Toufik Mansour 1. Department of Mathematics, Chalmers University of Technology, S Göteborg, Sweden

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1 COUNTING OCCURRENCES OF 32 IN AN EVEN PERMUTATION Toufik Mansour Department of Mathematics, Chalmers University of Technology, S-4296 Göteborg, Sweden Abstract We study the generating function for the number of even or odd) permutations on n letters containing exactly r 0 occurrences of 32 It is shown that finding this function for a given r amounts to a routine check of all permutations in S 2r 2000 Mathematics Subject Classification: Primary 05A05, 05A5; Secondary 05C90 Introduction Let [n] = {, 2,, n} and S n denote the set of all permutations of [n] We shall view permutations in S n as words with n distinct letters in [n] A pattern is a permutation σ S k, and an occurrence of σ in a permutation π = π π 2 π n S n is a subsequence of π that is order equivalent to σ For example, an occurrence of 32 is a subsequence π i π j π k i < j < k n) of π such that π i < π k < π j We denote by τπ) the number of occurrences of τ in π, and we denote by s r σn) the number of permutations π S n such that σπ) = r In the last decade much attention has been paid to the problem of finding the numbers s r σn) for a fixed r 0 and a given pattern τ see [, 2, 3, 5, 6, 7, 0,, 4, 5, 6, 7, 8, 9]) Most of the authors consider only the case r = 0, thus studying permutations avoiding a given pattern Only a few papers consider the case r > 0, usually restricting themselves to patterns of length 3 Using two simple involutions reverse and complement) on S n it is immediate that with respect to being equidistributed, the six patterns of length three fall into the two classes {23, 32} and {32, 23, 23, 32} Noonan [3] proved that s 23n) = 3 ) 2n n n 3 A general approach to the problem was suggested by Noonan and Zeilberger [4]; they gave another proof of Noonan s result, and conjectured that and s 2 23n) = 59n2 + 7n n2n )n + 5) s 32n) = ) 2n 3 n 3 ) 2n n 4 Research financed by EC s IHRP Programme, within the Research Training Network Algebraic Combinatorics in Europe, grant HPRN-CT

2 2 COUNTING OCCURRENCES OF 32 IN AN EVEN PERMUTATION The first conjecture was proved by Fulmek [8] and the latter conjecture was proved by Bóna in [6] A conjecture of Noonan and Zeilberger states that s r σn) is P -recursive in n for any r and τ It was proved by Bóna [4] for σ = 32 Mansour and Vainshtein [] suggested a new approach to this problem in the case σ = 32, which allows one to get an explicit expression for s r 32n) for any given r More precisely, they presented an algorithm that computes the generating function n 0 sr 32n)x n for any r 0 To get the result for a given r, the algorithm performs certain routine checks for each element of the symmetric group S 2r The algorithm has been implemented in C, and yields explicit results for r 6 Let π be any permutation The number of inversions of π is given by i π = {i, j) : π i > π j, i < j} The signature of π is given by signπ) = ) iπ We say π is an even permutation [respectively; odd permutation] if signπ) = [respectively; signπ) = ] We denote by E n [respectively; O n ] the set of all even [respectively; odd] permutations in S n Clearly, E n = O n = 2n! for all n 2 The following lemma holds immediately by definitions Lemma For any permutation π, signπ) = ) 2π) We denote by e r σn) [respectively; o r σn)] the number of even [respectively; odd] permutations π E n [respectively; π O n ] such that σπ) = r Apparently, for the first time the relation between even odd) permutations and pattern avoidance problem was suggested by Simion and Schmidet in [6] for σ S 3 In particularly, Simion