REFINED RESTRICTED PERMUTATIONS AVOIDING SUBSETS OF PATTERNS OF LENGTH THREE

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1 REFINED RESTRICTED PERMUTATIONS AVOIDING SUBSETS OF PATTERNS OF LENGTH THREE Toufik Mansour LaBRI, Université Bordeaux 1, 51 cours de la Libération 405 Talence Cedex, France Aaron Robertson 1 Department of Mathematics, Colgate University, Hamilton, NY 14 aaron@math.colgate.edu Abstract Define Sn(T k ) to be the set of permutations of {1,,...,n} with exactly k fixed points which avoid all patterns in T S m. We enumerate Sn(T k ), T S, for all T and 0 k n. 1 Introduction Let π S n be a permutation of {1,,...,n} written in one-line notation. Let α S m. We say that π contains the pattern α if there exist indices i 1 <i < <i m such that π i1 π i...π im is equivalent to α, where we define equivalence as follows. Define π ij = {x : π ix π ij, 1 x m}. If α = π i1 π i...π im then we say that α and π i1 π i...π im are equivalent. Define S k n(t ), T S m, to be the set of permutations with k fixed points which avoid all patterns in T and let s k n(t )= S k n(t ). In recent paper [RSZ], the authors began the study of S k n(t ) for T S, T = 1. In this paper we enumerate S k n(t ) for all T S with T. Thus, all that remains to enumerate are S k n(1) and S k n(1) for all 0 k n. We start by giving two bijections which preserve the parameter number of fixed points. Let I : S n S n be the group theoretical inverse. Let R : S n S n be the reversal bijection, i.e. R(π 1 π...π n )=π n π n 1...π 1, and let C : S n S n be the complement bijection, i.e. C(π 1 π...π n )=(n +1 π 1 )(n +1 π )...(n +1 π n ). 1 Homepage: aaron/ 000 Mathematics Subject Classification: 05A15, 8R15 1

2 It is easy to see that the bijections I and RC preserve the number of fixed points in a permutation. This reduces the number of cases that we have to consider. In the sequel, we will be using the following notation. Let I S be the characteristic function, i.e. I S =1ifS is true and I S =0ifS is false. We also define s j n(t )=0forj<0orj>n for any T S. The Case T = Using I and RC we see that we have the following cases. (1) {1, 1} = {{1, 1}} () {1, 1} = {{1, 1}, {1, 1}} () {1, 1} = {{1, 1}, {1, 1}} (4) {1, 1} = {{1, 1}} (5) {1, 1} = {{1, 1}, {1, 1}, {1, 1}, {1, 1}} () {1, 1} = {{1, 1}, {1, 1}} (7) {1, 1} = {{1, 1}} (8) {1, 1} = {{1, 1}, {1, 1}} Theorem.1 {s 0 n(1, 1)} n 0 =1, 0, 1,, 4, 0, 0,..., {s 1 n(1, 1)} n 0 =0, 1, 0,, 0, 0,..., {s n(1, 1)} n 0 =0, 0, 1, 0, 0..., and s k n(1, 1) = 0 for all k n. Proof. Obvious. Theorem. For n 1, s k n(1, 1) = 0 for k n, 4 n 1 + if i =0 s n+i(1, 1) =, s 1 n+i(1, 1) = 0 if i =1 4 n 1 if i =0 s 0 n+i(1, 1) =. (4 n 1) if i =1 (4 n 1 1) if i =0 4 n + if i =1, and

