Research Article Generalized Pattern Avoidance Condition for the Wreath Product of Cyclic Groups with Symmetric Groups

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1 Kitaev, Sergey and Remmel, Jeffrey and Riehl, Manda (203) Generalized pattern avoidance condition for the wreath product of cyclic groups with symmetric groups. ISRN Combinatorics, 203., This version is available at Strathprints is designed to allow users to access the research output of the University of Strathclyde. Unless otherwise explicitly stated on the manuscript, Copyright and Moral Rights for the papers on this site are retained by the individual authors and/or other copyright owners. Please check the manuscript for details of any other licences that may have been applied. You may not engage in further distribution of the material for any profitmaking activities or any commercial gain. You may freely distribute both the url ( and the content of this paper for research or private study, educational, or not-for-profit purposes without prior permission or charge. Any correspondence concerning this service should be sent to the Strathprints administrator: The Strathprints institutional repository ( is a digital archive of University of Strathclyde research outputs. It has been developed to disseminate open access research outputs, expose data about those outputs, and enable the management and persistent access to Strathclyde's intellectual output.

2 ISRN Combinatorics Volume 203, Article ID , 7 pages Research Article Generalized Pattern Avoidance Condition for the Wreath Product of Cyclic Groups with Symmetric Groups Sergey Kitaev, Jerey Remmel, 2 and Manda Riehl 3 Department of Computer and Information Sceinces, University of Strathclyde Glasgow, Livingstone Tower, 26 Richmond Street, Glasgow G XH, UK 2 Department of Mathematics, University of California, San Diego, La Jolla, CA 92093, USA 3 Department of Mathematics, University of Wisconsin Eau Claire, Eau Claire, WI , USA Correspondence should be addressed to Sergey Kitaev; sergey.kitaev@gmail.com Received 30 September 202; Accepted 7 October 202 Academic Editors: Y. Simsek and Y. Zhang Copyright 203 Sergey Kitaev et al. is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. 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.. 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 []. is the group of signed permutations where there are signs, = 0,, 2,,, where is a primitive th root of unity. We can think of the elements as pairs = where = and = { }. For ease of notation, if = 2 where {0 } for =, then we simply write = where = 2. Moreover, we think of the elements of = 2 as the colors of the corresponding elements of the underlying permutation. We dene a concept for matchings in words over a nite alphabet [ = {0 }. Given a word = [, let red be the word found by replacing the th largest integer that appears in by. For example, if = ,thenred = Given a word [ such that red =, wesaythataword [ has a -match starting at position provided red =. Let -mch be the number of -matches in the word. Similarly, we say that occurs inaword if there exist < < such that red =. We say that avoids if there are no occurrences of in. ereareanumberofpapersonpatternmatchingand pattern avoidance in [2 5]. We now present a selection of the previously studied denitions for pattern matching and avoidance, so that the reader may see how ours diers from and/or extends those in the literature. For example, the following pattern matching condition was studied in [3 5]. Denition Let, be a subset of, and. () One says that has an exact occurrence of (resp., ) if there are < 2 < < such that red = (resp., red ). (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

3 2 ISRN Combinatorics -match in ( starting at position ) if (red( = ( (resp., (red(. 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 }. enmansourandwest[5]provedthat = 2 2 = = 3 = = 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. enition Let ( where red( =, a subset of where for all (, red( =, and (. () () One says that ( has an occurrence of ( (resp., ) if there are < 2 < < such that (red( red( = ( (resp., (red( red( ). (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 (red( red( = ( (resp., (red( red(. For example, suppose that ( = ( 2 and ( = ( en there are no exact occurrences or exact matches of ( in (. However,therearetwo occurrences of (, one in positions 2 and 4 and one in positions 3 and 4. us there are two occurrences of ( in ( andthereisa( -match in ( starting at position 3. Finally, a more general-matching condition which generalizes both occurrences and matches and exact occurrences and exact matches in was introduced in []. enition Let (, be a subset of, = ( be a sequence of subsets of [, and (. () One says that ( has an occurrence of ( (resp., ( ) ifthereare < 2 < < such that (red( red( = ( red( (resp., (red( red( is equal to ( red( for some ( in ) and for =. (2) One says that ( avoids ( (resp., ( ) if there are no occurrences of ( (resp., ) in (. (3) Onesaysthatthereisa( -match in ( starting at position (resp., ( -match in ( starting at position ) if (red(, red( = ( red( (resp., (red( red( is equal to ( red( for some ( in ) and for =. us a ( -occurrence (or a ( -match) where = and = ({ } { } is just an exact occurrence (resp., an exact match) of (. Similarly, a ( occurrence (or a ( -match) where =, red( =, 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 (. emainresultof[] was to derive a generating function for the distribution of ( -matches where ( is any element of 2.Tostatethemainresultof[], we need to dene some notation. First we dene the -analogues of

