Centrosymmetric words avoiding 3-letter permutation patterns

Transcription

1 Centrosymmetric words avoiding 3-letter permutation patterns Luca S. Ferrari Dipartimento di Matematica, Università di Bologna Piazza di Porta San Donato, Bologna, Italy Mathematics Subject Classification: 05A05, 05A5 Abstract A word is centrosymmetric if it is invariant under the reverse-complement map. In this paper, we give enumerative results on k-ary centrosymmetric words of length n avoiding a pattern of length 3 with no repeated letters. Introduction Asequenceofpositiveintegers... n (for instance, a permutation or a word) is said to avoid a pattern =... s if it does not contain any subsequence i i... i s which is order isomorphic to. In the last two decades, the study of pattern avoidance has interested several research fields in combinatorics: many results have been found on permutations, which is still the most advanced area, and a considerable work has been made on words since 998. Burstein [4] gave for the first time an explicit formula for the number of words avoiding asetofpermutationpatternsins 3 ; later, Burstein and Mansour [5] extended the results to patterns with repeated letters. Many improvement and generalizations have been made in the last years. In 006 Firro and Mansour [7] presented a new method, the scanning-element algorithm (that we exploit in Section 5) to obtain an easier proof of Burstein s results for the pattern 3; in the same year, Mansour [] applied the block decomposition method for the pattern 3. Brändén and Mansour [3] used finite automata theory to give a combinatorial explanation for the number of words avoiding patterns of length three; furthermore, Jelínek and Mansour [9] determined all the Wilf-equivalence classes of subsequence patterns of length at most six. In this paper, we focus on centrosymmetric words, namely, words that are invariant under the reverse-complement map, or, equivalently, whose corresponding tableaux via the Robinson-Schensted-Knuth algorithm are invariant under Schützenberger s involution (see [8], [0] and [3] for more details). We find the generating functions for the number

2 ! Figure. The words 8387 W 5,8, 6465 W 5,{,4,5,6,8} and 553 W 5,6 are all order isomorphic. Their common renormalization, as well as the representative of their equivalence class, is 443 W 5,4. of centrosymmetric words avoiding a pattern of length 3 with no repeated letters, and, for the pattern 3, even the explicit formulae. In the special case of centrosymmetric surjective words, we also reobtain some of the results described in Egge s paper [6]. Preliminaries Definition. A word of length n over a k-letter alphabet (also said a k-ary word of length n, oraword of type (n, k)) is any map : {,,...,n}!a k, where A k is a set of positive integers such that A k = k. We denote the set of all words of length n over A k by W n,ak ;inparticular,whena k = {,,...,k} we use the symbol W n,k instead of W n,{,,...,k}. A word in W n,ak will be represented either by the one-line notation... n, or by the usual graphical representation, as in figure (). Definition. Let W n be the set of all words of length n: W n = [ k W n,ak. We say that two words, W n are order isomorphic if, for every apple a, b apple n, we have a b () a b for any relation {<, =,>}.

3 Figure. The word W 8,7 is centrosymmetric while W 8,7 is not (figures above). Moreover, two words containing the same letters can be or not centrosymmetric depending on the alphabet: the word W 7,7 is centrosymmetric while W 7,8 is not (figures below). Roughly speaking, two words are order isomorphic if and only if, ignoring the row numbering, their graphical representations are identical, or, alternatively, one can be obtained from the other by erasing or adding a certain number of empty rows. Observe that each word in a given equivalence class of order isomorphic words contains the same number d of di erent letters. Hence, it is natural to give the following definition. Definition.3 Given a word W n containing exactly d di erent letters, we call renormalization of the only word ( ) W n,d which is order isomorphic to. This word will also be taken as the representative of the equivalence class that contains. Definition.4 A word W n,ak is said to contain a pattern W s,t, where s apple n and t apple k, if there exist indices apple i apple applei s apple n such that the subsequence i... i s is order isomorphic to =... s. Otherwise, is said to avoid. For example, the word 3536 avoids the pattern 3 but contains, since the subsequence 336 is order isomorphic to. We denote the set of words in W n,ak which avoid by the symbol W n,ak ( ). Definition.5 A word W n,k is centrosymmetric if, for every i =,,...,n, i + n+ i = k +. In this case, we say that i is the conjugate letter of n+ i. 3

