Enumeration of snakes and cycle-alternating permutations
aa r X i v : . [ m a t h . C O ] N ov ENUMERATION OF SNAKES ANDCYCLE-ALTERNATING PERMUTATIONS
MATTHIEU JOSUAT-VERG`ES
To the memory of Vladimir Arnol’d
Abstract.
Springer numbers are an analog of Euler numbers for the group of signedpermutations. Arnol’d showed that they count some objects called snakes, thatgeneralize alternating permutations. Hoffman established a link between Springernumbers, snakes, and some polynomials related with the successive derivatives oftrigonometric functions.The goal of this article is to give further combinatorial properties of derivativepolynomials, in terms of snakes and other objects: cycle-alternating permutations,weighted Dyck or Motzkin paths, increasing trees and forests. We obtain the gen-erating functions, in terms of trigonometric functions for exponential ones and interms of J -fractions for ordinary ones. We also define natural q -analogs, make a linkwith normal ordering problems and combinatorial theory of differential equations. Introduction
It is well-known that the Euler numbers E n defined by ∞ X n =0 E n z n n ! = tan z + sec z (1)count alternating permutations in S n , i.e. σ such that σ > σ < σ > . . . σ n . Thestudy of these, as well as other classes of permutations counted by E n , is a vast topicin enumerative combinatorics, see the survey of Stanley [21]. By a construction dueto Springer [20], there is an integer K ( W ) defined for any Coxeter group W suchthat K ( S n ) = E n . As for the groups of signed permutations, Arnol’d [1] introducedsome particular kind of signed permutations called snakes, counted by the numbers K ( S Bn ) and K ( S Dn ). Thus snakes can be considered as a “signed analog” of alternatingpermutations. Algorithmically, Arnol’d [1] gives a method to compute these integerswith recurrences organized in triangular arrays similar to the Seidel-Entriger triangleof Euler numbers E n (see for example [11]).In this article, we are mostly interested in the numbers S n = K ( S Bn ), which werepresviously considered by Glaisher [9, §§
109 and 119] in another context. Springer [20]shows that they satisfy ∞ X n =0 S n z n n ! = 1cos z − sin z , (2)and we call S n the n th Springer number (although there is in theory a Springer numberassociated with each Coxeter group, this name without further specification usually Date : May 28, 2018.2000
Mathematics Subject Classification.
Primary: 05A15, 05A19. Secondary: 11B83.Supported by the French National Research Agency ANR, grant ANR08-JCJC-0011, and the AustrianScience Foundation FWF, START grant Y463. refer to type B ). Another link between snakes and trigonometric functions has beenestablished by Hoffman [13], who studies polynomials associated with the successivederivatives of tan and sec, i.e. P n ( t ) and Q n ( t ) such that:d n d x n tan x = P n (tan x ) , d n d x n sec x = Q n (tan x ) sec x. (3)Among various other results, Hoffman proves that Q n (1) = S n , and that P n (1) = 2 n E n counts some objects called β -snakes that are a superset of type B snakes. A firstquestion we can ask is to find the meaning of the parameter t in snakes.The main goal of this article is to give combinatorial models of these derivativepolynomials P n and Q n (as well as another sequence Q ( a ) n related the with the a thpower of sec), in terms of several objects: • Snakes (see Definition 3.2). Besides type B snakes of Arnol’d and β -snakes ofHoffman [13], we introduce another variant. • Cycle-alternating permutations (see Definition 3.7). These are essentially the imageof snakes via Foata’s fundamental transform [17]. • Weighted Dyck prefixes and weighted Motzkin paths (see Section 3). These aretwo different generalizations of some weighted Dyck paths counted by E n . • Increasing trees and forests (see Section 4). The number E n +1 counts increasingcomplete binary trees, and E n “almost complete” ones, our trees and forests area generalization of these.Various exponential and ordinary generating functions are proved combinatorially us-ing these objects. They will present a phenomenon similar to the case of Euler numbers E n : they are given in terms of trigonometric functions for the exponential generatingfunctions, and in terms of continued fractions for the ordinary ones. For example,in Theorem 3.17, we prove bijectively a continued fraction for the (formal) Laplacetransform of (cos z − t sin z ) − a . While this result was known analytically since longago [22], we show here that it can be fully understood on the combinatorial point ofview.This article is organized as follows. In Section 2, we give first properties of thederivative polynomials, such as recurrence relations, generating functions, we alsointroduce a natural q -analog and link this with the normal ordering problem. InSection 3, we give combinatorial models of the derivative polynomials in terms ofsubset of signed permutations, and obtain the generating functions. In Section 4, wegive combinatorial models of the derivative polynomials in terms of increasing treesand forests in two different ways: via the combinatorial theory of differential equationsand via the normal ordering problem.2. Derivative polynomials
Definitions.
Generalizing Q n ( t ) defined in (3), we consider Q ( a ) n ( t ) such thatd n d x n sec a x = Q ( a ) n (tan x ) sec a x. (4)Besides Q n which is the case a = 1 in Q ( a ) n , of particular interest will be the case a = 2denoted by R n = Q (2) n . Since sec x is the derivative of tan x , it follows that there isthe simple relation P n +1 ( t ) = (1 + t ) R n ( t ). NUMERATION OF SNAKES AND CYCLE-ALTERNATING PERMUTATIONS 3
By differentiation of (3) and (4), one obtains the recurrence relations: P n +1 = (1 + t ) P ′ n , Q ( a ) n +1 = (1 + t ) dd t Q ( a ) n + atQ n , (5)together with P = t and Q ( a )0 = 1. Let us give other elementary properties which arepartly taken from [13]. Proposition 2.1.
The polynomial Q ( a ) n has the same parity as n , i.e. Q ( a ) n ( − t ) =( − n Q ( a ) n ( t ) , and P n has different parity. We have: P n (1) = 2 n E n , Q n (1) = S n , R n (1) = 2 n E n +1 . (6) (Recall that Q n = Q (1) n and R n = Q (2) n .) The generating functions are: ∞ X n =0 P n ( t ) z n n ! = sin z + t cos z cos z − t sin z , ∞ X n =0 Q ( a ) n ( t ) z n n ! = 1(cos z − t sin z ) a . (7) Proof.
The generating function of { Q ( a ) n } n ≥ is not present in the reference [13], butcan be obtained in the same way as in the particular case a = 1 (which is in [13]). Bya Taylor expansion and a trigonometric addition formula, we have: ∞ X n =0 Q ( a ) n (tan u ) sec a u z n n ! = sec a ( u + z ) = sec a u (cos z − tan u sin z ) a , (8)We can divide on both sides by sec a u , let t = tan u , and the result follows.These exponential generating functions will be obtained combinatorially in Section 3using classes of signed permutations. It is interesting to note that P ( z, t ) = P P n z n /n !is a M¨obius transformation as a function of t , in such a way that z ( t P ( z, t ))is a group homomorphism from R to the elliptic M¨obius transformations of C fixing i and − i . This has an explanation through the fact that P ( z, t ) = tan(arctan t + z ) , (9)and consequently, as observed in [13]: P ( z, P ( z ′ , t )) = P ( z + z ′ , t ) , (10)which is the concrete way to say that z ( t P ( z, t )) is a group homomorphism.2.2. Operators and q -analogs. Let D and U be operators acting on polynomials inthe variable t by D ( t n ) = [ n ] q t n − , U ( t n ) = t n +1 , (11)where [ n ] q = − q n − q . The first one is known as the q -derivative or Jackson derivative ,and an important relation is DU − qU D = I , where I is the identity. Definition 2.2.
