Abstract

We continue the study of the generalized pattern avoidance condition for , the wreath product of the cyclic group with the symmetric group , initiated in the work by Kitaev et al., In press. Among our results, there are a number of (multivariable) generating functions both for consecutive and nonconsecutive patterns, as well as a bijective proof for a new sequence counted by the Catalan numbers.

1. Introduction

The goal of this paper is to continue the study of pattern-matching conditions on the wreath product of the cyclic group and the symmetric group initiated in [1]. is the group of signed permutations where there are signs, , , ,, , where is a primitive th root of unity. We can think of the elements as pairs where and . For ease of notation, if where for , then we simply write where . Moreover, we think of the elements of as the colors of the corresponding elements of the underlying permutation .

We define a concept for matchings in words over a finite alphabet . Given a word , let be the word found by replacing the largest integer that appears in by . For example, if , then . Given a word such that , we say that a word has a -match starting at position provided . Let -mch be the number of -matches in the word . Similarly, we say that occurs in a word if there exist such that . We say that avoids if there are no occurrences of in .

There are a number of papers on pattern matching and pattern avoidance in [2–5]. We now present a selection of the previously studied definitions for pattern matching and avoidance, so that the reader may see how ours differs from and/or extends those in the literature. For example, the following pattern matching condition was studied in [3–5].

Definition 1. Let , be a subset of , and . (1)One says that has an exact occurrence of (resp., ) if there are such that (resp., ).(2)One says that   avoids an exact occurrence of (resp., ) if there are no exact occurrences of (resp., ) in .(3)One says that there is an exact -match in   starting at position (resp., exact -match in   starting at position ) if (resp., .
That is, an exact occurrence or an exact match of in an element is just an ordinary occurrence or match of in where the corresponding signs agree exactly. For example, Mansour [4] proved via recursion that, for any , the number of elements in which avoid exact occurrences of is . This generalized a result of Simion [6] who proved the same result for the hyperoctahedral group . Similarly, Mansour and West [5] determined the number of permutations in that avoid all possible exact occurrences of 2 or 3 element sets of patterns from . For example, let be the number of that avoid all exact occurrences of the patterns in the set , the number of that avoid all exact occurrences of the patterns in the set , and the number of that avoid all exact occurrences of the patterns in the set . Then Mansour and West [5] proved that where is the th Fibonacci number.

An alternative matching condition arises when we drop the requirement of the exact matching of signs and replace it by the condition that the two sequences of signs match in the sense of words described above. That is, we shall consider the following matching conditions.

Definition 2. Let where , a subset of where for all , , and . (1)One says that has an occurrence of (resp., ) if there are   such that , (resp., ).(2)One says that   avoids (rep., ) if there are no occurrences of (resp., ) in .(3)One says that there is a -match in  starting at position (resp., -match in    starting at position ) if (resp., .

For example, suppose that and . Then there are no exact occurrences or exact matches of in . However, there are two occurrences of , one in positions 2 and 4 and one in positions 3 and 4. Thus there are two occurrences of in and there is a -match in starting at position .

Finally, a more general-matching condition which generalizes both occurrences and matches and exact occurrences and exact matches in was introduced in [1].

Definition 3. Let , be a subset of , be a sequence of subsets of , and . (1)One says that has an occurrence of (resp., ) if there are such that (resp., is equal to for some in ) and for .(2)One says that   avoids (resp., ) if there are no occurrences of (resp., ) in . (3)One says that there is a -match in starting at position (resp., -match in starting at position ) if , red (resp., is equal to for some in ) and for .

Thus a -occurrence (or a -match) where and is just an exact occurrence (resp., an exact match) of . Similarly, a -occurrence (or a -match) where , , and such that for is just an occurrence (or match) of .

Suppose we are given , , where for , and . We let ()mch (resp., ()Emch) denote the number of -matches (resp., exact -matches) in . We let mch (resp., Emch) denote the number -matches (resp., exact -matches) in . We let ()mch be the number of -matches in and ()mch the number of -matches in .

The main result of [1] was to derive a generating function for the distribution of -matches where is any element of . To state the main result of [1], we need to define some notation. First we define the -analogues of , , , and by respectively. We define the -analogs of , , , and by , , , and , respectively.

