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

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1 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, Jeffrey 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 CC kk SS, the wreath product of the cyclic group CC kk with the symmetric group SS, 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 CC kk SS of the cyclic group CC kk and the symmetric group SS initiated in []. CC kk SS is the group of kk n signed permutations where there are kk signs, = ωω 0, ωω, ωω 2,, ωω kkkk, where ωω is a primitive kkth root of unity. We can think of the elements CC kk SS as pairs γγ γ γγγγ γγγ where σσ σ σσ σσ SS and εεεεε εε {, ωωω ω ω ωω kkkk }. For ease of notation, if εεε (ωω ww,ωω ww 2,,ωω ww ) where ww ii {0,, kk k kk for ii i ii, then we simply write γγ γ γγγγ γγγ where ww w ww ww 2 ww. Moreover, we think of the elements of ww w ww ww 2 ww as the colors of the corresponding elements of the underlying permutation σσ. We de ne a concept for matchings in words over a nite alphabet [kkk k kkk kk k k kk k kk. Given a word wwwww ww [kkk, let red(www be the word found by replacing the iith largest integer that appears in ww by. For example, if ww w ,thenred(www w w w w w w. Given a word uuu [kkk jj such that red(uuu u uu, wesaythatawordww w wwww has a uu-match starting at position ii provided red(ww ii ww iii )=uu. Let uu-mch(www be the number of uu-matches in the word ww. Similarly, we say that uu occurs inawordww if there exist ii < < ii jj such that red(ww ii ww iijj ) = uu. We say that ww avoids uu if there are no occurrences of uu in ww. ereareanumberofpapersonpatternmatchingand pattern avoidance in CC kk SS [2 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 5]. De nition Let (τττ τττ τ ττ kk SS jj, Υ be a subset of CC kk SS jj, and (σσσ σσσ σ σσ kk SS. () One says that (σσσ σσσ has an exact occurrence of (τττ τττ (resp., Υ) if there are ii < ii 2 < < ii jj such that (red(σσ ii σσ iijj ), ww ii ww iijj ) = (τττ τττ (resp., (red(σσ ii σσ iijj ), ww ii ww iijj ) Υ). (2) One says that (σσσ σσσ σ σσ kk SS avoids an exact occurrence of (τττ τττ (resp., Υ) if there are no exact occurrences of (τττ τττ (resp., Υ) in (σσσ σσσ. (3) One says that there is an exact (τττ τττ-match in (σσσ σσσ starting at position ii (resp., exact

2 2 ISRN Combinatorics Υ-match in (σσσ σσσ starting at position ii) if (red(σσ ii σσ σσ iii ), ww ii ww ww iii ) = (τττ τττ (resp., (red(σσ ii σσ σσ iii ), ww ii ww ww iii ) Υ). at is, an exact occurrence or an exact match of (τττ τττ τ CC kk SS jj in an element (σσσ σσσ σ σσ kk SS is just an ordinary occurrence or match of ττ in σσ where the corresponding signs agree exactly. For example, Mansour [4] proved via recursion that, for any (τττ τττ τ ττ kk SS 2, the number of elements in CC kk SS which avoid exact occurrences of (τττ τττ is jjjj jjj jjj j jjjj jj 2. is generalized a result of Simion [6] who proved the same result for the hyperoctahedral group CC 2 SS. Similarly, Mansour and West [5] determined the number of permutations in CC 2 SS that avoid all possible exact occurrences of 2 or 3 element sets of patterns from CC 2 SS 2. For example, let KK be the number of (σσσ σσσ σ CC 2 SS that avoid all exact occurrences of the patterns in the set {( 2, 0 0), ( 2, 0 ), (2, 0)}, KK 2 the number of (σσσ σσσ σ σσ 2 SS that avoid all exact occurrences of the patterns in the set {( 2, 0 ), ( 2, 0), (2, 0 )}, and KK 3 the number of (σσσ σσσ σ σσ 2 SS that avoid all exact occurrences of the patterns in the set {( 2, 0 0), ( 2, 0 ), (2, 0 0)}. enmansourandwest[5]provedthat KK =FF 2, KK 2 = n, jj jjjj KK 3 = n n n jj, jjjj where FF 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 (τττ τττ τ ττ kk SS jj where red(uuu u uu, Υ a subset of CC kk SS jj where for all (τττ τττ τ τ, red(uuu u uu, and (σσσ σσσ σ CC kk SS. () () One says that (σσσ σσσ has an occurrence of (τττ τττ (resp., Υ) if there are ii < ii 2 < < ii jj such that (red(σσ ii σσ iijj ), red(ww ii ww iijj )) = (τττ τττ (resp., (red(σσ ii σσ iijj ), red(ww ii ww iijj )) Υ). (2) One says that (σσσ σσσ σ σσ kk SS avoids (τττ τττ (rep., Υ) if there are no occurrences of (τττ τττ (resp., Υ) in (σσσ σσσ. (3) One says that there is a (τττ τττ-match in (σσσ σσσ starting at position ii (resp., Υ- match in (σσσ σσσ starting at position ii) if (red(σσ ii σσ σσ iii ), red(ww ii ww ww iii )) = (τττ τττ (resp., (red(σσ ii σσ σσ iii ), red(ww ii ww ww iii )) Υ). For example, suppose that (τττ τττ τ ττ ττ τ ττ and (σσσ σσσ σ ( 3 2 4, 2 2 2). en there are no exact occurrences or exact matches of (τττ τττ in (σσσ σσσ. However,therearetwo occurrences of (τττ τττ, one in positions 2 and 4 and one in positions 3 and 4. us there are two occurrences of (τττ τττ in (σσσ σσσ andthereisa(τττ τττ-match in (σσσ σσσ starting at position 3. Finally, a more general-matching condition which generalizes both occurrences and matches and exact occurrences and exact matches in CC kk SS was introduced in []. e nition Let (τττ τττ AAA A AA kk SS jj, Υ be a subset of CC kk SS jj, AA A AAA,,AA jj ) be a sequence of subsets of [kkk, and (σσσ σσσ σ CC kk SS. () One says that (σσσ σσσ has an occurrence of (τττ τττ AAA (resp., (Υ, AAA) ifthereare ii < ii 2 < < ii jj such that (red(σσ ii σσ iijj ), red(ww ii ww iijj )) = (τττ red(uuuu (resp., (red(σσ ii σσ iijj ), red(ww ii ww iijj )) is equal to (τττ red(uuuu for some (τττ τττ in Υ) and ww iiss AA ss for sssssssss. (2) One says that (σσσ σσσ σ σσ kk SS avoids (τττ τττ AAA (resp., (Υ, AAA) if there are no occurrences of (τττ τττ (resp., Υ) in (σσσ σσσ. (3) Onesaysthatthereisa(τττ τττ AAA-match in (σσσ σσσ starting at position ii (resp., (Υ, AAA-match in (σσσ σσσ starting at position ii) if (red(σσ ii σσ σσ iii ), red(ww ii ww ww iii )) = (τττ red(uuuu (resp., (red(σσ ii σσ σσ iii ), red(ww ii ww ww iii )) is equal to (τττ red(uuuu for some (τττ τττ in Υ) and ww i AA ssss for sssssssssss. us a (τττ τττ AAA-occurrence (or a (τττ τττ AAA-match) where uuuuu uu jj and AA A AAAA },, {uu jj }) is just an exact occurrence (resp., an exact match) of (τττ τττ. Similarly, a (τττ τττ AAAoccurrence (or a (τττ τττ AAA-match) where uu u uu uu jj, red(uuu u uu, and AA A AAA,,AA jj ) such that AA ii = [kkk for ii i ii is just an occurrence (or match) of (τττ τττ. Suppose we are given (τττ τττ τ ττ kk SS jj, Υ CC kk SS jj, AA A AAA,,AA jj ) where AA ii [kkk for ii i,,jj, and (σσσ σσσ σ σσ kk SS. We let (τττ ττ)-mch((σσσ σσσσ (resp., (τττ ττ)-emch((σσσ σσσσ) denote the number of (τττ τττmatches (resp., exact (τττ τττ-matches) in (σσσ σσσ. We let Υ-mch((σσσ σσσσ (resp., Υ-Emch((σσσ σσσσ) denote the number Υ-matches (resp., exact Υ-matches) in (σσσ σσσ. We let (τττ τττ AA)-mch((σσσ σσσσ be the number of (τττ τττ AAA-matches in (σσσ σσσ and (Υ, AA)-mch((σσσ σσσσ the number of (Υ, AAA-matches in (σσσ σσσ. emainresultof[] was to derive a generating function for the distribution of (τττ τττ AAA-matches where (τττ τττ τ ττ kk is any element of CC kk SS 2.Tostatethemainresultof[], we need to de ne some notation. First we de ne the ppp pp-analogues of