Our q -analogs of the polynomials Q n and R n are defined by: Q n ( t, q ) = ( D + U DU ) n , R n ( t, q ) = ( D + DU U ) n . (12)Of course, 1 should be seen as t since the operators act on polynomials in t . Proposition 2.3.
We have Q n ( − t, q ) = ( − n Q n ( t, q ) , R n ( − t, q ) = ( − n R n ( t, q ) , (13) moreover Q n ( t, q ) and R n ( t, q ) are polynomials in t and q with nonnegative coefficientssuch that Q n ( t,
1) = Q n ( t ) and R n ( t,
1) = R n ( t ) . MATTHIEU JOSUAT-VERG`ES
Proof.
From DU − qU D = I , we can write D + U DU = ( I + qU ) D + U, D + DU U = ( I + q U ) D + (1 + q ) U, (14)then from (12) and (14) it follows that Q n +1 ( t, q ) = (1 + qt ) D (cid:0) Q n ( t, q ) (cid:1) + tQ n ( t, q ) , (15) R n +1 ( t, q ) = (1 + q t ) D (cid:0) R n ( t, q ) (cid:1) + (1 + q ) tR n ( t, q ) , (16)which generalize the recurrences for Q n and R n ( a = 1 and a = 2 in (5)). Theelementary properties given in the proposition can be proved recursively using (15)and (16).The fact that Q n ( t, q ) and R n ( t, q ) have simple forms in terms of D and U makes alink with normal ordering [2]. Given an expression f ( D, U ) of D and U , the problemis to find some coefficients c i,j such that we have the normal form : f ( D, U ) = X i,j ≥ c i,j U i D j . (17)When f ( D, U ) is polynomial, using the commutation relation DU − qU D = I we canalways find such coefficients and only finitely of them are non-zero. For example, thisis what we have done in (14). In Section 4, we will use some general results on normalordering to give combinatorial models of P n and Q n (when q = 1).What is interesting about normal ordering is that having the coefficients c i,j givemore insight on the quantity f ( D, U ). For example, consider the case of Q n ( t, q ) andlet f ( D, U ) = ( D + U DU ) n , then we have Q n ( t, q ) = ( D + U DU ) n X i,j ≥ c i,j U i D j X i ≥ c i, t i , (18) i.e. we can directly obtain the coefficients of Q n ( t, q ) through this normal form.2.3. Continued fractions.
Let D and U be the matrices of operators D and U inthe basis { t i } i ∈ N . Moreover, let W be the row vector ( t i ) i ∈ N and V be the columnvector ( δ i ) i ∈ N where we use Kronecker’s delta. We have W = (1 , t, t , . . . ), and: D = q (0)0 [2] q , U = , V = . (19)From the previous definitions, it follows that: Q n ( t, q ) = W ( D + U D U ) n V , R n ( t, q ) = W ( D + D U U ) n V , (20) D U − qU D = I, W U = tW , D V = 0 . (21)The point of writing this is that from (20) we only need the relations in (21) tocalculate Q n ( t, q ) and R n ( t, q ). Even more, if D , U , W , and V are a second setof matrices and vectors satisfying relations similar to (21), we also have Q n ( t, q ) = W ( D + U D U ) n V and R n ( t, q ) = W ( D + D U U ) n V . Indeed, when we havethe relation (17) for a given f ( D, U ), the coefficients c i,j are obtained only using the NUMERATION OF SNAKES AND CYCLE-ALTERNATING PERMUTATIONS 5 commutation relation, so the same identity also holds with either ( D , U ) or ( D , U ).We can take W = (1 , , , . . . ), V = V , D = D , and: U = t (0)1 tq tq (0) . . . . . . . (22)Then we have indeed D U − qU D = I , W U = tW , and D V = 0. What is niceabout this second set of matrices and vectors is that it enables us to make a link withcontinued fractions. Definition 2.4.
For any two sequences { b h } h ≥ and { λ h } h ≥ , let J ( b h , λ h ) denote J ( b h , λ h ) = 11 − b z − λ z − b z − λ z . . . . (23)We will always use z as variable and h as index of the two sequences so that thereshould be no ambiguity in the notation.These are called J -fractions, or Jacobi continued fractions. They are linked withmoments of formal orthogonal polynomials, but let us give the results for Q n ( t, q ) and R n ( t, q ) before more details about this. Proposition 2.5.
We have: ∞ X n =0 Q n ( t, q ) z n = J ( b Qh , λ Qh ) , ∞ X n =0 R n ( t, q ) z n = J ( b Rh , λ Rh ) , (24) where: b Qh = tq h ([ h ] q + [ h + 1] q ) , b Rh = tq h (1 + q )[ h + 1] q , (25) λ Qh = (1 + t q h − )[ h ] q , λ Rh = (1 + t q h )[ h ] q [ h + 1] q . (26) Proof.
First, observe that for any matrix M = ( m i,j ) i,j ∈ N , the product W M n V isthe upper-left coefficient ( M n ) , of M n . We can obtain this coefficient W M n V thefollowing way: W M n V = X i ,...,i n − ≥ m ,i m i ,i . . . m i n − ,i n − m i n − , . (27)When the matrix is tridiagonal, we can restrict the sum to indices such that | i j − i j +1 | ≤
1, so that these indices are the successive heights in a Motzkin path. Then (27) showsthat W M n V is the generating function of Motzkin paths of length n with someweights given by the coefficients of M , and using a classical argument [8] it followsthat P ∞ n =0 ( W M n V ) z n = J ( m h,h , m h − ,h m h,h − ).In the present case, it suffices to check that D + U D U and D + D U U aretridiagonal and calculate explicitly their coefficients to obtain the result. See also (34)and (36) below for a more general result.In particular, it follows from the previous proposition that we have: Q n (0 , q ) = E n ( q ) , R n (0 , q ) = E n +1 ( q ) , (28) MATTHIEU JOSUAT-VERG`ES where E n ( q ) and E n +1 ( q ) are respectively the q -secant and q -tangent numbers de-fined by Han, Randrianarivony, Zeng [12] using continued fractions. Remark 2.6.
From the fact that Q n (1) = S n and R n (1) = 2 n E n +1 , the previousproposition implies ∞ X n =0 S n z n = J (2 h + 1 , h ) , ∞ X n =0 E n +1 z n = J ( h + 1 , h ( h +1)2 ) . (29)We found only one reference mentioning the latter continued fraction, Sloane’s OEIS[19], but there is little doubt it can be proved by classical methods. For example,using a theorem of Stieltjes and Rogers [10, Theorem 5.2.10], the continued fractioncan probably be obtained through an addition formula satisfied by the exponentialgenerating function of { E n +1 } n ≥ , this function being the derivative of tan z + sec z ,explicitly (1 − sin z ) − . Combinatorially, the result can be proved using Andr´e trees [7,Section 5] and the bijection of Fran¸con and Viennot [10, Chapter 5]. More generally,this shows that if we add a parameter x , then J ( h + 1 , x h ( h +1)2 ) is the generatingfunction of the Andr´e polynomials defined by Foata and Sch¨utzenberger [7].An important property of J -fractions is the link with moments of (formal) orthog-onal polynomials, and we refer to [8, 23] for the relevant combinatorial facts. We con-sider here the continuous dual q -Hahn polynomials p n ( x ; a, b, c | q ), or p n ( x ) for short.In terms of the Askey-Wilson polynomials which depend on one other parameter d , p n ( x ) is just the specialization d = 0 (see [15] for the definitions of these classicalsequences, but our notations differ, in particular because of a rescaling x → x/ xp n ( x ) = p n +1 ( x ) + (cid:0) a + a − A n − C n (cid:1) p n ( x ) + A n − C n p n − ( x ) , (30)together with p − ( x ) = 0 and p ( x ) = 1, where A n = a (1 − abq n )(1 − acq n ) and C n = a (1 − q n )(1 − bcq n − ) . (31)The p n ( x ) are orthogonal with respect to the scalar product ( f, g ) L ( f g ) where L is the linear form such that ∞ X n =0 L ( x n ) z n = J (cid:0) a + a − A h − C h , A h − C h (cid:1) . (32)The quantity L ( x n ) is called the n th moment of the orthogonal sequence { p n ( x ) } n ≥ .Let µ n ( a, b, c ) denote this n th moment of the continuous dual q -Hahn polynomials p n ( x ). An elementary calculation shows the following: Proposition 2.7.
