aa r X i v : . [ m a t h . C O ] D ec Cyclically consecutive permutation avoidance
Richard Ehrenborg
Abstract
We give an explicit formula for the number of permutations avoiding cyclically a consecutivepattern in terms of the spectrum of the associated operator of the consecutive pattern. As anexample, the number of cyclically consecutive 123-avoiding permutations in S n is given by n !times the convergent series P ∞ k = −∞ (cid:16) √ π ( k +1 / (cid:17) n for n ≥ The Euler number E n enumerates the number of alternating permutations in the symmetricgroup S n . The Euler number has the following classical expression E n = n ! · · (cid:18) π (cid:19) n +1 · ∞ X k = −∞ k + 1) n +1 . (1.1)Simons and Yao [8] were the first to the author’s knowledge to give a probabilistic–geometric argu-ment for this identity. See also [5, Section 4] and [7]. The underlying idea is that to obtain a randompermutation π where each permutation is equally likely, pick a random point x = ( x , x , . . . , x n )from the cube [0 , n . Next let the permutation π be the standardization of the vector x , that is, π i < π j is equivalent to x i < x j for all indexes i = j . Geometrically, this corresponds to pick thepermutation associated with the simplex containing the point x in the standard triangulation of thecube [0 , n . Hence the probability that the point is alternating, that is, · · · < x n − > x n − < x n isgiven by E n /n !. By considering the distribution of the last coordinate f n of the point x one obtainsthe recursion f n = T ( f n − ) where T is the operator given by T ( f ) = R x f (1 − t ) dt . This recursionis straightforward to solve using Fourier series. Integrating this series yields equation (1.1).Ehrenborg, Kitaev and Perry [4] continued this work by studying consecutive pattern avoidanceusing operators on the space L ([0 , m ). The largest eigenvalue of the associated operator yields theasymptotic behavior for the number of permutations. In fact, more knowledge of the eigenvaluesgives a better asymptotic expansion; see [4, Theorem 1.1]. However, a complete description ofthe eigenvalues is hard to find. It is is only known in three cases. First, consecutive 123 , n !; see Theorem 3.4.We end with open questions for further research. For a vector x = ( x , . . . , x k ) of k distinct real numbers, define Π( x ) to be standardization of thevector x , that is, the unique permutation σ = ( σ , . . . , σ k ) in S k such that for all indices 1 ≤ i < j ≤ k the inequality x i < x j is equivalent to σ i < σ j . Let S be a set of permutations in the symmetricgroup S m +1 . We say that a permutation π in S n avoids the set S consecutively if there is no index1 ≤ j ≤ n − m such that Π( π j , π j +1 , . . . , π j + m ) ∈ S . Similarly, a permutation π ∈ S n avoids theset S cyclically consecutively if there is no index 1 ≤ j ≤ n such that Π( π j , π j +1 , . . . , π j + m ) ∈ S ,where the indexes are modulo n .We consider the more general situation of weighted enumeration of consecutive patterns; see [3].Let wt be a real-valued weight function on the symmetric group S m +1 . Similarly, let wt , wt betwo real-valued weight functions on the symmetric group S m . We call wt and wt the initialand final weight function, respectively. We extend these three weight functions to the symmetricgroup S n for n ≥ m by definingWt( π ) = wt (Π( π , π , . . . , π m )) · n − m Y i =1 wt(Π( π i , π i +1 , . . . , π i + m )) · wt (Π( π n − m +1 , π n − m +2 , . . . , π n )) . Let α n be the sum of all the weights of permutations in S n , that is, α n = X π ∈ S n Wt( π ) . This framework can be used to study consecutive pattern avoidance by defining the weightwt( σ ) = ( σ S, σ ∈ S. Furthermore, let both the initial and final weight functions be the constant function . Then for n ≥ m , α n is the number of permutations in S n that avoid the set S .Define the function χ on the ( m + 1)-dimensional cube [0 , m +1 by χ ( x , x , . . . , x m +1 ) =wt(Π( x , x , . . . , x m +1 )). The associated linear operator T on the space L ([0 , m ) is defined by T ( f )( x , . . . , x m ) = Z χ ( t, x , . . . , x m ) · f ( t, x , . . . , x m − ) dt. (2.1)Similar to χ define the two functions κ and µ on the m -dimensional unit cube [0 , m by κ ( x ) =wt (Π( x )) and µ ( x ) = wt (Π( x )). Then the quantity α n is given by the inner product (cid:0) T n − m ( κ ) , µ (cid:1) = α n /n ! . α n one has to obtain the spectrum of the operator T .We now turn our attention to the power T n where n ≥ m . Begin by expanding χ by the product χ n ( x , x , . . . , x n ) = n − m Y i =1 χ ( x i , x i +1 , . . . , x i + m ) . Define the kernal K n ( x, y ) for x, y ∈ [0 , m for x = ( x , . . . , x m ) and y = ( y , . . . , y m ) by theintegral K n ( x, y ) = Z [0 , n − m χ n + m ( x , . . . , x m , t , . . . , t n − m , y , . . . , y m ) dt · · · dt n − m . Since T n f ( y ) = Z [0 , m K n ( x, y ) dx. (2.2)we conclude that T n is a Hilbert–Schmidt operator, and hence T n is compact. Define the cyclic weight of a permutation π in S n , where n ≥ m + 1, by the productWt c ( π ) = n Y i =1 wt(Π( π i , π i +1 , . . . , π i + m )) , where the indexes are modulo n . Similar to α n , we consider the sum of cyclic weights to be β n = X π ∈ S n Wt c ( π ) . We will write β n ( S ) when the weighting function is associated with avoiding a certain pattern S . Lemma 3.1.