3 ISRN Combinatorics 3, n, kk, and aa,,aa mm by en Kitaev et al. [] proved that, for any Υ CC kk SS 2 and AA A AAA,AA 2 ) where AA,AA 2 [kkk, [] ppppp = pp qq ppppp =pp +pp qq + + pppp +qq, [] ppppp!= [] ppppp [] ppppp [2] ppppp [] ppppp, kk = ppppp = aa,,aa mm ppppp [] ppppp! [kk] ppppp! [n] ppppp!, [] ppppp! aa ppppp! aa mm ppppp!, (2) tt pp coinv(σσσ qq inv(σσσ xx (Υ, AAA-mch((σσσσσσσ zz (ww) [] ppppp! (σσσσσ) CC kk SS = zz 0 + +zz kkkk ttttt [] ppppp! (xxxx) mmmm (Υ, AAAAAA ppp ppp pp 0,,zz kkkk. (5) tt respectively. We de ne the qq-analogs of, n, kk, and aa,,aa mm by [n,qq, [n,qq!, kk,qq, and aa,,aa mm,qq, respectively. Next suppose that Υ CC kk SS 2 and AA A AAA,AA 2 ) where AA,AA 2 [kkk. enwewillsaythat(σσσ σσσ σ σσ kk SS is a maximum packing for (Υ, AAA if (σσσ σσσ has (Υ, AAA-matches starting at positions, 2,, n n. We let MPP (Υ, AAAAAA denote the set (σσσ σσσ σ σσ kk SS of maximum packings for (Υ, AAA. Given any word wwwww ww [kkk, we let zzzzzz z z zz ww ii. For any σσ σ σσ σσ SS, we let inv(σσσ (resp., coinv(σσσ) equal the number of pairs (iii iii such that and σσ ii >σσ jj (resp., σσ ii < σσ jj ). We then de ne mmmm (Υ, AAAAAA ppp ppp pp 0,,zz kkkk = pp coinv(σσσ qq inv(σσσ zz (ww). (σσσσσ) MPP (Υ, AAAAAA We will also be interested in the specializations (3) Duane and Remmel [7] gave the generating function for the number of (τττ τττ-matches and exact (τττ τττ-matches for a certain collection of (τττ τττ τ ττ jj SS where jjjjarecalled minimal overlapping patterns. Given a word uu u uuu uu u u uuu } jj such that red(uuu u uu and ττ τ ττ jj, Duane and Remmel say that (τττ τττ has the CC kk SS -minimal overlapping property if the smallest ii such that there exists a (σσσ σσσ σ σσ kk SS ii with (τττ ττ)-mch((σσσ σσσσ σ σ is 2jj j j. is means that in a kkcolored permutation (σσσ σσσ, two (τττ τττ-matches can share at most one pair of letters, and this pair of letters must occur at the end of the rst (τττ τττ-match and at the start of the second (τττ τττ-match. Similarly, we say that (τττ τττ has the CC kk SS -exact match minimal overlapping property if the smallest ii such that there exists a (σσσ σσσ σ σσ kk SS ii with (τττ ττ)-emch((σσσ σσσσ σ σ is 2jj j j. Now if (τττ τττ has the CC kk SS -minimal overlapping property, then the shortest kk-colored permutations (σσσ σσσ, such that (τττ ττ)-mch((σσσ σσσσ σ σσ have length n n n n. We let MPP kk (τττττττττττττττττ equalthesetofkk-colored permutations, (σσσ σσσ σ σσ kk SS,suchthat(τττ ττ)-mch(σσσ σσσ σ σσ. Werefertothekk-colored permutations in MPP kk (τττττττττττττττττ as maximum packings for (τττ τττ. We let mmmm (Υ, AAAAAA ppppppppppp = pp coinv(σσσ qq inv(σσσ, (σσσσσ) MPP (Υ, AAAAAA mmmm (Υ AAAAAA ppp ppp pp ppp p p ppkkkk (4) mmmm kk (τττττ),jjjj+ = MPPkk (τττττ),jjjj+, mmmm kk (τττττ),jjjj+ ppp ppp pp = (σσσσσ) MPP kk (τττττ),jjjj+ pp coinv(σσσ qq inv(σσσ rr ww, (6) = pp coinv(σσσ qq inv(σσσ rr wwww, (σσσσσ) MPP (Υ, AAAAAA mmmm kk (τττττ),jjjj+ ppp ppp pp 0,,zz kkkk = pp coinv(σσσ qq inv(σσσ zz (ww). (σσσσσ) MPP kk (τττττ),jjjj+ where for any word ww w ww ww [kkk, wwww w ww + +ww. In the special case where Υ = {(τττ ττττ where (τττ τττ τ ττ kk SS 2, we shall denote mmmm (Υ, AAAAAA (ppp ppp pp 0,,zz kkkk ) as mmmm (ττττττ AAAAAA (ppp ppp pp 0,,zz kkkk ). Similarly, we let EMPP kk (τττττττττττττττττ denote the set of kkcolored permutations, (σσσ σσσ σ σσ kk SS,suchthat

4 4 ISRN Combinatorics (τττ ττ)-emch(σσσ σσσ σ σσ, and refer to the kk-colored permutations in EMPP kk (τττττττττττττττττ as exact match maximum packings for (τττ τττ. We let eeeeee kk (τττττ),jjjj+ = EMPPkk (τττττ),jjjj+, eeeeee kk (τττττ),jjjj+ ppp ppp pp = (σσσσσ) EMPP kk (τττττ),jjjj+ eeeeee kk (τττττ),jjjj+ ppp ppp pp 0,,zz kkkk = (σσσσσ) EMPP kk (τττττ),jjjj+ pp coinv(σσσ qq inv(σσσ rr ww, pp coinv(σσσ qq inv(σσσ zz (ww). Duane and Remmel [7]provedthat(I)ifred(uuu u uu and (τττ τττ has the CC kk SS -minimal overlapping property, then tt xx (ττττττ-mch(σσσσσσ pp coinv(σσσ qq inv(σσσ zz (ww) n (σσσσσ) CC kk SS = / zz 0 + +zz kkkk tt + jjjj + ppppp! (xxxx) tt mmmm kk (τττττ),jjjj+ ppp ppp pp 0,,zz kkkk, and (II) if (τττ τττ τ ττ kk SS jj has the CC kk SS -exact match minimal overlapping property, then tt xx (ττττττ-emch(σσσσσσ pp coinv(σσσ qq inv(σσσ zz (ww) n (σσσσσ) CC kk SS = / zz 0 + +zz kkkk tt + jjjj + qq! (xxxx) tt (7) (8) eeeeee kk (τττττ),jjjj+ ppp ppp pp 0,,zz kkkk. (9) Duane and Remmel s proof of (I) and (II) works equally well for (Υ, AAA-matches. emaingoalofthispaperistoproveanumberofresults on the number of (τττ τττ AAA-avoiding elements of CC kk SS and to nd a generating function for the number of (τττ τττ AAAmatches 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 (τττ τττ AAA-avoiding elements of CC kk SS where (τττ τττ τ ττ kk SS 2. In Section 3, we prove a number of results on the number of (Υ, AAA-avoiding elements of CC kk SS and for certain subsets Υ of CC kk SS 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 SS that had no (jjjjj j jj j jj-matches. 2. (τττ τττ AAA-Avoiding Patterns Given (τττ τττ τ ττ kk SS jj and AA AAAA,,AA jj ) where AA ii [kkk for ii i ii, let AAAA (ττττττ AAA denote the number of (σσσ σσσ σ CC kk SS which avoid (τττ τττ AAA.InthespecialcasewhereAA ii = [kkk for i, we shall denote AAAA (ττττττ AAA as simply AAAA (ττττττ. us AAAA (ττττττ is the number (σσσ σσσ σ σσ kk SS which avoid (τττ τττ. Inthiscase,weshall ndformulasforaaaa (ττττττ AAA or AAAA (ττττττ where (τττ τττ τ ττ kk SS 2. ere are a number of natural maps for which the distribution of (τττ τττ AAA-occurrences and (τττ τττ AAA-matches remain invariant. at is, for any σσσσσ σσ SS, we de ne the reverse of σσ, σσ rr and the complement of σσ, σσ cc, respectively, by σσ rr =σσ σσ σσ, σσ cc = n n. (0) If wwwww ww [kkk, then we de ne the reverse of ww, ww rr and the complement of ww relative to [kkk, ww (cccccc, by ww rr =ww ww ww, ww (cccccc = kkkkkkk kkkkkkk. () If AA A AAA,,aa ss } [kkk, then we de ne AA (cccccc = {kkkkk aa,,kkkkkkk ss }. en if AA A AAA,,AA jj ) where AA ii [kkk, for i, then we de ne AA rr = AA jj,aa jjjj,,aa, (2) AA (cccccc = AA ccccc,,aaccccc jj. en consider the maps φφ aaaaa CC kk SS CC kk SS where φφ aaaaa ((σσσ σσσσ σ σσσ aa,ww bb ) for aaa aa a aaaa aaa aaa and bb b bbbb bbb bbbb bbbb where ii is the identity map; that is, σσ ii = σσ for all σσ σ σσ, ww ii =wwforall ww w wwww, and AA ii = AA for all AA A AAA,,AA ) with AA ii [kkk for i.itiseasytoseethata(σσσ σσσ has a (τττ τττ AAA-match or (τττ τττ AAA-occurrence if and only if (σσ aa,ww bb ) has a (ττ aa,uu bb, AA bb )-occurrence or (ττ aa,uu bb, AA bb )-match. It follows that if red(uuu u uu and (τττ τττ τ ττ kk SS 2, then we need only to compute AAAA (ττττττ for two patterns; namely, (τττ τττ τ ττ ττ τ ττ and (τττ τττ τ ττ ττ τ ττ. We start by considering the pattern ( 2, 0 0). Lemma 4. e number of elements in CC kk SS avoiding ( 2, 0 0) is given by ii + +ii kk = ii 0,,ii kk 0 2. ii,,ii kk (3)