Next suppose that and where . Then we will say that is a maximum packing for if has -matches starting at positions . We let denote the set of maximum packings for . Given any word , we let . For any , we let (resp., ) equal the number of pairs such that and (resp., ). We then define We will also be interested in the specializations where for any word , . In the special case where where , we shall denote as .

Then Kitaev et al. [1] proved that, for any and where ,

Duane and Remmel [7] gave the generating function for the number of -matches and exact -matches for a certain collection of where are called minimal overlapping patterns. Given a word such that and , Duane and Remmel say that has the -minimal overlapping property if the smallest such that there exists a with ()mch is . This means that in a -colored permutation , two -matches can share at most one pair of letters, and this pair of letters must occur at the end of the first -match and at the start of the second -match. Similarly, we say that has the -exact match minimal overlapping property if the smallest such that there exists a with ()Emch is . Now if has the -minimal overlapping property, then the shortest -colored permutations , such that ()mch have length . We let equal the set of -colored permutations, , such that ()mch. We refer to the -colored permutations in as maximum packings for . We let Similarly, we let denote the set of -colored permutations, , such that ()Emch, and refer to the -colored permutations in as exact match maximum packings for . We let Duane and Remmel [7] proved that (I) if and has the -minimal overlapping property, then and (II) if has the -exact match minimal overlapping property, then Duane and Remmel's proof of (I) and (II) works equally well for -matches.

The main goal of this paper is to prove a number of results on the number of -avoiding elements of and to find a generating function for the number of -matches for certain patterns which do not have the minimal overlapping property. That is, in Section 2, we shall prove a number of results about the number of -avoiding elements of where . In Section 3, we prove a number of results on the number of -avoiding elements of and for certain subsets of . Finally, in Section 4, we will prove an analog of a theorem of Mendes and Remmel which gave a generating function by the number of descents for the set of elements of that had no -matches.

2. -Avoiding Patterns

Given and where for , let denote the number of which avoid . In the special case where for , we shall denote as simply . Thus is the number which avoid . In this case, we shall find formulas for or where .

There are a number of natural maps for which the distribution of -occurrences and -matches remain invariant. That is, for any , we define the reverse of , and the complement of , , respectively, by If , then we define the reverse of , and the complement of relative to , , by If , then we define . Then if where , for , then we define Then consider the maps where for and where is the identity map; that is, for all , for all , and for all with for . It is easy to see that a has a -match or -occurrence if and only if has a -occurrence or -match.

It follows that if and , then we need only to compute for two patterns; namely, and .

We start by considering the pattern .

Lemma 4. The number of elements in avoiding is given by

Proof. We first observe that elements of different colors are independent in permutations avoiding , meaning that no two elements with different colors can form a prohibited configuration. Thus assuming we have elements of color , , we can choose how to place these colors to form a word in ways. Then we can choose the sets of elements from which will correspond to the colors in in ways. Finally, in order to construct a which avoids the prohibited pattern, we must place the elements of in the positions which are colored by in decreasing order.

The proof of Lemma 4 suggests an obvious generalization for patterns of the form where .

Theorem 5. Let . Then the number of elements in avoiding is given by where is the number of permutations in avoiding .

Proof. The proof here is essentially the same as the proof of Theorem 4, except that we can place elements of a permutation in of the same color in any of ways.

It is well known that the number of -permutations avoiding any pattern of length 3 is given by the th Catalan number . As a corollary to Theorem 5, we have that, for any pattern , the number of permutations in avoiding is given by

One can generalize Theorems 4 and 5 even further, by considering the distribution of patterns. Indeed, assuming that we know the number of -permutations containing occurrences of a pattern , we can write down the number of permutations in with occurrences of as For example, the distribution of the occurrences of the pattern is the same as the distribution of coinversions in permutations, so one can extract the numbers as the coefficients to in and substitute them in the last formula to get the distribution of the number of occurrences of in .

Next we consider the pattern in the case where .

The number of permutations in avoiding is shown in [6, page 19] to be equal to . However, in Theorem 6 below we provide an independent derivation of the exponential generating function in this case.

