AVOIDING PERMUTATIONS AND THE NARAYANA NUMBERS

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1 J Korean Math Soc 50 (2013, No 3, pp AVOIDING PERMUTATIONS AND THE NARAYANA NUMBERS Youngja Park and Seungkyung Park Abstract We study 132 avoiding permutations that also avoid (2r + 1(2r+2 12 but contain (2r 1(2r 12 pattern We find an identity between the number of these permutations and the Narayana number We also present relations between 132 avoiding permutations and polygon dissections Finally, a generalization of these permutations is obtained 1 Introduction Avoiding permutations of certain patterns have been studied over the years Especially, the number of avoiding permutations of pattern 132 with r descents is the Narayana number N(n, k (See [3] We study 132-avoiding permutations that avoid another pattern but contain some other pattern Let S n be the set of permutations on [n] := {1,2,,n} For π S n and τ S k, we say that π contains τ-pattern if there exists a subsequence 1 i 1 < i 2 < < i k n such that (π i1 π i2 π ik is in the same relative order as τ We say that π is τ-avoiding if such a subsequence does not exist The set of all τ-avoiding permutations in S n is denoted by S n (τ The set of all permutations containing τ in S n is denoted by S n (τ So S n (σ 1,σ 2 is the set of all permutations avoiding σ 1 but containing σ 2 in S n We consider the case of τ r = (2r 1(2r(2r 3(2r 2 12 pattern, and enumerate the number of permutations with i ascents in S n (132,τ r+1,τ r for r 1, and denote the set by Wn(r i We generalize this set with the pattern ((r 1d+1 (rd 1(rd (12 d by τ r,d and Wn(r,d, i the set of these permutations with i ascents in S n (132,τ r+1,d,τ r,d We found that if d = 2 in our formula, then the number is the same as in the following identity for d = 2 obtained by Sulanke([5]: d 1 ( ( 1 m+n+i n+i = ( dn+m k N(d,n,k, n n dn i=0 k 0 Received April 26, 2012; Revised September 14, Mathematics Subject Classification 05A15 Key words and phrases avoiding permutation, Narayana number, dissection number 529 c 2013 The Korean Mathematical Society

2 530 YOUNGJA PARK AND SEUNGKYUNG PARK where N(d, n, k is the d-dimensional Narayana Numbers He used Stanley s P-partitions theory and MacMahon s work Callan [1] also found the same identity by counting Dyck paths according to the number of plateaus But for d 3, our numbers are different from theirs In Section 2, we provecombinatoriallythat N(n,i+1 = i r=1 wi n (r, where Wn i(r = wi n (r by Algorithms A, B, C, and D on avoiding permutations Although we use the four algorithms, they give some applications on themselves Moreover, we can generalize Wn i (r further using those algorithms to get Wn(r,d i If d = 2, Wn(r,2 i = Wn(r, i which is the same as the results of Sulanke and Callan We discuss this in Section 5 Section 3 deals with the polygon dissection numbers derived by Algorithms A D In Section 4, we give the number of the avoiding permutations that have exactly one pattern of τ r 2 Avoiding permutations with the specific restrictions We call i a descent (resp, ascent of a permutation π = π 1 π n if π i > π i+1 (resp, π i < π i+1 for i = 1,2,,n 1 and count an extra descent at the end of each permutation We denote by des(π the number of descents in π An ascending run of a permutation is an increasing sequence of consecutive entries Thus ifπ hask descents, then it hask ascendingruns Ifa i+1 a i+j is an ascending run, then we say it is of length j Specially, we call an ascending run with length more than 1 in 132 avoiding permutation a long ascending run Note that any permutation can be decomposed into ascending runs, each of which is denoted by I j For example, if π = ,then π = has three ascending runs separated by bars, namely, I 1 = 56,I 2 = 3,and I 3 = 1247 of lengths 2, 1, and 4, respectively We represent by I j I k the concatenation of two runs of I j and I k Defineτ r :=(2r 1(2r(2r 3(2r 2 12forr 1 ThenS n (132,τ r+1,τ r is the set of all permutations that avoid the patterns 132 and τ r+1 but contain the pattern τ r Thus any permutation in S n (132,τ r+1,τ r must have at least r ascending runs of length at least 2 Let Wn(r i be the set of permutations π in S n (132,τ r+1,τ r satisfying des(π = n i Wewritewn(rfor W i n(r i LetS n,r (132bethesetofallpermutations that avoid the pattern 132 and have r descents Krattenthaler [3] showed that the number of such permutations is the Narayana number N(n,r If π Wn i(r, then π = I 1 I n i, where each I j is an ascending run Since π contains the pattern τ r, there must be r ascending runs, where the concatenation of any distinct pair of them contains a 3412 pattern We call these the basic ascending runs Notice that any ascending run of length 1 cannot be a basic ascending run, and that the first ascending run of length greater than 1 automatically becomes a basic ascending run The following theorem provides a closed formula for the number wn i (r using this fact