4 ISRN Combinatorics 3,,, and,, by en Kitaev et al. [] proved that, for any and =, ) where, [], [], = = , [], = [], [], [], [],, =, =,,, [], [], [],, [],,,, (2) coinv) inv), )-mch,)) ) 0[],,) = [], ), ),, 0,,. (5) respectively. We dene the -analogs of,,, and,, by [],, [],,,, and,,,, respectively. Next suppose that and =, ) where, []. enwewillsaythat, ) 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 inv) (resp., coinv)) equal the number of pairs, ) such that and > (resp., ). We then dene, ),, 0,, = coinv) inv) ).,), ), We will also be interested in the specializations (3) 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,,, 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. 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. Now if, ) has the -minimal overlapping property, then the shortest -colored permutations, ), such that (, )-mch, )) = have length ) +. We let,),)+ equalthesetof-colored permutations,, ) )+,suchthat(, )-mch, ) =. Werefertothe-colored permutations in,),)+ as maximum packings for, ). We let, ),,,, = coinv) inv),,), ), ),,,,, (4),),+ =,),+,,),+,, =,),),+ coinv) inv), (6) = coinv) inv),,), ),,),+, 0,, = coinv) inv) ).,),),+ where for any word = [], = + +. In the special case where = {, ) where, ), we shall denote, ),, 0,, ) as,, ),,, 0,, ). Similarly, we let,),)+ denote the set of colored permutations,, ) )+,suchthat

5 4 ISRN Combinatorics ( )-Emch(, andrefertothe-colored permutations in (( as exact match maximum packings for (. We let ( ( ( ( ( ( 0 ( ( coinv( inv( coinv( inv( (. Duane and Remmel [7]provedthat(I)ifred( and ( has the -minimal overlapping property, then (-mch( coinv( inv( ( 0 ( / 0 ( ( ( 0 and (II) if ( has the -exact match minimal overlapping property, then (-Emch( coinv( inv( ( 0 ( / 0 ( ( (7) (8) ( 0. (9) Duane and Remmel s proof of (I) and (II) works equally well for ( -matches. emaingoalofthispaperistoproveanumberofresults 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 ( -avoiding elements of and for certain subsets of 2. Finally, in Section 4, wewillproveananalogofatheorem 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 (.Inthespecialcasewhere [ for, we shall denote ( as simply (. us ( is the number ( which avoid (. Inthiscase,weshallndformulasfor ( or ( where ( 2. ere are a number of natural maps for which the distribution of ( -occurrences and ( -matches remain invariant. at is, for any, we dene the reverse of, and the complement of,, respectively, by. (0) If [, then we dene the reverse of, and the complement of relative to [, (, by (. () If } [, then we dene ( }. en if ( where [, for, then we dene (. (2) en consider the maps where (( ( for } and ( } where is the identity map; that is, for all, for all [, and for all ( with [ for.itiseasytoseethata( 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, ( ( and ( ( 2 0. We start by considering the pattern ( Lemma 4. e number of elements in avoiding ( is given by (3)