Theorem 6. The exponential generating function for the number of elements in avoiding the pattern is given by

Proof. Let and If contains the element 1 colored by the color 1, then there are no restrictions for placing this element, thereby giving possibilities. On the other hand, if contains the element 1 colored by the color 0 in position and is to avoid , it must be the case that every element to the right of 1 is colored with the color 0 and these elements can be arranged in any of ways. Moreover, it immediately follows that no instance of occurrence of exists where the first element is to the left of 1 and the second is to the right of 1. Thus it follows that avoids if and only if there is no occurrence of in . Then in the case where 1 is colored by 0 and is in position , we have possibilities where the binomial coefficient is responsible for choosing the elements to the left of 1 and placing the remaining elements to the right of 1. To summarize, we obtain Multiplying both parts of the equation above by and summing over all , we have so that It follows that Integrating and using the fact that , we see that which implies that

This type of argument can be generalized. That is, we have the following theorem.

Theorem 7. Let and , where . Let denote the number of which avoid , and Then if , If , then

Proof. Note that if contains the element 1 colored with a color from , then there are no restrictions for placing this element, thereby giving possibilities. On the other hand, if contains the element 1 colored by the color in position where , then if is to avoid , it must be the case that every element to the right of 1 must be colored with a color from and these elements can be arranged in any ways. Moreover, it immediately follows that no instance of a occurrence of exists where the first element is to the left of 1 and the second is to the right of 1. Then it follows that avoids if and only if there is no occurrence of in . Thus in the case where 1 is colored by a color from and is in position , we have possibilities where the binomial coefficient is responsible for choosing the elements to the left of 1 and placing the remaining elements to the right of 1. Hence Multiplying both parts of the equation above by and summing over all , we have so that for all , It follows that
Now suppose that . Then integrating (31) and using the fact that and the fact that , we see that which implies that which proves (26).
Next suppose that . Then (31) becomes Integrating (34) and using the fact that , we see that Thus which proves (27).

Next consider the special case where and so . Then clearly, It follows from the form of these generating functions that . Moreover, in the case where , it is easy to see that our definitions imply that the number of which avoid is the same as the number of which avoid . Then where is the number of avoiding . This is also easy to see combinatorially. Namely, if which avoids , then if is the result of replacing each occurrence of in by 0 and each occurrence of in by 1, then will be an element of which avoids . Vice versa, if avoids and arises from by replacing each by some element from and each 1 by some element from , then will avoid . In fact, if we are in the case where so that , then by the same argument, we can get an arbitrary which avoids by starting out with an which avoids and replacing each by some element from and each 1 by some element from . Considering the form of the generating function for when , this suggests the following theorem which can easily be proved by the same method as we used to prove Theorem 7.

Theorem 8. Given , let equal the number of 's in and equal the number of 1’s in . Let denote the set of which avoids . Then

3. Avoidance for Sets of Patterns

In this section, we shall prove a variety of results for the number of elements of that avoid certain sets of patterns of length 2.

If is a set of permutations of , we let where is the number of permutations such that avoids . If and where for , then we let denote the number of which avoid . We then let In the special case where for all , and for , we shall write instead of . That is, is the number of which avoid . Then we let

We start with a few simple results on sets of patterns where the avoidance of forces certain natural conditions on the possible sets of signs for elements of .

Lemma 9. (1) If  , then for all and .
(2) If  , then for all and .
(3) If  , , , then for all and .
(4) If  , , , then for all and .

Proof. For (1), it is easy to see that avoids if and only if all the signs are pairwise distinct. Therefore, there are ways to pick the signs, and then you have ways to arrange those signs and ways to pick .
For (2), it is easy to see that avoids if and only if . Therefore, there are ways to pick the signs in this case and there are ways to pick .
For (3), it is easy to see that avoids if and only if . Therefore, there are ways to pick the signs in this case and there are ways to pick .
For (4), it is easy to see that avoids if and only if . Therefore, there are ways to pick the signs in this case and there are ways to pick .

For , we let Then we have the following.

Theorem 10. Let be any set of permutations of . Then if , for all .