3 AVOIDING PERMUTATIONS AND THE NARAYANA NUMBERS 531 Theorem 21 For i r 1, wn(r i = N(i,r ( n r+i, where N(i,r = ( i r 1 is the Narayana number 1 i( i r Proof Algorithm A shows φ(π S i,r (132 and Algorithms B, C, and D show the map φ is ( n r+i -to-one function To define a map φ : W i n (r S i,r (132 for a permutation π = a 1 a n = I 1 I n i Wn(r i with the descent set {j 1,,j n i }, we apply the following algorithm to π and the output φ(π becomes an 132-avoiding permutation with r descents If I j is a long ascending run and I k the next long ascending run, and if I j I k does not contain a 3412 pattern, then we choose the first entry of I k as d j for j = 1,2,,n i 1 Otherwise, choose the last element from each of the long ascending runs Let {d 1,d 2,,d k,,d n i } be the chosen entries Notice that d n i = a n We now delete d k s according to the following algorithm: Algorithm A STEP 1 Delete d n i = a n STEP 2 For each k = 1,2,,n i 1, (a Subtract 1 from each entry before d k, and subtract 1 from each entry greater than d k after d k (b If there exist a s before d k and a t after d k with a s = a t, then replace a t by the number d k 1 Then remove d k from the ascending run Now we show that φ(π S i,r (132 and that the map is a ( n r+i -to-one function Since we remove n i entries from π, clearly, φ(π is a permutation on i letters Moreover, for each k = 1,2,,n i 1, the permutation after Step 2 is still 132-avoiding Notice that after the one run of Step 2, every entry is distinct If (b of Step 2 does not occur, then clearly 132-avoiding is preserved For the case of (b, every number appearing between d k 2 and d k 1 is less than d k 2, which means that it preserves 132-avoiding after Step (b Notice also that all length 1 ascending runs in π are removed by the algorithm So ignore those runs and only consider the long ascending runs Recall that the first long ascending run is the first basic ascending run If the next ascending run is a basic ascending run, then removing the last entry in the first one does not change the descent But if it is a nonbasic ascending run, which means that there is no 3412-pattern in the concatenation of the two, then removing the first element by Step 2 deletes the descent So the next ascending run is merged to the first one, which becomes a basic ascending run with a bigger size As we proceed with the algorithm, the number of basic ascending runs does not change Thus when the algorithm is completed, we obtain a permutation on i letters with r descents In other words, φ(π S i,r (132

4 532 YOUNGJA PARK AND SEUNGKYUNG PARK Secondly, we show that the map is a ( n r+i -to-one function Let φ(π = σ = b 1 b i S i,r (132 and the descent set of σ be Des(σ = {j 1,,j r },j r = i We will insert r entries into σ to obtain σ = c 1 c i+r S i+r,r (132, where Des(σ = {j 1 +1,,j r +r} Let σ = I 1 I r, where I k s are ascending runs and let a k,lk be the last entry of I k Algorithm B STEP 1 For each k = r 1,r 2,,1, (a Insert the number a k,lk +1 into I k as its last entry (b Add 1 to each entry in I j including a k,lk +1 for 1 j k, and add 1 to each entry in I j, j > k, that is greater than or equal to a k,lk +2 We denote by the output of this step (c If an entry a s,i s in I s is the same as an entry a t,i t in I t with k s < t, then replace a t,i t by a s,1 s 1, where a s,1 s is the first entry of I s Repeat this process until all entries in I j, j k, become distinct STEP 2 Attach the unused number from [i+r] = {1,2,,i+r} to the end of the last ascending run I r, which becomes I r = c i+r The output σ = c 1 c i+r = I1 I r is a permutation on i + r letters that avoids 132-pattern Before Step 2, since every number between 1 and the largest element a r,lr of Ir is used from the construction, the unused number must be greater than a r,lr Thus attaching the unused one to the end of Ir does not produce a descent Therefore, the number of descents of σ is still r By (a, (b of Step1 and Step 2, each Ij 2,1 j r By (c of Step 1, each Ij,1 j r, has at least two elements less than the first element of I j 1 So it has r basic ascending runs Therefore, σ has τ r pattern and σ avoids τ r+1 Now we insert m(0 m n i r numbers into the basic ascending runs of σ = I1 Ir from right to left, creating m descents, to obtain σ in Algorithm C Algorithm C For each m of 0,1,,n i r, find all possible positions that are spaces betweena k,2k anda k,lk fromeachik withrepetitionallowedfork = 1,2,,r 1 For k = r, include one more space, the last space Suppose that we select t k positions for each k = r,r 1,,1 and k t k = m with t k 0 We insert t k numbers into the selected positions Ik For each m = 0,1,,n i r and for k = r,r 1,,1, STEP 1 For the chosen t k positions from the possible spaces from I k, add the number t k to each entry in σ that is greater than or equal to a k,1k STEP 2 Insert t k numbers of a k,1k + (t k 1,,a k,1k + 1,a k,1k into the positions selected from right to left, keeping this decreasing order