6 ISRN Combinatorics 5 Proof. e rst observe that elements of dierent colors are independent in permutations avoiding ( 2, 0 0), meaning that no two elements with dierent colors can form a prohibited conguration. us assuming we have elements of color, 0,,,wecanchoosehowtoplace these colors to form a word in,, ways. en we can choose the sets of elements 0,, from {,,,, which will correspond to the colors 0,,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. e proof of Lemma 4 suggests an obvious generalization forpatternsoftheform(, 0 ) where. eorem 5. Let. en the number of elements in avoiding (, 0 ) is given by 0,, 0 2,,, where isthenumberofpermutationsin avoiding. 2 (4) Proof. eproofhereisessentiallythesameastheproof of eorem 4, except that we can place elements of a permutation in ofthesamecolorinanyof ways. Itiswellknownthatthenumberof-permutations avoiding any pattern of length 3 is given by the th Catalan number (/()) 2.Asacorollarytoeorem 5, we have that, for any pattern 3, the number of permutations in avoiding (, 0 0 0) isgivenby 0,, ,, (5) 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,wecanwritedownthe number of permutations in with occurrences of (, 0 ) as 0,, 0 0,, 0, 2, 2,.,, 2 2 (6) For example, the distribution of the occurrences of the pattern 2 isthesameasthedistributionofcoinversions in permutations, so one can extract the numbers, as the coecients to in ( ) and substitute them in the last formula to get the distribution of the number of occurrences of ( 2, 0 0) in. Next we consider the pattern ( 2, 0 ) in the case where 2. e number of permutations in 2 avoiding ( 2, 0 ) is shown in [6,page9]tobeequalto 0 2. However, in eorem 6 below we provide an independent derivation of the exponential generating function in this case. eorem 6. e exponential generating function for the number of elements in 2 avoiding the pattern ( 2, 0 ) is given by Proof. Let ( 2,0 ) and () /(). (7) () 0. (8) If (, ) 2 contains the element colored by the color, then there are no restrictions for placing this element, thereby giving ( ) possibilities. On the other hand, if (, ) contains the element colored by the color 0 in position and (, ) is to avoid ( 2, 0 ), it must be the casethateveryelementtotherightofiscoloredwiththe color 0 and these elements can be arranged in any of ( ) ways. Moreover, it immediately follows that no instance of occurrence of ( 2, 0 ) exists where the rst element is totheleofandthesecondistotherightof.usit follows that (, ) avoids ( 2, 0 ) if and only if there is no occurrence of ( 2, 0 ) in (, ). en in the case where is colored by 0 and is in position, we have ( ) ( ) / possibilities where the binomial coecient is responsible for choosing the elements to the le of and placing the remaining elements to the right of. To summarize, we obtain (). (9) Multiplying both parts of the equation above by / and summing over all 0, we have so that It follows that () () () 0 (20) (). (2) ln ( ()) () () () 2. (22) Integrating and using the fact that (0),weseethat ln ( ()) ln () ln () (23)

7 6 ISRN Combinatorics which implies that Multiplying both parts of the equation above by and summing over all 0, we have () = (). (24) is type of argument can be generalized. at is, we have the following theorem. eorem 7. Let and = ({0 { ), where. Let ( ) = ( 0 ) denotethenumberof( ) which avoid ( 0 ), and en if, () = ( ) 0. (25) () () = () (). (26) If =, then () = (()) (). (27) Proof. Note that if ( ) contains the element 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 colored by the color in position where {0, then if ( ) is to avoid ( 0 ), itmustbethecasethateveryelementtothe right of 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 elementistotheleofandthesecondistotheright of. en it follows that ( ) avoids ( 0 ) if and only if there is no occurrence of ( 0 ) in ( ).usinthecasewhereiscoloredby a color from {0 and is in position,wehave ( ) ( ) ( ) = ( ) possibilities where the binomial coecient is responsible for choosing the elements to the le of and placing the remaining elements to the right of. Hence ( ) 0 = () ( ) 0 so that for all, It follows that () ( ) 0 () = () () =0 ( ) (29) () () (). (30) ln () = () () = () (() )(). (3) Now suppose that. en integrating (3) and using the fact that (0 ) = and the fact that ()(ln( ) ln( )) = ( )( )( ), we see that ln () = ln (() ) which implies that () (ln ( ) ln (() )) () () = () () which proves (26). Next suppose that =. en (3) becomes ln () = () () = () (() ). (32) (33) (34) Integrating (34) and using the fact that (0 ) =, we see that ln () = ln (() ) ( ) = ()() ( ) ( ). =0 (28) () = ln (() ) (). (35)