Proof. It is easy to see that if and avoids , then it must be the case that . So suppose that where . Then clearly the number of such that avoids is That is, the binomial coefficient allows us to choose the elements of that correspond to the constant segment in . Then we only have to arrange the elements of so that it avoids which can be done in ways. Thus it follows that or, equivalently, that which implies (44).

We immediately have the following corollary.

Corollary 11. For any , the number of elements of which avoid or is .

Proof. Let and . Then clearly, for so that for . But then by Theorem 10, so that for .

We can derive theorems analogous to Theorem 10 for the other sign conditions in Lemma 9. That is, suppose that is any set of patterns from . Then we let where is the set of all permutations of . Then we have the following.

Theorem 12. For any , let Then, for all , (1) for all , (2) for all ,(3) for all .

Proof. For (1), note that for to avoid , we must have for some . Then avoids if and only if avoids . Thus .
For (2), note that for to avoid , we must have where . Then for any of the strictly increasing words , avoids if and only if avoids . Thus .
For (3), note that, for to avoid , we must have where the letters of are pairwise distinct. Then for any of the words which have pairwise distinct letters, avoids if and only if avoids . Thus .

Next we will prove two more results about for other sets of patterns that contain .

Theorem 13. Let Then (1) for all and , (2) for all and .

Proof. For (1), note that as in the proof of Theorem 10, if and avoids , then it must be the case that .
Now suppose that where . Assume that also avoids and is the set of elements of that correspond to the signs in . Then it follows that all the elements of must be larger than all the elements of , all the elements of must be larger than all the elements of , and so forth. Thus consists of the largest elements of consists of next largest elements of , and so forth. Then we can arrange the elements of in any order in the positions corresponding to in to produce a which avoids . Hence there are elements of the form which avoid in . Therefore (1) immediately follows.
For (2), observe that if in addition such also avoids , then we must place the elements of in increasing order. Hence is the number of solutions of with which is well known to be .

Theorem 14. Let . Then for , where and for .

Proof. We shall classify the elements which avoid by the number of elements which follow in . Now if so that ends in , the fact that avoids both and means that all the signs must be the same. That is, must be of the form for some . But since must also avoid , must be the identity. Thus there are choices for such .
Now suppose there are elements following in . Then again the fact that avoids both and means that all the signs in corresponding to (where ) must be the same, say that sign is . The fact that avoids means that (i) must be in increasing order and (ii) all the signs corresponding to must be different from . But then it follows from the fact that avoids both and that all the elements in must be greater than all the elements in . Hence there are such elements if and there are no such elements if . Thus from (ii) it follows that since is the only element of which avoids . For , we have
In the case , (52) becomes and in the case , (52) becomes In general, assuming that it follows that

We next consider simultaneous avoidance of the patterns and .

Theorem 15. Let , and . Then

Proof. Fix and suppose that is an element of which avoids both and . Now if is constant, then clearly can be arbitrary so that there are such elements.

Next assume that is not constant so there is an such that where for and for . We claim that must be the largest elements of . If not, then let be the largest element which is not in . Thus and there must be at least one with such that . Now it cannot be since otherwise would be an occurrence of either or . Hence it must be the case that and for some . But then no matter what color we choose for , either or would be an occurrence of either or . We can continue this reasoning to show that for any , the elements of corresponding to the block in must be strictly larger than the elements of corresponding to the block in . This given, it follows that we can arrange the elements of corresponding to a block in in any way that we want and we will always produce a pair that avoids both and . Thus for such a , we have ways to choose the colors and ways to choose the permutation . It follows that which is equivalent to (57).

Next we want to consider the problem of avoiding where and . If , then we say that is a left-to-right minimum of if for all and we let denote the number of left-to-right minima of . Let denote the set of all which avoid . Then we let

Then we have the following theorem.

Theorem 16. For and , one has

Proof. Let . If contains the element 1 colored by the color in position where , then if is to avoid , it must be the case that every element to the right of 1 must be colored with a color from and these elements can be arranged in any of ways. Moreover, it immediately follows that no instance of an occurrence of exists where the first element is to the left of 1 and the second is to the right of 1. Thus it follows that avoids if and only if there is no occurrence of in . Thus it follows that Here the binomial coefficient is responsible for choosing the elements to the left of 1 and placing the remaining elements to the right of 1 and the factor of accounts for the fact that 1 is always a left-to-right minimum in and none of the elements following 1 can be a left-to-right minimum of . This shows that Multiplying both parts of the equation above by and summing over all , we have so that, for all , It follows that Integrating and using the fact that , we see that and hence,

Using the fact that we see that In particular, We note that, by our symmetry maps where we apply the identity map to and the composition of complement and reverse to and , one can show that It is easy to see that our definitions imply that so that

Next we consider the case where , , and .