4 Figure 3. The centrosymmetric word C 6,{,3,5,8,9,}. We denote the set of centrosymmetric words of length n over the alphabet {,,...,k} by the symbol C n,k. Needless to say, a word is centrosymmetric if and only if its graphical representation is symmetric with respect to the center of the grid (see figure ). Observe that there are no centrosymmetric words with n odd and k even, since in this case there is no type (n, k) thatsatisfies (n+)/ = k +. Inall theother cases, the number of k-ary centrosymmetric words of length n is C n,k = k b n c. We need the following generalization of centrosymmetry. Definition.6 Let A k = {a,a,...,a k } be a k-letter alphabet, with a <a < <a k, and let =... n W n,ak, with i = a j(i) for every i =,...,n. The word is said to be centrosymmetric if, for every i =,...,n, j(i)+j(n + i) =k +. We denote the set of centrosymmetric words in W n,ak by the symbol C n,ak. 3 Symmetries Definition 3. We call reverse and complement, respectively, the transformations 7 7 r : W n,k! W n,k c: W n,k! W n,k! r! c such that ( r ) i = n+ i and ( c ) i = k + i. The transformations r and c generate the group G = hr, ci = {id, r, c, rc}, whichis isomorphic to the group of symmetries of a rectangle D 4 (the dihedral group). The action of G on W n,k allows us to state that, for every W n,k, W s,t and ' G avoids () '( )avoids'( ). Hence, in order to find the cardinality of C n,k ( ) foreverypermutationpattern of length three it is su cient to study the sets C n,k (3) and C n,k (3). 4

5 We finally remark that centrosymmetric words are exactly those words which are invariant under the reverse-complement (rc) operation. Then, for every C n,k, avoids () avoids rc and hence C n,k ( ) = C n,k ( rc ) = C n,k (, rc ). () 4 Surjective words From definition (.), it is natural to define a surjective word of length n over a k-letter alphabet A k to be a surjective map : {,,...,n}!a k. In other terms, a surjective word is a word in W n,ak which contains all the letters of its alphabet. We denote the set of surjective words in W n,ak by S n,ak,andthesetofcentrosymmetricsurjectivewordsin W n,ak by CS n,ak. As usual, when A k = {,,...,k}, we use the symbols S n,k and CS n,k, respectively. Obviously, S n,ak 6=? if and only if n k. We also observe that the set S n of permutations of length n can be seen as the set S n,n of surjective words with length equal to the cardinality of the alphabet. Now, we focus on centrosymmetric words. Let c n,k be the number of centrosymmetric words of type (n, k) andcs n,k the number of centrosymmetric surjective words of the same type: c n,k = C n,k, cs n,k = CS n,k. Proposition 4. For m, h (i) c m,h = (ii) c m,h+ = the following relations hold: h i= (iii) c m+,h+ = h cs m,i i h i= h cs m,i + i h i=0 h i=0 h cs m+,i+ i h cs m,i+ i Proof. Let d be the number of distinct letters contained in a fixed word C n,k, and consider the graphical representation. In case (i) and (iii), is obtained from its renormalization ( ) CS n,d,whichhasthesametypeparitiesof,byaddingasuitable number (possibly zero) of pairs of empty rows, and relabelling the alphabet letters from tok. Topreservecentrosymmetry,everypairconsistsoftworowsinsertedatthesame distance from the center. In the remaining case (ii), the renormalization ( ) CS n,d of an even-odd centrosymmetric word (i.e. a word with even length over an odd cardinality alphabet) can either be even-even, or even-odd. In both cases, to obtain from ( )we have to add again some pairs of empty rows, while in the first case a further central empty row is needed. 5 ()

6 ! Figure 4. Two words in C 6,7 obtained from their renormalizations by adding the central row and another pair (figure above) or two pairs of rows (figure below).! By definition, a word contains a given pattern whenever its renormalization ( ) contains. This implies that relations () hold also for restricted words. Hence, to get the desired enumeration of C n,k ( ) it is su cient to obtain the number CS n,k ( ) of centrosymmetric surjective words avoiding, andthenuse(). Obviously, the same can be done for the related generating functions. Define and let c n,k = C n,k ( ) and cs n,k = CS n,k ( ), C (x, y) = n 0 cn,k x n y k be the ordinary generating function for the number of centrosymmetric words avoiding the pattern. Definingtheseries C ee(x, y) = n 0 n even C eo(x, y) = n 0 n even C oo(x, y) = n 0 n odd k even k odd k odd c n,k x n y k, CS ee(x, y) = n 0 n even c n,k x n y k, CS eo(x, y) = n 0 n even c n,k x n y k, CS oo(x, y) = n 0 n odd k even k odd k odd cs n,k x n y k, cs n,k x n y k, cs n,k x n y k, (3) 6