5 ISRN Combinatorics 5 Proof. e rst observe that elements of different colors are independent in permutations avoiding ( 2, 0 0), meaning that no two elements with different colors can form a prohibited con guration. us assuming we have ii jjjj elements of color jj, jj j jjjjjj j j,wecanchoosehowtoplace these colors to form a word ww in ii,,ii kk ways. en we can choose the sets of elements CC 0,,CC kkkk from {,, n ii,,ii kk which will correspond to the colors 0,,kkkkin σσ in ways. Finally, in order to construct a (σσσ σσσ which avoids the prohibited pattern, we must place the elements of CC ii in the positions which are colored by ii in decreasing order. e proof of Lemma 4 suggests an obvious generalization forpatternsoftheform(τττ τ jj ) where τττττ jj. eorem 5. Let τττττ jj. en the number of elements in CC kk SS avoiding (τττ τ jj ) is given by ii + +ii kk = ii 0,,ii kk 0 AA ii AA ii2 AA iikk, ii,,ii kk where AA isthenumberofpermutationsinss avoiding ττ. 2 (4) Proof. eproofhereisessentiallythesameastheproof of eorem 4, except that we can place ii elements of a permutation in CC kk SS ofthesamecolorinanyofaa ii ways. Itiswellknownthatthenumberof-permutations avoiding any pattern of length 3 is given by the th Catalan number CC = (/( 2.Asacorollaryto eorem 5, we have that, for any pattern τττττ 3, the number of permutations in CC kk SS avoiding (τττ τ τ ττ isgivenby ii + +ii kk = ii 0,,ii kk 0 2ii ii 2ii 2 ii 2 2ii kk ii kk ii +ii 2 + ii kk +. ii,,ii kk (5) One can generalize eorems 4 and 5 even further, by considering the distribution of patterns. Indeed, assuming that we know the number AA i of ii-permutations containing mm occurrences of a pattern ττ τ ττ jj,wecanwritedownthe number of permutations in CC kk SS with mm occurrences of (τττ τ jj ) as ii + +ii kk = mm + +mm kk =mm ii 0,,ii kk 0 mm 0,,mm kk 0 AA ii,mm AA ii2,mm 2 AA iikk,mm kk. ii,,ii kk 2 2 (6) For example, the distribution of the occurrences of the pattern 2 isthesameasthedistributionofcoinversions in permutations, so one can extract the numbers AA i as the coefficients to qq mm in ii ssss (+qqqqqqqssss ) and substitute them in the last formula to get the distribution of the number of occurrences of ( 2, 0 0) in CC kk SS. Next we consider the pattern ( 2, 0 ) in the case where kkkk. e number of permutations in CC 2 SS avoiding ( 2, 0 ) is shown in [6,page9]tobeequalto jjjj jjj jj 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 CC 2 SS avoiding the pattern ( 2, 0 ) is given by Proof. Let AA = AAAA ( 2,0 ) and AA (tt) = eettttttttt tt. (7) tt AA (tt) =AA n. (8) If (σσσ σσσ σ σσ 2 SS contains the element colored by the color, then there are no restrictions for placing this element, thereby giving ( n possibilities. On the other hand, if (σσσ σσσ contains the element colored by the color 0 in position and (σσσ σσσ is to avoid ( 2, 0 ), it must be the casethateveryelementtotherightofiscoloredwiththe color 0 and these elements can be arranged in any of ( n ways. Moreover, it immediately follows that no instance of occurrence of ( 2, 0 ) exists where the rst element is tothele ofandthesecondistotherightof. usit follows that (σσσ σσσ avoids ( 2, 0 ) if and only if there is no occurrence of ( 2, 0 ) in (σσ σσ ii,ww ww ii ). en in the case where is colored by 0 and is in position, we have ( n ( ii ) AA ii = n ii /iii possibilities where the binomial coefficient is responsible for choosing the elements to the le of and placing the remaining elements to the right of. To summarize, we obtain AA AA = () AA + n ii iii. (9) Multiplying both parts of the equation above by tt /n and summing over all, we have so that It follows that tt AA n = tt AA n tt + n + AA (tt) =AA(tt) + tttt (tt) + tt AA ii iii (20) AA (tt) tt. (2) dd dddd ln (AA (tt)) = AA (tt) AA (tt) = tt + ( tt) 2. (22) Integrating and using the fact that AAAAA A A,weseethat ln (AA (tt)) = ln ( tt) + tt = ln ( tt) + tt tt (23)

6 6 ISRN Combinatorics which implies that Multiplying both parts of the equation above by tt /n and summing over all, we have AA (tt) = eettttttttt tt. (24) is type of argument can be generalized. at is, we have the following theorem. eorem 7. Let rr r rr r rr r r and AA rrrrrrrr = ({0,, rr r rrr rrrr r r rr r rrr, where kkkk. Let AAAAAA AAA AAA AAA A AAAA ( 2,0,AA rrrrrrrr) denotethenumberof(σσσ σσσ σ σσ kk SS which avoid ( 2, 0, AA rrrrrrrr ), and en if kkkkkkkk, AA rrrrrrrr (tt) =AA (n n n ) tt n. (25) rrrrrrrrrrrrr AA rrrrrrrr (tt) = (kkkkk) tt ssss (kkkkk) tt. (26) If kkkkkkkk, then AA rrrrrrrrrrr (tt) = eerrrrrrrrrrrrrrrrrr (kkkkk) tt. (27) Proof. Note that if (σσσ σσσ σ σσ kk SS contains the element colored with a color from {rrr r r rr r rr, then there are no restrictions for placing this element, thereby giving ( n n n n possibilities. On the other hand, if (σσσ σσσ contains the element colored by the color aa in position where aa a aaa a a aa a aa, then if (σσσ σσσ is to avoid ( 2, 0, AAA, itmustbethecasethateveryelementtothe right of must be colored with a color from {0,, ss s } and these elements can be arranged in any ( n ways. Moreover, it immediately follows that no instance of a occurrence of ( 2, 0, AA rrrrrrrr ) exists where the rst elementistothele ofandthesecondistotheright of. en it follows that (σσσ σσσ avoids ( 2, 0, AA rrrrrrrr ) if and only if there is no occurrence of ( 2, 0, AA rrrrrrrr ) in (σσ σσ ii,ww ww ii ). usinthecasewhereiscoloredby a color from {0,, rr r rr and is in position,wehave ss n ( n ( ii ) AAAAAA AAA AAA AAA A AAA AA n AAAAAA AAA AAA AAAAAAA possibilities where the binomial coefficient is responsible for choosing the elements to the le of and placing the remaining elements to the right of. Hence AA ( n n n ) tt n = (kkkkk) AA (n n n ) tt n so that for all kkkk, It follows that + (kkkkk) (n n n ) tt n +rrrr mmmm tt AA rrrrrrrr (tt) = (kkkkk) AA rrrrrrrr (tt) AA (iii iii iii ii) ss n, iii (29) + (kkkkk) tttt rrrrrrrr (tt) +rr AA rrrrrrrr (tt) ssss. (30) dd dddd ln AA rrrrrrrr (tt) = AA rrrrrrrr (tt) AA rrrrrrrr (tt) = kkkkk (kkkkk) tt + rr ( (kkkkk) tt)( ssss). (3) Now suppose that kk k kkkkk. en integrating (3) and using the fact that AAAAA AAA AAA AAA A A and the fact that (ddddddddddddd d ddddd d dddd d dddddd d ddd d dddddd d ddddddd d ddddd, we see that ln AA rrrrrrrr (tt) = ln ( (kkkkk) tt) + which implies that rr (kkkkkkkk) (ln ( ssss) ln ( (kkkkk) tt)), rrrrrrrrrrrrr AA rrrrrrrr (tt) = (kkkkk) tt ssss (kkkkk) tt which proves (26). Next suppose that kkkkkkkk. en (3) becomes dd dddd ln AA rrrrrrrrrrr (tt) = AA rrrrrrrrrrr (tt) AA rrrrrrrrrrr (tt) = jj (kkkkk) tt + rr ( (kkkkk) tt) 2. (32) (33) (34) Integrating (34) and using the fact that AAAAA AAA AA A AAA AAA A A, we see that ln AA rrrrrrrrrrr (tt) = ln ( (kkkkk) tt) AA ( n n n ) = (kkkkk)() AA (n n n ) AA (iii iii iii ii) + rrrrr ss n. iii (28) + rr kkkkk (kkkkk) tt = ln ( (kkkkk) tt) + rrrr (kkkkk) tt. (35)

