Abstract
We derive several multivariable generating functions for a generalized pattern-matching condition on the wreath product of the cyclic group and the symmetric group . In particular, we derive the generating functions for the number of matches that occur in elements of for any pattern of length 2 by applying appropriate homomorphisms from the ring of symmetric functions over an infinite number of variables to simple symmetric function identities. This allows us to derive several natural analogues of the distribution of rises relative to the product order on elements of . Our research leads to connections to many known objects/structures yet to be explained combinatorially.
1. Introduction
The goal of this paper is to study pattern-matching conditions on the wreath product of the cyclic group and the symmetric group . is the group of signed permutations where we allow signs of the form for some 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 .
Given a sequence of distinct integers, let red be the permutation found by replacing the th largest integer that appears in by . For example, if , then . Given a permutation in the symmetric group , we say a permutation has a -match starting at position provided . Let -mch be the number of -matches in the permutation . Similarly, we say that occurs in if there exist such that red . We say that avoids if there are no occurrences of in .
We can define similar notions for words over a finite alphabet . Given a word , let red be the word found by replacing the largest integer that appears in by . For example, if , then red. Given a word such that red, we say 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 red. We say that avoids if there are no occurrences of in .
There are a number of papers on pattern matching and pattern avoidance in [1–4]. For example, the following pattern matching condition was studied in [2–4].
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
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 [3] proved via recursion that for any , the number of elements in which avoid exact occurrences of is . This generalized a result of Simion [5] who proved the same result for the hyperoctahedral group . Similarly, Mansour and West [4] determined the number of permutations in that avoid all possible exact occurrences of 2 or 3 element sets of patterns of elements of . For example, let be the number of that avoid all exact occurrences of the patterns in the set , let be the number of that avoid all exact occurrences of the patterns in the set , and let be the number of that avoid all exact occurrences of the patterns in the set . Then Mansour and West [4] 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 will consider the following matching conditions.
Definition 2. Let where red, be a subset of where for all , red, and . (1)One says that has an occurrence of (resp., ) if there are such that (resp., ).(2)We say that avoids (resp., ) if there are no occurrences of (resp., ) in .(3)One says that there is a -match in starting at position (resp., -matchin 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, we will consider a more general matching condition which generalizes both occurrences and matches and exact occurrences and exact matches in .
Definition 3. Let , let be a subset of , and let 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)We say that there is a -match in starting at position (resp., -match in starting at position ) if , , is equal to for some in ) and for .
Thus, a -occurrence or -match where and is such that for is just an exact occurrence or exact match of . Similarly, a -occurrence or -match where , , and is 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 of -matches (resp., exact -matches) in . We let -mch be the number of -matches in and let -mch be the number of -matches in .
The main result of this paper is to derive a generating function for the distribution of -matches where is any element of . To state our main result, we first need some notation. We define the -analogues of , , , and by
respectively. We define the -analogues 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 which are 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 shall also be interested in the specializations where for any word , . In the special case, where where , we will denote as . For example, if , , and , then is a maximum packing for if and only if is the identity permutation and is such that . Thus, where for any power series , we write for the coefficient of in . Then it is easy to prove that If , , and , then is a maximum packing for if and only if is the identity permutation and for some . Thus, It then follows that Similarly if and , then is a maximum packing for if and only if is the identity permutation and is such that . Thus, We shall then prove that
The main result of this paper is to prove the following.
Theorem 4. For any and where ,
We note that (12) can be specialized to give natural analogues for generating functions of rises and descents in where we compare pairsusing the product order. That is, instead of thinking of an element of as a pair , we can think of it as a sequence of pairs . We then define a partial order on such pairs by the usual product order. That is, if and only if and . Then we define the following sets and statistics for elements : We shall refer to as the descent set of , as the weak descent set of , and as the strict descent set of . Similarly, we will refer to as the rise set of , as the weak rise set of , and as the strict rise set of . It is easy to see that if and only if there is a -match starting at position , if and only if there is a -match starting at position , and if and only if there is a -match starting at position where .
Similarly, if and only if there is a -match starting at position , if and only if there is a -match starting at position , and if and only if there is a -match starting at position where .
If , then we define the reverse of , by . Similarly, if , then we define . It is easy to see that Thus, we need to find the distributions for only one of the corresponding pairs. Substituting into (12) in these special cases will yield the following generating functions for rises, strict rises, and weak rises in .
