Generalized Pattern Avoidance Condition for theWreath Product of Cyclic Groups with Symmetric Groups

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Introduction
e goal of this paper is to continue the study of patternmatching 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, 1 =  0 , ,  2 ,…,  1 , where  is a primitive th root of unity.We can think of the elements   ≀   as pairs  =   where  =  1 ⋯   ∈   and  =  1 ⋯   ∈ {1  …   1 }  .For ease of notation, if  =   1    2  …      where   ∈ {0 …    1} for  = 1 …  , then we simply write  =   where  =  1  2 ⋯   .Moreover, we think of the elements of  =  1  2 ⋯   as the colors of the corresponding elements of the underlying permutation .
We de�ne a concept for matchings in words over a �nite alphabet [ = {0 1 …    1}.Given a word  =  1 ⋯   ∈ [  , let red be the word found by replacing the th largest integer that appears in  by   1.For example, if  = 2 7 2 4 7, then red = 0 2 0 1 2. Given a word  ∈ [  such that red = , we say that a word  ∈ [  has a -match starting at position  provided red  ⋯  1  = .Let -mch be the number of -matches in the word .
Similarly, we say that  occurs in a word  if there exist 1 ≤  1 < ⋯ <   ≤  such that red  1 ⋯     = .We say that  avoids  if there are no occurrences of  in .
ere are a number of papers on pattern matching and pattern avoidance in   ≀   [2][3][4][5].We now present a selection of the previously studied de�nitions 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][4][5].
at 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 (  ∈   ≀  2 , the number of elements in   ≀   which avoid exact occurrences of (  is ∑  =  (        2 .is generalized a result of Simion [6] who proved the same result for the hyperoctahedral group  2 ≀   .Similarly, Mansour and West [5] determined the number of permutations in  2 ≀   that avoid all possible exact occurrences of 2 or 3 element sets of patterns from  2 ≀  2 .For example, let    be the number of (  ∈  2 ≀   that avoid all exact occurrences of the patterns in the set {( 2   ( 2   (2   },  2   the number of (  ∈  2 ≀   that avoid all exact occurrences of the patterns in the set {( 2   ( 2   (2   }, and  3   the number of (  ∈  2 ≀   that avoid all exact occurrences of the patterns in the set {( 2   ( 2   (2   }.en Mansour and West [5] 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.at is, we shall consider the following matching conditions.�e�nition �� Let (  ∈   ≀   where red( = , Υ a subset of   ≀   where for all (  ∈ Υ, red( = , and (  ∈   ≀   . (1) One says that (  has an occurrence of (  (resp., Υ) if there are  ≤   <  2 < ⋯ <   ≤  such that (red(   ⋯     red(   ⋯     = (  (resp., (red(   ⋯     red(   ⋯     ∈ Υ).
(3) One says that there is a For example, suppose that (  = ( 2   and (  = ( 3 2 4  2 2 2.en 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. us there are two occurrences of (  in (  and there is a ( -match in (  starting at position 3.
e main result of [1] was to derive a generating function for the distribution of (   -matches where (  ∈   is any element of   ≀  2 .To state the main result of [1], we need to de�ne some notation.First we de�ne the  -analogues of We will also be interested in the specializations where for any word  =  . ( 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   {0, 1, … ,  − 1  such that red) =  and     , Duane and Remmel say that , ) has the   ≀   -minimal overlapping property if the smallest  such that there exists a , )    ≀   with (, )-mch, )) =  is  − 1. is 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 �rst , )-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  − 1.Now if , ) has the   ≀   -minimal overlapping property, then the shortest -colored permutations , ), such that (, )-mch, )) =  have length  − 1) + 1.We let ℳ  ,),−1)+1 equal the set of -colored permutations, , )    ≀  −1)+1 , such that (, )-mch, ) = .We refer to the -colored permutations in ℳ  ,),−1)+1 as maximum packings for , ).We let Similarly, we let ℰℳ  ,),−1)+1 denote the set of colored permutations, , )    ≀  −1)+1 , such that Duane and Remmel's proof of (I) and (II) works equally well for (Υ  -matches.e main goal of this paper is to prove a number of results on the number of (   -avoiding elements of   ≀   and to �nd a generating function for the number of (   matches for certain patterns which do not have the minimal overlapping property.at is, in Section 2, we shall prove a number of results about the number of (   -avoiding elements of   ≀   where (  ∈   ≀  2 .In Section 3, we prove a number of results on the number of (Υ  -avoiding elements of   ≀   and for certain subsets Υ of   ≀  2 .