7 ISRN Combinatorics 7 us which proves (27). AA rrrrrrrrrrr (tt) = eerrrrrrrrrrrrrrrrrr (kkkkk) tt (36) Next consider the special case where 2rrrrrand rrrrrso AA rrrrrrrrr = ({0,, rr r rrr rrrr r r rrr r rrr. en clearly, AA rrrrrrrrr (tt) = eerrrrrrrrrrrrr rrrr, AA,,2 (tt) = eettttttttt tt. (37) It follows from the form of these generating functions that AAAAAA AAA AAA AAAA A AA AAAAAA AA AA AA. Moreover, in the case where kkkk, it is easy to see that our de nitions imply that the num ber of (σσσ σσσ σ σσ 2 SS which avoid (( 2, 0 ), ({0}, {})) is the same as the number of (σσσ σσσ σ σσ 2 SS which avoid ( 2, 0 ). en AAAAAA AAA AAA AAAA A AA AA where AA is the number of (σσσ σσσ σ σσ 2 SS avoiding ( 2, 0 ). is is also easy to see combinatorially. Namely, if (σσσ σσσ σ σσ 2rr SS which avoids (( 2, 0 ), ({0, rrrrrr rrrr r r rrrrrrrr, then if ww is the result of replacing each occurrence of 0,,rrrrin ww by 0 and each occurrence of rrr r r rrrrr in ww by, then (σσσ σσ ) will be an element of CC 2 SS which avoids ( 2, 0 ). Vice versa, if (σσσ σσ ) CC 2 SS avoids ( 2, 0 ) and ww arises from ww by replacing each 0 by some element from {0,, rrrrr and each by some element from {rrr r r rrr r rr, then (σσσ σσσ will avoid (( 2, 0 ), ({0,, rr r rrr rrr r rr r r rrr r rrrr. In fact, if we are in the case where rrrrrsothat AA rrrrrrrr = ({0,, rr r }, {rr r rr r r rrrr, then by the same argument, we can get an arbitrary (σσσ σσσ σ σσ kk SS which avoids (( 2, 0 ), AA rrrrrrrr ) by starting out with an (σσσ σσ ) CC 2 SS which avoids ( 2, 0 ) and replacing each 0 by some element from {0,, rr r rr and eachbysomeelementfrom{rr r rr r r rr r rr. Considering the form of the generating function for AA rrrrrrrr (ttt when rrrrr, rrrrrrrrrrr AA rrrrrrrr (tt) = (kkkkk) tt rrrr (kkkkk) tt, (38) this suggests the following theorem which can easily be provedbythesamemethodasweusedtoprove eorem 7. eorem 8. Given (σσσ σσσ σ σσ 2 SS, let pos(σσσ σσσ equal the number of 0 s in ww and neg(σσσ σσσ equalthenumberof sinww. ( 2,0 ) Let AA denote the set of (σσσ σσσ σ σσ 2 SS which avoids ( 2, 0 ). en AA ttt ttt tt = n = tt ( 2,0 ) (σσσσσ) AA yyyyyyyyyy yyyy zzzz zzzz. 3. Avoidance for Sets of Patterns yy pos(σσσσσσ zz neg(σσσσσσ (39) In this section, we shall prove a variety of results for the number of elements of CC kk SS that avoid certain sets of patterns of length 2. If Υ SS jj is a set of permutations of SS jj, we let φφ Υ (tt) =AAAA Υ tt n, (40) where AAAA Υ is the number of permutations σσσσσ such that σσ avoids Υ. If Υ CC kk SS jj and AA A AAA,AA 2,,AA jj ) where AA ii [kkk for ii i ii, then we let AAAA (Υ, AAA n denote the number of (σσσ σσσ σ σσ kk SS which avoid (Υ, AAA. We then let θθ (Υ, AAA kk (tt) = AAAA (Υ, AAA tt n n. (4) In the special case where for all (σσσ σσσ σ σ, red(uuu u uu and AA ii = [kkk for ii i ii,weshallwriteaaaa Υ n instead of AAAA (Υ, AAA n. at is, AAAA Υ n is the number of (σσσ σσσ σ σσ kk SS which avoid Υ. en we let θθ Υ kk (tt) =AAAA Υ tt n n. (42) 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 CC kk SS. Lemma 9. () If Υ = {( 2, 0 0), (2, 0 0)}, then AAAA Υ n = kk n2 for all and kkkk. (2) If Υ 2 = {( 2, 0), (2, 0)}, then AAAA Υ 2 n = n n for all and kkkk. (3) If Υ 3 = {( 2, 0 0), ( 2, 0), (2, 0 0), (2, 0)}, then AAAA Υ 3 n = kk n for all and kkkk. (4) If Υ 4 = {( 2, 0 ), ( 2, 0), (2, 0 ), (2, 0)}, then AAAA Υ 3 n = kkkkk for all and kkkk. Proof. For(),itiseasytoseethat(σσσ σσσ σ σσ kk SS avoids Υ if and only if all the signs are pairwise distinct. erefore, there are kk ways to pick the signs, and then you have n ways to arrange those signs and n ways to pick σσ. For(2),itiseasytoseethat(σσσ σσσ σ σσ kk SS avoids Υ 2 if and only if 0 ww ww kkkk. erefore, there are n ways to pick the signs in this case and there are n ways to pick σσ. For(3),itiseasytoseethat(σσσ σσσ σ σσ kk SS avoids Υ 2 if and only if 0 ww < <ww kkkk. erefore, there are kk ways to pick the signs in this case and there are n ways to pick σσ. For(4),itiseasytoseethat(σσσ σσσ σ σσ kk SS avoids Υ 2 if andonlyif0 ww = =ww kkkk. erefore, there are kk ways to pick the signs in this case and there are n ways to pick σσ. For Υ SSSSS, we let enwehavethefollowing. Υ 0 = σσσ σ jj σσσσ. (43)