Theorem 5. Consider the following:
Other distribution results for -matches follow from these results. For example, if , then we define the complement of , by If , then we define the complement of , by We can then consider maps where for . Such maps will easily allow us to establish that the distribution of -matches is the same for various classes of ’s. For example, one can use such maps to show that the distributions of -matches, -matches, -matches, and -matches are all the same.
Another interesting case is when we let . In this case, we have a -match in starting at if and only if and . In that case, (12) can be specialized to prove the following.
Theorem 6. Consider the following:
Substituting for into (12) will yield the following generating function: Let Thus, -matches correspond to rises, -matches correspond to weak rises, and -matches correspond to strict rises. We shall find for and find . For example, we will show that the following generating function for the distribution of inversions, coinversions, and rises over is an immediate consequence of (12).
Lemma 7. Consider the following: which reduces to (15) when one sets .
We shall prove (12) by applying a ring homomorphism, defined on the ring of symmetric functions over infinitely many variables , to a simple symmetric function identity. There has been a long line of research, [6–15], which shows that a large number of generating functions for permutation statistics can be obtained by applying homomorphisms defined on the ring of symmetric functions over infinitely many variables to simple symmetric function identities. For example, the th elementary symmetric function, , and the th homogeneous symmetric function, , are defined by the generating functions We let where is the th power symmetric function. A partition of is a sequence such that and . We write if is a partition of , and we let denote the number of parts of . If , we set , , and . Let denote the space of homogeneous symmetric functions of degree over infinitely many variables so that . It is well known that , , and are all bases of . It follows that is an algebraically independent set of generators for and, hence, we can define a ring homomorphism where is a ring by simply specifying for all .
Now it is well known that A surprisingly large number of results on generating functions for various permutation statistics in the literature and large number of new generating functions can be derived by applying homomorphisms on to simple identities such as (24) and (25). We shall show that (12) can be proved by applying appropriate ring homomorphisms to identity (24).
The outline of this paper is as follows. In Section 2, we will provide the necessary background in symmetric functions that we will need to derive our generating functions. In Section 3, we will prove (12) and give a number of cases where we can compute or its specializations which will prove all of the formulas in the special cases described above. We shall also show that if denotes the number of such that has no -matches, then the sequence appears in the OIES [16] in several special cases. In Section 4, we will study as a function of , the size of the underlying cyclic group . We will show that in several cases, is a polynomial in whose coefficients have interesting combinatorial properties. Finally, in Section 5, we will discuss some related results and directions for future research.
2. Symmetric Functions
In this section, we give the necessary background on symmetric functions needed for our proofs of the generating functions.
Let denote the ring of symmetric functions over infinitely many variables with coefficients in the field of complex numbers . The th elementary symmetric function in the variables is defined by and the th homogeneous symmetric function in the variables is defined by Thus, 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 is an algebraically independent set of generators for , we can specify a ring homomorphism on by simply defining for all .
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 1.
Let denote the set of all -brick tabloids of shape and let . Through simple recursions stemming from (28), Eğecioğlu and Remmel proved in [17] that
We end this section with two lemmas that will be needed in later sections. Both of the lemmas follow from simple codings of a basic result of Carlitz [18] that where is the number of rearrangements of 1's and 0's. We start with a lemma from [19]. Fix a brick tabloid . Let denote the set of all fillings of the cells of with the numbers so that the numbers increase within each brick reading from left to right. We then think of each such filling as a permutation of by reading the numbers from left to right in each row. For example, Figure 2 pictures an element of whose corresponding permutation is .
Then the following lemma which is proved in [19] gives a combinatorial interpretation to .
Lemma 8. If is a brick tabloid in , then
Another well known combinatorial interpretation for ([13]) is that it is equal to the sum of the sizes of the partitions that are contained in an rectangle. Thus one has the following lemma.
Lemma 9. Consider the following:
3. Generating Functions
The main goal of this section is to prove (12). That is, we will prove the following theorem.