Theorem 17. Let where and , and . Then if , If , then and

Proof. Let so that
Now suppose that contains the element 1 colored by a color from . Then there are no restrictions for placing this element, thus giving possibilities. On the other hand, if contains the element 1 colored by the color in position where , then if is to avoid , it must be the case that every element to the right of 1 must be colored with a color from and these elements can be arranged in any of ways. Moreover, it immediately follows that no instance of an occurrence of exists where the first element is to the left of 1 and the second is to the right of 1. Then it follows that avoids if and only if there is no occurrence of in . So in the case where 1 is colored by a color from and is in position , we have possibilities where the binomial coefficient is responsible for choosing the elements to the left of 1 and placing the remaining elements to the right of 1. Thus Multiplying both parts of the equation above by and summing over all , we have so that, for all and , It follows that
Now suppose that . Then integrating (80) and using the fact that and that , we see that which implies that and proves (74).
Now if . Then and (80) becomes Integrating (83) and using the fact that , we see that Thus which proves (75).

Finally, we end this section by considering the elements of which avoid both and . In this case, we only have a result for the case when .

Theorem 18. The number of permutations in simultaneously avoiding and is given by the -th Catalan number .

Proof. We prove the statement by establishing a bijection between the objects in question of length and the Dyck paths of semilength known to be counted by the Catalan numbers (a Dyck path of semi-length is a lattice path from to with steps and that never goes below the -axis; Figure 4).
Suppose . Note that must avoid the pattern 1 2 3, as in an occurrence of such a pattern in , there are two letters of the same color leading to an occurrence of (1 2,0 0). Thus the structure of , as it is well known, is two decreasing sequences shuffled.
Subdivide into the so-called reverse irreducible components. A reverse irreducible component is a factor of of minimal length such that everything to the left (resp., right) of is greater (resp., smaller) than any element of . For example, the subdivision of is . The blocks of size 1 are called singletons. In the example above, 4 and 3 are singletons. It is easy to see that any singleton element in can have any color (either 0 or 1). We will now show that the color of each element of a nonsingleton block is uniquely determined.
Indeed, irreducibility of a single block means that two decreasing sequences in the structure of -avoiding permutations are the block's sequence of left-to-right minima and the block's sequence of right-to-left maxima which do not overlap. Thus for any left-to-right minimal element (except possibly the last element), one has an element greater than to the right of it inside the same block and vice versa, from which we conclude that must receive color 1, whereas must receive color 0 (otherwise a prohibited pattern will occur).
We are ready to describe our bijection. For a given , consider a matrix representation of , that is, an integer grid with the opposite corners in (0,0) and and with a dot in position for . We will give a description of a path (corresponding to ) from to involving only steps and that never goes above the line . Clearly, can be transformed to a Dyck path of length by taking a mirror image with respect to the line , rotating 45 degrees counterclockwise and making a parallel shift.
To build , set and , and do the following steps, letting begin at . Clearly, each reverse irreducible block of defines a square on the grid which is a matrix representation of the block. We call the reverse irreducible block of with the left-top corner at the current block.
Step 1. If the current block is not a singleton, go to Step 2. If the color of the element with -coordinate equal to is (resp., 1) travel around the current block counterclockwise (resp., clockwise) to get to the point . Note that touches the line if the color is 1. Set and , and proceed with Step 3.
Step 2. In Step 2 we follow a standard bijection between -avoiding permutations and Dyck paths that can be described as follows. Let be the point of the current block opposite to . Start going down from until the -coordinate of the current node gets less than the -coordinate of the dot with -coordinate equal . Start moving horizontally to the right and go as long as possible making sure that none of the dots are below the part of constructed so far and and . Suppose is the last point the procedure above can be done (i.e., we were traveling on the line and either and or there is a dot with -coordinate having -coordinate less than ). If and , proceed with Step 3; otherwise set and and go to Step 2. Note that in Step 2    never touches the line .
Step 3. If , make as many as it takes horizontal steps to get to the point and terminate; otherwise go to Step 1.
Returning to our example, , we have given the matrix diagram in Figure 1. We have outlined the reverse irreducible blocks in Figure 2. We start our path at . We travel down until we reach , when we are less than the y-coordinate of our first point . We then continue traveling right and down as described in Step 2. We travel clockwise around our singleton colored and counterclockwise around our singleton colored . Then we continue to the final reverse irreducible block and finish our path, given in Figure 3.