8 8 ISRN Combinatorics eorem 0. Let Υ SS jj be any set of permutations of SS jj. en if Γ=Υ 0 {( 2, 0), (2, 0)}, for all kkkk. θθ Γ kk (tt) = φφ Υ (tt) kk, (44) Proof. Itiseasytoseethatif(σσσ σσσ σ σσ kk SS and (σσσ σσσ avoids {( 2, 0), (2, 0)}, thenitmustbethecasethat 0 uu uu 2 uu kkkk. So suppose that wwww aa aa 2 (kk k kk aa kk where aa + +aa kk =. en clearly the number of σσσσσ such that (σσσ σσσ avoids Υ 0 is AAAA Υ aa,,aa aa AAAA Υ aa kk. (45) kk at is, the binomial coefficient allows us to choose the elements CC ii of {,, n thatcorrespondtotheconstant segment ii aa in ww. en we only have to arrange the elements of CC ii sothatitavoidsυ which can be done in AAAA Υ aa ways. usitfollowsthat AAAA Γ n = aa + +aa kk = aa ii 0 or, equivalently, that which implies (44). AAAA Γ n n aa,,aa kk AAAA Υ aa AAAA Υ aa kk, = aa + +aa kk = aa ii 0 kk We immediately have the following corollary. (46) AAAA Υ aa ii aa ii!, (47) Corollary. For any kkkk, the number of elements of CC kk SS which avoid Γ = {( 2, 0 0), ( 2, 0), (2, 0)} or Γ 2 = {(2, 0 0), ( 2, 0), (2, 0)} is kk. Proof. Let Υ = { 2} and Υ 2 = {2 }. en clearly, AAAA Υ ii = for ii i ii i so that φφ Υ ii (ttt t tttt for ii i ii i. But then by eorem 0, θθ Γ ii n (ttt t ttttt ) kk =ee kkkk so that AAAA Γ ii n =kk for ii i ii i. We can derive theorems analogous to eorem 0 for the other sign conditions in Lemma 9. at is, suppose that Υ is anysetofpatternsfromss jj. en we let Υ ii = τττ τ τ τ jjjj ττττ, Υ dd = τττ ττ) ττττττττττ jj, (48) where DD jj is the set of all permutations of {0,,, jjjjj. en we have the following. eorem 2. For any Υ SS jj, let Γ =Υ 0 {( 2, 0 ), ( 2, 0), (2,0 ), (2, 0)}, Γ 2 =Υ ii {( 2, 0), ( 2, 0 0), (2, 0), (2,0 0)}, Γ 3 =Υ dd {( 2, 0 0), (2, 0 0)}. en, for all kkkk, (49) () AAAA Γ n = kkkkkkυ for all, (2) AAAA Γ 2 n = kk AAAAΥ for all, (3) AAAA Γ 3 n = kk n Υ for all. Proof. For (), note that for (σσσ σσσ σ σσ kk SS to avoid {( 2, 0 ), ( 2, 0), (2, 0 ), (2, 0)}, we must have wwwww for some ii i iii i i ii i ii. en (σσσ σσ ) CC kk SS avoids Υ 0 ifandonlyifσσavoids Υ. us AAAA Γ n = kkkkkkυ. For (2), note that for (σσσ σσσ σ σσ kk SS to avoid {( 2, 0), ( 2, 0 0), (2, 0), (2, 0 0)}, we must have ww w ww ww where ww < < ww. en for any of the kk strictly increasing words ww w www w w ww w ww, (σσσ σσσ σ σσ kk SS avoids Υ ii if and only if σσ avoids Υ. us AAAA Γ 2 n = kk AAAAΥ. For (3), note that, for (σσσ σσσ σ σσ kk SS to avoid {( 2, 0 0), (2, 0 0)}, we must have wwwww ww where the letters of ww are pairwise distinct. en for any of the kk n words ww w www w w ww w ww which have pairwise distinct letters, (σσσ σσσ σ σσ kk SS avoids Υ dd ifandonlyifσσ avoids Υ. us AAAA Γ 3 n = kk n Υ. Next we will prove two more results about AAAA Υ n for other sets of patterns Υ that contain {( 2, 0), (2, 0)}. eorem 3. Let en Υ = {( 2,0 ), ( 2, 0), (2, 0)}, Υ 2 = {( 2,0 ), ( 2, 0), (2, 0), (2, 0 0)}. (50) () AAAA Υ n = aa + +aa kk = aa! aa kk! for all and kkkk, aa ii 0 (2) AAAA Υ n = n kkkk for all and kkkk. Proof. For (), note that as in the proof of eorem 0, if (σσσ σσσ σ σσ kk SS and (σσσ σσσ avoids {( 2, 0), (2, 0)}, thenitmustbethecasethat0 uu uu 2 uu kkkk. Now suppose that ww w w aa aa 2 (kk k kk aa kk where aa + +aa kk =. Assume that (σσσ σσσ also avoids ( 2, 0 ) and CC ii is the set of elements of σσ that correspond to the signs (ii i ii aa ii in ww. en it follows that all the elements of CC must be larger than all the elements of CC 2, all the elements of CC 2 must be larger than all the elements of CC 3,andsoforth. us CC consists of the aa largest elements of {,, 2 consists of next aa 2 largest elements of {,, n, and so forth. en we can arrange the elements of CC ii in any order in the positions corresponding to (ii i ii aa ii in ww to produce a (σσσ σσσ which avoids Υ. Hence there are aa! aa kk! elements of the form (σσσ σσσ which avoid Υ in CC kk SS. erefore () immediately follows. For (2), observe that if in addition such (σσσ σσσ also avoids (2, 0 0), then we must place the elements of CC ii in increasing order. Hence AAAA Υ 2 is the number of solutions of n aa + +aa kk = with aa ii 0 whichiswellknowntobe. n kkkk

9 ISRN Combinatorics 9 eorem 4. Let Υ = {( 2, 0 ), ( 2, 0), (2, 0 0)}. en for, AAAA Υ kkkk n =kkk AAAA Υ =, AAAA Υ = 2n jjjj (kk) jj jj for kk k kk (5) where (kkkk 0 =and (kkkk = kkkkk k kk k kkk k kk k kk for. Proof. We shall classify the elements (σσσ σσσ σ σσ kk SS which avoid Υ by the number of elements ss which follow in σσ. Now if ssssso that σσ ends in, thefactthat(σσσ σσσ avoids both ( 2, 0 ) and ( 2, 0) means that all the signs must be the same. at is, ww mustbeoftheformii for some ii i. But since (σσσ σσσ must also avoid (2, 0 0), σσσσ σ σ σσ must be the identity. us there are kk choices for such (σσσ σσσ. Now suppose there are ss elements following in σσ. en again the fact that (σσσ σσσ avoids both ( 2, 0 ) and ( 2, 0) means that all the signs in ww corresponding to σσ σσ n (where σσ n =)mustbethesame,saythatsign is ii. e fact that (σσσ σσσ avoids (2, 0 0) means that (i) σσ σσ n must be in increasing order and (ii) all the signs corresponding to σσ n σσ must be different from ii. But then it follows from the fact that (σσσ σσσ avoids both ( 2, 0 ) and ( 2, 0) that all the elements in σσ σσ n must be greater than all the elements in σσ n σσ. Hence there are kkkkkk Υ sssssss such elements if kk k k andtherearenosuch elements if kkkk. us from (ii) it follows that AAAA Υ =since ( 2 n n ) istheonlyelementofcc kk SS which avoids Υ. For kkkk,wehave AAAA Υ n =kkk ssss Inthecasekkkk,(52) becomes AAAA Υ =2+ ssss andinthecasekkkk,(52) becomes AAAA Υ =3+ ssss In general, assuming that AAAA Υ kkkk n =kkk kkkkkk Υ sssssss. (52) 2 = 2n (53) 3 (2ss) =3+(3 2). (54) 2 jjjj (kk) jj, jj (55) it follows that AAAA Υ n =kkkkk ssss =kkkkk ssss (kkkk) AAAA Υ sssss kkkk (kkkk)kkk =kkkkk(kkkk) kk kkkk + jjjj ssss jjjj (kkkk)(kk) jj ss jj ssss =kkkkk(kkkk) 2 2 kk + (kkkk) jj jj. jjjj (kk) jj ss jj (56) We next consider simultaneous avoidance of the patterns ( 2, 0) and ( 2, 0 ). eorem 5. Let Υ = {( 2, 0), ( 2, 0 )}, AA Υ kk (ttt t AAAA Υ ntt and CCCCCC C C n. en AA Υ kk (tt) = +CC(tt) (kkkk) CC (tt). (57) Proof. Fix kk and suppose that (σσσ σσσ is an element of CC kk SS which avoids both ( 2, 0) and ( 2, 0 ).Nowifwwwww is constant, then clearly σσ can be arbitrary so that there are n such elements. Next assume that ww isnotconstantsothereisanssss such that ww w ww aa iiaa 2 2 iiaa ss ss where aa rr for rr r rrrrrr and ii rr ii rrrr for rr r rrrrrr r r.weclaimthatσσ σσ aa must be the aa largest elements of {,, n. If not, then let nbe the largest element which is not in σσ σσ aa. us rr r rr andtheremustbeatleastoneσσ tt with tt t tt such that σσ tt < n. Now it caot be σσ aa + = n since otherwise (σσ tt σσ aa +,ii ii 2 ) would be an occurrence of either ( 2, 0) or ( 2, 0 ).Henceitmustbethecasethat σσ aa + <nand σσ pp =nfor some ppppp +. But then no matter what color we choose for ww pp, either (σσ tt σσ pp,ii ww pp ) or (σσ aa +σσ pp,ii 2 ww pp ) would be an occurrence of either ( 2, 0) or ( 2, 0 ).Wecancontinuethisreasoningtoshowthatfor any pppppppp, the elements of σσ corresponding to the block ii aa pp pp in ww must be strictly larger than the elements of σσ corresponding to the block ii aa qq qq in ww. is given, it follows that we can arrange the elements of σσ corresponding to a block ii aa pp pp in ww inanywaythatwewantandwewillalwaysproducea pair (σσσ σσσ that avoids both ( 2, 0) and ( 2, 0 ). us for such a ww, wehavekkkkk k kk ssss ways to choose the colors