and Schmidt [6] proved that ) e 0 2n 32n) = + ) ) n and o 0 2n 2n + ) n n + n )/2 32n) = ) n 2n + ) n n + n )/2 In this paper, as a consequence of [2], we suggest a new approach to this problem in the case of even or odd) permutations where σ = 32, which allows one to get an explicit expression for e r 32n) for any given r More precisely, we present an algorithm that computes the generating functions E r x) = n 0 er 32n)x n and O r x) = n 0 or 32n)x n for any r 0 To get the result for a given r, the algorithm performs certain routine checks for each element of the symmetric group S 2r The algorithm has been implemented in C, and yields explicit results for 0 r 6 2 Recall definitions and preliminary results To any π S n we assign a bipartite graph G π in the following way The vertices in one part of G π, denoted V, are the entries of π, and the vertices of the second part, denoted V 3, are the occurrences of 32 in π Entry i V is connected by an edge to occurrence j V 3 if i enters j For example, let π = , then π contains 5 occurrences of 32, and the graph G π is presented on Figure Figure Graph G π for π =

3 COUNTING OCCURRENCES OF 32 IN AN EVEN PERMUTATION 3 We denote by G n π the connected component of G π containing entry n Let πi ),, πi s ) be the entries of π belonging to G n π, and let σ = σ π S s be the corresponding permutation We say that πi ),, πi s ) is the kernel of π and denote it ker π, σ is called the shape of the kernel, or the kernel shape, s is called the size of the kernel, and the number of occurrences of 32 in ker π is called the capacity of the kernel For example, for π = as above, the kernel equals 4283, its shape is 4253, the size equals 5, and the capacity equals 4 Theorem 2 Mansour and Vainshtein [2, Theorem ]) Let π S n such that 32π) = r, then the size of the kernel of π is at most 2r + We say that ρ is a kernel permutation if it is the kernel shape for some permutation π Evidently ρ is a kernel permutation if and only if σ ρ = ρ Let ρ S s be an arbitrary kernel permutation We denote by Sρ) the set of all the permutations of all possible sizes whose kernel shape equals ρ For any π Sρ) we define the kernel cell decomposition as follows The number of cells in the decomposition equals ss + ) Let ker π = πi ),, πi s ); the cell C ml = C ml π) for l s + and m s is defined by C ml π) = {πj) i l < j < i l, πi ρ m )) < πj) < πi ρ m))}, where i 0 = 0, i s+ = n +, and π0) = 0 for any π If π coincides with ρ itself, then all the cells in the decomposition are empty An arbitrary permutation in Sρ) is obtained by filling in some of the cells in the cell decomposition A cell C is called infeasible if the existence of an entry a C would imply an occurrence of 32 that contains a and two other entries x, y ker π Clearly, all infeasible cells are empty for any π Sρ) All the remaining cells are called feasible; a feasible cell may, or may not, be empty Consider the permutation π = The kernel of π equals 3845, its shape is 423 The cell decomposition of π contains four feasible cells: C 3 = {2}, C 4 =, C 5 = {}, and C 4 = {6, 7} see Figure 2) All the other cells are infeasible; for example, C 32 is infeasible, since if a C 32, then aπ i 2 )π i 4 ) is an occurrence of 32 for any π whose kernel is of shape 423 m C 32 l Figure 2 Kernel cell decomposition for π S423) As another example, permutation π = belongs to the same class S423) Its kernel is , and the feasible cells are C 3 = {4, 6, 5}, C 4 = {3}, C 5 = {, 2}, C 4 = {, 0}