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.

(𝜏𝜏𝜏 𝜏𝜏𝜏 󵱂󵱂 𝐴𝐴𝜎-Avoiding Patterns
Given Proof.�e �rst observe that elements of different colors are independent in permutations avoiding (1 2, 0 0), meaning that no two elements with different colors can form a prohibited con�guration.us assuming we have  1 elements of color ,   0,  ,   1, we can choose how to place these colors to form a word  in    1 ,,   ways.en we can choose the sets of elements  0 ,  ,  1 from {1,  ,  which will correspond to the colors 0,  , 1 in  in    1 ,,   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.e proof of Lemma 4 suggests an obvious generalization for patterns of the form (, 0  ) where     .eorem 5. Let     .en the number of elements in   ≀   avoiding (, 0  ) is given by where   is the number of permutations in   avoiding .
Proof.e proof here is essentially the same as the proof of eorem 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    (1/(1))  2   .As a corollary to eorem 5, we have that, for any pattern    3 , the number of permutations in   ≀   avoiding (, 0 0 0) is given by One can generalize eorems 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 (, 0  ) as For example, the distribution of the occurrences of the pattern 1 2 is the same as the distribution of coinversions in permutations, so one can extract the numbers  , as the coefficients to   in ∏  1 (1    ⋯   1 ) and substitute them in the last formula to get the distribution of the number of occurrences of (1 2, 0 0) in   ≀   .
Next we consider the pattern (1 2, 0 1) in the case where   2.
e number of permutations in  2 ≀   avoiding (1 2, 0 1) is shown in [6,  , it must be the case that every element to the right of 1 must be colored with a color from {0 …     and these elements can be arranged in any (  ) ways.Moreover, it immediately follows that no instance of a occurrence of (  0     ) exists where the �rst element is to the le of 1 and the second is to the right of 1. en it follows that ( ) avoids (  0     ) if and only if there is no occurrence of (  0     ) in (  ⋯      ⋯   ).us in the case where 1 is colored by a color from {0 …     and is in position   , we have   (  ) (   ) (   ) =    (   ) possibilities where the binomial coefficient is responsible for choosing the elements to the le of 1 and placing the remaining elements to the right of 1 . (39)

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 We start with a few simple results on sets of patterns Υ where the avoidance of Υ forces certain natural conditions on the possible sets of signs for elements of   ≀   .Proof.For (1) which implies (44).
We immediately have the following corollary.
For (2), observe that if in addition such (, ) also avoids (2 1, 0 0), then we must place the elements of   in increasing order.Hence  Υ 2 , is the number of solutions of  1 + ⋯ +   =  with    0 which is well known to be  +1 1 .Proof.We shall classify the elements (, )    ≀   which avoid Υ by the number of elements  which follow  in .Now if  = 0 so that  ends in , the fact that (, ) avoids both (1 2, 0 1) and (1 2, 1 0) means that all the signs must be the same.at is,  must be of the form   for some   .But since (, ) must also avoid (2 1, 0 0),  = 1 2   must be the identity.us there are  choices for such (, ).Now suppose there are  elements following  in .en again the fact that (, ) avoids both (1 2, 0 1) and (1 2, 1 0) means that all the signs in  corresponding to  1    (where   = ) must be the same, say that sign is .e fact that (, ) avoids (2 1, 0 0) means that (i)  1    must be in increasing order and (ii) all the signs corresponding to  1    must be different from .But then it follows from the fact that (, ) avoids both (1 2, 0 1) and (1 2, 1 0) that all the elements in  1    must be greater than all the elements in  1    .Hence there are  Υ ,1 such elements if   2 and there are no such elements if  = 1.us from (ii) it follows that  Υ ,1 = 1 since (1 2  , 0  ) is the only element of   ≀   which avoids Υ.For   2, we have In the case  = 2, (52) becomes and in the case  = , (52) becomes In general, assuming that We next consider simultaneous avoidance of the patterns (1 2, 1 0) and (1 2, 0 1).