5 AVOIDING PERMUTATIONS AND THE NARAYANA NUMBERS 533 Thus there are (( ( i r+1 m = i r+m m ways to get σ Finally, we construct σ from σ to complete the proof We insert n i r m numbers in the fronts of Ik s with repetition allowed from right to left in such a way that every time we place a number a new descent must be created Among r +m ascending runs in σ, there are r basic ascending runs and we denote them by I k = a k,1k a k,2k a k,lk for 1 k r Define l k as l k = a k 1,1k 1 a k,1k t k 1 fork = 2,3,,r, andl 1 = (i+r+m+1 a 1,11 Suppose that u k numbers h 1,,h uk are placed in front of the kth basic ascending run Ik and that k u k = n i r m Then these numbers should satisfy the conditions a k,2k +u k j h j a k,2k +u k j+l k 1 and h 1 > h 2 > > h uk Suppose that we insert u k numbers in front of Ik for each k = 1,2,,r and let r k=1 u k = n i r m with u k 0 For each of k = r,r 1,,1, Algorithm D STEP 1 Find l k for all k = r,r 1,,1 STEP 2 For k = r,r 1,,1, (a For j = u k,u k 1,,1, find the range for h j (b Find all possible sequences with h 1 > h 2 > > h uk 1 > h uk, then place each of such sequences in front of I k in the order of h uk,h uk 1,,h 2,h 1 from right to left (c Each time h j is inserted, add 1 to each entry of σ that is greater than or equal to h j After completing Algorithms B through D, we obtain a permutation σ on n letters and by our construction, it is clear that σ Wn(r i Since k l k = i+r, the total number of ways of ( inserting n i r m numbers into ( i+r i+r positions with repetition allowed is n i r m = ( n m 1 i+r 1 By Vandermonde s, we get n i r m=0 ( i r +m m ( n m 1 i+r 1 = Thus the map φ is a ( n r+i -to-one function Therefore, wn i(r = N(i,r( ( n r+i n r i = 1 i i( i r r 1 ( n r+i n r i ( n r+i Example 22 Apply Algorithm A to the following permutations π to obtain φ(π π = = φ( π = = φ( Example 23 Clearly, S 2,2 (132 = {21} We apply Algorithm B through D to find all π s on n = 6 letters whose image is φ(π = σ = 21 Since 0 m n i r = = 2, we consider each of these three cases:

6 534 YOUNGJA PARK AND SEUNGKYUNG PARK First, apply AlgorithmBto σ = 2 1to obtain σ : σ = = σ Now, if m = 2, apply Algorithm C to σ to get σ : σ = = σ We have found a permutation in W6(2 2 so we do not need to apply Algorithm D For m = 1, by Algorithm C, σ = = σ By Algorithm D, to insert in front of the first basic ascending run 45, 5 h 1 6 since l 1 = i+r+m+1 a 1,11 = = 2 Similarly, to insert in front of the second basic ascending run 23, 3 h 1 4 since l 2 = a 1,11 a 2,12 0 = 4 2 = 2 Thus we will have four outputs of Algorithm D σ = = σ, or σ = = σ, σ = = σ, σ = = σ Finally, for m = 0, we have to insert n i r m = = 2 numbers according to Algorithm D into two positions of σ = in front of I 1 = 34(a, I 2 = 12(b, or one of each(c First, we find l k s: l 1 = 2 and l 2 = 2 Case (a: 5 h 1 6 and 4 h Case (b: 3 h 1 4 and 2 h Case (C: for inserting 2 h 1 3 in front of I 2 = 12, then for 5 h 1 6 in front of the first run, one at a time Thus we have W6 2 (2 = {563421,645231,546231,564231,563241,564312,564213,563214, , , , , , , } Corollary 24 i wn(r i = r=1 i ( n r+i N(i,r = N(n,i+1 n r i r=1