4 4 COUNTING OCCURRENCES OF 32 IN AN EVEN PERMUTATION Given a cell C ij in the kernel cell decomposition, all the kernel entries can be positioned with respect to C ij We say that x = πi k ) ker π lies below C ij if ρk) < i, and above C ij if ρk) i Similarly, x lies to the left of C ij if k < j, and to the right of C ij if k j As usual, we say that x lies to the southwest of C ij if it lies below C ij and to the left of it; the other three directions, northwest, southeast, and northeast, are defined similarly Let us define a partial order on the set of all feasible cells by saying that C ml C m l C ml if m m and l l Theorem 22 Mansour and Vainshtein [2]) i) is a linear order ii) Let C ml and C ml be two nonempty feasibly cells such that l < l Then for any pair of entries a C ml, b C ml, one has a > b iii) Let C ml and C m l be two nonempty feasibly cells such that m < m Then any entry a C ml lies to the right of any entry b C m l Let π be any permutation with a kernel permutation ρ, and assume that the feasible cells of the kernel cell decomposition associated with ρ are ordered linearly according to, C, C 2,, C fρ) Let d j be the size of C j For example, let π = with kernel permutation ρ = 423, as on Figure 2, then d = 2, d 2 =, d 3 = 0, and d 4 = We denote by l j ρ) the number of the entries of ρ that lie to the north-west from C j or lie to the south-east from C j For example, let ρ = 423, as on Figure 2, then l ρ) = 3, l 2 ρ) = 2, l 3 ρ) = 3, and l 4 ρ) = 4 Clearly, l ρ) = sρ) and l fρ) = sρ) for any nonempty kernel permutation ρ Lemma 23 For any permutation π with a kernel permutation ρ, ) fρ) i j fρ) signπ) = ) didj+ fρ) djljρ) signρ) signc j ) Proof To verify this formula, let us count the number of occurrences of the pattern 2 in π There four possibilities for an occurrence of 2 in π The first possibility is an occurrence occurs in one of the cells C j, so in this case there are fρ) 2Cj ) occurrences The second possibility is an occurrence occurs in ρ, so there are 2ρ) occurrences The third possibility is an occurrence of two elements which the first belongs to ρ and the second belongs to C i, so there are fρ) d jl j ρ) see Theorem 22) occurrences The fourth possibility is an occurrence of two elements which the first belongs to C i and the second belongs to C j where i < j Theorem 22 yields every entry of C i is greater than every entry of C j for all i < j), so there are i<j fρ) d id j occurrences Therefore, by Lemma we have equivalently, signπ) = ) signπ) = ) fρ) 2Cj) ) 2ρ) ) fρ) ) i j fρ) didj+ fρ) djljρ) djljρ) ) i<j fρ) didj, signρ) fρ) signcj ) We say the vector v = v, v 2,, v n ) is a binary vector if v i {0, } for all i, i n We denote the set of all binary vectors of length n by B n For any v B n, we define v = v + v v n For example, B 2 = {0, 0), 0, ),, 0),, )} and,, 0, 0, ) = 3 Let ρ be any kernel permutations, we denote by X ρ a [respectively; Y ρ a ] the set of all the binary vectors v B fρ) such that ) v +sρ) = a [respectively; ) v = a] For any v B fρ), we define z ρ v) = ) i<j fρ) vivj+ fρ) ljρ)vj signρ)

5 COUNTING OCCURRENCES OF 32 IN AN EVEN PERMUTATION 5 Let ρ be any kernel permutations and v = v, v 2,, v fρ) ) B fρ), we denote by Sρ; v) the set of all permutations of all sizes with kernel permutation ρ such that the corresponding cells C j satisfy ) dj = ) vj, in such a context v is called a length argument vector of ρ By definitions, the following result holds immediately Lemma 24 For any kernel permutation ρ, Sρ) = v B fρ) Sρ; v) Let ρ be any kernel permutations and let v = v, v 2,, v fρ) ), u = u, u 2,, u fρ) ) B fρ), we denote by Sρ; v, u) the set of all permutations in Sρ; v) such that the corresponding cells C j satisfy signc j ) = if and only if u j = 0, in such a context u is called a signature argument vector of ρ By Lemma 24, the following result holds immediately Lemma 25 For any kernel