We have Q n ( t, q ) = µ n ( i √ q, − i √ q, t )(1 − q ) n , R n ( t, q ) = µ n ( iq, − iq, t )(1 − q ) n . (33) Proof.
We just have to identify the ordinary generating functions of both sides in eachidentity. This is possible because we know in each case the explicit form of the J -fraction expansion, from (32) and Proposition 2.5. See also (34) and (35) below for amore general result.This very simple link between our polynomials and the moments µ n ( a, b, c ) is one ofthe properties indicating that Q n ( t, q ) and R n ( t, q ) are interesting q -analogs of Q n ( t ) NUMERATION OF SNAKES AND CYCLE-ALTERNATING PERMUTATIONS 7 and R n ( t ). Note that in view of (15) and (16), it is tempting to define a q -analog Q ( a ) n ( t, q ) of Q ( a ) n ( t ) by the recurrence: Q ( a ) n +1 ( t, q ) = (1 + q a t ) D (cid:16) Q ( a ) n ( t, q ) (cid:17) + [ a ] q tQ ( a ) n ( t, q ) , (34)together with Q ( a )0 ( t, q ) = 1. Though we will not study these apart the particular cases a = 1 and a = 2, it is worth mentioning that a natural generalization of the previousproposition holds, more precisely we have: Q ( a ) n ( t, q ) = µ n ( iq a/ , − iq a/ , t )(1 − q ) n , (35)which can be proved by calculating explicitly the matrix ( I + q a U ) D + [ a ] q U . Theentries in this matrix show that the generating function of { Q ( a ) n ( t, q ) } has a J -fractionexpansion with coefficients: b h = tq h (cid:0) [ h − a ] q + q a − [ h + 1] q (cid:1) , λ h = (1 + t q h − a )[ h ] q [ h − a ] q . (36)Indeed we have the same coefficients if we replace ( a, b, c ) with ( iq a/ , − iq a/ , t ) in (31)and (32). 3. Combinatorial models via signed permutations
Definition 3.1.
We will denote [ n ] = { . . . n } and J n K = {− n . . . − } ∪ { . . . n } . A signed permutation is a permutation π of J n K such that π ( − i ) = − π ( i ) for any i ∈ J n K .It will be denoted π = π . . . π n where π i = π ( i ), and S ± n is the set of all such π . Wewill also use the cycle notation, indicated by parenthesis: for example π = 3 , − , , , , , − , − , − −
4) in cycle notation.We consider here two classes of signed permutations: snakes and cycle-alternatingpermutations. These will give combinatorial models of the derivative polynomials. Asfor the q -analog, going through other objects (weighted Dyck prefixes) we can obtaina statistic on cycle-alternating permutations counted by the parameter q .3.1. Snakes.
There are several types of snakes to be distinguished, so let us first givethe definition:
Definition 3.2.
A signed permutation π = π . . . π n is a snake if π > π < π >. . . π n . We denote by S n ⊂ S ± n the set of snakes of size n . Let S n ⊂ S n be thesubset of π satisfying π >
0, and S n ⊂ S n be the subset of π satisfying π > − n π n < S n , and are usually calledsnakes of type B . He shows in particular that the cardinal of S n is the n th Springernumber S n . The objects in S n are called β -snakes by Hoffman [13], and he showedthat the cardinal of this set is P n (1) = 2 n E n .It is practical to take a convention for the values π and π n +1 . If π ∈ S n (respectively, π ∈ S n , π ∈ S n ) we define π = − ( n + 1) and π n +1 = ( − n ( n + 1) (respectively, π = 0 and π n +1 = ( − n ( n + 1), π = π n +1 = 0). This is not consistent withthe inclusions S n ⊂ S n ⊂ S n , so we can think of these sets to be disjoint to avoidambiguity in the sequel. Then an element of any of these three sets is a π ∈ S ± n suchthat π < π > π < π > . . . π n +1 . For example in the case of S n , the condition( − n π n < n is odd, π n > π n +1 = 0, and if n is even, π n < π n +1 = 0.For clarity, we will often write π = ( π ) , π , . . . , π n , ( π n +1 ). MATTHIEU JOSUAT-VERG`ES
Definition 3.3.
Let π be a snake in one of the sets S n , S n , or S n , and π , π n +1 asabove. We define a statistic sc( π ) as the number of sign changes through the values π . . . π n +1 , i.e. sc( π ) = { i | ≤ i ≤ n, π i π i +1 < } .Using this statistic, we have combinatorial models of the derivative polynomials. Theorem 3.4.
For any n ≥ , we have: P n ( t ) = X π ∈S n t sc( π ) , Q n ( t ) = X π ∈S n t sc( π ) , R n ( t ) = X π ∈S n +1 t sc( π ) . (37) Proof.