7 by inverse binomial transform (see e.g. [4]), relations () imply that the following relations hold:! Cee(x, y) = y CS y ee x, p, y! C eo(x, y) = y y CS ee x, C oo(x, y) = y p y + p y CS oo x, p y CS eo x,! y p. y! y p y y y, Hence, since C (x, y) =Cee(x, y)+ceo(x, y)+coo(x, y), we finally obtain that!! C (x, y) = y CS ee y x, p + p CS y y y eo x, p + y + p CS y y oo x, p y! y y. (4) 5 The pattern 3 In order to compute the cardinality of the set CS n,k (3) we exploit the scanningelement algorithm, presented in [7]. Denote by csn,k,i 3 the number of centrosymmetric surjective words avoiding 3 and starting with the letter i. Ofcourse,wehave cs 3 n,k = k i= cs 3 n,k,i. Observe that a word CS n,k starting with such that apple apple k containsthe pattern 3. In fact, by surjectivity there exists a letter x such that the subsequence x n is a 3-pattern. Hence, the preceding formula reduces to k csn,k 3 = csn,k,i. 3 i=d k e Now, our first goal is to find a recurrence relation for the terms csn,k,i 3. In order to do this, we consider the number csn,k,i,j 3 (with n 4andk 3) of centrosymmetric surjective words, that avoid 3, with first letter i and second letter j. Bythepreceding considerations, we have cs 3 n,k,i = k j=d k e 7 cs 3 n,k,i,j. (5)

8 The case j = k holds only in some special cases, that we present in the next result. Theorem 5. For n 4 and k 3, if csn,k,i,j 3 holds: (i) k apple i<j= k, (ii) k i = j k, (iii) k i>j, k (iv) n is even, i = k and j = k (v) n and k are even, i = k + and j = k., 6=0then only one of the following cases Moreover, for each one of the preceding cases we have the following recurrences: (i) csn,k,i,j = csn,k,i + csn,k,i, (ii) csn,k,i,j 3 3 = csn,k,i, ( cs 3 (iii) csn,k,i,j 3 = n,k, k csn 3,k,j if i = k +, j = k 3 + csn,k,j otherwise and n, k are even, (iv) cs 3 n,k,d k e,d k e = ( cs 3 n,k, k cs 3 n,k, k (v) cs 3 3 = cs. n,k, k +, k n,k, k if n and k are even if n is even and k is odd, Proof. Let =... n be a word in CS n,k (3) with = i and = j, respectively. Recall that i k and j k. Suppose i<j. If j 6= k, surjectivity implies that the sequence ijk is a 3-pattern: hence, j = k. In this case, the symbol k in the second position of and its conjugate inthe(n )-th position cannot form any 3 pattern. Hence, the number csn,k,i,j 3 is completely determined by the subsequence = 3... n n, whosesymbolscan belong either to the alphabet {,...,k }, or{,...,k}. Obviously,inthefirstcasewe can renormalize, soitsfirstletterbecomesi anditsalphabet{,...,k }. All these considerations prove recurrence (i). The case i = j is quite simple: when has two identical starting letters, it avoids 3 if and only if its subword =... n does. Furthermore, surjectivity is preserved between and, andhencerecurrence(ii) isproved. Now, suppose that i>j. In this case, the only fact that there is no 3-pattern with first letter j implies that no 3-pattern with first letter i occurs. The same is true for the conjugates of i and j, andhencethenumbercsn,k,i,j 3 is completely determined by the subword = 3... n. To preserve surjectivity, may either be over the alphabet {,...,k} r {, n}, or{,...,k}. Thisleadstothesecondrecurrenceof(iii). Observe 8