8 ISRN Combinatorics 7 us which proves (27). () = (()) () (36) Next consider the special case where 2=and =so 2 = ({0 { 2 2). en clearly, 2 () = () 2 () = (). (37) 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 denitions 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 0in by 0 and each occurrence of 2 in by, then ( ) will be an element of 2 which avoids ( 2 0 ). Vice versa, if ( ) 2 avoids ( 2 0 ) and arises from by replacing each 0 by some element from {0 and each by some element from { 2, then ( ) will avoid (( 2 0 ) ({0 { 2 )). In fact, if we are in the case where =sothat = ({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 eachbysomeelementfrom{. Considering the form of the generating function for () when, (2) () = () () (38) this suggests the following theorem which can easily be provedbythesamemethodasweusedtoproveeorem 7. eorem 8. Given ( ) 2, let pos( ) equal the number of 0 s in and neg( ) equalthenumberof sin. ( 20 ) Let denote the set of ( ) 2 which avoids ( 2 0 ). en = = 0 ( 20 ) () (). 3. Avoidance for Sets of Patterns pos() neg() (39) 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 () = 0 (40) where is the number of permutations such that avoids. If and = ( 2 ) where [ for =, then we let ( ) denote the number of ( ) which avoid ( ). We then let ( ) () = 0 ( ). (4) In the special case where for all ( ), red() = and = [ for =,weshallwrite instead of ( ). at is, is the number of ( ) which avoid. en we let () = 0. (42) 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. () If = {( 2 0 0) (2 0 0), then = 2 for all and. (2) If 2 = {( 2 0) (2 0), then 2 = for all and. (3) If 3 = {( 2 0 0) ( 2 0), (2 0 0), (2 0), then 3 = for all and. (4) If 4 = {( 2 0 ) ( 2 0), (2 0 ), (2 0), then 3 = for all and. Proof. For(),itiseasytoseethat( ) avoids if and only if all the signs are pairwise distinct. erefore, there are ways to pick the signs, and then you have ways to arrange those signs and ways to pick. For(2),itiseasytoseethat( ) avoids 2 if and only if 0. erefore, there are ways to pick the signs in this case and there are ways to pick. For(3),itiseasytoseethat( ) avoids 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),itiseasytoseethat( ) avoids 2 if andonlyif0 ==. erefore, there are ways to pick the signs in this case and there are ways to pick. For, we let enwehavethefollowing. 0 = 0. (43)

9 8 ISRN Combinatorics eorem 0. Let be any set of permutations of. en if = 0 {( 2, 0), (2, 0)}, for all. () = (), (44) Proof. Itiseasytoseethatif(, ) and (, ) avoids {( 2, 0), (2, 0)}, thenitmustbethecasethat 0 2. So suppose that =0 2 ( ) where ++ =. en clearly the number of such that (, ) avoids 0 is,,. (45) at is, the binomial coecient allows us to choose the elements of {,, } thatcorrespondtotheconstant segment + in. en we only have to arrange the elements of sothatitavoids which can be done in + ways. usitfollowsthat, = ++ = 0 or, equivalently, that which implies (44).,,,, = ++ = 0 = We immediately have the following corollary. (46), (47) Corollary. For any, the number of elements of which avoid = {( 2, 0 0), ( 2, 0), (2, 0)} or 2 = {(2, 0 0), ( 2, 0), (2, 0)} is. Proof. Let = { 2} and 2 = {2 }. en clearly, = for =, 2 so that () = for =, 2. But then by eorem 0,, () = ( ) = so that, = for =, 2. We can derive theorems analogous to eorem 0 for the other sign conditions in Lemma 9. at is, suppose that is anysetofpatternsfrom. en we let =, 0, = (, ),, (48) where is the set of all permutations of {0,,, }. en we have the following. eorem 2. For any, let = 0 {( 2, 0 ), ( 2, 0), (2,0 ), (2, 0)}, 2 = {( 2, 0), ( 2, 0 0), (2, 0), (2,0 0)}, 3 = {( 2, 0 0), (2, 0 0)}. en, for all, (49) (), = for all, (2) 2, = for all, (3) 3, = for all. Proof. For (), note that for (, ) to avoid {( 2, 0 ), ( 2, 0), (2, 0 ), (2, 0)}, we must have = for some {0,, }. en (, ) avoids 0 ifandonlyifavoids. us, =. For (2), note that for (, ) to avoid {( 2, 0), ( 2, 0 0), (2, 0), (2, 0 0)}, we must have = where < <. en for any of the strictly increasing words {0,, }, (, ) avoids if and only if avoids. us 2, =. For (3), note that, for (, ) to avoid {( 2, 0 0), (2, 0 0)}, we must have = where the letters of are pairwise distinct. en for any of the words {0,, } which have pairwise distinct letters, (, ) avoids ifandonlyif avoids. us 3, =. Next we will prove two more results about, for other sets of patterns that contain {( 2, 0), (2, 0)}. eorem 3. Let en = {( 2,0 ), ( 2, 0), (2, 0)}, 2 = {( 2,0 ), ( 2, 0), (2, 0), (2, 0 0)}. (50) (), = ++ = for all and, 0 (2), = + for all and. Proof. For (), note that as in the proof of eorem 0, if (, ) and (, ) avoids {( 2, 0), (2, 0)}, thenitmustbethecasethat0 2. Now suppose that = 0 2 ( ) where + + =. Assume that (, ) also avoids ( 2, 0 ) and is the set of elements of that correspond to the signs ( ) in. en it follows that all the elements of must be larger than all the elements of 2, all the elements of 2 must be larger than all the elements of 3,andsoforth.us consists of the largest elements of {,, }, 2 consists of next 2 largest elements of {,, }, and so forth. en 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. erefore () immediately follows. For (2), observe that if in addition such (, ) also avoids (2, 0 0), then we must place the elements of in increasing order. Hence 2 is the number of solutions of, ++ = with 0 whichiswellknowntobe. +