Theorem 10. Let and where . For all ,
Proof. Define a ring homomorphism by setting
Then we claim that
for all . That is,
Next we want to give a combinatorial interpretation to (36). By Lemma 8, for each brick tabloid , we can interpret as the sum of the weights of all fillings of with a permutation such that is increasing in each brick, and we weight with . If and covers cell , then we will interpret the factor as picking an element to put on top of . Next suppose that and covers cells . Let be the elements of in cells , respectively. Then we interpret as the numbers of ways of picking a maximum packing for where we weight by . We then reorder by the proper contribution after reordering which is to the coinversion count of the resulting permutation and to the inversion count of the resulting permutation. Finally, we interpret as all ways of picking a label of the cells of each brick except the final cell with either an or a . For completeness, we label the final cell of each brick with . For example, suppose that , , and . Thus is a maximum packing for only if is strictly decreasing and is a weakly increasing word over the alphabet . Then at the top of Figure 3, we have pictured the brick tabloid along with a permutation which is increasing within bricks. Below that, we have picked our choices of for the bricks for and choice of 1 for brick which is of length 1. These choices result in the filled brick tabloid pictured at the bottom of Figure 3. We have also specified a labeling of the cells with either , , or so that the last cell of each brick is labeled with 1 and the remaining cells are labeled with either of . We shall call all such objects created in this way filled labeled brick tabloids, and let denote the set of all filled labeled brick tabloids that arise in this way. Thus, a consists of a brick tabloid , an element , and a labeling of the cells of with elements from such that (1)if and covers cell , then is an arbitrary element of , (2)if and covers cells , then is an element of , (3)the final cell of each brick is labeled with 1,(4)each cell which is not a final cell of a brick is labeled with or .
We then define the weight of to be times the product of all the labels in and the sign of to be the product of all the labels in . For example, if is the filled labeled brick tabloid pictured at the bottom of Figure 3, then and . It follows that
Next we define a weight-preserving sign-reversing involution . To define , we scan the cells of from left to right looking for the leftmost cell such that either (i) is labeled with or (ii) is at the end of a brick , and the brick immediately following has the property that if together with cover cells , then is a maximum packing for . 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 . If neither case (i) or case (ii) applies, then we let . For example, if is the element of pictured in Figure 3, then is pictured in Figure 4.
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 it cannot be the case that there is -match starting at position in since otherwise we could combine and . For any such fixed point, we associate an element . For example, a fixed point of is pictured in Figure 5 where and .
It follows that if cell is at the end of a brick, then is not the start of a -match. However, if is a cell which is not at the end of a brick, then our definitions force to be the start of a -match. Since each such cell must be labeled with an , it follows that . Vice versa, if , then we can create a fixed point by having the bricks in end at cells of the form where is not the start of a -match, labeling each cell which is the start of a -match with , and labeling the remaining cells with . Thus, we have shown that
as desired.
Applying to the identity , we get
which proves (33).
To be able to use Theorem 10, we need to be able to compute . In fact, this is easy to do in most cases. Thus, we end this section by giving several such examples. We will examine the cases when (i) and , (ii) and , (iii) and , (iv) and , (v) and , (vi) and , (vii) and , (viii) and .
Example 11. Let and .
In this case for any , . It is easy to see that for , if and only if is the identity permutation and where . As we pointed out in the introduction, it follows that
Note that if and for , then and . Hence,
Here the last equality follows from Lemma 9. Thus, it follows that for all ,
Example 12. Let and .
In this case for any , . It is easy to see that for , if and only if is the identity permutation and where . As we pointed out in the introduction, it follows that
Hence, for all ,
Example 13. Let and .
In this case for any , . It is easy to see that for , if and only if is the identity permutation and where . As we pointed out in the introduction, it follows that
Hence,
Thus, for all ,
Example 14. Let and .
In this case, it is easy to see that for , if and only if and where for . In this case, one cannot find simple compact expressions for or . However, it is clear that . Thus, it follows that, for all ,
Example 15. Let and .
In this case, (,)-mch(, )-Emch where . There are two cases that we have to consider depending on whether equals or not. If , then for , is an element of if and only if and . Thus, . It follows that if , then
In particular,
If , then the only possible maximum packings for are when in which case is an element of if and only if . Thus, . It follows that if with , then
In particular,
Example 16. Let and .
In this case, it is easy to see that for all , if and only if and where . It follows that for all ,
Hence,
Setting in (55), we obtain that
Let denote the number of that have no -matches. Then setting in (56), we see that
One can easily calculate that the sequence starts out with
which is sequence A000629 in the OEIS and which also counts the number of necklaces of partitions of labeled beads. Similarly, the sequence starts out with
and the sequence starts out with
Neither of these two sequences appear in the OEIS.
Example 17. Let and .
In this case, it is easy to see that for all and , if and only if and is either of the form where and or where . Hence, for all ,
It follows that
Setting in (62), we obtain that
Let denote the number of that have no -matches. Then setting in (63), we see that
Note that it follows that for ,
It would be interesting to find a direct combinatorial proof of this recursion, which we leave as an open problem.