7 AVOIDING PERMUTATIONS AND THE NARAYANA NUMBERS 535 Table 1 w i n (r n r\i n r\i Proof Since wn i(r = N(i,r( n r+i, i i ( n r+i wn i (r = N(i,r n r i r=1 r=1 Moreover, i r=1 wi n(r = S n,n i (132,whereS n,n i (132isthesetof132avoiding permutations on n letters with n i descents This is known to be S n,n i (132 = N(n,n i = N(n,i+1 from [3] Remark 25 From the definition of wn i (r, we provide that n+r ( n+r +i S n+2r (132,τ r+1,τ r = N(i,r n+r i for r 1, n 0 In particular, we take the sequence of S n+2 (132,3412,12 = F 2(n+1 1 is {1,4,12,33,88,232,} for n 0,r = 1 because the sequence of S n+2 (132,3412 = S n+2 (132,3412,12 {n(n 1 21} = F 2(n+1 is {1,2,5,13,34,89,}(see[6], [4], wheren 0andF n isthefibonacci number i=r 3 An application of the algorithms We denote the set of permutations with j long ascending runs in S n (132 by P(n,j We write p n (j for P(n,j Let n = 2j+m(j,m 0 A permutation with j long ascending runs in 132 avoiding permutation has s(0 s j 1 ascending runs and j s basic ascending runs Suppose that π P(2j +m,j hast(0 t mascendingrunswithlength 1 Thenπ hasj+m tascents To obtain the number of permutations with j long ascending runs in 132 avoiding permutation, we use Algorithms A, B, C, and D

8 536 YOUNGJA PARK AND SEUNGKYUNG PARK Table 2 S n+2r (132,τ r+1,τ r n\r Proposition 31 Define p m (0 = 1 for j = 0 Then for j 1, [ m j 1 ( ] ( m t+s 2j +m p 2j+m (j = N(j +m t,j s, s t t=0 s=0 where N(i,r = i( 1 i ( i r r 1 is the Narayana number and i r 1 Proof Every long ascending run is merged to a basic ascending run by Algorithm A So π P(2j + m,j becomes φ(π S j+m t,j s (132, where j is the number of long ascending runs By Algorithm B, j s basic ascending runs are formed So we obtain a σ S 2j+m t s,j s (132 By making s long ascending runs but not basic ascending run, we get j long ascending runs in Algorithm C, where repetition is not allowed and the last space is excluded Thus there are ( ( j+m t (j s s = m t+s s choices Finally, by adding t ascending runs with length 1 at the front of each ascending run and the last space in Algorithm D, where l 1 = 2j + m t + 1 a 1,11,l j+1 = a j,1j,l k = a k 1,1k 1 a k,1k,(2 k j and j+1 k=1 u k = t, j+1 k=1 l k = 2j+m t+1 Also, h q = a k 1,1k 1 +u k q at the last space and the front of ascending run, not basic ascending run At the front of basic ascending run, a k,2k +u k q h q a k,2k +u k q+(l k 1 There are (( ( 2j+m t+1 = 2j+m choices Therefore, we have m t=0 [ j 1 s=0 N(j +m t,j s( m t+s s Now we consider the inner summation t t ] ( 2j+m t [ j 1 s=0 N(j +m t,j s( ] m t+s s This counts the permutations with j long ascending runs and not with ascending runs of length 1 in S n (132 Dyck n path, D n, is the set of paths on the square lattice with steps (1,1 and (1, 1 from (0,0 to (2n,0 that never falls below the x-axis We denote steps (1,1 and (1, 1 by U(upstep and D(downstep, respectively We call a two consecutive step UD a peak By the bijection between S n (132 and Dyck n paths D n in [3], we see that the above inner summation is equal to the number of paths with j UDDs and j peak UDs in D 2j+m t Callan [2] presented that dissections of a regular (n+2gon with n 1 k noncrossing diagonals correspond to k marked Dyck n paths in one-to-one fashion We insert UD to each peak at m t marked Dyck