3 Proof. Let π S n (1, 1). It is easy to prove by induction that we must have π = (π(1),π(),...,π(m)) where π(j) =(t j 1 1,t j 1,...,t j +1,t j 1 ) for some n = t 0 > t 1 >t > >t m = 0. We call π(j) aj-block of π. (See [Man] for more details.) Clearly s k n(1, 1) = 0 for k n since three fixed points yield a 1 pattern. Hence, we need only consider k = 0, 1,. However, using a result in [SiSc] and the fact that s k n(1, 1) = 0 for k n, we immediately obtain obtain for s k n(1, 1) for one of k =0, 1,, provided we know it for the other two values of k. Hence, we may prove the stated formulas by providing formulas for the following: 1. s n(1, 1). s 1 n+1(1, 1). s 0 n(1, 1). We start with s n(1, 1). Consider the graph of π, which consists of the points (i, π i ) Z. By the symmetries of the graph of π, we see that the number of intersections with the line y = x (i.e. the number of fixed points of π) is two if and only if m is odd for π =(π(1),π(),...,π(m)) and the intersections are in the m+1 -block of π. So, the number of elements in the m+1 -block must be even, as well as the total number of elements of π. Hence, we have s n(1, 1) = 0 for n odd. For the case n, let d be the number of elements in the m+1 -block. Since the other blocks cannot contain a fixed point, we have (s n d(1, 1)) permutations for d =1,,...,n. Using a result in [SiSc], this gives 1 + n 1 d=1 (n d 1 ) permutations. Hence, s n(1, 1) = 4n 1 +. Next, we look at s 1 n+1(1, 1) and again consider the graph of π; we have a situation similar to the case above. We must have the middle block containing an odd number of elements. Let this middle block have d + 1 elements. Using a result in [SiSc], this gives 1+ n 1 d=0 (n d 1 ) permutations. Hence, s 1 n+1(1, 1) = 4n +. Lastly, we consider s 0 n(1, 1). From the block decomposition given at the beginning of this proof and the fact that we do not have a fixed point, for π Sn(1, 0 1) we have π = π(1)π() where π() S n (1, 1) and π(1) is a permutation of n+1,n+,...,n which avoids both given patterns. Thus, using a result in [SiSc], s 0 n(1, 1) = (s n (1, 1)) = 4 n 1.

4 Theorem. For n, s k n(1, 1) = 0 for k n, n(n+) if n 0(mod ) 4 s n(1, 1) = n 1 4 if n 1, 5(mod ) (n+)(n+4) 4 if n, 4(mod ) n 9 4 if n (mod ) n if n 0(mod ), s 1 n(1, 1) = 7n 1n+9 4 if n 1, 5(mod ) n 4 if n, 4(mod ) 7n 1n+45 4 if n (mod ), and s 0 n(1, 1) = ( n ) +1 s 1 n (1, 1) s n(1, 1). Proof. Clearly s k n(1, 1) = 0 for k n since three fixed points yield a 1 pattern. Hence, we need only consider k =0, 1,. However, using a result in [SiSc] and the fact that s k n(1, 1) = 0 for k n, we immediately obtain the stated equation for s 0 n(1, 1) upon proving the formulas for k =1,. Hence, we need only consider k =1,. Let π S n (1, 1) with π i = n. Ifi 1 then we must have π =(i 1)(i ) 1n(n 1)(n ) i for any i n. This case contributes n 1 permutations to S 1 n(1, 1) for n odd and n to S n(1, 1) for n even. For i = 1, we have π = n(n 1) (n x + 1)y(y 1) 1(n x)(n x 1) (y + 1) for some 1 x n and 1 y n x 1, or we have π = n(n 1) 1. We now consider the cases k =1, for i =1. We have three potential intervals in which fixed points may reside: (A) n(n 1) (n x +1), (B) y(y 1) 1, or (C) (n x)(n x 1) (y +1) Hence, we have the following situations in which a fixed point appears. 4

5 A. For n odd we get one fixed point for any n+1 x n and 1 y n x 1, or if π = n(n 1) 1. The fixed point occurs at position n+1 B. For x + y odd we get one fixed point for any 1 x n and x +1 y n x 1. The fixed point occurs at position x+y+1. C. For n + y odd we get one fixed point for any 1 x n and 1 y n x 1. The fixed point occurs at position n+y+1. We now consider the case k =(i = 1). To have two distinct fixed points we must have both B and C both occurring. Hence, we get n(n ) if n 0(mod ) 4. n x=1 n x 1 y=x+1 n+y odd x+y odd 1= (n 1)(n+1) 4 if n 1, 5(mod ) (n 4)(n ) 4 if n, 4(mod ) (n )(n+) 4 if n (mod ) permutations with two fixed points in this case. We now consider the case k =1(i = 1). We may have only A occur, only B occur, or only C occur: Only A occurs: In this case we require n to be odd and we get 1+ n x= n+1 n x 1 y=1 1= (n )(n 1) 8 +1 permutations with one fixed point. 5