permutation ρ, Sρ) = Sρ; v) = v B fρ) For any a, b {0, } we define H r a, b) = v B fρ) u B fρ) Sρ; v, u) 2 E rx) + ) a E r x)), b = 0 2 O rx) + ) a O r x)), b = By definitions, the following result holds immediately Lemma 26 Let a, b {0, } The generating function for all permutations π such that 32π) = r, ) π = ) a, and signπ) = ) b is given by H r a, b) 3 Main Theorem The main result of this note can be formulated as follows Denote by K the set of all kernel permutations, and by K t the set of all kernel shapes for permutations in S t Let ρ be any kernel permutation, for any a, b {0, } and any r,, r fρ) we define L ρ r,,r fρ) a, b) = v X ρ ) a Theorem 3 Let r For any a, b {0, }, 3) H r a, b) = ρ K 2r+ u Y ρ ) b zρv) r + +r fρ) =r cρ), r j 0 fρ) H rj v j, u j ) L ρ r,,r fρ) a, b) Proof Let us fix a kernel permutation ρ K 2r+, a length argument vector v = v,, v fρ) ) X ) aρ), and a signature argument vector u = u,, u fρ) ) Y fρ) ) b z ρv) Recall that the kernel ρ of any π contains exactly cρ) occurrences of 32 The remaining r cρ) occurrences of 32 are distributed between the feasible cells of the kernel cell decomposition of π By Theorem 22, each occurrence of 32 belongs entirely to one feasible cell, and the occurrences of 32 in different cells do not influence one another

6 6 COUNTING OCCURRENCES OF 32 IN AN EVEN PERMUTATION Let π be any permutation such that 32π) = r, signπ) = ) b and ) π = ) a together with a kernel permutations ρ, length argument vector v, and signature argument vector u Then by Lemma 25, the cells C j satisfy the following conditions: ) v j = 0 if and only if d j is an even number, 2) u j = 0 if and only if signc j ) =, 3) ) v++v fρ)+sρ) = ) a, and 4) ) u++u fρ) zρ v) = ) b Therefore, by Lemma 26 this contribution gives x sρ) r + +r fρ) =r cρ), r j 0 fρ) H rj v j, u j ) Hence by Lemma 25 and [2, Theorem ], if summing over all the kernel permutations ρ K 2r+, length argument vectors v X ) aρ), and signature argument vectors u Y fρ) ) b z ρv) then we get the desired result Theorem 3 provides a finite algorithm for finding E r x) and O r x) for any given r 0, since we have to consider all permutations in S 2r+, and to perform certain routine operations with all shapes found so far Moreover, the amount of search can be decreased substantially due to the following proposition Proposition 32 The only kernel permutation of capacity r and size 2r + is ρ = 2r 2r + 2r 3 2r 2r 2j 3 2r 2j 4 2 Its parameters are given by sρ) = 2r +, cρ) = r, fρ) = r +2, signρ) =, and z ρ v,, v r+2 ) = ) +vr+2+ i<j r+2 vivj) Proof The first part of the proposition holds by [2, Prorposition] Besides, by using the form of ρ we get sρ) = 2r +, cρ) = r, fρ) = r + 2; signρ) =, and l j ρ) = 2r for all j =, 2,, r + and l r+2 ρ) = 2r + Therefore, z ρ v,, v r+2 ) = ) +vr+2+ i<j r+2 vivj) By this proposition, it suffices to search only permutations in S 2r Below we present several explicit calculations 3 The case r = 0 Let us start from the case r = 0 Observe that Theorem 3 remains valid for r = 0, provided the left hand side of Equation 3 for a = b = 0 is replaced by H r 0, 0) = 2 E rx) + E r x)) ; subtracting here accounts for the empty permutation So, we begin with finding kernel shapes for all permutations in S The only shape obtained is ρ =, and it is easy to see that sρ ) =, cρ ) = 0, fρ ) = 2, and X ρ ) = Y = {, 0), 0, )}, X ρ ) = Y = {0, 0),, )}, z ρ 0, 0) = z ρ, 0) = z ρ, ) = z ρ 0, ) =