9 AVOIDING PERMUTATIONS AND THE NARAYANA NUMBERS 537 j +m t path Then we insert UD to unmarked interior vertex after deleting the marked interior vertex, where we use interior vertices of down steps instead of up steps So we expect that a Dyck 2j + m t path with j UDDs and j peak UDs correspond to an m t marked Dyck path Thus we can represent dissection numbers in terms of Narayana number Corollary 32 j 1 ( m t+s d j 1 (j +m t+2 = N(j +m t,j s, s s=0 where j 1 is the number of noncrossing diagonals in regular (j+m t+2gon Therefore, Proposition 31 becomes: Corollary 33 For j 1, p 2j+m (j = m ( 2j +m d j 1 (j +m t+2 t t=0 In fact, since the strings of UUD and UDD are equidistributed, p n (j in S n (132 is equal to the number of Dyck n paths with j long ascents (A [4] 4 Permutations with τ r pattern exactly once in S n+2r 1 (132,τ r+1 We count the number of permutations that contain τ r pattern exactly once in S n+2r 1 (132,τ r+1 (n 1,r 2 by Algorithms A D We denote the set of permutations that contain τ r pattern exactly once in Wn+2r 1 i (r by W 1,i n+2r 1 (r LetEr n betheset ofpermutationsthatcontainsτ r patternexactly once in S n+2r 1 (132,τ r+1 We denote E n r by er n (n 1,r 2 Then er n = n+r 1 i=r W 1,i n+2r 1 (r Proposition 41 Let e r 1 = 1,w0 i t (1 = 1 and w0 i t (t = 0 (t 2 Then ( n+r 1 e r n = r 1 ( n 1+i wi t i r (t n+r 1 i = i=r n+r 1 i=r t=1 ( r 1 ( 2i 2t r N(i r,t 2i 2r t=1 ( n 1+i n+r 1 i Proof Since each Wn(r i is disjoint and Wn(r i = if i < r or n i < r, we know that each W 1,i n+2r 1 (r are disjoint and Er n = n+r 1 i=r W 1,i n+2r 1 (r Since π W 1,i n+2r 1 (r has τ r pattern exactly once, the length of the first basic ascending run is 2 and other r 1 basic ascending runs have lengths greater than orequalto2, whereonlythefirsttwoentriesofeachbasicascendingrunareless than the first entry of the preceding ascending run Also, non-basic ascending runs of lengths greater than or equal to 2 must occur after the second basic

10 538 YOUNGJA PARK AND SEUNGKYUNG PARK ascending run Otherwise, π would have τ r pattern more than once To obtain W 1,i n+2r 1 (r, we use Algorithms A D All elements except the first element in each ascending run with length greater than or equal to 2 in φ(π are larger than the elements before the ascending run Each ascending run with length more than or equal to 3 in φ(π has at least two elements less than the second elements after the ascending run We consider each ascending run with length greater than or equal to 3 except the first element Let φ(π have t ascending runs with length more than or equal to 3 If we reverse the order of ascending runs, we regard φ(π as a permutation in W i r i t (t So the number of φ(π is (t (1 t r 1 by the aforementioned reasons There are ((i r+1 m = choices for σ For σ, we change a k,2k of Ik in the range of h j into a k,2k + 1 and change l k 1 into l k 2 for exactly one τ r pattern So W ( i t i r i r+m m a k,2k +1+u t j h j a k,2k +1+u t j+l k 2,(1 j t Also, r k=1 (l k 1 = i Then we get (( ( i n+2r 1 i r m = n+r 2 m ( n+r 1 i m = n+r 2 m i 1 choices for σ By Vandermonde s, we get n+r 1 i( i r+m ( n+r 2 m ( m=0 m i 1 = n 1+i n+r 1 i Therefore, for n 2,r 2, E r n = e r n = = n+r 1 i=r ( n+r 1 r 1 i=r n+r 1 W 1,i n+2r 1 (r = t=1 ( r 1 ( n+i 1 w i r i t (t n+r 1 i i=r t=1 ( ( 2i 2t r n 1+i N(i r,t 2i 2r n+r 1 i In particular, if r = 2, n+1 ( n 1+i e 2 n = = F 2n 1 n+1 i i=2 n ( n+i = S n+1 (132,3412,12 = N(i,1 n i i=1 5 A generalization We generalize Theorem 21 to the case of any length d 3 Let us denote the pattern ((r 1d+1 (rd 1(rd (12 d by τ r,d and write Wn i(r,d for the set of these permutations with i ascents in S n (132,τ r+1,d,τ r,d, and Wn(r,d i = wn(r,d i We use Algorithms A, B, C, and D again Since π Wn i (r,d, there should be r ascending runs, where the concatenation of any distinct pair of them contains a (d + 1 2d(12 d pattern We call them the d-basic ascending runs Notice that any ascending run of length 1,,d 1 cannot be a basic ascending run, and that the first ascending run of length greater than d 1 automatically becomes a basic ascending run

