Three New Refined Arnold Families

The Springer numbers, introduced by Arnold, are generalizations of Euler numbers in the sense of Coxeter groups. They appear as the row sums of a double triangular array $(v_{n,k})$ of integers, $1\leq|k|\leq n$, defined recursively by a boustrophedon algorithm. We say a sequence of combinatorial objects $(X_{n,k})$ is an Arnold family if $X_{n,k}$ is counted by $v_{n,k}$. A polynomial refinement $V_{n,k}(t)$ of $v_{n,k}$, together with the combinatorial interpretations in several combinatorial structures was introduced by Eu and Fu recently. In this paper, we provide three new Arnold families of combinatorial objects, namely the cycle-up-down permutations, the valley signed permutations and Knuth's flip equivalences on permutations. We shall find corresponding statistics to realize the refined polynomial arrays.

1. Introduction 1.1.Springer numbers and Arnold triangle.Let W be a finite Coxeter group W with set of generators S. For w ∈ W , its descent set is defined by Des(w) := {s ∈ S : ℓ(w) < ℓ(ws)}, where ℓ(w) is the length function.For each J ⊂ S, Springer [15] considered the set D J := {w ∈ W : Des(w) = J} and define K(W ) := max J⊂S |D J |, which we called the Springer numbers of type W .It can be proved that in type A n−1 we have K(S n ) = E n , the Euler number which counts the number of alternating permutations σ = σ 1 σ 2 . . .σ n ∈ S n with σ 1 > σ 2 < σ 3 > • • • and therefore Springer numbers are generalizations of the Euler numbers.See [16,17] for more information on Euler numbers and alternating permutations.In 1877, Seidel defined the triangular array (E n,k ) for the calculation of E n by with E 1,1 = 1, E n,1 = 0(n ≥ 2), and showed that E n = k E n,k .The refinement E n,k of E n is called the Entringer number since Entringer [7] has proved that E n,k is the number of alternating permutations with first entry k.
The above idea of refinement can be generalized into type B n and D n .For the Coxeter group B n (D n , respectively) of type B n (D n , respectively) Arnold [2] proved that the Springer numbers can be read as the row sums of the double triangle generated recursively by the boustrophedon recurrence relations with v 1,1 = v 1,−1 = 1 and v n,−n = 0 for all n ≥ 2. Initial values are listed in Table 1.
We call v n,k the Arnold numbers.  .For simplicity we denote −i by ī.A signed permutation is usually denoted by its window notation σ = σ 1 σ 2 . . .σ n , where σ i = σ(i).Since B n and D n are respectively the permutation models of the type B n and D n Coxeter groups [3], just like S n is for type A n−1 , by an abuse of notation we let B n denote the set of signed permutations of [n], and let D n ⊂ B n be those signed permutations with |{i : σ i < 0}| being even.We may define three types of alternating signed permutations [2,10] as Arnold called those permutations in S 0 n and S D n the snakes of type B n and D n , respectively.Moreover, he proved that v n,k (v n,−k , respectively) counts the number of snakes of type B n (D n , respectively) with first entry k (−k, respectively).
Recently, Eu and Fu [8] gave a polynomial refinements of v n,k in the Arnold triangle as follows.For 1 ≤ |k| ≤ n, define the polynomials V n,k = V n,k (t) by and V n,−n = 0 (n ≥ 2).Note that v n,k = V n,k (1).The first few polynomials V n,k (t) are listed in Table 2.
We call V n,k the Arnold-Hoffman polynomials since it is proved in [8] that where the polynomials P n , Q n defined by d n dx n tan(x) = P n (tan(x)) and are introduced by Hoffman [9].Following [14], a sequence of combinatorial objects Furthermore, it is a refined Arnold family if we can find a statistic over (X n,k ) such that V n,k is the enumerator with respect to this statistic.For example, for signed permutations, it is proved by Josuat-Vergés [10] that V n,k (t) for k > 0 (k < 0, respectively) counts the type B n snakes (D n , respectively) with respect to the statistic 'change of sign'.Recently in [8] Eu and Fu came up with several new refined Arnold families, among them the signed increasing 1-2 trees, signed André permutations, signed Simsun permutations of type I and type II, and complete increasing binary trees.

Structures investigated and Main Theorems
Our motivation comes from the fact that there are still other combinatorial structures counted by Euler numbers or Springer numbers, but their complete profiles (or even corresponding definitions in types B n or D n ) in the sense of refined Arnold families are missing.In this paper we investigate three of them, namely the cycle-up-down permutations, the valley signed permutations and Knuth's flip equivalence classes on permutations, and complete their profiles.