We check that the right-hand sides satisfy the recurrences in (5) (case a = 1 or a = 2), but we only detail the case of R n (case a = 2), the other ones being similar.We have R = 1, and this corresponds to the snake (0) , , (0). Also R = 2 t , and thiscorresponds to (0) , , − , (0) and (0) , , − , (0). Suppose the result is proved for R n − ,then to prove it for R n we distinguish three kinds of elements in S n +1 (the conventionis π = π n +2 = 0). We will denote by π − i the number π i − π i > π i + 1 if π i < • First, suppose that | π n +1 | = 1, hence π n +1 = ( − n . Let π ′ = π − . . . π − n . De-pending on the parity of n , we have either π n < π n +1 = 1 or π n > π n +1 = − − n π n <
0, on the other hand that π n π n +1 < π ′ ∈ S n , and sc( π ′ ) = sc( π ) −
1. The map π π ′ is bijective, and with therecurrence hypothesis, it comes that the set of π ∈ S n +1 with | π n +1 | = 1 is countedby tR n − . • Secondly, suppose that π = 1, and let π ′ = − π − , . . . , − π − n +1 . From π > π , weobtain π <
0, hence − π − >
0. It follows that π ′ ∈ S n . Moreover, since π π < π ′ ) = sc( π ) −
1. With the recurrence hypothesis, it comes that theset of π ∈ S n +1 with π = 1 is also counted by tR n − . • Thirdly, suppose that there is j ∈ { . . . n } such that | π j | = 1. We have either π j − > π j < π j +1 or π j − < π j > π j +1 , and it follows that π j − and π j +1 havethe same sign. We will obtain the term R ′ n − in the subcase where π j have alsothe same sign as π j − and π j +1 , and the term t R ′ n − otherwise. Let us prove thefirst subcase, the second will follow since it suffices to consider snakes of the firstsubcase where π j is replaced with − π j .Let π ′ = π − , . . . , π − j − , − π − j +1 , . . . , − π − n +1 . Then we can check that π ( π ′ , j )is a bijection between π of the first subcase, and couples ( π ′ , j ) where π ′ ∈ S n and j ∈ { . . . n } is such that π ′ j − π ′ j <
0. At the level of generating function, choosinga sign change in π ′ is done by differentiation with respect to t , and this explainswhy π of the first subcase are counted by R ′ n − .Adding the above terms, we obtain (1 + t ) R ′ n − + 2 tR n − = R n . This completes therecurrence. The difference in the case of Q n (respectively, P n ) is roughly that only thesecond and third points (respectively, only the third point) are to be considered.The exponential generating function of { Q n ( t ) } n ≥ can be directly obtained fromsnakes in the form ∞ X n =0 Q n ( t ) z n n ! = sec z − t tan z . (38)To prove this, let π ∈ S n , and consider its unique factorization π = f . . . f k , whereeach f k is a maximal non-empty factor such that all its entries have the same sign. NUMERATION OF SNAKES AND CYCLE-ALTERNATING PERMUTATIONS 9
Each of these factor has odd length, except possibly the last one. So it is convenientto think that if the last factor f k has odd size, there is also an empty factor f k +1 at theend, and consider the new factorization π = f . . . f j (with j = k or j = k + 1). Notethat sc( π ) = j −
1. The factorization proves (38), because π is built by assemblingalternating permutations of odd size f . . . f j − , and an alternating permutation ofeven size f j .Using the snakes, we can also directly obtain that ∞ X n =0 R n ( t ) z n n ! = ∞ X n =0 Q n ( t ) z n n ! ! . (39)Indeed, let π = (0) , π . . . π n +1 , (0) be among snakes counted by R n ( t ), i.e. π ∈ S n +1 .There is j ∈ { . . . n +1 } such that | π j | = n +1. We define π ′ = (0) , π . . . π j − , ( π j ) and π ′′ = (0) , ( − n π n +1 , . . . , ( − n π j +1 , (( − n π j ). After some relabelling, these definetwo snakes in S j − and S n − j +1 . We have sc( π ) = sc( π ′ ) + sc( π ′′ ), and π is built byassembling π ′ and π ′′ , so this proves (39).Another interesting question concerning snakes is the following. For a given per-mutation σ = σ . . . σ n ∈ S n , the problem is to choose signs ǫ = ǫ . . . ǫ n ∈ {± } n such that π = ǫ σ , . . . , ǫ n σ n is a snake. Arnol’d [1] described the possible choicessuch that π ∈ S n in terms of ascent and descent sequences in the permutation. Let i ∈ { . . . n − } , then a case-by-case argument from [1] show that: • if σ i − < σ i < σ i +1 , then ǫ i = ǫ i +1 , • if σ i − > σ i > σ i +1 , then ǫ i − = ǫ i , • if σ i − > σ i < σ i +1 , then ǫ i − = ǫ i +1 . (40)Following this, we can answer the problem of building a snake from a permutation interms of some permutations statistics. We can apply this argument to the three sets S n , S n , and S n , essentially by varying the conventions on σ (0) and σ ( n + 1). Definition 3.5.
Let σ = σ . . . σ n ∈ S n , we make the convention that σ = σ n +1 = n + 1. We also need to consider other conventions, so let S n and S n be “copies” ofthe set S n , where: • If σ ∈ S n , we set σ = 0 and σ n +1 = n + 1. • If σ ∈ S n , we set σ = σ n +1 = 0.Let σ be in one of the sets S n , S n , or S n . An integer i ∈ [ n ] is a valley of σ if σ i − > σ i < σ i +1 , a peak if σ i − < σ i > σ i +1 , a double descent if σ i − > σ i > σ i +1 ,and a double ascent if σ i − < σ i < σ i +1 . Let va( σ ) denote the number of valleys in σ , pk( σ ) denote the number of peaks in σ , and dda( σ ) denote the number of doubledescents and double ascents in σ .For example, we see 1 , , ∈ S as (0) , , , , (4). Then 3 is a valley. On the otherhand, we see 1 , , ∈ S as (0) , , , , (0). Then 3 is a double descent. Proposition 3.6.
When n ≥ , we have: P n ( t ) = X σ ∈ S n t dda( σ ) (1 + t ) va( σ ) = X σ ∈ S n t dda( σ ) (1 + t ) pk( σ ) , (41) Q n ( t ) = X σ ∈ S n t dda( σ ) (1 + t ) va( σ ) , R n ( t ) = X σ ∈ S n +1 t dda( σ ) (1 + t ) va( σ ) . (42) The last two identities are also true when n = 0 . Proof.
This follows from Theorem 3.4 and (40), which now can be extended to thecases i = 1 or i = n using the conventions on σ and σ n +1 (note that we also needto consider ǫ and ǫ n +1 ). Let us first prove the case of P n , σ ∈ S n and π ∈ S n .We have ǫ = − π = − ( n + 1). The first two rules in (40)show that to each double ascent or double descent we can associate a sign change in π , and that the only remaining choices to be done concern the valleys. At a valley σ i − > σ i < σ i +1 , we can have either ǫ i − = ǫ i = ǫ i +1 , or ǫ i − = − ǫ i = ǫ i +1 , i.e. Q n and R n are proved similarly, exceptthat we don’t consider ǫ and we have ǫ = 1.An important property of the statistics va( σ ) and dda( σ ) is that they can be followedthrough the bijection of Fran¸con and Viennot (see for example the book [10]). Omittingdetails, this gives combinatorial proof of: ∞ X n =0 Q n ( t ) z n = J (cid:0) (2 h + 1) t, (1 + t ) h (cid:1) , (43) ∞ X n =0 R n ( t ) z n = J (cid:0) (2 h + 2) t, (1 + t ) h ( h + 1) (cid:1) , (44)which is the particular case q = 1 of Proposition 2.5. Unfortunately, it seems difficultto follow the parameter q through this bijection, and obtain nice statistics on S n and S n +1 corresponding to Q n ( t, q ) and R n ( t, q ).3.2. Cycle-alternating permutations.
From the combinatorial interpretation interms of snakes, we can derive other ones using simple bijections. We need somegeneral definitions concerning signed permutations. Let π ∈ S ± n . Definition 3.7.
Let neg( π ) be the number of i ∈ [ n ] such that π i <
0. For any orbit O of π , let − O = {− x | x ∈ O } . A cycle of π is an unordered pair of orbits { O , O } suchthat − O = O . It is called a one-orbit cycle if O = O and two-orbit cycle otherwise.(This is known to be a natural notion of cycle since there is a decomposition of anysigned permutation into a product of disjoint cycles.) For any signed permutation π ,its arch diagram is defined as follows: draw on the horizontal axis 2 n nodes labelled bythe integers in J n K in increasing order, then for any i draw an arch from i to π ( i ) suchthat the arch is above the horizontal axis if i ≤ π ( i ) and below the axis otherwise.See Figure 1 for an example of an arch diagram. Since the labels from − n to n are in increasing order we do not need to explicitly write them. See also Figure 2further for another example. We did not specify what happens with the fixed pointsin the arch diagram. Although this will not be really important in the sequel, we canchoose to put a loop b at each positive fixed point and a loop b at each negativefixed point. This way, the arch diagram of a signed permutation is always centrallysymmetric. bbbbbbb b b b b b b b Figure 1.
The arch diagram of π = 2 , − , , , − , , NUMERATION OF SNAKES AND CYCLE-ALTERNATING PERMUTATIONS 11
Definition 3.8.
Let π ∈ S ± n . An integer i ∈ J n K is a cycle peak of π if π − ( i ) < i >π ( i ), and it is a cycle valley of π if π − ( i ) > i < π ( i ). The signed permutation π is cycle-alternating if every i ∈ J n K is either a cycle peak or a cycle valley. Let C n ⊂ S ± n denote the subset of cycle-alternating signed permutations, and C ◦ n ⊂ C n be the subsetof π ∈ C n with only one cycle. Lemma 3.9.