6 Only B occurs: In this case we get + ( ) n+ 4 n(n ) if n 0(mod ) n 1 x=1 1+ n x 1 y=n x x+y odd n 1+ n x 1 x= n 1 +1 y=x x+y odd n x=1 n x 1 y=x+1 x+y odd n+y even 1= + ( ) n+5 4 if n 1(mod ) (n 15)(n 1) + ( ) n+4 4 n(n ) (n 9)(n ) + ( ) n+ 4 (n 1)(n+) if n (mod ) if n (mod ) + ( ) n+8 4 if n 4(mod ) (n 5)(n ) + ( ) n+1 4 if n 5(mod ) permutations with one fixed point. Only C occurs: In this case we get + ( ) n+ 4 n(n ) if n 0(mod ) n 1 x=1 x y=1 n+y odd 1+ n n x 1 x= n 1 +1 y=1 n+y odd 1+ n x=1 n x 1 y=x+1 x+y even n+y odd 1= + ( ) n+5 4 if n 1(mod ) (n 15)(n 1) + ( ) n+4 4 n(n ) (n 9)(n ) + ( ) n+ 4 (n 1)(n+) if n (mod ) if n (mod ) + ( ) n+8 4 if n 4(mod ) (n 5)(n ) + ( ) n+1 4 if n 5(mod ) permutations with one fixed point. Summing over all cases yields the stated formula. Theorem.4 For n 1, s n 1 n (1, 1) = 0, s n n(1, 1) = 1, 5 4 n 1 if i =0 s 0 n+i(1, 1) = (4 n 1) if i =1,

7 s k n+i(1, 1) = s k+1 n+i (1, 1) = 4 n k 1 if i =0 0 if i =1 0 if i =0 4 n k 1 if i =1 for 1 k n 1, and for 0 k n 1. Proof. The proof is very similar to the proof of Theorem. and we provide a sketch of this proof. It is easy to prove by induction that π =(π(1),π(),...,π(m)) where π(j) =(t j +1,t j +,...,t j 1 1,t j 1 ) for some 0 = t 0 <t 1 <t < <t m = n. We call π(j) aj-block of π. (See [Man] for more details.) Recall that the graph of π consists of the points (i, π i ) Z. For k fixed points, we see that the middle block of π =(π(1),...,π(m)) must be of length k and must contain all k fixed points and that n must be even. Hence, we have s k n = (s n k (1, 1)) =4 n k 1 and s k n+1 = 0 for 1 k n 1. Similarly, for k + 1 fixed points, we have s k+1 n+1 =(s n k (1, 1)) =4 n k 1 and s k+1 n =0 for 0 k n 1. Using a result in [SiSc] gives the stated formula for s 0 n(1, 1). Theorem.5 For n, s 0 n(1, 1) = 1 (n 1 +( 1) n ), s k n(1, 1) = (n k 1 + ( 1) n k ) for 1 k n, s n 1 n (1, 1) = 0, and s n n(1, 1) = 1. Proof. Let π S n (1, 1). Clearly, we must have either π 1 = n or π n = n. For the case π n = n we have s k 1 n 1(1, 1) permutations. Hence, we need only consider π 1 = n. Write π =(n, π(1), 1,π()). We must have π(1) be a decreasing sequence and π() be an increasing sequence. Hence, we cannot have a fixed point in π() and we can have at most one fixed point in π(1). Thus, for k the case π 1 = n is empty giving s k n(1, 1) = s k 1 n 1(1, 1) for k. It remains to consider k =0, 1 for π =(n, π(1), 1,π()). We consider k = 0 first (the case k = 1 will follow immediately). We may place the elements,,...,n 1in n ways such that π(1) is decreasing and π() is increasing. To see this, insert the elements,,...,n 1 in reverse order at 1 s position and slide the element all the way to the left or all the way to right. However, we must subtract those permutations which contain a fixed point, which may only occur in π(1). 7