9 that, for particular values of i and j only one of the alphabets holds. In the even-even case, if i = k +andj = k the subword can only be over {,...,k}: thisprovesthe first recurrence of (iii). The last case j = k holds if and only if i = k (with n even) or i = k + (with both n and k even), otherwise a 3-pattern occurs. For the same reason, there cannot be any letter equal to i or its conjugate in the subword, hencethe only {,...,k} r {, n} alphabet is allowed for. Thisyieldsrecurrences(iv) and(v). Theorem (5.) and relation (5), for n 4, k 3(nand k not odd and even, respectively) and for every i such that k apple i apple k, yieldthefollowingrecurrencerelationfor csn,k,i 3 : csn,k,i 3 = cs cs n,k,d + cs k e n,k,i+ where and n,k,d k e + d k e+appleiapplek + i d k e+ j=d k e+ (cs3 n,k,i + cs 3 i A = n,k,i )+ (cs 3 n,k,i + cs 3 n,k,i )+ i=d k if A holds 0 otherwise 8 >< cs 3 if n and k are even n,k, k '(n, k) = 0 if n is even and k is odd >: if n and k are odd. cs 3 n,k, k Now, the base cases of the recurrence are needed. First of all, we set e '(n, k), (6) cs 3 0,0,0 =, cs 3 n,0,0 =0 and cs 3 0,k,i =0 (7) for every n, k andforeveryi such that apple i apple k. If k = or k = it is straightforward that csn,, 3 = 8 n, (8) n csn,, 3 = csn,, 3 if n, n even = (9) 0 if n, n odd. Moreover, we recall that, for every i, csn,k,i 3 =0ifk>n. Hence, the only remaining base cases necessary for the recurrence are cs 3 3,3, = cs 3 3,3, =0 and cs 3 3,3,3 =. Now we are ready to use the scanning-element algorithm. We will show its application only for the even-even case, since the remaining two cases can be treated analogously. First of all, we consider P 3 ee (x, y, v) = n 0 n even 9 k even cs 3 x n y k v k n,k, k.

10 Recurrence (6), for every n, k 4andfori = k,yields cs 3 n,k, k =cs 3 n,k, k +cs 3 n,k, k Hence, multiplying by x n y k v k, summing over all even n, k 4 and solving the resulting equation we get Pee 3 (x, y, v) = x x y v x x y v. (0) Now, define CS 3 n,k (v) = k cs 3 n,k,i v i. i= k and CS 3 ee (x, y, v) = n 0 n even k even k i= k cs 3 n,k,i x n y k v i = n 0 n even k even CS 3 n,k (v) x n y k. Multiplying recurrence (6) by v i and summing over all i from k to k we obtain CSn,k 3(v) = v v Observe that cs 3 n,k,k = CS 3 n v k+ 3 CSn v,k() + v 3 CSn,k v (v),k(v) v k+ 3 CSn,k v () + cs 3 v k n,k, k csn 3,k,k v k. k cs 3 n 4,k,i + k () cs 3 n 4,k,i = CS 3 n 4,k() + CS 3 n 4,k (). () i= k i= k Substituting () in (), multiplying () by x n y v k and summing over all possible even n, k 4weget v x y v x v( v) vx vx y = v v where and obviously x 4 CSee x, 3 y v,v = x 4 y CS 3 ee (x, y, ) + x y v Pee 3 x, y v,v (0) = v x v x y v x v x y CS 3 ee (x, y, ) = CSee 3 (x, y). 0 Pee x, 3 y v,v +(x ), (3)

11 Now, we apply the kernel method (for further details and examples, see [] and []). The coe cient of CSee 3 x, y,v in (3) vanishes for v where v ± (x, y) = x ± p (x, y), ( x ) (x, y) = 4x ( x )( + y ). Hence, substituting v = v (x, y) in(3) andsolvingtheresultingequationwefinally obtain that CSee 3 (x, y) = y + p (x, y) ( + y ) p (x, y). (4) The same method, used for the even-odd and the odd-odd case, yields CSeo 3 (x, y) = y( p x )( (x, y)) ( x )( + y ) p (x, y) and (5) CS 3 oo (x, y) = y( p (x, y)) x( x )( + y ). (6) The desired generating function C 3 (x, y) isnowimmediatelyobtainedby(4): C 3 (x, y) = y y + y x( x) y( x y) p +y x x( x) p y p y x. Moreover, evaluating the coe cients of the series expansions of the generating functions (4), (5) and (6), we get the following explicit formulae for the number of surjective centrosymmetric words avoiding 3: cs 3 m,h = cs 3 m,h+ = cs 3 m+,h+ = m i=0 m i= m i=0 i i ( ) m i i m i i m ( ) m i ( ) m i i + i, h m, h, i, m i h m, h 0, i i i, i m i h m 0, h 0. As already observed, the set S n,n corresponds to the set S n of permutations of length n. We recall that, as Egge already proved in [6] (Theorems. and.7), centrosymmetric permutations avoiding 3 are counted by the central binomial coe cients and the Catalan numbers, according to the parity of their length (for a bijective proof, see also []). We submit that this result can be reobtained from the preceding formulae by setting m = h in the first and last of (7). (7)