Let π ∈ C ◦ n . Then n is even (respectively, odd) if and only if π has atwo-orbit cycle (repectively, a one-orbit cycle).Proof. In a cycle-alternating permutation, each orbit has even cardinal because thereis an alternance between cycle peaks and cycle valleys, so if there are two oppositeorbits of size n , necessarily n is even.In the cycle-alternating permutation π , − n is a cycle valley and n is a cycle peak.If there is one orbit, it can be written π = ( n, i , . . . , i n − , − n, − i , . . . , − i n − ) in cyclenotation. Because of the alternance of cycles peaks and cycle valleys, it follows that n is odd. Theorem 3.10.
We have : P n ( t ) = X π ∈C ◦ n +1 t neg( π ) , Q n ( t ) = X π ∈C n t neg( π ) , R n ( t ) = X π ∈C ◦ n +2 π > t neg( σ ) . (45) Proof.
There are simple bijections between snakes and cycle-alternating permutationsto prove this from Theorem 3.4.Let us begin with the case of P n , so let π ∈ S n . Suppose first that n is odd,hence π = π n +1 = − ( n + 1). We can read π in cycle notation and consider thetwo-orbit cycle π ′ = ( π , π , . . . , π n )( − π , − π , . . . , − π n ). Then π ′ ∈ C ◦ n +1 . When n is even, we have − π = π n +1 = n + 1, and the bijection is defined by taking π ′ =( π , π , . . . , π n , − π , − π , . . . , − π n ). The map π π ′ is clearly invertible, and usingthe previous lemma, we have a bijection between S n and C ◦ n +1 . Let i ∈ { , . . . , n } , then π i π i +1 < π ′ ( | π i | ) = −| π i +1 | <
0, and it follows that sc( π ) = neg( π ′ ).This proves the first equality of the theorem.The case of R n is slightly different. In this proof, we will denote by π + i the number π i + 1 if π i >
0, and π i − π i <
0. Let π ∈ S n +1 . If n + 1 is odd, hence π n +1 >
0, wedefine π ′ = (1 , π +1 , . . . , π + n +1 )( − , − π +1 , . . . , − π + n +1 ). If n + 1 is even, hence π n +1 < π ′ = (1 , π +1 , . . . , π + n +1 , − , − π +1 , . . . , − π + n +1 ). In each case, it is easily checkedthat this defines a bijection from S n +1 to the subset of π ∈ C ◦ n +2 satisfying π >
0, sothat a change of sign in π correspond to a i > π ′ ( i ) < Q n requires more attention. Let π ∈ S n , we need to split it into blocksthat will correspond to the cycles of an element π ′ ∈ C n . This can be done by avariant of Foata’s fundamental transform [17]. Let a < · · · < a k be the left-to-rightmaxima of | π | . . . | π n | ∈ S n (recall that u is a left-to-right maximum of σ ∈ S n when σ ( i ) < σ ( u ) for any i < u ). We consider the factors of π , f = π a . . . π a − , f = π a . . . π a − , up to f k = π a k . . . π n (we can take the convention a k +1 = n +1). Weform π ′ by putting together cycles, so that if f i has odd length it gives a one-orbit cycle( π a i . . . π a i +1 − , − π a i . . . − π a i +1 − ), and if it has even length it gives a two-orbit cycle( π a i . . . π a i +1 − )( − π a i . . . − π a i +1 − ). The definition of a i shows that π a i is greater than π a i +1 . . . π a i +1 − , and consequently these cycles are indeed alternating. The inversebijection is easily deduced: we can write π ′ ∈ C n as a product of cycles π ′ = c . . . c k ,where the cycles are sorted so that their maximal entries are increasing. To each c i we associate a word f i so that c i is either ( f i , − f i ) or ( f i )( − f i ), and such that the first letterof f i is the maximum entry of c i . Then we can form the snake π = f , ǫ f , . . . , ǫ k f k ,where the signs ǫ , . . . , ǫ k ∈ {± } are the unique ones such that π is alternating.It is in order to give examples of the previous two bijections. If we start froma snake π = (0) , , − , − , − , , − , (7) ∈ S , the left-to-right maxima of | π | are1 , ,
6, and this gives π ′ = (4 , − , − , − , , − , , − , − ∈ C . As for theother direction, let π ′ = (5 , − , , − , , − , − − , , − − , ∈ C . It is aproduct of three cycles so that the snake π is formed by putting together three words5 , − ,
3, and 6 , −
1, and 7 , −
4. We need to change the signs of the second and thirdwords to obtain a snake, which is π = (0) , , − , , − , , − , , ( − π ) = neg( π ′ ), butthis does not immediately follows the construction and need to be proved now. Let a, b >
0, such that π ′ ( a ) = − b , we can associate a sign change in π to each suchpair ( a, b ). The two integers are in the same cycle c i . If the word f i is of the form . . . , a, − b, . . . or . . . , − a, b, . . . , then either the factor a, − b or − a, b appears in theword ǫ i f i , and consequently appears in π as a sign change. The other possibility isthat f i = b, . . . , − a which only occur when f i has even length, and f i = b, . . . , a whichonly occur when f i has odd length. If b (respectively − b ) appear in π , it is greater(respectively, smaller) than his neighbor entries, and after examining a few cases itcomes that the same is true for a . It follows that there is a sign change in π betweenthe last entry of ǫ i f i and its right neighbor. This completes the proof.Using cycle-alternating permutations, there is a bijective proof of the fact that P n +1 = (1 + t ) R n . This can be written X π ∈C ◦ n +2 t neg( π ) = (1 + t ) X σ ∈C ◦ n +2 σ > t neg( σ ) . (46)Now, consider the conjugation by the transposition (1 , − i.e. the map that sends π to (1 , − π (1 , − C ◦ n +2 which proves combi-natorially (46), as is easily seen on the arch diagrams. Indeed, in this representationthe involution exchanges the dot labelled 1 with the dot labelled −
1, see Figure 2. Wehave π (1) > π ′ = (1 , − π (1 , −
1) does not satisfy π ′ (1) > π (1) > π ′ ) = neg( π ) + 2. This proves (46). Note that it alsois possible to prove P n +1 = (1 + t ) R n on the snakes, using Proposition 3.6, but thisis more tedious. bbbbb b b b b b bbbbb b b b b b Figure 2.
The arch diagram of the cycle-alternating permutation π = (1, 4, − − − − −
4, 5, 2, 3) ∈ C ◦ , and of its conjugate(1 , − π (1 , − { P n } n ≥ , knowing the one of { Q n } n ≥ . Indeed, a cycle-alternatingpermutation is an assembly of cycle-alternating cycles, in the sense of combinatorial NUMERATION OF SNAKES AND CYCLE-ALTERNATING PERMUTATIONS 13 species, and this gives immediately: ∞ X n =0 P n ( t ) z n +1 ( n + 1)! = log ∞ X n =0 Q n ( t ) z n n ! ! = log (cid:18) z − t sin z (cid:19) , (47)and after differentiation: ∞ X n =0 P n ( t ) z n n ! = sin z + t cos z cos z − t sin z . (48)3.3. The q -analogs. It is possible to give combinatorial interpretations of Q n ( t, q )and R n ( t, q ) using cycle-alternating permutations, and the notion of crossing for signedpermutations which was defined in [5]. But in the case of cycle-alternating permuta-tions, there is a more simple equivalent definition which is the following. Definition 3.11. A crossing of π ∈ C n is a pair ( i, j ) ∈ J n K such that i < j < π ( i ) <π ( j ). Equivalently, it is the number of intersections of two arches above the horizontalaxis in the arch diagram of π . We denote by cr( π ) the numbers of crossings in π .Because of the symmetry in the arch diagram, cr( π ) is also half the total numberof intersections between two arches. For example, the two permutations in Figure 2have respectively 2 and 3 crossings. This kind of statistic can be followed throughbijections between permutations and paths [4], but let us first show that Q n ( t, q ) and R n ( t, q ) are related with some weighted Dyck prefixes (a Dyck prefix being similar toa Dyck path except that the final height can be non-zero). Definition 3.12.