8 Let f be a fixed point and write π(1) = (σ(1),f,σ()). We have ( ) n f 1 f choices for the elements in σ(1). Once these are chosen, there are then f ways to place the elements,,...,f 1intoσ() and π() such that σ() is decreasing and π() is increasing. Thus, we have n+1 ( ) n f 1 s 0 n(1, 1) = n f. f From here, it is easy to check that s 0 n(1, 1) = s 0 n 1(1, 1) + s 0 n (1, 1), which gives s 0 n(1, 1) = 1 (n 1 +( 1) n ) for s 0 1(1, 1) = 0 and s 0 (1, 1) = 1. f= For the case k = 1, the above argument shows that s 1 n(1, 1) = s 0 n 1(1, 1) + n+1 f= ( n f 1 f ) f. From here, it is easy to check that s 1 n(1, 1) = s 1 n 1(1, 1) + s 1 n (1, 1), which gives s 1 n(1, 1) = (n +( 1) n 1 ) for s 1 1(1, 1) = 1 and s 1 (1, 1) = 0. Remark. s 0 n(1, 1) = J n where J n is the n th Jacobsthal number. Theorem. For n 1, s k n(1, 1) = n k 1 for 0 k n 1 and s n n(1, 1) = 1. Proof. Let π S n (1, 1). It is easy to see that in order to avoid both 1 and 1 we have either π =(π,n) where π S n 1 (1, 1) or π =(j + 1)(j +) (n 1)n1...j for some 1 j n 1. Adjusting for the number of fixed points we see that n 1 s k n(1, 1) = s k 1 n 1(1, 1) + I k=0. j=1 A straightforward induction on k for 0 k<nfinishes the proof. As a corollary of Theorem. we rederive a result given in [SiSc]. Corollary.7 s n (1, 1) = ( n ) +1 Proof. Summing over 0 k n we obtain the result immediately. 8

9 Theorem.8 For n 1, ( ( n k n+k ) ( n+k ) ) s k n k + n k for n + k even n(1, 1) = 0 otherwise Proof. For π S n (1, 1) it is easy to see that we must have π = π(1)n(n 1) j with π(1) Sj 1(1, k 1) S k 1 j 1 (1, 1) (depending on the parity of n+j) for some 1 j n. Hence, we have s k n(1, 1) = n 1 j=1 n j odd s k j 1(1, 1) + n j=1 n j even s k 1 j 1 (1, 1). From here we see that s k n(1, 1) = s k n (1, 1) + s k 1 n 1(1, 1). Hence, using initial conditions it is easy to check that the formula given is correct. Theorem.9 Let G k (x) be the generating function for {s k n(1, 1)} n 0. Then G k (x) = x k (1 x) k+1 (1 x x ) k+1. In particular, s 0 n(1, 1) = F n, for n, where {F n } n 0 is the Fibonacci sequence initialized by F 0 = F 1 =1. Proof. Let π S n (1, 1). It is easy to see that we must have π = (π,σ) where π S k I j=1 n j (1, 1) and σ = n(n j + 1)(n j +) (n 1) for some 1 j n. Hence, we have s k n(1, 1) = s k 1 n 1(1, 1) + n j= sk n j(1, 1) for 0 k n. This implies that s k n(1, 1) = s k 1 n 1(1, 1) + s k n (1, 1) + s k n 1(1, 1) s k 1 n (1, 1). For k =0wehaves 0 n(1, 1) = s 0 n 1(1, 1) + s 0 n (1, 1), from which it follows that G 0 (x) = 1 x. For k>0wehaveg 1 x x k (x) = x(1 x) G 1 x x k 1 (x). From here all of the results follow. The following corollary rederives a result in [SiSc]. Corollary.10 n 0 s n(1, 1)x n = 1 x 1 x, i.e. s n(1, 1) = n 1 for n 1. Proof. Summing k 0 G k(x) gives the desired result. We now make the following definition. 9