One can easily calculate that the sequence starts out with
which is sequence A000522 in the OEIS having multiple combinatorial interpretations including the number of permutations with no such that , that is, in the Babson-Steingrimsson notation, the number of in which avoid the pattern (the same pattern is denoted in the recently introduced notation [20]).
The sequence starts out with
which is sequence A010844 of the OEIS. This sequence does not have a combinatorial interpretation listed in the OEIS so that we have obtained a simple combinatorial interpretation of this sequence. However, if is the th term of this sequence where , then is equal to times the permanent of the matrix with on the main diagonal and 1s everywhere else.
The sequence starts out with
is sequence A010844 of the OEIS. This sequence does not have a combinatorial interpretation listed in the OEIS. Thus, we have obtained a new combinatorial interpretation of this sequence. In this case, if is the th term of the sequence where , then is equal to times the permanent of the matrix with on the main diagonal and 1's everywhere else.
Similarly, the sequences for , and appear in the OEIS as sequences as A056545, A056546, and A056547, respectively.
Example 18. Let and .
In this case, it is easy to see that for . That is, for , we must have and where and for . Now if , our choices for are either or where or so that . Now if , our choices for are where , so that .
It follows that
Setting in (69), we obtain that for
Let denote the number of that have no -matches. Then setting in (70), we see that
One can easily calculate the initial terms of the sequences . For example, the sequence starts out
The sequence starts out
The sequence starts out
However, none of these sequences appear in the OEIS.
4. Expressions for as Polynomials in and Connections to Other Objects
Recall that for any and where , is the number of such that has no -matches. We start out by considering the special case where and . In that case, we will simply write for . Thus, is the number of permutations such that has no -matches. By setting in (33), we see that
We can easily compute the initial sequences of values for these generating functions using any computer algebra system such as Mathematica or Maple. In this section, we will demonstrate these values in several cases.
4.1.
For example, if , equals the number of such that . Table 1 gives initial values of .
Several of these sequences appear in the OEIS [16]. In fact, we can easily calculate as a polynomial in . For example, we have
We point out that forms the familiar sequence of pentagonal numbers (A000326 in [16]). Other previously documented sequences appearing in Table 1 include the structured octagonal antiprism numbers (A100184 in [16]) for , as well as (A000522 in [16]), for which there are many known combinatorial interpretations, including the total number of arrangements of all subsets of .
We conjecture that for and , is always of the form where is a polynomial of degree whose leading coefficient is positive and such that signs of the remaining coefficients alternate. Now we can prove that is always of the form where is a polynomial of degree and the term of degree 1 in is . That is, for any , if we set and in the proof of Theorem 10 and use the fact that , we see that where Here we let and for . But then It is easy to see that the right-hand side of (79) is a polynomial of degree , and the lowest degree term comes from the term corresponding to which is of the form . On the other hand, the highest-degree term arises when we pick the highest power of from the products which is clearly just . It follows that the highest power of in is just
We can give a combinatorial interpretation to That is, we will show that counts the number of pairs of permutations such that if and , then there is no such that and . That is, is the number of pairs that have no common rises. To prove our claim, we can give a combinatorial interpretation to the right-hand side of (81). That is, if we start with a brick tabloid , then by Lemma 8, we can interpret as all ways to fill the cells of with pairs of permutations such that both and are increasing in each brick. Then we interpret as labeling each cell of which is not at the end of brick with and labeling each cell which is at the end of the brick with 1. We let denote the set of all filled brick tabloids constructed in this way. For example, Figure 6 pictures a typical element of . If , we define the sign of , , as the product of labels at the top of the cells of .
We now can define a sign-reversing involution on .
To define , suppose that and and are the two permutations in with on the bottom and on the top. Then we scan the cells of from left to right looking for the leftmost cell such that either (i) is labeled with or (ii) is at the end of a brick , and the brick immediately following has the property that both and are strictly increasing in all the cells corresponding to and . In case (i), 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 changing the label of cell from to . In case (ii), is the result of replacing the bricks and in by a single brick and changing the label of cell from to . If neither case (i) or case (ii) applies, then we let . For example, if is the element of pictured in Figure 6, then is 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 all bricks in must be of size 1 and . Moreover, if and are two consecutive bricks in and is the last cell of , then it cannot be the case that and where and are the two permutations in since otherwise we could combine and . Thus, is a pair of permutations in with no common rises. Vice versa, if is a pair of permutations in with no common rises, we can create a fixed point of by having be the bottom permutation of , having be the top permutation of , and having all bricks be of size 1. For example, Figure 8 pictures a fixed point of .