Let P n be the set of weighted Dyck prefixes of length n , where • each step ր between heights h and h +1 has a weight q i for some i ∈ { , . . . h +1 } , • each step ց between heights h and h + 1 has a weight q i for some i ∈ { , . . . h } .Let P ′ n ⊂ P n be the subset of paths p such that there is no step ր starting at height h with the maximal weight q h +1 . For any p ∈ P n , let fh( p ) be its final height, and letw( p ) be its total weight, i.e. the product of the weights of each step. Proposition 3.13.
We have: Q n ( t, q ) = X p ∈P ′ n t fh( p ) q w( p ) , R n ( t, q ) = X p ∈P n t fh( p ) q w( p ) . (49) Proof.
In the proof of Proposition 2.5, we have used the matrices D + U D U and D + D U U , thought of as transfer-matrices for walks in the non-negative integers.We can do something similar with the matrices D + U D U = q (0)[1] q q [2] q , D + D U U = q (0)[2] q q [3] q . (50)Recall that we have (20), where W = (1 , t, t , . . . ). In terms of paths, the coefficientsin the matrices and vectors have the following meaning: from V = (1 , , . . . ) ∗ , we onlyconsider paths starting at height 0, and from W = (1 , t, t , . . . ) there is the weight t fh( p ) for each path p . The coefficients ( h + 1 , h ) in the matrices give the possibleweights on steps ր from height h to h + 1, the coefficients ( h, h + 1) in the matricesgive the possible weights on steps ց from height h + 1 to h . The result follows. Proposition 3.14.
There is a bijection
Ψ : C n → P ′ n such that neg( π ) = fh( p ) and cr( π ) = w( p ) if the image of π is p .Proof. We can use the the bijection Ψ
F Z from [4] between permutations and weightedMotzkin paths. To do this, we identify S ± n to a subset of S n via the order-preservingbijection J n K → [2 n ]. This subset is characterized by the fact that the arrow diagramsare centrally symmetric, and via Ψ F Z from [4], it is in bijection with some weightedMotzkin paths that are vertically symmetric. Cycle-alternating permutations corre-pond to the case where there is no horizontal step, i.e. they are in bijection with someweighted Dyck paths that are vertically symmetric. Keeping the first half of theseDyck paths gives the desired bijection.For convenience, we rephrase here explicitly the bijection of the previous proposition.Let π ∈ C n , then we define a Ψ( π ) ∈ P ′ n such that, for any j ∈ {− n . . . − } : • the ( n + 1 + j )th step is ր if π − ( j ) > j < π ( j ) and ց if π − ( j ) < j > π ( j ), • if the ( n + 1 + j )th step is ր , it has a weight q k where k is the number of i suchthat ( i, j ) is a crossing, i.e. i < j < π i < π j , • if the ( n + 1 + j )th step is ց , it has a weight q k where k is the number of i suchthat ( − i, − j ) is a crossing, i.e. − i < − j < − π i < − π j .See Figure 3 for an example. b b b b b b b bbbbbbb → q q q q q q q Figure 3.
The bijection Ψ from C n to P ′ n in the case of π =3 , , − , , , , Q n ( t, q ) and R n ( t, q ) usingthe notion of crossing. Proposition 3.15.
We have Q n ( t, q ) = X π ∈C n t neg( π ) q cr( π ) , R n ( t, q ) = X π ∈C n +1 π n +1 < t neg( π ) − q cr( π ) . (51) Alternatively, the condition π n +1 < in the second identity can be replaced with π − ( n + 1) < .Proof. The first identity follows from Proposition 3.13 and the bijection Ψ between C n and P ′ n , but the second one is not as immediate.First, note that the bijection π π − stabilizes the set C n and preserves the numberof crossings, since it is just an horizontal symmetry on the arch diagrams. This provesthe fact that we can replace π n +1 < π − ( n + 1) < P ′′ n +1 ⊂ P ′ n +1 of elements p such that there is nostep ց from height h + 1 to h with weight q h (note that this condition implies thereis no return to height 0). There is an obvious bijection from P ′′ n +1 to P n , becauseremoving the first step in p ∈ P ′′ n +1 gives a path which is in P n with respect to the NUMERATION OF SNAKES AND CYCLE-ALTERNATING PERMUTATIONS 15 shifted origin (1 , R n ( t, q ) = X p ∈P ′′ n +1 t fh( p ) − q w( p ) . (52)It remains to show that the image of P ′′ n +1 via Ψ − is precisely the set of π ∈ C n +1 such that π − ( n + 1) <
0. Essentially, this follows from the fact that there are somesteps in the path which characterize the right-to-left minima in π , and we can useLemma 3.2.2 from [14]. In our case, the subset P ′′ n +1 ⊂ P ′ n +1 corresponds to a subsetof S n +2 (via the identification S ± n +1 → S n +2 ), more precisely to the subset of σ having no right-to-left minima among 1 . . . n + 1, i.e. σ − (1) > n + 1. In terms of π ∈ S ± n +1 , this precisely means that π − ( − n − > Remark 3.16.
Independently from our work, Chen, Fan and Jia [3] gave a bijectionbetween snakes of type B and Dyck prefixes such that there are h + 1 possible choiceson each step between height h and h + 1, and their statistic α ( π ) (see Theorem 4.5)is the same as our statistic sc( π ), though defined differently. We can have such abijection by composing two we have presented, but their bijection is not the same asours and does not seem to be directly related. To advocate our bijection S n → P ′ n ,note that going through the set C n we just had to adapt two known bijections: Foata’sfundamental transform [17] (in the direction S n → C n ), then Corteel’s bijection Ψ F Z [4] originally given by Foata and Zeilberger (in the direction C n → P ′ n ).3.4. Another variant of Ψ F Z . With another adaptation of this bijection, we canlink C n and some weighted Motzkin path, so that we obtain a combinatorial proof ofa J -fraction generalizing (43) and (44) in Theorem 3.17 below. As a consequence ofTheorem 3.10, we have: Q ( a ) n ( t ) = X π ∈C n a cyc( π ) t neg( π ) (53)where cyc( π ) is the number of cycles, indeed since there is a “cycle structure” in C n it suffices to take the a th power of the exponential generating function to havea parameter a counting cyc( π ). With the recurrence for Q ( a ) n ( t ) in (5) and with thesame method as in the case of Q n and R n , we can prove a continued fraction for theordinary generating function of { Q ( a ) n ( t ) } n ≥ . We prove here bijectively this continuedfraction, by going from C n to weighted Motzkin paths. To this end, we use anotherrepresentation of a signed permutation, the signed arch diagram . Let π ∈ S ± n , thisdiagram is obtained the following way: draw n dots labelled by 1 . . . n from left toright on the horizontal axis, then for each i ∈ [ n ], draw an arch from i to | π i | , andlabel this arrow with a + if π i > − if π i <
0. We understand that as inthe previous arch notation, the arch is above the axis if i ≤ | π i | and below otherwise.See the left part of Figure 4 for an example. (In the case | π i | = i , the arch is a loopattached to the dot i .) A case-by-case study shows that π is cycle-alternating if andonly if it avoids certain configurations in the signed arch diagram, which are listed inthe right part of Figure 4 ( ± being either + or − ). These forbidden configurations areof course reminiscent of (40).Once we know the forbidden configurations, it is possible to encode the signed archdiagram of π ∈ C n by a weighted Motzkin path of n steps, in a way similar to Ψ F Z from [4]. The diagram of π is scanned from left to right, and the path is built suchthat: b b b b b b b + − − + −− + b + ± b + ± b + − b − + b + Figure 4.