7 COUNTING OCCURRENCES OF 32 IN AN EVEN PERMUTATION 7 Therefore, Equation 3 for a = b = 0 gives 32) 2 E 0x) + E 0 x)) = = xh 0, 0)H 0 0, 0) + xh 0, )H 0 0, ) + xh 0, 0)H 0 0, ) + xh 0, )H 0 0, 0), Equation 3 for a = and b = 0 gives 33) 2 E 0x) E 0 x)) = xh 2 0 0, 0) + xh 2 0 0, ) + xh 2 0, 0) + xh 2 0, ), Equation 3 for a = 0 and b = gives 34) 2 O 0x) + O 0 x)) = = xh 0, )H 0 0, 0) + xh 0, 0)H 0 0, ) + xh 0 0, 0)H 0, 0) + xh 0 0, )H 0, ), and Equation 3 for a = b = gives 35) 2 O 0x) O 0 x)) = 2xH 0 0, )H 0 0, 0) + 2xH 0, )H 0, 0) Out present aim to find explicitly E 0 x) and O 0 x), thus we need the following notation We define for all r 0 Clearly, M r x) = E r x) O r x) and F r x) = E r x) + O r x) H r 0, 0) H r 0, ) = 2 M rx) + M r x)), H r, 0) H r, ) = 2 M rx) M r x)), H r 0, 0) + H r 0, ) = 2 F rx) + F r x)), H r, 0) + H r, ) = 2 F rx) F r x)), for all r 0 Therefore, by subtracting respectively; adding) Equation 34 and Equation 32, and by subtracting respectively; adding) Equation 35 and Equation 33 we get { M0 x) + M 0 x) = 2 M 0 x) M 0 x) = xm 2 0 x) + M 2 0 x)) and { F0 x) + F 0 x) = 2 + xf 2 0 x) F 2 0 x)) F 0 x) F 0 x) = xf 2 0 x) + F 2 0 x)) Hence, M 0 x) = + 4x 2 2x and F 0 x) = 4x 2x Theorem 33 i) The generating function for the number of even permutations avoiding 32 is given by see [6]) E 0 x) = 4x + + ) 4x 2 2 2x 2x ii) The generating function for the number of odd permutations avoiding 32 is given by see [6]) O 0 x) = 4x ) 4x 2 2 2x 2x iii) The generating function for the number of permutations avoiding 32 is given by see [9]) F 0 x) = 4x 2x

8 8 COUNTING OCCURRENCES OF 32 IN AN EVEN PERMUTATION 32 The Case r = Since permutations in S 2 do not exhibit kernel shapes distinct from ρ, the only possible new shape is the exceptional one, ρ 2 = 32 Calculation of the parameters of ρ 2 gives sρ 2 ) = 3, cρ 2 ) =, fρ 2 ) = 3, X ρ 2 ) = Y = {, 0, 0), 0,, 0), 0, 0, ),,, )}, X ρ 2 ) = Y = {0, 0, 0),,, 0),, 0, ),,, 0)}, and z ρ2 0, 0, 0) = z ρ2, 0, 0) = z ρ2 0,, 0) = z ρ2,, 0) =, z ρ2 0, 0, ) = z ρ2, 0, ) = z ρ2 0,, ) = z ρ2,, ) = Therefore, by Theorem 3 we have 2H 0, 0) H 0, )) = = M x) + M x) = x3 2 M 0 x) M 0 x))m0 2 x) + M0 2 x)) 2H, 0) H, )) = = M x) M x) = 2xM 0 x)m x) + M 0 x)m x)) x3 2 M 0 x) + M 0 x))m0 2 x) + M0 2 x))), Using the expression for M 0 x) see the case r = 0) we get M x) = 2 + 3x + 2x2 ) + 3x 4x2 + 4x 3 4x 2 ) /2 2 Similarly, if considering the expressions for H 0, 0) + H 0, ) and H, 0) + H, ) we get F x) = 3x x ) + 4x) /2 2 2 Theorem 34 i) The generating function for the number of even permutations containing 32 exactly once is given by E x) = 2 2x x2 ) + 3x 4 4x) /2 + 3x 4x2 + 4x 3 4x 2 ) /2 4 ii) The generating function for the number of odd permutations containing 32 exactly once is given by O x) = 2 x + x2 ) + 3x 4x) /2 3x 4x2 + 4x 3 4x 2 ) /2 4 4 iii) The generating function for the number of permutations containing 32 exactly once is given by see [6]) F x) = 3x x ) + 4x) / The case r = 2 We have to check the kernel shapes of permutations in S 4 Exhaustive search adds four new shapes to the previous list; these are 243, 342, 423, and 243; besides, there is the exceptional 3542 S 5 Calculation of the parameters s, c, f, z, X a, Y a is straightforward, and we get Theorem 35 i) The generating function for the number of even permutations containing 32 exactly twice is given by E 2 x) = 2 xx3 + 3x 2 4x ) + 4 2x4 4x x 2 5x + 2) 4x) 3/2 4 6x7 48x 6 76x x x 3 2x 2 5x + 2) 4x 2 ) 3/2