11 AVOIDING PERMUTATIONS AND THE NARAYANA NUMBERS 539 Proposition 51 For d 3, w i n(r,d = i (d 1r j=0 w i r j i (r,d 1 ( n j r+i n j r i Proof By Algorithm A, π Wn(r,d i becomes φ(π W i r j i (r,d 1, where j is the number of ascending runs with length greater than or equal 2 which are not d-basic ascending runs So 0 j Min(i (d 1r, n i r By Algorithm B, we obtain a σ S i+r+j,r+j (132,τ r+1,d,τ r,d from σ W i r j i (r,d 1 We use Algorithms C and D to increase n i r j descents By Vandermonde s, we get n i r j ( ( ( i r j +m n m 1 n r j +i = m i+r+j 1 n r j i Therefore, m=0 i (d 1r wn i (r,d = j=0 w i r j i (r,d 1 ( n j r+i n j r i We know that for fixed n,i, and d, the sum of w i n(r,d for all r 0 is also a Narayana number Table 3 w i n (r,3 n r\i n r\i For the case when d = 3 and r = 0, S n+3r (132,τ r+1,3,τ r,3 = S n (132,123 = 2 n (A on OEIS For the case when d = 4 and r = 0, S n+3r (132,τ r+1,4,τ r,4 = S n (132,1234 = F 2n (A on OEIS, where F n is the Fibonacci number These two cases present well-known numbers, but all other numbers of r 1 and d 3 appear to be new

12 540 YOUNGJA PARK AND SEUNGKYUNG PARK Table 4 w i n (r,4 n r\i n r\i Table 5 S n+3r (132,τ r+1,3,τ r,3,s n+4r (132,τ r+1,4,τ r,4 d = 3 d = 4 n\r n\r By successively plugging in the initial condition w i n (r,2 = N(i,r( n r+i, which is Theorem 21, we obtain the following: Corollary 52 Let α k = j 1 + +j k,j l 0,1 k d 2,α 0 = 0 [ wn(r,d i = N(i (d 2r α d 2,r α d 2 i (d 1r d 2 q=1 ( 2(i qr αd 2 +α d 2 α q 1 In particular, when d = 3 we have: Corollary 53 i 2r wn i (r,3 = j=0 2(i qr α q ( 2i 2r j N(i r j,r j ( ] n α1 r+i n α 1 r i ( n j r+i n j r i Acknowledgements We would like to thank anonymous referee for his/her helpful comments

13 AVOIDING PERMUTATIONS AND THE NARAYANA NUMBERS 541 References [1] D Callan, Kreweras Narayana Number Identity Has a Simple Dyck Path Interpretation, (2005, preprint [2], Polygon Dissections and marked Dyck Paths, (2005, preprint [3] C Krattenthaler, Permutations with restricted patterns and Dyck paths, Adv in Appl Math 27 (2001, no 2-3, [4] N J A Sloane, On-line Encyclopedia of Integer Sequences, attcom/ njas/sequences/indexhtml [5] R A Sulanke, Generalizing Narayana and Schröder Numbers to Higher Dimensions, The Electronic Journal of Combinatorics 11 (2004, Art R54 [6] J West, Generating trees and forbidden subsequences, Discrete Math 157 (1996, no 1-3, Youngja Park Department of Mathematics Yonsei University Seoul , Korea address: ojpark@yonseiackr Seungkyung Park Department of Mathematics Yonsei University Seoul , Korea address: sparky@yonseiackr