10 ISRN Combinatorics 9 eorem 4. Let = {( 2, 0 ), ( 2, 0), (2, 0 0)}. en for,, =, =,,2 = 2, =2 () for, (5) where () 0 =and () = ( ) ( ) for. Proof. We shall classify the elements (, ) which avoid by the number of elements which follow in. Now if =0so that ends in, thefactthat(, ) avoids both ( 2, 0 ) and ( 2, 0) means that all the signs must be the same. at is, mustbeoftheform for some. But since (, ) must also avoid (2, 0 0), = 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 ( 2, 0 ) and ( 2, 0) means that all the signs in corresponding to (where =)mustbethesame,saythatsign is. e fact that (, ) avoids (2, 0 0) means that (i) must be in increasing order and (ii) all the signs corresponding to must be dierent from. But then it follows from the fact that (, ) avoids both ( 2, 0 ) and ( 2, 0) that all the elements in must be greater than all the elements in. Hence there are, such elements if 2 andtherearenosuch elements if =. us from (ii) it follows that, =since ( 2, 0 ) istheonlyelementof which avoids. For 2,wehave, = = Inthecase=2,(52) becomes,2 =2 = andinthecase=,(52) becomes, = = In general, assuming that, =,. (52) 2 = 2, (53) (2) =(2). (54) 2 =2 (), (55) it follows that, = = = = (), () =() =2 = =2 ()() = =() 2 2 (). = () (56) We next consider simultaneous avoidance of the patterns ( 2, 0) and ( 2, 0 ). eorem 5. Let = {( 2, 0), ( 2, 0 )}, () = 0, and () =. en () = () () (). (57) Proof. Fix and suppose that (, ) is an element of which avoids both ( 2, 0) and ( 2, 0 ).Nowif= is constant, then clearly can be arbitrary so that there are such elements. Next assume that isnotconstantsothereisan2 such that = 2 2 where for =,, and for =,,.Weclaimthat must be the largest elements of {,, }. If not, then let be the largest element which is not in. us andtheremustbeatleastone with such that. Now it cannot be = since otherwise (, 2 ) would be an occurrence of either ( 2, 0) or ( 2, 0 ).Henceitmustbethecasethat and =for some. But then no matter what color we choose for, either (, ) or (, 2 ) would be an occurrence of either ( 2, 0) or ( 2, 0 ).Wecancontinuethisreasoningtoshowthatfor any, 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 inanywaythatwewantandwewillalwaysproducea pair (, ) that avoids both ( 2, 0) and ( 2, 0 ). us for such a, wehave( ) ways to choose the colors

11 0 ISRN Combinatorics,, and! 2!! ways to choose the permutation. It follows that, =! =2 = >0 ()! 2!!, (58) which is equivalent to (57). Next we want to consider the problem of avoiding (, ) where = {( 2, 0 0), ( 2, 0 )} and = (, { }). If =,thenwesaythat is a le-to-right minimum of if > for all andwelet lrmin() denote the number of le-to-right minima of. Let, denote the set of all (, ) which avoid (, ). en we let (, ) = lrmin(). 0! (59) (,), en we have the following theorem. eorem 6. For = {( 2, 0 0), ( 2, 0 )} and = (, { }),onehas () (, ) = ( ). (60) Proof. Let, () = (,), lrmin(). If (, ) contains the element colored by the color in position where {0,, }, then if (, ) istoavoid(, ), it must be the casethateveryelementtotherightofmustbecoloredwith a color from {0,, 2} and these elements can be arranged in any of ( )! ways. Moreover, it immediately follows that no instance of an occurrence of (, ) exists where the rst elementistotheleofandthesecondistotherightof. usitfollowsthat(, ) avoids (, ) ifandonlyifthereis no occurrence of (, ) in (, ). us it follows that lrmin () =() (,),, = ()!, () =! (), ().! (6) Here the binomial coecient is responsible for choosing the elements to the le of and placing the remaining elements to therightofandthefactorof accounts for the fact that is always a le-to-right minimum in and none of the elements following can be a le-to-right minimum of. is shows that, (), () =! (). (62)! =0 Multiplying both parts of the equation above by! and summing over all, we have, () 0! =, () (), (63) 0! so that, for all 2, It follows that =0 (, ) = (, ) (). (64) ln (, ) = (). (65) Integrating and using the fact that (, 0) =,weseethat and hence, (, ) = ln (() ), (66) Using the fact that we see that In particular, () (, ) = (). (67) () = () () 0!, (68), () = () = ()( ()) ( ()()). ({( 2,0 0),( 2,0 )},(,{})), =(2 )(32) ( ()). (69) (70) Wenotethat,byoursymmetrymapswhereweapplythe identity map to and the composition of complement and reverse to and,onecanshowthat ({( 2,0 0),( 2,0 )},(,{})), ({( 2, ),( 2,2 )},({0},)) =,. It is easy to see that our denitions imply that ({( 2,0 0),( 2,0 )},({0},)), ({( 2, ),( 2,2 )},({0},)) =,, (7) (72)