Proof.Fix  and suppose that (, ) is an element of   ≀   which avoids both (1 2, 1 0) and (1 2, 0 1).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   2 where    1 for  = 1,  ,  and   ≠  1 for  = 1,  ,   1.We claim that  1    1 must be the  1 largest elements of {1,  , }.If not, then let    be the largest element which is not in  1    1 .us    1  1 and there must be at least one   with    such that      .Now it cannot be   1 1 =    since otherwise (    1 1 ,  1  2 ) would be an occurrence of either (1 2, 1 0) or (1 2, 0 1).Hence it must be the case that   1 1     and   =    for some    1  1.But then no matter what color we choose for   , either (    ,  1   ) or (  1 1   ,  2   ) would be an occurrence of either (1 2, 1 0) or (1 2, 0 1).We can continue this reasoning to show that for any 1      , the elements of  corresponding to the block     in  must be strictly larger than the elements of  corresponding to the block     in .is 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 (1 2, 1 0) and (1 2, 0 1).us for such a , we have (  1) 1 ways to choose the colors  1 , … ,   and  1 ! 2 !⋯   !ways to choose the permutation .It follows that which is equivalent to (57).
Subdivide  into the so-called reverse irreducible components.A reverse irreducible component is a factor  of  of minimal length such that everything to the le (resp., right) of  is greater (resp., smaller) than any element of .For example, the subdivision of    is   . e 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 le-to-right minima and the block's sequence of right-to-le maxima which do not overlap.us for any le-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 shi.
To build , set    and    + , and do the following steps, letting  begin at  .Clearly, each reverse irreducible block of  de�nes a square on the grid which is a matrix representation of the block.We call the reverse irreducible block of  with the le-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  8.We travel down until we reach  , when we are  less than the ycoordinate of our �rst 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 .en we continue to the �nal reverse irreducible block and �nish our path, given in Figure 3.

Pattern Matching Results for Pairs of
Length ≥ Goulden and Jackson [8] proved the following theorem.eorem 19 (see [8]).e main goal of this section is to generalize eorem 20.at is, let  )    )   where   .en we shall consider three sets of patterns for   ≀   : (a)  )  0  ), (b)  )         0), We can also �nd a generating function for   by considering the following argument.Take a rearrangement that is counted by   .Now remove the �nal number of the rearrangement.Either (a) we have removed a 0, (b) we have removed a 1.
In case (a), we can count the number of such rearrangements by  11 , except that we have overcounted by those rearrangements counted by  11 which end in   1 zeroes.We can correct this by considering removing the last 1 zeroes of the over-counted rearrangements, leaving those of length    with    zeros ending in a 1, of which there are  1 .In case (b), we can count the number of such sequences by  1 .us we obtain the following relationship: =  11   1 +  1 . (105) Next, we multiply by     and sum over all , from which we obtain a generating function with a �nite sum of small  values in the numerator and a quotient of 1     +    +1 .

F 4 :
e Dyck path corresponding to           0  0  0).Note that the path only touches the line        aer a singleton colored .
(  ∈   ≀   and    (   …     where   ⊆ [ for    …  , let  (   denote the number of (  ∈   ≀   which avoid (   .In the special case where    [ for    …  , we shall denote  (   as simply  ( .us  ( is the number (  ∈   ≀  which avoid ( .In this case, we shall �nd formulas for  (   or  ( where (  ∈   ≀  2 .ereare a number of natural maps for which the distribution of (   -occurrences and (   -matches remain invariant.atis, for any     ⋯   ∈   , we de�ne the reverse of ,   and the complement of ,   , respectively, by        ⋯              ⋯        .…          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 red(   and (  ∈   ≀  2 , then we need only to compute  ( for two patterns; namely, (   ( 2 0 0 and (   ( 2 0 .We start by considering the pattern ( 2 0 0.