10 0 ISRN Combinatorics ii,,ii ss and aa!aa 2! aa kk! ways to choose the permutation σσ. It follows that AAAA Υ n = kkkkk k ssss aa + +aa ss = aa ii >0 kk(kkkk) ssss aa!aa 2! aa ss!, (58) which is equivalent to (57). Next we want to consider the problem of avoiding (Υ, BBB where Υ = {( 2, 0 0), ( 2, 0 )} and BB B BBBBBB BBB B BBB. If σσ σ σσ σσ SS,thenwesaythatσσ jj is a le -to-right minimum of σσ if σσ ii >σσ jj for all iandwelet lrmin(σσσ denote the number of le -to-right minima of σσ. Let BB n denote the set of all (σσσ σσσ σ σσ kk SS which avoid (Υ, BBB. en we let tt BB kk (xxx xx) = xx lrmin(σσσ. n (59) (σσσσσ) BB n en we have the following theorem. eorem 6. For Υ = {( 2, 0 0), ( 2, 0 )} and BB B ([kkkk kkk k kkk,onehas kkkkkkkkkkk BB kk (xxx xx) = (kk k kkkk. (60) Proof. Let BB n (xxx x x (σσσσσσσσσn xx lrmin(σσσ. If (σσσ σσσ contains the element colored by the color aa in position where aaa {0,, kk k kk, then if (σσσ σσσ istoavoid(υ, BBB, it must be the casethateveryelementtotherightofmustbecoloredwith a color from {0,, kkkkk and these elements can be arranged in any of ( n ways. Moreover, it immediately follows that no instance of an occurrence of (Υ, BBB exists where the rst elementistothele ofandthesecondistotherightof. usitfollowsthat(σσσ σσσ avoids (Υ, BBB ifandonlyifthereis no occurrence of (Υ, BBB in (σσ σσ ii,ww ww ii ). us it follows that xx lrmin (σσσ =xx(kkkk) n (σσσσσ) BB n,σσ = (n)! ii BB i (xx) = n (kkkk)n BB i (xx). iii (6) Here the binomial coefficient is responsible for choosing the elements to the le of and placing the remaining elements to therightofandthefactorofxx accounts for the fact that is always a le -to-right minimum in σσ and none of the elements following can be a le -to-right minimum of σσ. is shows that BB i (xx) BB n (xx) = kkkkkkk (kkkk) n. (62) iii Multiplying both parts of the equation above by tt /n and summing over all, we have BB n (xx) tt n = kkkk kk tt BB i (xx) (kkkk) n, (63) mmmm iii so that, for all kkkk, It follows that BB kk (ttt tt) = kkkk BB kk (ttt tt) (kkkk) tt. (64) ln BB kkkk kk (xxx xx) = (kkkk) tt. (65) Integrating and using the fact that BB kk (xxx xx x x,weseethat and hence, BB kk (xxx xx) = kkkk ln ( (kkkk) tt), (66) kkkk Using the fact that we see that In particular, kkkkkkkkkkk BB kk (xxx xx) = (kkkk) tt. (67) ( zz) aa = aa (aaaa) (aaaaaaa) zz n, (68) BB n (xx) = kkkk kkkk kkkk kkkk + kkkk kkkk + kkkk) = (kkkk)(kkkk k (kkkk)) (kkkk k ()(kkkk)). ({( 2,0 0),( 2,0 )},([kkkkkkkkkkkk AAAA n =kk(2kk k k)(3kkkk) ( n ()). (69) (70) Wenotethat,byoursymmetrymapswhereweapplythe identity map to σσ and the composition of complement and reverse to ww and AA,onecanshowthat ({( 2,0 0),( 2,0 )},([kkkkkkkkkkkk AAAA n ({( 2,kkkk kkkkkkkk kkkkkk kkkkkkkkkkkkkkkkkk = AAAA n. It is easy to see that our de nitions imply that ({( 2,0 0),( 2,0 )},({0},[kkkkk AAAA n ({( 2,kkkk kkkkkkkk kkkkkk kkkkkkkkkkkkkkkkkk = AAAA n, (7) (72)

11 ISRN Combinatorics so that ({( 2,0 0),( 2,0 )},({0},[kkkkk AAAA n =kk(2kk k k)(3kkkk) ( n ()). (73) Multiplying both parts of the equation above by tt /n and summing over all, we have CC n n tt n = kkkkk CC n n tt n Next we consider the case where < jj j jj, CC jj = ({0,, jj j jjj jjj j jj j j jj j jjj, and Υ = {( 2, 0 0), ( 2, 0 )}. eorem 7. Let CC jj = ({0,, jj j jjj jjj j jj j j jj j jjj where kkkkand <jjjjj, and Υ = {( 2, 0 0), ( 2, 0 )}. en if kkkkkkkkkk, θθ (Υ, jjjjjjjjjjjjj CC jj ) (jj j jjjj kk (tt) = kkkkktt (kkkkkkkk. (74) If kkkkkkkkkk, then kkkkkkkkand Proof. Let CCCCCC CCC CCC C CCCC (Υ, CC jj ) n θθ (Υ, CC jj ) 2jjjj (tt) = eejjjjjjjjjjjjjjjjj jjjjtt. (75) so that CC kkkkk (tt) =CCn n tt n =θθ(υ,cc jj ) kk (tt). (76) Now suppose that (σσσ σσσ σ σσ kk SS contains the element colored by a color from {jjj j j jj j jj. en there are no restrictions for placing this element, thus giving ( n possibilities. On the other hand, if (σσσ σσσ contains the element colored by the color aa in position where aa a aaa a a aaaaa, then if (σσσ σσσ is to avoid (Υ, CC jj ),itmustbethecasethat every element to the right of must be colored with a color from {0,, jj j jj and these elements can be arranged in any of ( n ways. Moreover, it immediately follows that no instance of an occurrence of (Υ, CC jj ) exists where the rst elementistothele ofandthesecondistotherightof. en it follows that (σσσ σσσ avoids (Υ, CC jj ) if and only if there is no occurrence of (Υ, CC jj ) in (σσ σσ ii,ww ww ii ).Sointhe casewhereiscoloredbyacolorfrom{0,, jj j jj and is in position, we have (jj j jj n ( n ( ii ) CCCCCC CCC CCC C CCC CCC C ) n CCCCCC CCC CCCCCCC possibilities where the binomial coefficient is responsible for choosing the elements to the le of and placing the remaining elements to the right of. us CC n n = kkkkk () CC n n CC iii iii ii + jjjjj jjjj n. iii (77) + kkkkk n n tt n +jjjj mmmm tt so that, for all kkkkand <jjjjj, CC kkkkk (tt) = kkkkk CC kkkkk (tt) It follows that CC iii iii ii jjjj n, iii (78) + kkkkktttt CC kkkkk (tt) (79) kkkkk (tt) +jj jjjjtt. dd dddd ln CC kkkkk (tt) = CC kkkkk (tt) CC kkkkk (tt) = kkkkktt + jj kkkkk tt jjjj tt. (80) Now suppose that kkkkkkkkkk. en integrating (80) and using the fact that CCCCC CCC CCC C C and that (ddddddddddddd d ddddd d ln( yyyyyy y yyy y yyyyyy y yyyyy yy y yyyyy,weseethat ln CC kkkkk (tt) = ln kkkkk tt which implies that CC kkkkk (tt) = + jj kkkkk jjjj ln jjjj ln kkkkk tt jjjjjjjjjjjjj jjjjtt kkkkktt kkkkktt (8) (82) and proves (74). Now if kkkkkkkkkk. en kk k kkk k k and (80) becomes dd dddd ln CC 2jjjjjjj (tt) = CC 2jjjjjjj (tt) CC 2jjjjjjj (tt) = jjjj jjjjtt + jj jjjj tt 2. (83)

12 2 ISRN Combinatorics Integrating (83) and using the fact that CCCCC CCC C CC CCC C C, we see that us ln CC 2jjjjjjj (tt) = ln jjjj tt which proves (75). + jj jjjj jjjj tt = ln jjjj tt + jjjj jjjj tt. (84) CC 2jjjjjjj (tt) = eejjjjjjjjjjjjjjjjj jjjj tt, (85) Finally, we end this section by considering the elements of CC kk SS which avoid both ( 2, 0 0) and ( 2, 0 ).Inthis case,weonlyhavearesultforthecasewhenkkkk. eorem 8. enumberofpermutationsincc 2 SS simultaneously avoiding ( 2, 0 0) and ( 2, 0 ) isgivenbythe ( n -th Catalan number CC = (/( n n 2. Proof. We prove the statement by establishing a bijection between the objects in question of length and the Dyck paths of semilength ( n knowntobecountedbythe Catalan numbers (a Dyck path of semi-length is a lattice path from (0, 0) to (2n with steps (, ) and (, ) that never goes below the xx-axis; Figure 4). Suppose (σσσ σσσ σ σσ 2 SS. Note that σσ must avoid the pattern 2 3, as in an occurrence of such a pattern in σσ, there are two letters of the same color leading to an occurrence of ( 2,0 0). us the structure of σσ, as it is well known, is two decreasing sequences shuffled. Subdivide σσ into the so-called reverse irreducible components. A reverse irreducible component is a factor FF of σσ of minimal length such that everything to the le (resp., right) of FF is greater (resp., smaller) than any element of FF. For example, the subdivision of σσ σ σσσσσσσ is σσ σ σσσσσσσσσσ. 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 σσ 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 ( 2 3)-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 xx (except possibly the last element), one has an element yy greater than xx to the right of it inside the same block and vice versa, from which we conclude that xx must receive color, whereas yy must receive color 0 (otherwise a prohibited pattern will occur). We are ready to describe our bijection. For a given (σσσ σσσ σ σσ 2 SS, consider a matrix representation of σσ, that is, an integer grid with the opposite corners in (0,0) and (n n and with a dot in position (iii ii ii (/2)) for iii. We will give a description of a path PP (corresponding to (σσσ σσσ) from (0, n to ( n involving only steps (0, ) and (, 0) that never goes above the line yy y yyyyyyyy. Clearly, PP canbetransformedtoadyckpathoflength(n by taking a mirror image with respect to the line yy y yyyyyyyy, rotating 45 degrees counterclockwise and making a parallel shi. To build PP, set ii ii i and jjjjjjjj, and do the following steps, letting PP begin at (iii iii. Clearly, each reverse irreducible block of σσ de nes a square on the grid which is a matrix representation of the block. We call the reverse irreducible block of σσ with the le -top corner at (iii iii the current block. Step. If the current block is not a singleton, go to Step 2. If the color of the element with xx-coordinate equal to ii i iii is 0 (resp., ) travel around the current block counterclockwise (resp., clockwise) to get to the point (ii. Note that PP touches the line yyyyyyyyyyyif the color is. Set and jjjjjjjj, and proceed with Step 3. Step 2. In Step 2 we follow a standard bijection between ( 2 3)-avoidingpermutationsandDyckpathsthatcanbe described as follows. Let (kkk kk be the point of the current block opposite to (iii iii. Start going down from (iii iii until the yy-coordinate of the current node gets /2 less than the yycoordinate of the dot with xx-coordinate equal ii i iii. Start moving horizontally to the right and go as long as possible makingsurethatnoneofthedotsarebelowthepartofpp constructed so far and iand jjjj. Suppose (ii,jj ) is the last point the procedure above can be done (i.e., we were traveling on the line yyyyy and either ii =kkand jj =lor thereisadotwithxx-coordinate ii +(/2) having yy-coordinate less than jj ). If ii ii ii and jj jj j, proceed with Step 3; otherwise set ii ii ii and jj jj jj andgotostep2.notethatin Step 2 PP never touches the line yyyyyyyyyyy. Step 3. If jjjj, make as many as it takes horizontal steps to gettothepoint( n and terminate; otherwise go to Step. Returning to our example, σσ σ σσ σ σ σ σ σ σσ 0 0 0), we have given the matrix diagram in Figure. We have outlined the reverse irreducible blocks in Figure 2. We start our path at (0, 8). We travel down until we reach (0, 6), whenweare/2 less than the y- coordinate of our rst point (/2, 6(/2)). 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 0. en we continue to the nal reverse irreducible block and nish our path, given in Figure Pattern Matching Results for Pairs of Length 2 Goulden and Jackson [8] proved the following theorem.