Figure 1 
Matrix representation of .

Figure 2 
Matrix representation of with reverse irreducible blocks outlined.

Figure 3 
The Dyck path corresponding to added to Figure 2.

Figure 4 
The Dyck path corresponding to . Note that the path only touches the line after a singleton colored .

4. Pattern Matching Results for Pairs of Length

Goulden and Jackson [8] proved the following theorem.

Theorem 19 (see [8]). If , where , then

Later Mendes and Remmel [9] refined this result by proving the following.

Theorem 20. When , where is the number of rearrangements of zeroes and ones such that zeroes never appear consecutively.

The reason that Theorem 20 is a refinement of Theorem 19 is that Mendes and Remmel [9] proved that so that setting in (87) yields (86).

The main goal of this section is to generalize Theorem 20. That is, let where . Then we shall consider three sets of patterns for :(a),(b), (c) which equals the set of all such that , is weakly decreasing, and . In [1], Kitaev et al. defined three different types of descents. That is, for elements , they defined We shall refer to as the descent set of , as the weak descent set of , and as the strict descent set of .

Theorem 21. Let where . (1) If , then (2) If , then
(3) If   equals the set of all such that , is weakly decreasing, and , then

Proof. Our proof is an adaptation of the proof that Mendes and Remmel [9] used to prove Theorem 20. The basic idea is to show that (90), (91), and (92) arise by applying appropriate ring homomorphisms, defined on the ring of symmetric functions over infinitely many variables , to a simple symmetric function identity.
The th elementary symmetric function, , and the th homogenous symmetric function, , are defined by the generating functions Clearly,
Let be an integer partition; that is, is a finite sequence of weakly increasing positive integers. Let denote the number of parts of . If the sum of these integers is , we say that is a partition of and write . For any partition , let . The well-known fundamental theorem of symmetric functions says that is a basis for or that is an algebraically independent set of generators for . Similarly, if we define , then is also a basis for .
Since the elementary symmetric functions and the homogeneous symmetric functions are both bases for , it makes sense to talk about the coefficient of the homogeneous symmetric functions when written in terms of the elementary symmetric function basis. These coefficients have been shown to equal the sizes of certain sets of combinatorial objects up to a sign. A brick tabloid of shape and type is a filling of a row of squares of cells with bricks of lengths such that bricks do not overlap. One brick tabloid of shape and type is displayed In Figure 5.
Let denote the set of all -brick tabloids of shape and let . Through simple recursions stemming from (94), Eğecioğlu and Remmel [10] proved that
We now consider three different ring homomorphisms which map to where is the field of rational numbers. Since is an algebraically independent set of generators for , we can define a ring homomorphism by simply specifying its value on for all . For , define
Our first goal is to prove the following facts. (A) If , then (B)If , then (C) If consists of the set of all such , is weakly decreasing, and , then
Let , , and . Then for ,
We wish to give a combinatorial interpretation of right-hand side of (100). First we select a partition and then we select a brick tabloid . Next we interpret the multinomial coefficient as the number of ways to pick sets of sizes , respectively, which partition . We then place these elements in in the bottom of the cells of the brick in decreasing order reading from left to right. Next if , then which we interpret as the ways of picking an element and placing at the top of each cell of . If , then which we interpret as the ways of picking a strictly decreasing sequence from and placing on top of the th cell of . If , then which we interpret as the ways of picking a weakly decreasing sequence from and placing on top of the th cell of . Finally, we interpret the term as follows. First, we pick a sequence which consists of    's and 's for some such that there is no consecutive sequence of in . Then if , we label the th cell of with , and if , we label the th cell of with . Finally, we label the last cell of with . Thus for any given , we will have cells labeled with which accounts for the sign . The only term in (100) that we have not accounted for is the term which we use to change that last cell of each brick from to . We shall call an object created in this way a filled labeled brick tabloid and we let denote the set of all filled labeled brick tabloids that arise in this way for interpreting . Thus a consists of a brick tabloid , a permutation , a sequence , and a labeling of the cells of with elements from such that the following conditions hold. (a) is strictly decreasing in each brick.