It follows that is the number of pairs with no common rises so that the leading coefficient of is .
4.2.
For , equals the number of such that . Table 2 gives initial values of .
In fact, we can easily calculate as a polynomial in . For example, we have We point out that forms the familiar sequence of the second pentagonal numbers (A005449 in [16]). None of the other rows or columns in Table 2 matched any previously known sequences in [16].
We conjecture that for and , is always of the form where is a polynomial of degree with positive coefficients. In fact, we see that the coefficients of and are the same up to a sign for all . we can prove this. That is, for any , if we set and in the proof of Theorem 10 and use the fact that , we see that where Here we let and for . But then
Since for any , , it is easy to see that the right-hand side of (86) is obtained from the right-hand side of (79) by replacing by and multiplying by . Thus, conjecture that has positive coefficients is equivalent to the conjecture that the signs of the coefficients of alternate.
4.3.
For , equals the number of such that . Table 3 gives initial values of .
In fact, we can easily calculate as a polynomial in . For example, we have We point out that forms the familiar sequence of hexagonal numbers (A000384 in [16]). Additionally, matches the sequence counting the number of necklaces on the set of labeled beads (A000629 in [16]). In fact, in this case, we can give a completely combinatorial interpretation of the coefficients of as a polynomial in . Let denote the set of ordered set partitions of . For any set partition , let denote the number of parts of . Then we claim that so that the coefficient of in is equal to where is the Stirling number of the second kind which is the number of set partitions of into parts. That is, for any , if we set and in Theorem 10 and use the fact that for all , then we see that where But then Since counts the number of ordered set partitions such that , it is easy to see that the right-hand side of (91) equals the right-hand side of (88).
4.4.
Recall that . Table 4 gives initial values of .
In fact, we can easily calculate as a polynomial in . For example, we have In this case, we will show that is just the Eulerian polynomial That is, for any , if we set and in the proof of Theorem 10 and the fact that , we see that where But then Next we want to give a combinatorial interpretation to (96). For any brick tabloid , we can interpret as the set of all fillings of with a permutation such that is increasing in each brick. We then interpret as all ways of picking a label of the cells of each brick except the final cell with either an or a and letting the label of the last cell of each brick be . We let denote the set of all filled labeled brick tabloids that arise in this way. Thus, a consists of a brick tabloid , a permutation , and a labeling of the cells of with elements from such that(1) is strictly increasing in each brick, (2)the final cell of each brick is labeled with ,(3)each cell which is not a final cell of a brick is labeled with or . 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 9 pictures such a composite object where and .
Thus,
Next we define a weight-preserving sign-reversing involution . To define , we scan the cells of from left to right looking for the leftmost cell such that either (i) is labeled with or (ii) is at the end of a brick , and the brick immediately following has the property that is strictly increasing in all the cells corresponding to and . 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 . If neither case (i) or case (ii) applies, then we let . For example, if is the element of pictured in Figure 9, then is pictured in Figure 10.
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 it cannot be the case that since otherwise we could combine and . For any such fixed point, we associate an element . For example, a fixed point of is pictured in Figure 11 where It follows that if cell is at the end of a brick which is not the last brick, then . However, if is a cell which is not at the end of a brick, then our definitions force . Since each such cell must be labeled with an , it follows that where the comes from the fact that the last cell of the last brick is also labeled with . Vice versa, if , then we can create a fixed point by having the bricks in end at cells of the form where and labeling each such cell with , labeling the last cell with , and labeling the remaining cells with . Thus, we have shown that as desired.
4.5.
Next we consider another case where we can show that as a polynomial in , has coefficients whose signs alternate. That is, it follows from (64) that We can easily calculate as a polynomial in . For example, we have
Thus, we see that the signs of the polynomial seem to alternate and that the leading term seems to be . We can prove both of these facts. Taking the coefficient of on both sides of (101), we see that
Now we can give a combinatorial interpretation to the right-hand side of (103) as follows. First for each , we interpret as the number of ways to pick a sequence of pairwise distinct elements from . Then we interpret as picking of the elements which we indicate by putting an above each of the elements picked. Finally, we interpret as placing above the elements in the sequence. We let denote the set of all labeled sequences constructed in this way. For example, if , , and , then the following labeled sequence would be in : Given a labeled sequence , we let the sign of , , be the product of the labels in . We let . Then it is easy to see that
Given a sequence , we say that the minimal sequence of is the longest string such that for all , and for . The minimal sequence for might be empty. For example, it could be the last label or it could be that . We let denote the length of the minimal sequence of . For example, the minimal sequence of is empty so that . The minimal sequence of is so that .