12 Now, using relations () we finally obtain the formulae for the number of centrosymmetric words avoiding 3: c 3 m,h = c 3 m,h+ = c 3 m+,h+ = m i=0 m i= m i=0 i i ( ) m i i m i ( ) m i m +h i i ( ) m i i + i i i + h h i i + h i m i h i i + h m i h, m, h,, m, h 0,, m 0, h 0. (8) 6 The pattern 3 Theorem 6. A word W n,k, with n, k 6, n and k not odd and even respectively, is a centrosymmetric surjective word that avoids 3 if and only if is of one of the following types: (i) = k, where is a centrosymmetric surjective word of length n, that avoids 3, either over the alphabet {,...,k }, or {,...,k}; (ii) = 0, where (a) =... j is a binary word over the alphabet {k,k}, with = k and j = k, such that it is the longest prefix of of this kind; (b) is a (possibly empty) centrosymmetric surjective word, that avoids 3, either over the alphabet {3,...,k }, or {,...,k }; (c) 0 = ( ) rc ; (iii) = 0, where (a) =... j is a surjective weakly increasing word over the alphabet {,..., j}, with k +apple apple k and j = k, such that it is the longest prefix of of this kind; (b) is a (possibly empty) centrosymmetric surjective word, that avoids 3, either over the alphabet {k+,..., } (if > k +), or {k+,..., }; (c) 0 = ( ) rc ; (iv) is a weakly increasing centrosymmetric surjective word over the alphabet {,...,k}. Proof. It is easily checked that a word either of type (i), (ii), (iii) or(iv) isacentrosymmetric surjective word that avoids 3. To prove the converse, let =... n be a centrosymmetric surjective word over the alphabet {,...,k} and avoiding 3. First

13 (a) Type (i) (b) Type (ii) (c) Type (iii) (d) Type (iv) Figure 5. Some examples of the four types of words descripted in Theorem (6.). 3

14 of all, observe that we can t have apple apple k,otherwise n = k + and hence the sequence k n would be an occurrence of 3. Therefore, we analyse the remaining cases for. (i) If = k then, by centrosymmetry, we have n =. Hence, the subword 3... n is a surjective word either over the alphabet {,...,k } or {,...,k} that avoids 3. (ii) If = k, let j be the rightmost position such that j = k. Then, the prefix =... j contains only the letters k andk: otherwise,itwouldcontainan occurrence of the pattern 3, and hence, by centrosymmetry, would contain the pattern 3. It is immediately checked that j apple n and that = 0,where 0 = ( ) rc and, if j 6= n, is a centrosymmetric surjective word either over the alphabet {3,...,k }, or{,...,k }. Ifj = n, is necessarily the empty word. (iii) If k +apple apple k, then is necessarily the first letter of a weakly increasing subword =... j, wherej apple n is the rightmost position such that j = k. If not, would contain a pattern that yields a 3 by the surjectivity of.moreover, in order to avoid both 3 and 3 (see relation ()), we have i apple for every i>j. Hence, = 0 0,where = ( ) rc and, if j 6= n, is a centrosymmetric surjective word, that avoids 3, either over the alphabet {k +,..., } (if k + apple, i.e. > k +), or {k +,..., }. As in the previous case, if j = n, is necessarily the empty word. (iv) If =then n = k, so,inordertoavoidboth3and3,thesubword... n cannot contain any pattern. Therefore, is weakly increasing. Theorem (6.) allows us to determine a recurrence relation for the number csn,k 3 of k-ary centrosymmetric surjective words of length n that avoid 3. In order to do this, it is convenient to analyse the first values of n and k. We recall that csn,k 3 =0ifn is odd and k is even, and we set cs 3 0,0 =, cs 3 n,0 =0 and cs 3 0,k =0 8n, k. For k =ork =,wehave cs 3 n, = 8 n, (9) n csn, 3 if n, n even = (0) 0 if n, n odd. If k =3andn 3, a word CS n,3 (3) can be either of type (i), (ii) or(iv) of Theorem (6.). If is of type (i) thenitscentralsubword belongs either to CS n,3 (3), or CS n, (3); then, there are csn 3,3 +ofsuchwords. 4