(b)(1)If , then is constant in each brick. (2)If , then is strictly decreasing in each brick. (3)If , then is weakly decreasing in each brick. (c) The final cell of each brick is labeled with 1. (d) Each cell which is not a final cell of a brick is labeled with or and there is no consecutive string of cells of length labeled within a brick. We then define the weight of to be the product of all the labels in and the sign of to be the product of all the labels in .
For example, if , , , and , then Figure 6 pictures a filled labeled brick tabloid at the top, a a filled labeled brick tabloid in the middle, and a filled labeled brick tabloid at the bottom. In each case, and .
Next we define sign-reversing, weight-preserving involutions on . To define , we scan the cells of from left to right looking for the leftmost cell such that either (i) is labeled with , (ii) is at the end of a brick and the brick immediately following has the property that is strictly decreasing in all the cells corresponding to and and is constant on all the cells corresponding to and if , is strictly decreasing on all the cells corresponding to and if , is weakly decreasing on all the cells corresponding to and if .
In case (i), where is the result of replacing the brick in containing by two bricks and where contains the cell plus all the cells in to the left of and contains all the cells of to the right of , , , and is the labeling that results from by changing the label of cell from to . In case (ii), where is the result of replacing the bricks and in by a single brick , , and is the labeling that results from by changing the label of cell from to . Note that since the last cell in each brick is labeled with a 1, when we combine the two bricks in case (ii), we cannot create a run of consecutive cells which are labeled . If neither case (i) or case (ii) applies, then we let . For example, if we consider the 's in pictured in Figure 6, then their corresponding images are pictured in Figure 7.
It is easy to see that is a weight-preserving, sign-reversing involution, and hence shows that
Thus we must examine the fixed points of . First there can be no labels in so that . Moreover, if and are two consecutive bricks in and is the last cell of , then (1)If , it cannot be the case that and since otherwise we could combine and , (2)if , it cannot be the case that and since otherwise we could combine and , and (3)if , it cannot be the case that and since otherwise we could combine and . Thus if is the last cell of a brick, then if , if , and if . However, the label on a cell of any brick must be if is not the last cell of the brick and must be 1 if is the last cell of a brick. Moreover, if is not the last cell of a brick, then and if , if , if . Thus if is not the last cell of a brick, then if , if , and if . It follows that and if , if , and if . Then Therefore (A), (B), and (C) hold.
We can now prove (90) by applying to (94). That is, we have and , and then
Similarly, (91) can be proved by applying to (94) and (92) can be proved by applying to (94).


Figure 5 
A brick tabloid of shape and type .

Figure 6 
Filled labeled brick tabloids in .

Figure 7 
for the s in Figure 6.

In fact, we can calculate in several cases. For instance, routine combinatorial techniques are able to show that is whenever . When , we find that

We can also find a generating function for by considering the following argument. Take a rearrangement that is counted by . Now remove the final number of the rearrangement. Either(a)we have removed a , (b)we have removed a .

In case (a), we can count the number of such rearrangements by , except that we have overcounted by those rearrangements counted by which end in zeroes. We can correct this by considering removing the last zeroes of the over-counted rearrangements, leaving those of length with zeros ending in a , of which there are . In case (b), we can count the number of such sequences by . Thus we obtain the following relationship:

Next, we multiply by and sum over all , from which we obtain a generating function with a finite sum of small values in the numerator and a quotient of .

Acknowledgment

J. Remmel was partially supported by NSF Grant DMS 0654060 and M. Riehl was supported by UWEC’s Office of Research and Sponsored Programs.