Now suppose that is such that , , and its minimal sequence is empty. Let be the elements of . Then we define the minimal sequence chain of to be where and for . Note that the signs of the elements of a minimal sequence chain of alternate. Thus, if the minimal sequence chain of is of even length, then is odd, and the sum of the signs of elements in the minimal sequence chain of is 0. If the minimal sequence chain of is of odd length, then is even, and the sum of the signs of elements in the minimal sequence chain of is . Clearly every element of is an element of a unique minimal sequence chain for some whose minimal sequence is empty so that Thus, we have shown that This shows that as a polynomial in , is sign alternating starting with the leading term . That is, by (109), But the elements of have no labels since every element of has an empty minimal sequences and, hence, .
We can also compute several other coefficients of as a polynomial in . For example, we can explicitly compute the coefficient of in . Note that Now an element has exactly one label which is . If , then the last element of is automatically its minimal sequence so that it is not counted in . Thus, counts all such that which is clearly .
We claim that where is the number of derangements of , that is, the number of permutations with no fixed points. It is well known that for . But by (105), One can also use our combinatorial interpretation of to give a combinatorial proof of this fact. That is, it is well known that , for . In our case, we know that Thus, we must show that . Now if , then all the labels in are so that we can identify each element with just a permutation . Now consider a whose minimal sequence has even length. If , then clearly the minimal sequence of has the same length as the minimal sequence of . Vice versa, given , let be the permutation of such that . Clearly for , is a permutation in whose minimal sequences have the same length as the minimal sequence of . It follows that
Next suppose that is such that , and its minimal sequence has even length. Then we claim that has a minimal sequence whose length has the same parity as the length of the minimal sequence of . That is, if is an element of the minimal sequence of , then it must be the case that is the identity so that has minimal sequence of length . If is not in the minimal sequence of , then it cannot be the case that is in the minimal sequence of so that the lengths of the minimal sequences of and are the same. Vice versa, given , let be the permutation of such that . Clearly for , is a permutation in whose lengths of the minimal sequences have the same parity as the length of the minimal sequence of . It follows that Hence, satisfies the same recursion as the recursion for .
Finally, it is not difficult to see that To prove this, we will construct a bijection which has the property that for all , and have the same parity. Now suppose we are given an where and . Then is must be the case that there is a unique such that . Let . That is, arises from by inserting just before the element of that is marked with and adding 1 to all the elements of and removing the and replacing it with and labeling 1 with . It is easy to see that is a bijection. Thus, all we need to do is to show that for all , and have the same parity. The minimal sequence of does not contain . In , the minimal sequence does not contain since . Now if the minimal sequence of does not contain , then it is easy to see that the minimal sequence of does not contain which means that the lengths of the minimal sequences of and are the same. So suppose that the minimal sequence of is . Then there are two cases. Namely, either (i) in which case the minimal sequence of is or (ii) in which case the minimal sequence of is . Hence in all cases, the lengths of the minimal sequences of and have the same parity. Thus, shows that which by (109) establishes (117).
5. Further Research
An obvious question is to ask whether the results of this paper can be extended to patterns of length . There are some partial results in this direction due to Duane and Remmel [21]. 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 in 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 equal the set of -colored permutations, , such that -Emch, and refer to the -colored permutations in as exact match maximum packings for . We let Then Duane and Remmel [21] proved that (I) if and has the -minimal overlapping property, then
(II) if has the -exact match minimal overlapping property, then Duane and Remmel’s proof of (I) and (II) works equally well for -matches.
There are, however, several examples in the literature of generating functions for the number of -matches in permutations where does not have the minimal overlapping property, and we have found some analogues of such results for . In addition, there are many formulas in the literature for the number of permutations that avoid certain patterns in permutations. We have found several examples of such formulas for the matching conditions for discussed in this paper. Such results will appear in subsequent papers.
Acknowledgments
J. Remmel was partially supported by NSF Grant DMS 0654060. M. Riehl was partially supported by a grant from the Office of Research and Sponsored Programs, UWEC.