2.1.
Cycle-up-down permutations.A permutation σ ∈ S n can be written in the standard cycle notation, which means that in each cycle the first entry is the smallest entry of this cycle, and these smallest first entries among the cycles are increasing.For example, σ = (1, 8, 5)(2, 4)(3)(6, 9, 7) is a standard cycle notation.From now on, when we mention a cycle notation it is always standard.A permutation σ ∈ S n is called cycle-up-down if, in its standard cycle notation, each cycle (σ i 1 , σ i 2 , . . ., σ ir ) is reverse alternating, i.e., Denote CUD n the set of cycle-up-down permutations of length n.They are first defined and investigated by Deutsch and Elizalde [5].
To our ends we define types B n and D n counterparts of the cycle-up-down permutations.A signed permutation σ ∈ B n is a signed cycle-up-down permutation of type B n if it simultaneously meets the following conditions.
In other words, (CUD together with the statistic npk is a realization of the refined Arnold family.The proof will be put in Section 3.

2.2.
Valley signed permutations.Valley signed permutations are first defined by Josuat-Vergés, Novelli, and Thibon [11] and are only defined in types B n and D n .For σ ∈ S n , an entry n be the set of valley signed permutations of type B n and  Define the statistic neg of σ ∈ VS (B)   n (or VS (D)  n ) by neg(σ) := #{i : σ i < 0} and we obtain another realization of V n,k (t).
In other words, (VS n,1 ) together with the statistic neg, is a realization of the refined Arnold family.The proof will be put in Section 4. We extend the concept of flip equivalent to other types.For a signed permutation σ = σ 1 . . .σ k σ k+1 . . .σ n in its window notation, the signed permutation σ k . . .σ 1 σ k+1 . . .σ n is said to be a flip from σ if Note that from this definition the numbers of flip equivalences classes on signed permutations are not Springer numbers, but certain subsets are if we adapt the following.For a signed word σ = σ 1 σ 2 . . .σ m such that |σ i | ̸ = |σ j | for all 1 ≤ i, j ≤ m, define the statistic signed maximum smax(σ) of σ by writing σ as σ = σ L σσ R , where |σ| = min(|σ|) and computing smax(σ) recursively: Similarly, let FL (D) n be the set of flip equivalence classes of signed permutations in B n whose signed maximums are negative, and define We will see that both are counted by Springer numbers of types B n and D n .Moreover, define the statistic signed peak spk for σ ∈ B n by . With this we obtain the third realization of the refined Arnold family.
In other words, (FL n,1 ) together with the statistic spk forms a refined Arnold family.
2.4.complete increasing binary trees with n labelled nodes.A (plane) binary tree is complete if every non-leaf node has two children.Let T n be the set of complete binary trees such that there are n nodes labelled by integers from 1 to n but some leaves can be left unlabelled (called empty leaves), and along the path from the root to any leaf the labels are increasing.These trees are first introduced by [10], and in the rest of the paper, by complete increasing binary trees we always mean the trees from the set T n .Figure 1 shows the sixteen complete increasing binary trees with three labelled nodes.Recently a realization of the refined Arnold family in terms of complete increasing binary trees is found by Eu and Fu [8].Given τ ∈ T n , the rightmost path of τ is a sequence of nodes (v 1 , v 2 , . . ., v d ) where v 1 is the root and v i+1 is the right child of v i for all i = 1, 2, . . ., d − 1.The node v d is called the rightmost leaf of τ .Let T • n (T * n , respectively) denote the subset of T n consists of the trees whose rightmost leaf is empty (labelled, respectively).The label of v ) be the subset of trees whose rightmost label is k.Let emp(τ ) be the number of empty leaves in τ , we have the following.
It is then natural to seek bijection between complete increasing binary trees to cycle-up-down permutations, valley signed permutations, and flip equivalence classes.
The main results of this paper are twofold.Firstly, we will give definitions (if needed) of types B n and D n versions of the above structures and find suitable statistics such that each is a new realization of the refined Arnold family.Secondly, as the set of complete increasing binary trees is the key structure for linking other Arnold families together (see [8]), we will also give bijections between these new combinatorial objects with the complete increasing binary trees.
The rest of the paper is organized as follows.In Section 3 we give a realization of the refined Arnold family in terms of cycle-up-down permutations and in Section 4 in terms of valley signed permutations.The interlude Sections 5 (resp.Section 6) are devoted to bijections between cycle-up-down permutations (resp.valley signed permutations) with complete increasing binary trees.In Section 7 we investigate Knuth's flip equivalence classes on signed permutations.

Proofs of Main Theorem I
To prove Theorem 2.1, by referring to the recurrences of V n,k (t) we need to prove the following three recurrences: We investigate them in the propositions below.