(10)If     ⋯   ∈ [  , then we de�ne the reverse of ,   and the complement of  relative to [, ( , by        ⋯     (         ⋯        .(11) If      …    } ⊆ [, then we de�ne  (         …        }. en if    (   …     where   ⊆ [, for    …  , then we de�ne (12) en consider the maps   ∶   ≀   →   ≀   where   ((   (      for   ∈   } and  ∈   ( } where  is the identity map; that is, Lemma 4. e number of elements in   ≀   avoiding ( 2 0 0 is given by page 19] to be equal to ∑  0   Let  ≤  ≤  ≤    and    = ({0 …     { …    ), where   .Let (   ) =  ( 0   ) denote the number of ( )    ≀   which avoid (  0     ), and possibilities.On the other hand, if (, ) contains the element 1 colored by the color 0 in position   1 and (, ) is to avoid (1 2, 0 1), 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 (1 2, 0 1) exists where the �rst element is to the le of 1 and the second is to the right of 1. us it follows that (, ) avoids (1 2, 0 1) if and only if there is no occurrence of (1 2, 0 1) in ( 1 ⋯   ,  1 ⋯   ).en in the case where 1 is colored by 0 and is in position   1, we have (  ) (   )       / possibilities where the Integrating and using the fact that (0)  1, we see that ln ( ())   ln (1  ) 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   {0 …    , then if ( ) is to avoid (  0   ) . Hence (     ) = (  ) (  )  (   )Multiplying both parts of the equation above by    and summing over all   0, we have Now suppose that     .en integrating (31) and using the fact that (0   ) =  and the fact that ()(ln(  )  ln(  )) = (  )(  )(  ), It follows from the form of these generating functions that (   2) =   (   2).Moreover, in the case where  = 2, it is easy to see that our de�nitions imply that the num� ber of ( )   2 ≀   which avoid (( 2 0 ) ({0 {)) is the same as the number of ( )   2 ≀   which avoid ( 2 0 ).en (   2) =     where   is the number of ( )   2 ≀   avoiding ( 2 0 ).is is also easy to see combinatorially.Namely, if ( )   2 ≀   which avoids (( 2 0 ) ({0 …  { …  2)), then if  ′ is the result of replacing each occurrence of 0 …   in  by 0 and each occurrence of  …  2 in  by 1, then (  ′ ) will be an element of  2 ≀   which avoids ( 2 0 ).Vice versa, if (  ′ )   2 ≀   avoids ( 2 0 ) and  arises from  In fact, if we are in the case where  =  so that    = ({0 …     {   …  ), then by the same argument, we can get an arbitrary ( )    ≀   which avoids (( 2 0 )    ) by starting out with an (  ′ )   2 ≀   which avoids ( 2 0 ) and replacing each 0 by some element from {0 …     and each 1 by some element from {   …    .Considering the form of the generating function for   () when   , ′ by replacing each 0 by some element from {0 …   and each 1 by some element from { …  2  , then ( ) will avoid (( 2 0 ) ({0 …     {   …  2  )).eorem 8. Given ( )   2 ≀   , let pos( ) equal the number of 0's in  and neg( ) equal the number of 1's in .
Υ ⊆   is a set of permutations of   , we let , it is easy to see that ( )    ≀   avoids Υ  if and only if all the signs are pairwise distinct.erefore,thereareways 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 Υ 2 if and only if 0 ≤   ≤ ⋯ ≤   ≤   .erefore,thereare     ways to pick the  signs in this case and there are  ways to pick .For (3), it is easy to see that ( )    ≀   avoids Υ 2 if and only if 0 ≤   < ⋯ <   ≤   .erefore,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 Υ 2 if and only if 0 ≤   = ⋯ =   ≤   .erefore, there are  ways to pick the  signs in this case and there are  ways to pick .the binomial coefficient allows us to choose the elements   of {1, … , } that correspond to the constant segment   +1 in .en we only have to arrange the elements of   so that it avoids Υ which can be done in  Υ for some   {0, … ,   1}.en (,   )    ≀   avoids Υ 0 if and only if  avoids Υ. us 1, 0 0)}, we must have  =  1 ⋯   where  1 < ⋯ <   .en for any of the     strictly increasing words   {0, … ,   1}, (, )    ≀   avoids Υ  if and only if  avoids Υ. us  , note that, for (, )    ≀   to avoid {(1 2, 0 0), (2 1, 0 0)}, we must have  =  1 ⋯   where the letters of  are pairwise distinct.en for any of the      words   {0, … ,   1} which have pairwise distinct letters, (, )    ≀   avoids Υ  if and only if  avoids Υ. us 1, ,  =    (  1)  , , 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   .eorem 18. e number of permutations in   ≀   simultaneously avoiding     and     is given by the  + -th Catalan number  +   +   + + .
, where   , then where ℛ  is the number of rearrangements of  zeroes and    ones such that  zeroes never appear consecutively.ereason that eorem 20 is a re�nement of eorem 19 is that Mendes and Remmel[9]proved that