The signed arch diagram of π = 2 , − , − , , − , − ,
7, andthe forbidden configurations in the signed arch diagram of a cycle-alternating permutation. • If the i th node in π is b then the i th step of p if ր , moreover this step has label+ (resp. − ) if the two arches starting from the i th node have label + (resp. − ). • If the i th node in π is b then the i th step of p if → . Moreover this step has label( j ) if the left strand in the i th node is connected to the j th strand of the “partial”signed arch diagram (say, from bottom to top). See for example the left part ofFigure 5 where there are three possible choices to connect this kind of node. Wecan see that the possible labels are (1) , . . . , ( h ) where h is the starting height ofthe i th step in the path. • Similarly if the i th node in π is b then the i th step of p if → , and this step haslabel ( j ′ ) if the i th node is connected on the left to the j th strand of the “partial”signed arch diagram (say, from top to bottom). • If the i th node in π is b − then the i th step is → with label (0). • If the i th node in π is b then the i th step is ց with a label ( j, k ) where j and k respectively encode where the upper and lower strands are connected, as in thecase of b and b above. b b b b b b b b + − + − −−− ++++ . . . → + (1) + (2) − (2 ′ )(2,1) + . . . Figure 5.
Construction of the bijection from the signed arch diagramof π ∈ C n to a labelled Motzkin path.Quite a few bijections of this kind are known so we will not give further details. Asfor the parameters a and t , they are taken into account the following way: • For each arch with label − linking i and j where i < j , we give a weight t to the i th step in the path. Thus there is a weight t on the step ր with label − , and aweight t on each step → . • For each i which is the maximal element of a cycle, we give a weight a to the i thstep. Thus each step → with weight (0) has a weight a , since these correspond tonodes b − or cycles of the form ( i, − i ). Besides, there is a weight a on some of thestep ց . More precisely, there are h possible labels on a step ց starting at height h , and h of them have this weight a . Indeed, for each choice of where to connectthe upper strand of the node b there is a unique choice of where to connect thelower strand so that a new cycle is “closed”. NUMERATION OF SNAKES AND CYCLE-ALTERNATING PERMUTATIONS 17
All in all, using the classical correspondence between weighted Motzkin paths and J -fractions, this bijection proves the following theorem. Theorem 3.17.
Let Q ( a ) n ( t ) be the polynomials in a and t with exponential generatingfunction (cos z − t sin z ) − a , or defined combinatorially by (53) . Then, using the notationin Definition 2.4, we have: ∞ X n =0 Q ( a ) n ( t ) z n = J (cid:0) (2 h + a ) t, h ( h − a )(1 + t ) (cid:1) . (54)Apart the combinatorial side, this result was already known by Stieltjes [22]. Besideshis analytical proof, it is also possible to obtain the continued fraction by showing thatthe exponential generating function satisfies some particular kind of addition formulawhich permits to make use of a theorem of Stieltjes and Rogers, see for example[10, Theorem 5.2.10]. We thank Philippe Flajolet for communicating these facts andreferences. 4. Increasing trees and forests
So far, we have only used recurrence relations and bijections to derive our newmodels of the derivative polynomials. Some more elaborate methods apply pretty wellin our case to give combinatorial models in terms of increasing trees and forests. Wefirst present how to use the combinatorial theory of differential equations, as exposedby Leroux and Viennot in [16], and secondly how to use recent results of B lasiak andFlajolet related with operators and normal ordering [8].4.1.
Increasing trees via differential equations.
An archetype example in thecombinatorial theory of differential equation is the one of the tangent and secantfunctions (see [16]), and it has become a classical method to show that Euler numbers E n count some increasing trees, i.e. labelled trees where labels are increasing from theroot to the leaves. Here we have a system of two differential equations similar to theone of tangent and secant, except that an initial condition is given by the parameter t instead of 0. This will give rise to more general trees where we allow some leaves withno label having a weight t . Lemma 4.1.
Let f = P P n z n /n ! and g = P Q n z n /n ! , then we have: ( f ′ = 1 + f f (0) = t,g ′ = f g g (0) = 1 . (55) Proof.
Of course, this can be checked on the closed form given in (7). Also, f ′ = 1 + f can be directly checked on the snakes as in the case of alternating permutations [21],and g ′ = f g follows from f = (log g ) ′ previously seen.It is adequate to rewrite the equations in the following way: ( f = t + z + R f ,g = 1 + R f g. (56)Let us begin with the case of f ( z ). From f = t + z + R f , and proceeding as in[16], f ( z ) counts increasing trees that are recursively produced by the following rules,starting from an isolated node marked by f : • a node marked by f can become a leaf with no label (this corresponds to the term t in f = t + z + R f , f ( z ), so these leaves will have a weight t ), • a node marked by f can become a leaf with an integer label, this label being thesmallest integer that does not already appear in the tree (this corresponds to theterm z in f = t + z + R f ), • a node marked by f can become an internal node having an integer label andtwo (ordered) sons marked by f (this corresponds to the last term R f in f = t + z + R f ). As before, the integer label is the smallest integer that does notalready appear in the tree.More precisely, the coefficient of z n in f ( z ) counts these trees having n integer labels.See Figure 6 for an example of tree produced by these rules. f → f f → f f f → f f → f → Figure 6.
Tree produced via the equation f = t + z + R f .The case of g ( z ) is quite similar. Starting from an isolated node marked by g , eachnode marked by g can become either a leaf with no label, or an internal node havingtwo sons respectively marked by f and g . Note that we need the production rules for f to build a tree counted by g , and note also that g can produce an empty leaf whichhas no weight t , contrary to the empty leaves produced by f .The two kinds of tree can also be given a non-recursive definition. Definition 4.2.
Let T n be the set of complete binary trees, such that: • except some leaves that are empty, the nodes are labelled by integers so that each i ∈ [ n ] appears exactly once, • labels are increasing from the root to the leaves.Let em( T ) be the number of empty leaves of T ∈ T n , and let T ∗ n ⊂ T n be the subsetof trees such that the rightmost leaf is empty.The production rules for f (respectively, g ) can be checked on the set T n (respectively T ∗ n ), so that the result of the above discussion is the following. Theorem 4.3.
We have: P n ( t ) = X T ∈T n t em( T ) , Q n ( t ) = X T ∈T ∗ n t em( T ) − . (57)4.2. Increasing trees via normal ordering.
Some recent results of B lasiak andFlajolet [2] directly apply to this context and also give models of P n ( t ) and Q ( a ) n ( t ) interms of increasing trees and forests. Let D be the derivation with respect to t and U the multiplication by t ( i.e. we are in the particular q = 1 of operators definedin (11)). There holds DU − U D = I . A general idea in [2] is that the coefficients c i,j in the normal form of f ( D, U ) as in (17), at least for certain particular formsof f ( D, U ), naturally counts some labelled directed graphs which are produced byconnecting some “gates”. In the present case, when f ( D, U ) is ( D + U DU ) n for P n and ( D + aU + U U D ) n for Q ( a ) n , and we obtain some increasing trees and forests. NUMERATION OF SNAKES AND CYCLE-ALTERNATING PERMUTATIONS 19
The theorem of this subsection is a direct application of a main result of Flajolet andB lasiak [2, Theorem 1], so as a proof we will roughly explain some ideas leading to thedefinition below and refer to [2] for more details.