15 If is of type (ii) thenthecentralsubwordisnecessarily =..., so we have b n c n = b n c n = of such words. It is easily checked that there are n words Hence, the preceding considerations lead to the recurrence csn,3 3 = csn 3,3 + b l n c n m +, of type (iv). which yields cs 3 n,3 = 8 >< >: n + n if n is even n + n + n 3 if n is odd. If n, k 4, denote by csn,k (i), csn,k (ii), csn,k (iii) and csn,k (iv) the number of elements of CS n,k (3) of type (i), (ii), (iii) and(iv), respectively (see Theorem 6.). Of course, we have cs 3 n,k = cs 3 n,k (i) + cs 3 n,k (ii) + cs 3 n,k (iii) + cs 3 n,k (iv). We deduce a recurrence formula for each one of the sequences cs 3 n,k ( ). (i) It is immediately verified that cs 3 n,k (i) = cs 3 n,k + cs 3 n,k. (ii) Fix the length n of the subword. There are n of such subwords, and each one matches with csn 3 3 n,k + csn n,k 4 surjective subwords. Hence, cs 3 n,k (ii) = b n c n = n csn 3 n,k + csn 3 n,k 4. (iii) Of course, csn,k 3 (iii) 6=0ifandonlyifn 6andk 5. Fix the length n and the number of symbols k of the subword. We have 3 apple n apple n and 3 apple k apple min{n, k }.Thereare n of such subwords, and each of them n k is associated with: csn 3 3 n,k k + + csn n,k k subwords, ifk 6andk apple k the only subword = k+ k+ k+ 5 ; k+,ifk 5isoddandk =.

16 Hence, for n while, for n, k 6, cs 3 n,k (iii) = = 6andk =5,wehavek =3andthen b n c n =3 cs 3 n,5(iii) = min{n,b k c} k =3 c k+ b n c n =3 n n k n n 3 n = 3, cs 3 n n,k k + + cs 3 n n,k k b n n + (k odd) n k = n =d k e b n c min{n,b k c} n csn 3 n,k k + + csn 3 n,k k n k n =3 k =3 n + (k odd). (iv) The number of weakly increasing centrosymmetric surjective words is trivially n csn,k 3 (iv) = k. Hence, for n, k 4, we conclude that b n c csn,k 3 = csn 3,k + csn 3,k + b n c + (n 6^k 6) n = min{n,b k c} n n k n =3 k =3 n n + (k 5^kodd) + k. k+ n csn 3 n,k + csn 3 n,k 4 + cs 3 n n,k k + + cs 3 n n,k k Further computations in the special cases k =4andk =5yield 8 >< csn,5 3 = >: cs 3 n,4 = n + n n if n is even, n, n n 4 + n4 +4n 3 00n 08n (n +7) n 5 + n4 58n 384n if n is even, n if n is odd, n

17 n/ Table. The first values for csn,k 3. The values for 6 apple k apple n apple 9 have been computed directly. Routine calculations yield the following recurrence. Formula 6. The number csn,k 3 of k-ary centrosymmetric surjective words of length n that avoid 3, with n 0 and k 6, is given by the following recurrence relation: cs 3 n,k = 6cs 3 n,k +cs 3 n,k 4cs 3 n 4,k 9cs 3 n 4,k +6csn 3 6,k+ 5cs 3 n 6,k + cs 3 n 6,k 4 9cs 3 n 8,k cs 3 n 8,k cs 3 n 8,k 4+ cs 3 n 0,k +3cs 3 n 0,k cs 3 n 0,k 6. Multiplying the above equation by x n y k,summingoverallevenn and k (with n 0 and k 6) and solving the resulting equation we obtain that CS 3 ee (x, y) = Similarly, CS 3 eo (x, y) = and 6x +4x 4 6x 6 +9x 8 x 0 + +x 4 y 3x 6 y +3x 8 y x 0 y x 6 y 4 + x 8 y 4 + x 0 y 4 ( x + x 4 + x 4 y )( 4x +5x 4 x 6 x y +4x 4 y x 6 y + x 6 y 4 ). () x y ( + x y 5x +9x 4 x 4 y 7x 6 +x 8 x 8 y 4 + x 8 y ) ( x + x 4 + x 4 y )( 4x +5x 4 x 6 x y +4x 4 y x 6 y + x 6 y 4 ) () CS 3 oo (x, y) = xy ( 3x +x 4 ) 4x +5x 4 x 6 x y +4x 4 y x 6 y + x 6 y 4. (3) Even in this case, in the series expansions of () and (3) the coe cients of the monomials associated to words of length equal to their alphabet cardinality agree with Egge s result (see [6]). Hence, for every m 0wehave cs 3 m,m = m and cs 3 m+,m+ = m. 7