The sum of the cyclic weights β n is given by the integral β n = n ! · Z [0 , n χ n + m ( x , x , . . . , x n , x , . . . , x m ) dx · · · dx n . Proof.
We evaluate the integral by partitioning the n -dimensional cube by the standard triangula-tion and noting that the function χ n + m is constant on each of the n ! simplexes, that is, Z [0 , n χ n + m ( x , x , . . . , x n , x , . . . , x m ) dx · · · dx n = X π ∈ S n Z x ∈ [0 , n Π( x )= π χ n + m ( x , x , . . . , x n , x , . . . , x m ) dx · · · dx n = X π ∈ S n Z x ∈ [0 , n Π( x )= π Wt c ( π ) dx · · · dx n = 1 n ! · X π ∈ S n Wt c ( π ) . heorem 3.2. Let ( λ k ) ≤ k ≤ K , where K ≤ ∞ , be a list of the eigenvalues of the operator T (including multiplicities), where T has the form given in equation (2.1) . Let n ≥ m + 1 . Assumethat the series P k λ nk converges absolutely. Then the sum of the cyclic weights is given by β n = n ! · K X k =1 λ nk . Proof.
Begin to observe that T n is an Hilbert-Schmidt operator. Since the sum P k | λ k | n convergesthen the operator T n is a trace class operator; see Section XI.8, Exercise 49 in [1]. The trace of theoperator is given by the convergent sum tr( T n ) = K X k =1 λ nk . But the trace of T n is also given by the following integral; see part (c) of the above mentionedexercise in [1]:tr( T n ) = Z [0 , m K ( x, x ) dx = Z [0 , n χ n + m ( x , . . . , x m , t , . . . , t n − m , x , . . . , x m ) dx · · · dx m dt dt n − m = β n /n ! , proving the result.We give two examples. Theorem 3.3.
The number of cyclically consecutive -avoiding permutations in the symmetricgroup S n for n ≥ is given by β n (123) = n ! · ∞ X k = −∞ √ π ( k + 1 / ! n . Proof.
Here we let the weight function wt on S be given by wt( σ ) = 1 − δ σ, , where δ denotes theKronecker delta. The eigenvalues are given by λ k = √ π ( k +1 / where k ranges over all the integers;see [4, Proposition 5.1]. Note for n ≥ P ∞ k = −∞ | λ k | n converges. By Theorem 3.2 theresult follows in the case n ≥
3. The case n = 2 is can be verified by a separate calculation.A different approach to prove this theorem is to cyclically shift the permutations such that π n = n . Now we are looking for 123-avoiding permutations π · · · π n − in S n − . However, we alsorequire that the last two entries yields a descent, since otherwise the three entries π n − π n − π n would have the 123 pattern. The number of such permutations in S n − can be calculated by( n − · ( T n − m ( κ ) , µ ) where κ is the constant function and µ is the 0 , µ ( x, y ) = 1 if x > y and µ ( x, y ) = 0 if x < y . However, Theorem 2.1 in [3] only gives an asymptotic expansion4or the desired quantity. Lastly, one has to do an analytic argument to show that the asymptoticexpansion actually gives a convergent series.As our next example, we consider the weighted example from [3, Section 6]. Let bb ( π ) and bb c ( π ) denote the number of double descents of the permutation π , respectively, the number ofcyclically double descents of the permutation π , that is, bb ( π ) = { i ∈ { , , . . . , n − } : π i < π i +1 < π i +2 } , bb c ( π ) = { i ∈ { , , . . . , n } : π i < π i +1 < π i +2 } , where the indexes are modulo n . Theorem 3.4.
The sum of bb c ( π ) over all permutations π in S n which cyclically do not have adouble ascent is given by n ! , that is, X π ∈ S nπ cyclically -avoiding bb c ( π ) = n ! . Proof.
Here the weight function wt on S be given by wt(132) = wt(213) = wt(231) = wt(312) = 1,wt(123) = 0 but wt(321) = 2. The associated operator has only one eigenvalue namely 1; see [3,Theorem 6.1]. Hence the result follows directly for n ≥
3. The three cases n ≤ Corollary 3.5.
The sum of bb ( π ) over all permutations in S n such that π = n , there is no doubleascent and which end with a descent, that is, π n − > π n , is given by ( n − .Proof. The weight of a permutation in the previous theorem does not change when it is cyclicallyshifted. Hence shift the permutation such that π = n and the sum is given by n ! /n .As an example, there are 9 permutations in S that are 123-avoiding, begin with the element 5and end with a descent. In lexicographic order they are 51432, 52143, 52431, 53142, 53241, 53421,54132, 54231, and 54321. Note that the 8 first permutations all have one double descent and thelast permutation has three double descents yielding the sum 8 · + 1 · = 4!.It is tempting to use [4, Proposition 7.2] to obtain a result for the number of cyclically con-secutive 123 , , Are there other operators of the form (2.1) where we can determine the spectrum? This is not astraightforward question since the spectrum for 213-avoiding permutations is given by the equationerf (cid:18) √ · λ (cid:19) = r π , where erf denotes the error function. Is is always true that the sum P k | λ k | m is absolutely con-vergent? In other words, is the operator T m always trace class? Does the generating function P n ≥ β n · z n /n ! have a nicer form than the generating function P n ≥ α n · z n /n !; see [2, 6].5t is interesting to note that Theorem 3.3 gives a exact expression for the number of cyclicallyconsecutive 123-avoiding permutations. However, for consecutive 123-avoiding permutations thecorresponding result [4, Theorem 5.4] only yields an asymptotic expansion.Since this result of Theorem 3.4 is purely combinatorial, it is natural to ask for a bijective proof. Acknowledgments
The author is grateful to Margaret Readdy for her comments on an earlier version of this paper.The author was partially supported by National Security Agency grant H98230-13-1-0280.
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