10 Definition.1 Let S 1,S S m. If s k n(s 1 )=s k n(s ) for all 0 k n for all n m we say that S 1 and S are Super-Wilf equivalent. Furthermore, we say that S 1 and S are in the same Super-Wilf class. Remark. Using the above definition and results from [RSZ] we see that 1, 1, and 1 are Super-Wilf equivalent, 1 and 1 are Super-Wilf equivalent, and the eight cases given at the beginning of Section provide eight Super-Wilf classes. The Case T = Using I and RC we see that we have the following cases, except that case (1) is grouped together because both 1 and 1 are to be avoided, which is not possible for n 5. (1) {1, 1, 1} = {{1, 1, 1}, {1, 1, 1}, {1, 1, 1}, {1, 1, 1}} () {1, 1, 1} = {{1, 1, 1}} () {1, 1, 1} = {{1, 1, 1}, {1, 1, 1}, {1, 1, 1}, {1, 1, 1}} (4) {1, 1, 1} = {{1, 1, 1}} (5) {1, 1, 1} = {{1, 1, 1}, {1, 1, 1}} () {1, 1, 1} = {{1, 1, 1}} (7) {1, 1, 1} = {{1, 1, 1}, {1, 1, 1}} (8) {1, 1, 1} = {{1, 1, 1}, {1, 1, 1}, {1, 1, 1}, {1, 1, 1}} (9) {1, 1, 1} = {{1, 1, 1}} Theorem.1 Let α {1, 1, 1, 1}. Then {s 0 n(1, 1)} n 0 =1, 0, 1,, 1, 0, 0,... for α 1, 1 {s 0 n(1, 1)} n 0 =1, 0, 1, 1, 1, 0, 0,... for α = 1 or α = 1 {s 1 n(1,α,1)} n 0 =0, 1, 0,, 0, 0,... for α 1, 1 {s 1 n(1,α,1)} n 0 =0, 1, 0, 1, 0, 0,... for α = 1 or α = 1 {s n(1,α,1)} n 0 =0, 0, 1, 0, 0..., for any α, and s k n(1,α,1) = 0 for any α, for all k n. Proof. Obvious. 10

11 Theorem. For n, s k n(1, 1, 1) = 0 for k n, s n(1, 1, 1) = F n I neven, s 1 n(1, 1, 1) = F n 1 I n odd, and s 0 n(1, 1, 1) = F n F n F n F n 1 if n is even if n is odd where {F n } n 0 is the Fibonacci sequence initialized by F 0 = F 1 =1., Proof. Clearly, s k n(1, 1, 1) = 0 for k since three fixed points are a 1 pattern. We may then use a result from from [SiSc] to get s 0 n(1, 1, 1) = s n (1, 1, 1) s 1 n(1, 1, 1) s n(1, 1, 1) = F n s 1 n(1, 1, 1) s n(1, 1, 1). Hence, we only need consider k =1,. We start with k =. The fixed points must be adjacent in order to avoid all of 1, 1, 1. Let the fixed points be j and j + 1. Write π =(π(1),j,j +1,π()). Then π(1) is on the elements {n j +,...,n}. To avoid the 1 pattern, we require n j + j + and n j +1 j + 1. Hence, j = n (so that n must be even). Thus, since it is not possible to have fixed points in π(1) or π() we have s n(1, 1, 1) = (s n (1, 1, 1)) I neven, which, using a result from [SiSc], gives the stated formula. Now consider k = 1. Let j be the fixed point. Note that j 1,n in order to avoid all of 1, 1, 1. Write π =(π(1),j,π()). In order to avoid 1 and 1, there exists at most one x π(1) with x<j. Assume, for a contradiction, that such an x exists. Then we must have x = π j 1 in order to avoid the 1 pattern. Next, for z π() we must have z<jelse xjz is a 1 pattern. This forces x = j 1 to be a fixed point, a contradiction. Hence, no such x exists so that for all i π(1) we have i>j. In order to avoid 1 and 1, there exists at most one y π() with y>j. Assume, for a contradiction, that such a y exists. Then we must have y = π j+1 in order to avoid the 1 pattern. Furthermore, for all i π(1) we must have i>jso as not to have a 1 pattern with ijy. This forces y = j +1tobe a fixed point, a contradiction. Hence, no such y exists. Since we have i π(1), i>jand k π(), k<j, and for j to be a fixed point we require n to be odd so that j = n+1, we get s 1 n(1, 1, 1) = (s n 1 (1, 1, 1)) I n odd, which, using a result from [SiSc], gives the stated formula. Theorem. For n, s 0 n(1, 1, 1) = n, s1 n(1, 1, 1) = n +( 1)n+1, s n(1, 1, 1) = 1 (1 + ( 1)n ), and s k n(1, 1, 1) = 0 for k n. Proof. Let π S n (1, 1, 1). It is easy to see that we must have π = n(n 1) (n j +1)(n j 1)(n j ) 1(n j) for some 0 j n 1. From here the results follow easily. 11