12 ISRN Combinatorics so that ({( 2,0 0),( 2,0 )},({0},[)), =(2 )(32) ( ()). (73) Multiplying both parts of the equation above by and summing over all, we have,, 0 =,, 0 Next we consider the case where < <, = ({0,, }, {,, }), and = {( 2, 0 0), ( 2, 0 )}. eorem 7. Let = ({0,, }, {,, }) where 3and <<, and = {( 2, 0 0), ( 2, 0 )}. en if, (, (2) ) ( ) () = (). (74) If =, then =2and Proof. Let (,, ) = (, ), (, ) 2 () = (()). (75) so that, () =,, 0 =(, ) (). (76) Now suppose that (, ) contains the element colored by a color from {,, }. en there are no restrictions for placing this element, thus giving ( ) possibilities. On the other hand, if (, ) contains the element colored by the color in position where {0,, }, then if (, ) is to avoid (, ),itmustbethecasethat every element to the right of must be colored with a color from {0,, 2} and these elements can be arranged in any of ( ) ways. Moreover, it immediately follows that no instance of an occurrence of (, ) exists where the rst elementistotheleofandthesecondistotherightof. en it follows that (, ) avoids (, ) if and only if there is no occurrence of (, ) in (, ).Sointhe casewhereiscoloredbyacolorfrom{0,, } and is in position, we have ( ) ( ) ( ) (,, ) = ( ) (,, ) possibilities where the binomial coecient is responsible for choosing the elements to the le of and placing the remaining elements to the right of. us,, = (),,,,. =0 (77),, 0 so that, for all 3and <<,, () =, () It follows that =0,,, (78), () (79), (). ln, () =, (), () =. (80) Now suppose that. en integrating (80) and using the fact that (0,, ) = and that ()(ln( ) ln( )) = ( )( ) ( ),weseethat ln, () = ln which implies that, () = ln ln (2) (8) (82) and proves (74). Now if =. en = 2 and (80) becomes ln 2, () = 2, () 2, () = 2. (83)

13 2 ISRN Combinatorics Integrating (83) and using the fact that, we see that us ln ln which proves (75). + ln +. (84) (85) Finally, we end this section by considering the elements of which avoid both and.inthis case,weonlyhavearesultforthecasewhen. eorem 8. enumberofpermutationsin simultaneously avoiding and isgivenbythe + -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 + knowntobecountedbythe 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 2 3, as in an occurrence of such a pattern in, there are two letters of the same color leading to an occurrence of ( 2,0 0). us the structure of, as it is well known, is two decreasing sequences shued. 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 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 ). 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, 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. Wewillgiveadescriptionofapath (corresponding to ) from + to + involving only steps and that never goes above the line ++. Clearly, canbetransformedtoadyckpathoflength+ 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 denes 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. If the current block is not a singleton, go to Step 2. If the color of the element with -coordinate equal to + is (resp., ) travel around the current block counterclockwise (resp., clockwise) to get to the point +. Note that touches the line ++if the color is. Set + and, and proceed with Step 3. Step 2. In Step 2 we follow a standard bijection between -avoidingpermutationsanddyckpathsthatcanbe 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 makingsurethatnoneofthedotsarebelowthepartof 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 thereisadotwith-coordinate + having -coordinate less than ). If and, proceed with Step 3; otherwise set and andgotostep2.notethatin Step 2 never touches the line ++. Step 3. If, make as many as it takes horizontal steps to gettothepoint + and terminate; otherwise go to Step. Returning to our example,, we have given the matrix diagram in Figure. We have outlined the reverse irreducible blocks in Figure 2. We start our path at 8. We travel down until we reach, whenweare less than the y- coordinate 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 Pattern Matching Results for Pairs of Length Goulden and Jackson [8] proved the following theorem.