13 ISRN Combinatorics FIGURE : Matrix representation of σσ σ σσ σ σ σ σ σ σσ 0 0 0). FIGURE 4: e Dyck path corresponding to σσ σ σσ σ σ σ 3 2, 0 0 0).Notethatthepathonlytouchesthe line yyyyyyyyyyya er a singleton colored. eorem 9 (see [8]). If τττττττ τ, where jjjj, then 0 tt n σσ σσσ ττ-mch (σσ) =0 = tt jjjj (jjjjjj tt jjjjjj (jjjj j jjj. (86) 0 Later Mendes and Remmel [9] re ned this result by proving the following. FIGURE 2: Matrix representation of σσ σ σσ σ σ σ σ σ σσ 0 0 0)with reverse irreducible blocks outlined. 0 eorem 20. When τττττττ τ, tt n σσσσσ ττ-mch (σσ)=0 = xx des(σσσ + tt /n n ( ) ii R xx ii, (87) where R is the number of rearrangements of ii zeroes and n ones such that jj zeroes never appear consecutively. 0 e reason that eorem 20 is a re nement of eorem 9 is that Mendes and Remmel [9]provedthat 0 0 ( ) ii R = ( ) if jj divides n ( ) if jj divides n 0 otherwise, (88) so that setting xxxxin (87) yields (86). e main goal of this section is to generalize eorem 20. at is, let ττ (jjj = jjjjj j jj j j where jjjj. en we shall consider three sets of patterns for CC kk SS : FIGURE 3: e Dyck path corresponding to σσ σ ( , 0 0 0)added to Figure 2. (a) (ττ (jjj,0 jj ), (b) (ττ (jjj,jjjj jjjjjjjj,

14 4 ISRN Combinatorics (c) Υ which equals the set of all (ττ (jjj, www such that www {0,, kkkkk jj, ww is weakly decreasing, and red(www w ww. In [], itaev et al. de ned three different types of descents. at is, for elements (σσσ σσσ σ σσ kk SS, they de ned Des ((σσσ σσ)) = ii iii ii >σσ,ww ii ww, des ((σσσ σσ)) = Des ((σσσ σσ)), WDes ((σσσ σσ)) = ii iii ii >σσ,ww ii =ww, wdes ((σσσ σσ)) = WDes ((σσσ σσ)), SDes ((σσσ σσ)) = ii iii ii >σσ,ww ii >ww, sdes ((σσσ σσ)) = SDes ((σσσ σσ)). (89) We shall refer to Des((σσσ σσσσ as the descent set of (σσσ σσσ, WDes((σσσ σσσσ as the weak descent set of (σσσ σσσ, and SDes((σσσ σσσσ as the strict descent set of (σσσ σσσ. eorem 2. Let ττ (jjj =jjjj jwhere jjjj. () If uuuu jj, then tt n (σσσσσ) CC kk SS (τττττ)-mch ((σσσσσ))=0 tt wdes ((σσσσσσσ xx = + n kkkk ( ) ii R xx ii (2) If uuuuuuu uuuuuu u, then tt n (σσσσσ) CC kk SS (τττττ)-mch ((σσσσσ))=0 tt sdes ((σσσσσσσ xx = + n kk ( ) ii R xx ii.. (90) (9) (3) If Υ equals the set of all (ττ (jjj, www such that ww w {0,, kk k kk jj, ww is weakly decreasing, and red(www w ww, then tt n (σσσσσ) CC kk SS Υ-mch ((σσσσσ))=0 tt xx des((σσσσσσσ = + n kkkkkkk ( ) ii R xx ii. (92) Proof. Our proof is an adaptation of the proof that Mendes and Remmel [9]usedtoprove eorem 20. e basic idea is to show that(90), (9), and (92) arise by applying appropriate ring homomorphisms, de ned on the ring Λ of symmetric functions over in nitely many variables xx,xx 2,, to a simple symmetric function identity. FIGURE 5: A brick tabloid of shape ( 2) and type (,, 2, 3, 5). e th elementary symmetric function, ee,andtheth homogenous symmetric function, h,arede nedbythe generating functions Clearly, EE (tt) =ee tt =+xx ii tt, HH (tt) =h tt = xx ii tt. HH (tt) = ii ii (93) EE ( tt). (94) Let λλλλλλ,,λλ l ) be an integer partition; that is, λλ is a nite sequence of weakly increasing positive integers. Let l(λλλ λ λ denote the number of parts of λλ.ifthesumofthese integers is,wesaythatλλisapartition of and write λλλλλ. For any partition λλ λ λλλ,,λλ l ), let ee λλ =ee λλ ee λλl. ewellknown fundamental theorem of symmetric functions says that {ee λλ is a partition} is a basis for Λ or that {ee 0,ee, } is an algebraically independent set of generators for Λ. Similarly, if we de ne h λλ =h λλ h λλl, then {h λλ is a partition} is also a basis for Λ. Since the elementary symmetric functions ee λλ and the homogeneous symmetric functions h λλ are both bases for Λ, it makes sense to talk about the coefficient of the homogeneous symmetric functions when written in terms of the elementary symmetric function basis. ese coefficients have been shown to equal the sizes of certain sets of combinatorial objects up to a sign. A brick tabloid of shape (n and type λλ λ λλλ,,λλ kk ) isa llingofarowofsquares of cells with bricks of lengths λλ,,λλ kk such that bricks do not overlap. One brick tabloid of shape (2) and type (,, 2, 3, 5) is displayed In Figure 5. Let B λλλλλ denote the set of all λλ-brick tabloids of shape (n and let BB λλλλλ = B λλλλλ. rough simple recursions stemming from (94), Egecio glu and Remmel [0] proved that h =( ) BB λλλλλ ee λλ. (95) λλλλλ We now consider three different ring homomorphisms which map Λ to Q[xxx where Q is the eld of rational numbers. Since ee 0,ee, is an algebraically independent set of generators for Λ, we can de ne a ring homomorphism by simply specifying its value on ee for all. For, de ne Γ ee = ( ) n Γ 2 ee = ( ) n Γ 3 ee = ( ) n kkkk( ) n R xx ii, kk ( ) n R xx ii, kkkkkkk ( ) n R xx ii. (96)