12 Theorem.4 For n, s 0 n(1, 1, 1) = s n(1, 1, 1) = n (1 I n odd), s 1 n(1, 1, 1) = n(1 I neven ), and s k n(1, 1, 1) = 0 for k n. Proof. Let π S n (1, 1, 1). It is easy to see that we must have π = j(j 1) 1n(n 1) (j + 1). From here the results follow easily. Theorem.5 For n, s 0 n(1, 1, 1) = n I n odd +( n +1)I neven, s 1 n(1, 1, 1) = n I n odd, and s k n(1, 1, 1) = I n=k for k n. Proof. Let π S n (1, 1, 1). It is easy to see that we must have π = n(n 1) (n j +1)1 (n j) for some 1 j n. From here the results follow easily. Theorem. For n, s 0 n(1, 1, 1) = n 1 and s k n(1, 1, 1) = I n=k for 1 k n. Proof. Let π S n (1, 1, 1). Then we must have π = j(j +1) n1 (j 1) for some 1 j n. From here the results follow easily. Theorem.7 For n, s 0 n(1, 1, 1) = 1 (1+( 1)n ), and for k 1, { 1+( 1) s k n if n>k n (1, 1, 1) = 1 if n =k, and { 1+( 1) s k 1 n+1 if n>k 1 n (1, 1, 1) = 1 if n =k 1. Proof. Let π S n (1, 1, 1). It is easy to see that we must have π = j(j 1) 1(j + 1)(j +) n for some 1 j n. From here the results follow easily. Theorem.8 For n, 1 if 0 k n s k n(1, 1, 1) = 0 if k = n 1 1 if k = n. Proof. Let π S n (1, 1, 1). It is easy to see that we must have π = j1 (j 1)(j + 1) n for some 1 j n. From here the results follow easily. 1

13 Theorem.9 For n and 0 k n, s k n(1, 1, 1) = ( ) n+k In+k even. k Proof. Let π Sn(1, k 1, 1). Then π must be of the form (1,π )or(, 1,π ) where π Sn 1(1, k 1 1, 1) and π Sn 1(1, k 1, 1). This gives us s k n(1, 1, 1) = s k 1 n 1(1, 1, 1) + s k n (1, 1, 1). A straightforward induction on n + k finishes the proof. 4 The Cases T 4 The cases T 4 are easy and in fact s k n(t ) {0, 1, } for all T S, T 4, for all n 1 and 0 k n. Acknowledgment We thank an anonymous referee for a careful reading and helpful corrections. References [Man] T. Mansour, Permutations Avoiding a Pattern from S k and at Least Two Patterns from S, Ars Combinatoria (001), [RSZ] A. Robertson, D. Saracino, and D. Zeilberger, Refined Restricted Permutations, arxiv: math.co/000 [SiSc] R. Simion and F. Schmidt, Restricted Permutations, European Journal of Combinatorics (1985),