14 ISRN Combinatorics F : Matrix representation of 0 0 0). F 4: e Dyck path corresponding to 0 0 0).Notethatthepathonlytouchesthe line aer a singleton colored. eorem 9 (see [8]). If, where, then 0 -mch ) 0 0 0) 0 ). (86) 0 Later Mendes and Remmel [9] rened this result by proving the following. F 2: Matrix representation of 0 0 0)with reverse irreducible blocks outlined. 0 eorem 20. When, 0 -mch )0 des) / 0 ) (87) where is the number of rearrangements of zeroes and ones such that zeroes never appear consecutively. 0 e reason that eorem 20 is a renement of eorem 9 is that Mendes and Remmel [9]provedthat 0 0 ) 0 ) if divides ) if divides 0 otherwise (88) so that setting in (87) yields (86). e main goal of this section is to generalize eorem 20. at is, let ) ) where. en we shall consider three sets of patterns for : F 3: e Dyck path corresponding to 0 0 0)added to Figure 2. (a) ) 0 ), (b) ) 0),

15 4 ISRN Combinatorics (c) which equals the set of all ( (, such that {0,,, is weakly decreasing, and red(. In [], itaev et al. dened three dierent types of descents. at is, for elements (,, they dened Des ((, >,, des ((, Des ((,, WDes ((, >,, wdes ((, WDes ((,, SDes ((, >, >, sdes ((, SDes ((,. (89) We shall refer to Des((, as the descent set of (,, WDes((, as the weak descent set of (,, and SDes((, as the strict descent set of (,. eorem 2. Let ( where. () If 0, then 0 (, (,-mch ((,0 wdes ((, (,, 0 (2) If, then 0 (, (,-mch ((,0 sdes ((, (,, 0.. (90) (9) (3) If equals the set of all ( (, such that {0,,, is weakly decreasing, and red(, then 0 (, -mch ((,0 des((, (,, 0. (92) Proof. Our proof is an adaptation of the proof that Mendes and Remmel [9]usedtoproveeorem 20. e basic idea is to show that(90), (9), and (92) arise by applying appropriate ring homomorphisms, dened on the ring of symmetric functions over innitely many variables,,, to a simple symmetric function identity. F 5: A brick tabloid of shape ( and type (,,,, 5. e th elementary symmetric function,,andtheth homogenous symmetric function,,aredenedbythe generating functions Clearly, (, 0 ( 0. ( (93) (. (94) Let (,, be an integer partition; that is, is a nite sequence of weakly increasing positive integers. Let ( denote the number of parts of.ifthesumofthese integers is,wesaythatisapartition of and write. For any partition (,,, let.ewellknown fundamental theorem of symmetric functions says that { is a partition is a basis for or that { 0,, is an algebraically independent set of generators for. Similarly, if we dene, then { is a partition 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 coecient of the homogeneous symmetric functions when written in terms of the elementary symmetric function basis. ese coecients have been shown to equal the sizes of certain sets of combinatorial objects up to a sign. A brick tabloid of shape ( and type (,, isallingofarowof squares of cells with bricks of lengths,, such that bricks do not overlap. One brick tabloid of shape ( and type (,,,, 5 is displayed In Figure 5. Let, denote the set of all -brick tabloids of shape ( and let,,. rough simple recursions stemming from (94), Egecio glu and Remmel [0] proved that ( (,. (95) We now consider three dierent ring homomorphisms which map to [ where is the eld of rational numbers. Since 0,, is an algebraically independent set of generators for, we can dene a ring homomorphism by simply specifying its value on for all 0. For, dene ( ( ( (,,, 0 0(,,, 0 (,,. (96)