15 ISRN Combinatorics 5 ur rst goal is to prove the following facts. (A) If uuuu jj, then n n h = (σσσσσ) CC kk SS (τττττ)-mch ((σσσσσ))=0 (B) If uuuuuuu uuuuuu u, then n n 2 h = (σσσσσ) CC kk SS (τττττ)-mch((σσσσσ))=0 xx wdes((σσσσσσσ. xx sdes((σσσσσσσ. (97) (98) (C) If Υ consists of the set of all (ττ (jjj, www such ww w {0,, kkkkk jj, ww is weakly decreasing, and red(www w ww, then n n 3 h = (σσσσσ) CC kk SS Υ-mch ((σσσσσ))=0 xx des((σσσσσσσ. (99) Let fffff fff f ff, fffff fff f kk, and fffff fff f kkkkkkk. en for ii i iii ii ii, n n ii h = n ( ) BB λλλλλ Γ ii ee λλ λλλλλ = n λλλλλ = ( ) l(λλλ BB λλλλλ mmmm ( ) λλ mm λλ mm! ffiii ii mm ( ) λλmm ii R λλmm,iiixx ii λλλλλbbbbb,,bb l(λλλ B λλλλλ l(λλλ ) l(λλλ bb,,bb l(λλλ ff iii ii mm ( ) bbmm ii R bbmm,iiixx ii. mmmm (00) We wish to give a combinatorial interpretation of righthand side of (00). First we select a partition λλλλλand then we select a brick tabloid BB B BBB,,bb l(λλλ ) B λλλλλ. Next we interpret the multinomial coefficient bb,,bb l(λλλ as the number of ways to pick sets SS,,SS l(λλλ of sizes bb,,bb l(λλλ, respectively, which partition {,, n. We then place these elements in SS mm inthebottomofthecellsofthebrickbb mm in decreasing order reading from le to right. Next if ii i i, then fffff ff mm )=kkwhich we interpret as the ways of picking an element ii i iii i i ii i ii and placing ii at the top of each cell of bb mm. If ii i i, then fffff ff mm ) = bb kk mm which we interpret as the ways of picking a strictly decreasing sequence ss s ss > > ss bbmm from {0,, kk k kk and placing ss jj on top of the jjthcellofbb mm. If, then fffff ff mm )= kkkkk mm bb mm which we interpret as the ways of picking a weakly decreasing sequence ss s ss ss bbii from {0,, kk k kk and FIGURE 6: Filled labeled brick tabloids in FF 2,ii. placing ss jj on top of the jjth cell of bb ii. Finally, we interpret the term ( ( ) bb mm ii R bbmm,iiixx ii ) as follows. First, we pick a sequence ttttt tt bbmm which consists of ii ii s and bb mm i s for some ii such that there is no consecutive sequence of jj j j jjjjj in tt. en if tt ss =xx, we label the ssth cell of bb mm with xx, and if tt ss = 0, we label the ssth cell of bb mm with. Finally, we label the last cell of bb mm with. usforanygiven ii, we will have bb mm iicells labeled with which accounts for the sign ( ) bb ii. e only term in (00)thatwehavenot accounted for is the term ( ) l(λλλ 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 F n denote the set of all lled labeled brick tabloids that arise in this way for interpreting n n ii (h ). us a CCCC n consists of a brick tabloid TT, a permutation σσσσσ, a sequence ww w www w w ww w ww, and a labeling LL ofthecellsoftt with elements from {xxx xx xxx such that the following conditions hold. (a) σσ is strictly decreasing in each brick. (b) () If, then ww is constant in each brick. (2) If ii i i, then ww is strictly decreasing in each brick. (3) If ii i i, then ww 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 xx or andthereisnoconsecutivestringofcells of length jjjjlabeled xx within a brick. We then de ne the weight wwwwww of CC to be the product of all the xx labels in LL and the sign sgn(ccc of CC tobetheproduct of all the labels in LL. For example, if n, kkkk, jjjj, and TT T TTT TT TT, then Figure 6 pictures a lled labeled brick tabloid CCCC 2, at the top, a a lled labeled brick tabloid CCCC 2,2 in the middle, and a lled labeled brick tabloid CCCC 2,3 atthebottom.in each case, wwwwww w ww 6 and sgn(ccc C CC. Next we de ne sign-reversing, weight-preserving involutions II ii on FF n. o de ne II ii (CCC, wescanthecellsofcc C

16 6 ISRN Combinatorics FIGURE 7: II ii (CCC for the CCs in Figure 6. (TTT TTT TTT TTT from le to right looking for the le most cell tt such that either (i) tt is labeled with, (ii) tt is at the end of a brick bb jj and the brick bb jjjj immediately following bb jj has the property that σσ is strictly decreasing in all the cells corresponding to bb jj and bb jjjj and ww isconstantonallthecells corresponding to bb jj and bb jjjj if, ww is strictly decreasing on all the cells corresponding to bb jj and bb jjjj if, ww is weakly decreasing on all the cells corresponding to bb jj and bb jjjj if. In case (i), II ii (CCC C CCC,σσ,ww,LL ) where TT is the result of replacing the brick bb in TT containing tt by two bricks bb and bb where bb contains the cell tt plus all the cells in bb to the le of tt and bb contains all the cells of bb totherightof tt, σσ σ σσ, ww w ww, and LL is the labeling that results from LL by changing the label of cell tt from to. In case (ii), II ii (CCC C CCC,σσ,rr,LL ) where TT is the result of replacing the bricks bb jj and bb jjjj in TT by a single brick bbb bb b bb, wwwww, and LL is the labeling that results from LL by changing the label of cell tt 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 caot create a run of jjjjconsecutive cells which are labeled xx. If neither case (i) or case (ii) applies, then we let II ii (CCC C CC. For example, if we consider the CC s in F 2,ii pictured in Figure 6, then their corresponding images II ii (CCC are pictured in Figure 7. It is easy to see that II ii is a weight-preserving, signreversing involution, and hence II ii shows that n n ii h = sgn (CC) ww (CC). CCCC n,ii ii (CC)=CC (0) us we must examine the xed points CC C CCCC CCC CCC CCC of II ii.firsttherecanbeno labels in LL so that sgn(ccc C C. Moreover, if bb jj and bb jjjj are two consecutive bricks in TT and tt is the last cell of bb jj, then () If,itcaotbethecasethatσσ tt >σσ tttt and ww tt = ww tttt since otherwise we could combine bb jj and bb jjjj, (2) if, it caot be the case that σσ tt >σσ tttt and ww tt > ww tttt since otherwise we could combine bb jj and bb jjjj, and (3) if, it caot be the case that σσ tt >σσ tttt and ww tt ww tttt since otherwise we could combine bb jj and bb jjjj. us if ss is the last cell of a brick, then ssswdes(σσσ σσσ if, ssssdes(σσσ σσσ if, and sssdes(σσσ σσσ if. However, the label on a cell tt ofanybrickbb jj must be xx if tt is not the last cellofthebrickandmustbeiftt is the last cell of a brick. Moreover, if tt is not the last cell of a brick, then σσ tt >σσ tttt and ww tt =ww tttt if, ww tt >ww tttt if, ww tt ww tttt if. us if tt isnotthelastcellofabrick,thentttwdes(σσσ σσσ if, tttsdes(σσσ σσσ if, and tttdes(σσσ σσσ if. It follows that sgn(ccc C C and wwwwww w wdes((σσσ σσσσ if, wwwwww w sdes((σσσ σσσσ if ii i i, and wwwwww w wdes((σσσ σσσσ if. en n n h = n n 2 h = n n 3 h = (σσσσσ) CC kk SS (τττττ)-mch((σσσσσ))=0 (σσσσσ) CC kk SS (τττττ)-mch((σσσσσ))=0 (σσσσσ) CC kk SS (τττττ)-mch((σσσσσ))=0 xx wdes((σσσσσσσ xx sdes((σσσσσσσ if ii i ii if ii i ii xx des((σσσσσσσ if ii i ii (02) erefore (A), (B), and (C) hold. We can now prove (90) by applying Γ to (94). at is, we have uuuu jj and ττ τ ττ ττττττ τ,andthen tt n (σσσσσ) CC kk SS (τττττ)-mch(=)0 =Γ xx wdes((σσσσσσσ h tt = Γ ee ( tt) = + ( tt) kk( ) /n n ( ) n R xx ii = + kkkk /n n ( ) n R xx ii. (03) Similarly, (9) can be proved by applying Γ 2 to (94) and (92) can be proved by applying Γ 3 to (94). In fact, we can calculate RR in several cases. For instance, routine combinatorial techniques are able to show that RR n is n ii whenever 0 ii i ii. When jjjj, we nd that RR n = i pppp n. (04) pppppppppppppp pppppp

17 ISRN Combinatorics 7 We can also nd a generating function for RR by considering the following argument. Take a rearrangement that is counted by RR. Now remove the nal number of the rearrangement. Either [0] Ö. Eğecioğlu and J. B. Remmel, Brick tabloids and the coection 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 RR, except that we have overcounted by those rearrangements counted by RR which end in jjjj zeroes. We can correct this by considering removing the last jjjj zeroes of the over-counted rearrangements, leaving those of length nwith izeros ending in a, of which there are RR. In case (b), we can count the number of such sequences by RR. us we obtain the following relationship: RR =RR RR +RR. (05) Next, we multiply by qq ii xx and sum over all, from which we obtain a generating function with a nite sum of small RR values in the numerator and a quotient of xxxxxxxxxx ii xx jjjj. Acknowledgment J. Remmel was partially supported by NSF Grant DMS and M. Riehl was supported by UWEC s Office of Research and Sponsored Programs. References [] S. Kitaev, A. Niedermaier, J. B. Remmel, and A. Riehl, Generalized pattern matching conditions for CC kk SS, 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, Aals 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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