18 Using relation (4), we finally obtain that the generating function for the number of centrosymmetric words that avoid 3 is the rational function C 3 (x, y) = (x, y) (x, y), where (x, y) = and (x, y) = 6x 6 x 0 +8x 4 y 3 5x 4 y 7x 6 y 3 +9x 8 6x 4x 4 y +4x 4 + xy +45x 6 y +8x y 3y 43x 6 y 4 4x 8 y +x 8 y 4 +5x 0 y 3x 0 y 4 7x 7 y 8x y 4 +3y 4 +40x 4 y 4 8x 5 y 7 +5x 3 y 7 xy 7 +4x 7 y 7 +3xy 5 +5x 5 y 5 5x 3 y 5 4x 9 y 3 +x 9 y 5 +5x 3 y 3 3xy 3 5x 7 y 5 +9x 5 y +8x 7 y 3 +x 9 y 5x 3 y 6x 5 y 3 3x 4 y 6 7x 8 y 6 +6x y 6 y 6 3x 0 y 3 x y 3 +7x 6 y 5 +3x 8 y 3 + x 6 y 7 6x 8 y 5 x y 5 x 4 y 5 x 4 y 7 +x y 7 +4x 6 y 6 +9x 6 y 7x 8 y + x y +x 0 y ( y )( x + x 4 +x y y ) ( x 6 +3x 6 y 6x 4 y + x 4 y 4 +5x 4 +6x y x y 4 4x + y 4 y ). References [] M. Barnabei, F. Bonetti and M. Silimbani, The Eulerian numbers on restricted centrosymmetric permutations, Pure Mathematics and Applications, :99-8, 00. [] C. Banderier, M. Bousquet-Mélou, A. Denise, P. Flajolet, D. Gardy and D. Gouyou- Beauchamps, Generating functions for generating trees, DiscreteMathematics,46(- 3):9-55, 00. [3] P. Brändén and T. Mansour, Finite automata and pattern avoidance in words, Journal of Combinatorial Theory, Series A, 0():7-45, 005. [4] A. Burstein, Enumeration of words with forbidden patterns, Ph.D. Thesis, University of Pennsylvania, 998. [5] A. Burstein and T. Mansour, Words restricted by patterns with at most distinct letters, Electronic Journal of Combinatorics, 9():#R3, 00/03. [6] E. S. Egge, Restricted symmetric permutations, Annals of Combinatorics, : , 007. [7] G. Firro and T. Mansour, Three-letter-pattern-avoiding permutations and functional equations, Electronic Journal of Combinatorics, 3():#R5,

19 [8] William Fulton, Young Tableaux, with Applications to Representation Theory and Geometry, Cambridge University Press, 997. [9] V. Jelínek and T. Mansour, Wilf-equivalence on k-ary words, compositions, and parking functions, Electronic Journal of Combinatorics, 6():#R58, 009. [0] D. E. Knuth, Permutations, matrices and generalized Young tableaux, Pacific Journal of Mathematics, 34:709-77, 970. [] T. Mansour, Restricted 3-avoiding k-ary words, Chebyshev polynomials, and continued fractions, Advances in Applied Mathematics, 36():75-93, 006. [] H. Prodinger, The kernel method: a collection of examples, SéminaireLotharingien de Combinatoire, 50:B50f, 004. [3] M. P. Schützenberger, Quelques remarques sur une construction de Schensted, Mathematica Scandinavica, :7-8, 963. [4] N. J. A. Sloane and S. Plou e, The Encyclopedia of Integer Sequences, Academic Press,