elements that have a directed edge into a.Hence mp 2n+1,P (5) = 2.

Computing GM P n,τ (i) (x)
In this section, we shall study the problem of computing GMP n,τ (i) (x) for n 1 and i = 1, . . ., 5. First it is easy to see that for any i, GMP 1, That is, there is only one generalized maximum packing of length 1 which consists of a block of length 1 and weight 1.Similarly, there is only one generalized maximum packing of length 2 which consists of a block of length 2 and weight 1.There is only one generalized maximum packing of length 3, namely, 123 where 12 is a block of length 2 and 3 is a block of length one.Thus GMP 3,τ (i) = w(123) = −1.There are two generalized maximum packings of length 4, namely, 1234 which consists of two blocks of length 2 and has weight −1 and τ (i) which consists of a single block with weight x − 1.Thus GMP 4,τ (i) = x − 2.
In general, we do not know how to find closed formulas for GMP n,τ (i) (x) as function of n, but for i ∈ {1, 2, 4} there are simple recursions for computing GMP n,τ (i) (x).The key to our ability to develop recursions for GMP n,τ (i) (x) in the case where i ∈ {1, 2, 4} is due to the fact that τ (1) , τ (2) , and τ (4) either start with 1 or end with 4. This will allow us to develop recursions based on either the length of the first block or the length of the last block in a generalized maximum packing.Neither τ (3) = 2413 nor τ (5) = 3412 start with 1 or end with 4 and we have not been able to find any simple recursions for GMP n,τ (3) or GMP n,τ (5) .
The easiest case is for τ (1) = 1324 where we have the following theorem.
Proof.It is easy to see from the form of the graphs G 2n,P (1) that any maximum packing σ ∈ MP 2n,τ (1) must start with 1 and end with 2n.By the definition of a generalized maximum packing whose block structure is B 1 . . .B k , the last element of B i must be smaller than the first element of B i+1 for all i < k.Thus all the elements of B i are smaller than any element in B i+1 for all i < k.Now suppose that n 3 and σ = σ 1 . . .σ 2n ∈ GMP 2n,τ (1) .There are two possibilities.
Case 1. σ consists of a single block.
In this case σ is a maximum packing of τ (1) and w(σ Case 2. σ has block structure B 1 . . .B s where s 2. If B 1 is of length 2, then B 1 = 12 and has weight −1 = −C 0 and red(B 2 . . .B s ) is a generalized maximum packing for τ (1) of length 2n − 2. If B 1 has length 2k where k 2, then B 1 = 1 . . .2k is a maximum packing for τ (1) of length 2k which has weight −(x−1) k−1 and red(B 2 . . .B s ) is a generalized maximum packing of length 2n − 2k.Then there are C k−1 choices for B 1 .Hence the contribution of the permutations in case 2 to GMP 2n,τ (1) (x) is Thus for n 3, (28) holds.
Here is the list of the first few values of GMP 2n,τ (1) (x).
For (1), our formula obviously holds for m = 1 and m = 2. Now if n > 2 and we assume that G 2m,τ (1) (0) = (−1) m−1 C m for m < n, then by (28), we have For (2), GMP 4,τ (1) (x)| x = 1 so our formula holds for n = 2.For n > 2, we have that Using induction and some simplifications with binomial coefficients, one can show that this is equivalent to the following identity: One can then verify that the right-hand side of ( 29) is equal to 2n n−2 by induction using a series of routine manipulations and some simple identities for binomial coefficients.
To prove (32) and (33), note that Similarly, one can compute that Unfortunately, we have not been able to prove similar results for τ (i) where i 2 because in these cases, we have not been able to find explicit formulas for GMP n,τ Next we consider recursions for GMP n,τ (2) (x).
Here is the list of the first few values of GMP 2n,τ (3) (x). 3. for all 1 j k − 1, the last element of B j is less than the first element of B j+1 .
Then we can define the generalized maximum packing polynomials GMP 2n,Υ (x) and GMP 2n+1,Υ (x) in the same manner that we defined GMP 2n,τ (i) (x) and GMP 2n+1,τ (i) (x).If Υ ⊆ A 4 , the proof of Theorem 1 goes through without change if we replace maximum packings for τ with maximum packings for Υ and generalized maximum packings for τ by generalized maximum packings for Υ throughout the proof.Thus we have the following theorem.Proof.It is easy to see that mp 2n,D equals the number of F ∈ F 2,n such that for each i < n, there is either a P (2) -match or a P (5) -match starting in column i.Let F r be the reverse of F .That is, the first row of F r is F (1, n), F (1, n − 1) . . ., F (1, 1) and the second row of F r is F (2, n), F (2, n − 1) . . ., F (2, 1), reading from left to right.For example, (P (3) ) r = P (1) = 3 4 1 2 and (P (5) ) r = 2 4 1 3 .
It is easy to see that F ∈ F 2,n has the property that for each i < n, there is either a P (2) -match or a P (5) -match starting at column i if and only if F r ∈ F 2,n has the property that for each i < n, there is either a (P (2) ) r -match or a (P (5) ) r -match starting at column i.But note that (P (3) ) r and (P (5) ) r are the two standard tableaux of shape (2,2).Thus F r has the property that for each i < n, there is either a (P (2) ) r -match or a (P (5) ) r -match starting at column i if and only if F r is a standard tableau of shape (n, n).But it follows from the Frame-Robinson-Thrall hook formula [10] for the number of standard tableaux Unfortunately elements of MP 2n,D do not end in 1 or 2n so that there does not seem to be any way to develop simple recursions for GMP 2n,D (x) or GMP 2n+1,D (x).Nevertheless, J. Harmse [12] computed the following initial values of GMP n,D (x) (3) = C n−1 and the electronic journal of combinatorics 21(3) (2014), #P3.2 mp 2n,τ (2) = mp 2n,τ (4) = mp 2n,τ (5) = 1, where C n = 1 n+1 2n n is the nth Catalan number.

1 .
T = (d 1 , . . ., d k ) is a brick tabloid of shape (2n + 2) where each brick d j has even length, the electronic journal of combinatorics 21(3) (2014), #P3.2 2. α is a permutation of S 2n+1 such that in each brick d j with j < k, the elements in brick d j reduce to a permutation in GMP d j ,τ and the elements in brick d k fill the first d k − 1 cells of brick d k and reduce to a permutation in GMP d k −1,τ , and 3. L : {1, . . ., 2n} → Q[x] is the labeling of the cells of T described above.For example, in Figure 6, T = (2, 8, 6), α = 1 4 3 7 5 10 9 12 11 13 2 8 6 14 15, and L is the labeling where all the cells which do not have an explicit label in them are assumed to have label 1.

Figure 8 :
Figure 8: Computing the number of P -matches for elements in F 6,3 .

Figure 15 :
Figure 15: The graphs G n,D and G + n,D .