16 ISRN Combinatorics 5 ur rst goal is to prove the following facts. (A) If, then () ()-mch (()) (B) If, then () ()-mch(()) wdes(()). sdes(()). (97) (98) (C) If consists of the set of all ( () ) such {, is weakly decreasing, and red(), then 3 () -mch (()) des(()). (99) Let ( ), ( ), and (3 ). en for { 3, () () () () () () () () () () () () (). (00) We wish to give a combinatorial interpretation of righthand side of (00). First we select a partition and then we select a brick tabloid ( () ). Next we interpret the multinomial coecient () as the number of ways to pick sets () of sizes (), respectively, which partition {. We then place these elements in inthebottomofthecellsofthebrick in decreasing order reading from le 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 thcellof. If 3, then (3 ) which we interpret as the ways of picking a weakly decreasing sequence from { and F 6: Filled labeled brick tabloids in. 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. en 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.usforanygiven, we will have cells labeled with which accounts for the sign (). e only term in (00)thatwehavenot 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 and we let denote the set of all lled labeled brick tabloids that arise in this way for interpreting ( ). us a consists of a brick tabloid, a permutation, a sequence {, and a labeling ofthecellsof with elements from { such that the following conditions hold. (a) is strictly decreasing in each brick. (b) () If, then is constant in each brick. (2) If, then is strictly decreasing in each brick. (3) If 3, then is weakly decreasing in each brick. (c) e nal cell of each brick is labeled with. (d) ach cell which is not a nal cell of a brick is labeled with or andthereisnoconsecutivestringofcells of length labeled within a brick. We then dene the weight () of to be the product of all the labels in and the sign sgn() of tobetheproduct of all the labels in. For example, if,,, and ( ), then Figure 6 pictures a lled labeled brick tabloid at the top, a a lled labeled brick tabloid in the middle, and a lled labeled brick tabloid 3 atthebottom.in each case, () and sgn(). Next we dene sign-reversing, weight-preserving involutions on. o dene (), wescanthecellsof

17 6 ISRN Combinatorics F 7: ( for the s in Figure 6. ( from le to right looking for the lemost 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 isconstantonallthecells 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 le of and contains all the cells of totherightof,,, 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, 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, signreversing involution, and hence shows that sgn ( (. ( (0) us we must examine the xed points ( of.firsttherecanbeno labels in so that sgn(. Moreover, if and are two consecutive bricks in and is the last cell of, then () If,itcannotbethecasethat > 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. us if is the last cell of a brick, then WDes( if, SDes( if, and Des( if. However, the label on a cell ofanybrick must be if is not the last cellofthebrickandmustbeif is the last cell of a brick. Moreover, if is not the last cell of a brick, then > and if, > if, if. us if isnotthelastcellofabrick,thenwdes( if, SDes( if, and Des( if. It follows that sgn( and ( wdes(( if, ( sdes(( if, and ( wdes(( if. en ( (-mch((0 ( (-mch((0 ( (-mch((0 wdes(( sdes(( if if des(( if. (02) erefore (A), (B), and (C) hold. We can now prove (90) by applying to (94). at is, we have 0 and,andthen 0 ( (-mch(0 0 wdes(( 0 ( ( ( / 0 ( / 0 (. (03) Similarly, (9) can be proved by applying to (94) and (92) can be proved by applying to (94). In fact, we can calculate in several cases. For instance, routine combinatorial techniques are able to show that is whenever 0 /. When, we nd that / 0. (04)

18 ISRN Combinatorics 7 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 [0] Ö. Eeciolu and J. B. Remmel, Brick tabloids and the connection matrices between bases of symmetric functions, Discrete Applied Mathematics, vol. 34, no. 3, pp , 99. (a) we have removed a 0, (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. us we obtain the following relationship: = +. (05) 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 + +. Acknowledgment J. Remmel was partially supported by NSF Grant DMS and M. Riehl was supported by UWEC s Oce of Research and Sponsored Programs. References [] S. Kitaev, A. Niedermaier, J. B. Remmel, and A. Riehl, Generalized pattern matching conditions for, ISRN Combinatorics, vol. 203, Article ID , 20 pages, 203. [2] E. S. Egge, Restricted colored permutations and Chebyshev polynomials, Discrete Mathematics, vol. 307, no. 4, pp , [3] T. Mansour, Pattern avoidance in coloured permutations, Séminaire Lotharingien de Combinatoire, vol. 46, article B46g, p. 2, 200. [4] T. Mansour, Coloured permutations containing and avoiding certain patterns, Annals of Combinatorics, vol. 7, no. 3, pp , [5] T. Mansour and J. West, Avoiding 2-letter signed patterns, Séminaire Lotharingien de Combinatoire, vol. 49, article B49a, p., [6] R. Simion, Combinatorial statistics on type-b analogues of noncrossing partitions and restricted permutations, Electronic Journal of Combinatorics, vol. 7, no., pp. 27, [7] A. Duane and J. Remmel, Minimal overlapping patterns in colored permutations, Electronic Journal of Combinatorics, vol. 8, no. 2, p. 25, 20. [8] I. P. Goulden and D. M. Jackson, Combinatorial Enumeration, A Wiley-Interscience Publication, John Wiley & Sons, New York, NY, USA, 983. [9] A. Mendes and J. Remmel, Permutations and words counted by consecutive patterns, Advances in Applied Mathematics, vol. 37, no. 4, pp , 2006.

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