Definition 4.4.
Let F n be the set of plane rooted forests satisfying the followingconditions: • each root has exactly one child, and each of the other internal nodes has exactlytwo (ordered) children, • there are n nodes labelled by integers from 1 to n , but some leaves can be non-labelled (these are called empty leaves), and labels are increasing from each rootdown to the leaves.Note that the trees forming a forest are unordered. Let U n ⊂ F n be the subset oftrees, i.e. forests with one connected component. For any tree or forest T , let em( T )be the number of empty leaves, and let cc( T ) be its number of connected components.For example, there are 11 elements in F and they are: ,
12 3 ,
13 2 , ,
12 3 ,
13 2 , , ,
12 3 , ,
12 3 . Theorem 4.5.
We have: P n ( t ) = X T ∈U n +1 t em( T ) , Q ( a ) n ( t ) = X T ∈F n a cc( T ) t em( T ) . (58)For example, the forests in F given above illustrate Q = 6 t + 5 t . The last fourelements of the list are the trees, and illustrate P = 2 t + 2 t . By counting with aweight 2 on each connected component, we obtain from this list that R = 24 t + 16 t . Proof.
From the definition of derivative polynomials in terms of D and U , and therelation DU − U D = I , we have: P n ( t ) = ( D + U U D ) n t, Q ( a ) n ( t ) = ( D + aU + U U D ) n . (59)Let us consider the case of Q ( a ) n . Let f n ( D, U ) = ( D + aU + U U D ) n and c n,i,j thecoefficient of U i D j in its normal form, as in (17). Then from [2, Theorem 1], c n,i,j counts labelled diagrams obtained by connecting three kinds of “gates”, one for eachterm in D + aU + U U D . More precisely to each term U k D ℓ we associate a gateconsisting of one node with k outgoing strands and ℓ ingoing strands. See the left partof Figure 7. Then, c n,i,j count labelled diagrams obtained by connecting n of thesegates such that: • i outgoing strands and j ingoing strands are not connected, • all other strands are connected, so that each ingoing strand is connected with anoutgoing strand and these form a directed edge, • the gates labelled by the integers in { . . . n } , and labels are increasing when wefollow a directed edge, • at each node, the ingoing strands on one side and the outgoing strands on anotherside are ordered, • there is a weight a at each gate corresponding to the term U .We have Q ( a ) n ( t ) = ( D + aU + U U D ) n (cid:16) X i,j ≥ c i,j U i D j (cid:17) X i ≥ c i, t i (60)so that we can only consider diagrams with no unconnected ingoing strand, and t counts the unconnected outgoing strands.The labelled diagrams described by the above rules are essentially the same aselements in F n : it suffices to add an empty leaf at each unconnected outgoing strandto see the equivalence. It is clear that the node corresponding to the term U willappear exactly once in each connected component of the labelled diagrams, so theparameter a counts indeed the connected components.In the case of P n , we can also consider f n ( D, U ) = ( D + U U D ) n U and c n,i,j thecoefficient of U i D j in its normal form, as in (17). This case is somewhat differentsince f n ( D, U ) is not the n th power of some expression, but similar arguments applyas well: the labelled diagrams that appear have n + 1 gates, the gate labelled 1 is oftype U , all other gates are of type D or U U D . These labelled diagrams are the sameas elements in U n +1 , as in the previous case we just have to add an empty leaf to eachunconnected outgoing strand to see the equivalence. As previously said, we refer to[2] for precisions about this proof. i i i Figure 7.
The “gates” corresponding to terms in D + aU + DU U ,and a labelled diagram obtained by connecting them.There is a simple bijection between T n and U n +1 : given T ∈ T n , relabel the nodes by i i + 1, then add a new node with label 1 on top of the root. There is also a simplebijection between T ∗ n and F n : let T ∈ T ∗ n , remove the rightmost leaf, as well as alledges in the path from the root to the rightmost leaf, then the remaining componentsform the desired forest. See Figure 8 for an example.
12 35 6 4 →
12 35 6 4 →
12 35 6 4
Figure 8.
The bijection from T ∗ n to F n .Thus, Theorems 4.3 and 4.5 are essentially equivalent although obtained by differentmethods. It is also in order to give a bijection between U n and C ◦ n , and by applying thisbijection componentwise it will give a bijection between F n and C n . It is more practicalto give the bijection from T n to S n (recall that we already have simple bijections NUMERATION OF SNAKES AND CYCLE-ALTERNATING PERMUTATIONS 21 T n → U n +1 and S n → C ◦ n +1 ), so let T ∈ T n . Consider the “reading word” of this tree(it can be defined by w ( T ) = w ( T ) i w ( T ) if the tree T has a root labelled i , left son T and right son T ). This word contains integers from 1 to n , and some letters (say ◦ ) to indicate the empty leaves. The first step is to replace each i with n + 1 − i in thisword. To obtain the snake, replace each integer i by ( − j +1 i where j is the numberof ◦ before i in the word, then remove all ◦ . See Figure 9 for an example.
123 4 → , , ◦ , , ◦ , , ◦ → , , ◦ , , ◦ , , ◦ → ( − , − , − , , − , (5) Figure 9.
The bijection from T n to S n .Actually, this bijection can be highlighted by Proposition 3.6 and the discussionleading to it. Indeed, it is just a variant of a classical bijection between permutationsand unary-binary increasing trees [23], where double ascents and double descents cor-respond to nodes having only one child, and valleys correspond to leaves. Via thesebijections, removing the empty leaves of T ∈ T n is the same as taking the absolutevalue of a snake S ∈ S n . In the other direction, T seen as a unary-binary tree togetherwith the data of the empty leaves, is the same as a permutation together with a choiceof signs making it into a snake. Conclusion
There is still a wide range of results on alternating permutations that might have acounterpart in the case of snakes, for example one could ask if Andr´e permutations andsimsun permutations [21], both counted by E n , have natural generalizations in signedpermutations (as a first answer, see [18] for a definition of Andr´e signed permutation ,counted by S n ). Another problem is to give more combinatorial meaning to the Arnold-Seidel triangles [1, 6] used to compute the integers S n (see [11] for recent work on thissubject in the case of integers E n ). As for q -analogs, we have generalized the numbers E n ( q ) defined in [12] but there is another q -analog of E n related with inversions andmajor index [21] that might also be extended to snakes.Though we only discussed the type B ones, Arnol’d [1] also defined some type D snakes counted by the Springer number K ( D n ). Besides, Hoffman showed that K ( D n )is equal to P n (1) − Q n (1), so that it is also the number of π ∈ S n such that π <
0. Asusual with type D , it is much more difficult to obtain enumerative or bijective resultswith either of these two families, they are not even a subset of the group of even-signed permutation D n as one could expect. Having said that, one possible directiontowards a better understanding of these objects could be the geometric definition ofthe Springer number K ( W ) in terms of hyperplane arrangements [1]. For example,in this context it would be very interesting to give the polynomial Q n ( t ) a geometricmeaning that would refine Q n (1) = K ( S Bn ). Acknowledgement
Part of this research was done during a visit of the LABRI in Bordeaux, and I thankall the Bordelais for welcoming me and for various suggestions concerning this work.
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