9 COUNTING OCCURRENCES OF 32 IN AN EVEN PERMUTATION 9 ii) The generating function for the number of odd permutations containing 32 exactly once is given by O 2 x) = 2 x4 + 3x 3 5x 2 4x + 2) + 4 2x4 4x x 2 5x + 2) 4x) 3/ x7 48x 6 76x x x 3 2x 2 5x + 2) 4x 2 ) 3/2 iii) The generating function for the number of permutations containing 32 exactly twice is given by see [2]) F 2 x) = 2 x2 + 3x 2) + 2 2x4 4x x 2 5x + 2) 4x) 3/2 34 The cases r = 3, 4, 5, 6 Let r = 3, 4, 5, 6; exhaustive search in S 6, S 8, S 0, and S 2 reveals 20, 04, 503, and 2576 new nonexceptional kernel shapes, respectively, and we get Theorem 36 Let r = 3, 4, 5, 6, then M r x) = A r x) + B r x) 4x 2 ) r+/2) and F r x) = C r x) + D r x) 4x) r+/2) 2 2 where A 3 x) = 2x 6 + 0x 5 24x 4 30x x 2 + 7x 2, A 4 x) = 2x 8 + 4x 7 46x 6 90x 5 + 7x x 3 42x 2 8x +, A 5 x) = 2x 0 + 8x 9 76x 8 98x x x 5 355x 4 7x x 2 + 0x 2, A 6 x) = 256x 3 446x 2 68x + 94x 0 40x x x 7 702x 6 430x x x 3 88x 2 5x + 4, B 3 x) = 64x 320x 0 800x x x 7 972x 6 524x x x 3 43x 2 7x + 2, B 4 x) = 256x x x 3 320x x x x x x x x 5 775x 4 97x x 2 + 8x, B 5 x) = 024x 9 926x x x x x x x x x x x x x x x x 3 98x 2 0x + 2, B 6 x) = x x x x x x x x x x x x x x x x x x x x x 4 632x x 2 + 5x 4,

10 0 COUNTING OCCURRENCES OF 32 IN AN EVEN PERMUTATION C 3 x) = 2x 3 5x 2 + 7x 2, C 4 x) = 5x 4 7x 3 + 2x 2 + 8x 3, C 5 x) = 4x 5 7x 4 + x 3 6x 2 + 4x 2, and C 6 x) = 42x 6 44x 5 + 5x 4 + 4x 3 20x 2 + 9x 4, D 3 x) = 22x 6 06x x 4 302x x 2 27x + 2, D 4 x) = 2x x x 7 754x x x 4, D 5 x) = 50x 2568x x x x x x x x x 2 50x + 2, D 6 x) = 4x x x x x x x x x x x x x 2 07x + 4 Moreover, for r = 3, 4, 5, 6, E r x) = A r x) + C r x) + D r x) 4x) r+/2 + B r x) 4x 2 ) r+/2) 4 and O r x) = 4 A r x) C r x) + D r x) 4x) r+/2 B r x) 4x 2 ) r+/2) First of all, let us simplify the expression where a = 0,, r j 0 for all j 4 Further results and open questions L ρ r,,r fρ) a, 0) L ρ r,,r fρ) a, ), Lemma 4 Let v {0, } n be any vector, and let a {, } Then n H r v j, x j ) n n H r v j, y j ) = a g r j), x Y a y Y a where g r j) = H r v j, 0) H r v j, ) = 2 M rx) + ) vj M r x)) for all j Proof Let us define an order on the set B n, we say the vector v < u if there exists j such that u j + v j =, and u i = v i for all i j We say the 2 n vectors u,, u 2n of B n are satisfy the l-property if 0 = 0, 0,, 0) = u < u 2 < < u 2n, and we say the 2 n vectors u,, u 2n are satisfy the c-property if the vectors u,, u m),, u 2m,, u 2m m )

11 COUNTING OCCURRENCES OF 32 IN AN EVEN PERMUTATION are satisfy the l-property for all m =, 2,, n c-property by For example, the vectors of B 3 are satisfy the 0, 0, 0) <, 0, 0) <,, 0) < 0,, 0) < 0,, ) <,, ) <, 0, ) < 0, 0, ) First of all, let us prove by induction on n that there exists an order of the vectors of B n with c- property, for all n For n = the c-property holds with 0) < ) Suppose that there exists an order of the vector of B m with the c-property Let v j = u j,, uj m, 0) for all j =, 2,, 2 m and v 2m +j = u 2m + j,, u 2m + j m, ) for all j =, 2,, 2 m By definitions, v = 0, 0,, 0) and v < < v 2m+, so the l-property holds for m + Hence, by induction on m we get that there exists an order of the vector of B n with the c-property Now we are ready to prove the lemma Without loss of generality we can assume that 0, 0,, 0) Ya n which means a = ); otherwise it is enough to replace a by a Let x,, x 2n all the vectors of B n with the c-property Using 0, 0,, 0) Ya n and x 2i Y a n for all i =, 2,, 2 n Therefore, for all i =, 2,, 2 n, 2 n i= n ) H r v j, x 2i j ) n H r v j, x 2i j ) together with the c-property we get that x 2i Y n a = 2n i= = g r ) 2n 2 ) i g r ) n i= n j=2 H r ṽ j, y 2i j H r v j, x 2i j ) ) n ) = ) H r ṽ j, yj 2i), where y p = x 2p 2,, x2p n ) for all p =, 2,, 2 n, and ṽ = v 2, v 3,, v n ) Using the c-property for x,, x 2n we get that the vectors y,, y 2n are satisfy the c-property in B n Hence by induction on n by definitions the lemma holds for n = ), we get that the expression equals to a n g rj) As a remark, the vector 0, 0,, 0) Y ρ z ρv) if and only if z ρv) = for any kernel permutation ρ and vector v Therefore, by Theorem 3 and Lemma 4 we get the following result Theorem 42 Let a {0, } and r 0 Then 2 M rx) + ) a M r x)) δ r+a,0 = = ρ K 2r+ x sρ) r,,r fρ) =r cρ), r j 0 v X ) a ρ) ) 2 fρ) fρ) z ρ v) M rj x) + ) vj M rj x)) As a remark, the above theorem yields two equations for a = 0 and a = ) that are linear on M r x) and M r x) So, Theorem 42 provides a finite algorithm for finding M r x) for any given r 0, since we have to consider all permutations in S 2r+, and to perform certain routine operations with all shapes found so far Moreover, the amount of search can be decreased substantially due to the following proposition which holds immediately by Proposition 32 and Theorem 42 Proposition 43 Let r, a {0, }, and ρ = 2r 2r + 2r 3 2r 2r 2j 3 2r 2j 4 2 Then the expression x sρ) fρ) 2 fρ) z ρ v) M rj x) + ) vj M rj x)) r,,r fρ) =r cρ), r j 0 v X ) a ρ)

12 2 COUNTING OCCURRENCES OF 32 IN AN EVEN PERMUTATION is given by [r+2)/2] j=a ) r + 2 ) j a+ 2 r 2 x 2r+ M 0 x) M 0 x)) j M 0 x) + M 0 x)) r+2 j 2j + a By this proposition, it is sufficient to search only permutations in S 2r Besides, using Theorem 42 and the case r = 0 together with induction on r we get the following result Theorem 44 M r x) is a rational function on x and 4x 2 for any r 0 In view of our explicit results, we have even a stronger conjecture Conjecture 45 For any r, there exist polynomials A r x), B r x), C r x), and D r x) with integer coefficients such that E r x) = 4 A rx) + 4 B rx) 4x) r+/2 + 4 C rx) 4x 2 ) r+/2, O r x) = 4 D rx) + 4 B rx) 4x) r+/2 4 C rx) 4x 2 ) r+/2, where the coefficients of the polynomial A r x) + D r x) are divided by 2 References [] N Alon and E Friedgut On the number of permutations avoiding a given pattern J Combin Theory Ser A, 89):33 40, 2000 [2] M D Atkinson Restricted permutations Discrete Math, 95-3):27 38, 999 [3] M Bóna Exact enumeration of 342-avoiding permutations: a close link with labeled trees and planar maps J Combin Theory Ser A, 802): , 997 [4] M Bóna The number of permutations with exactly r 32-subsequences is P -recursive in the size! Adv in Appl Math, 84):50 522, 997 [5] M Bóna Permutations avoiding certain patterns: the case of length 4 and some generalizations Discrete Math, 75-3):55 67, 997 [6] M Bóna Permutations with one or two 32-subsequences Discrete Math, 8-3): , 998 [7] T Chow and J West Forbidden subsequences and Chebyshev polynomials Discrete Math, 204-3):9 28, 999 [8] M Fulmek Enumeration of permutations containing a presribed number of occurrences of a pattern of length three Adv Appl Math, to appear, mathco/02092 [9] DE Knuth The Art of Computer Programming, 2nd ed Addison Wesley, Reading, MA 973) [0] T Mansour Permutations containing and avoiding certain patterns In Formal power series and algebraic combinatorics Moscow, 2000), Springer, Berlin, 2000 [] T Mansour and A Vainshtein Restricted 32-avoiding permutations Adv Appl Math, 26: , 200 [2] T Mansour and A Vainshtein Counting occurrences of 32 in a permutation Adv Appl Math, 282):85 95, 2002 [3] J Noonan The number of permutations containing exactly one increasing subsequence of length three Discrete Math, 52-3):307 33, 996 [4] J Noonan and D Zeilberger The enumeration of permutations with a prescribed number of forbidden patterns Adv in Appl Math, 74):38 407, 996 [5] A Robertson Permutations containing and avoiding 23 and 32 patterns Discrete Math Theor Comput Sci, 34):5 54 electronic), 999 [6] R Simion and F W Schmidt Restricted permutations European J Combin, 64): , 985 [7] Z Stankova Forbidden subsequences Discrete Math, 32-3):29 36, 994 [8] Z Stankova Classification of forbidden subsequences of length 4 European J Combin, 75):50 57, 996 [9] J West Generating trees and the Catalan and Schröder numbers Discrete Math, 46-3): , 995