Proof of Main Theorem II
In this section we prove Theorem 2.2.For 1 We are going to show that both have the same recurrence relations as V n,k .
where (iii) We construct a bijection

Then we have ψ(σ) ∈ VS
where n,k+1 and neg(ψ(σ)) = neg(σ), which gives us the term V S n,1 are empty for all n ≥ 2, we have for n ≥ 2. Together with Proposition 4.1, the proof is completed.□

Complete increasing binary trees and Cycle-up-down permutations
Recall that T n is the set of complete increasing binary trees with n labelled nodes and T • n (T * n , respectively) are those trees with empty (resp.labelled) rightmost leaf.In this section, we shall establish the bijections The subscript C indicates that it is for cycle-up-down permutations.
5.1.Type B n .We roughly describe the flowchart of the desired bijection.We map an up-down sequence to a non-plane complete increasing tree by Algorithm 1.By Algorithm 2 we map each cycle of σ ∈ CUD (B)   n to a complete increasing binary tree with the help of Algorithm 1.By combing corresponding trees from all cycles we reach our desired bijection.Our running example is σ = (1, 3, 2)(4)(5, 6)(7, 9, 8) ∈ CU D (B) 9,7 .We need two operations first.(i) Double bracket decomposition: Decompose a sequence a 1 , a 2 , . . ., a n of distinct integer into ((a 1 , . . ., a i−1 )a i (a i+1 , . . ., a n )), where a i is the smallest entry.
(ii) Complement: Replace the i-th smallest entry of a sequence A = a 1 , a 2 , . . ., a n of distinct integers by the i-th greatest entry for each 1 ≤ i ≤ n.Denote the resulting sequence by A c .Algorithm 1 is actually adapted from [6], which maps a sequence of distinct integers into a non-plane complete increasing binary tree.
Set T (A) to be an empty tree.Apply the following procedures recursively.(i) If max(A) is on the left of min(A) : Let A := A c and go to (ii) (ii) If max(A) is on the right of min(A) : Apply double bracketing decomposition on A and write A = ((A L )a i (A R )).Construct a tree rooted at a i and define the children of a i to be two nodes labelled by min(A L ) and min(A R ) (If any of the sequence is empty, the child is defined to be empty).(iii) Apply the above procedure recursively on A L and A R .Output : A complete increasing binary (non-plane) tree T (A).Note that the leaves of T (A) are all empty.

From a sequence A to the complete increasing binary tree T (A).
We say a permutation σ = σ 1 , σ 2 , . . ., σ |A| of elements of a set A of distinct integers up-down if σ 1 < σ 2 > σ 3 < • • • .Let UD A be the set of up-down permutations and UD A,1 ⊆ UD A consisting of those with σ 1 = min(σ).The following is an easy consequence of Algorithm 1.
Corollary 5.1.Algorithm 1 induces a bijection between (i) UD A and the set of complete increasing (non-plane) tree with nodes labelled by elements of A such that all leaves are empty.(ii) UD A,1 and the set of complete increasing (non-plane) tree with nodes labelled by elements of A such that all leaves are empty and the root has an empty child.

.}. On the other hand, for a finite set A of positive integers we let T (1)
A denote the set of complete increasing binary plane trees with empty leaves, satisfying that (i) the nodes are labelled using the integers in A, and (ii) the rightmost labelled node is the root.The goal of Algorithm 2 is to map a cycle C i to a tree τ ∈ T -If a i,j < 0, let the labelled child be the right child of v and the empty node be the left child of v. -If a i,j > 0, let the labelled child be the left child of v and the empty node be the right child of v. (iii) If the node v with label |a i,j | contains two labelled children in T (A): -If a i,j < 0, let the child with smaller label be the right child of v and the child with larger label be the left child of v. -If a i,j > 0, let the child with smaller label be the left child of v and the child with larger label be the right child of v. Output : The resulting tree τ i .Note that it is easy to see that τ i ∈ T (1) int(C i ) .For example, the cycle C = (2, 6, 3, 7, 4, 9, 8) maps to a tree in T (1) int(C) as in Figure 3.

4.
Map each cycle to a non-plane complete increasing binary tree.
(ii) For C m = (k, k) : Let τ m be the tree of single node with label k.
In this way we obtain a sequence of trees τ 1 , τ 2 , . . ., τ m−1 such that τ i ∈ T int(C i ) for 1 ≤ i ≤ m − 1 and a single node tree τ m labeled by k.Since for 1 ≤ i ≤ m − 1 the right child of root of τ i is empty, we replace the empty right child of the root of τ i by the tree τ i+1 and get a tree τ ∈ T * n,k .This results in a bijection between CUD and their corresponding non-plane trees are given in Figure 7.By applying Algorithm 2 we obtain plane trees τ 1 , τ 2 , and τ 3 , shown in Figure 8 together with the tree τ 4 with a single node corresponding to the cycle (7,7).Finally, the tree τ ∈ T * 9,7 is constructed from τ 1 , τ 2 , τ 3 , and τ 4 , as shown in Figure 9.The following bonus is straightforward from Theorem 5.2 and Theorem 5.3 and we omit the proof.: σ → τ , the labels on the rightmost path of τ is exactly the set of minimum entries of cycles of |σ|.

Complete increasing binary Trees and Valley signed permutations
In this section, we give the bijections between valley signed permutations and complete increasing binary trees.
Let σ i = min(σ) and σ L = σ 1 , . . ., σ i−1 and σ R = σ i+1 , . . ., σ n .Construct a tree recursively by setting σ i as the root of T and T (σ L ) (resp.T (σ R )) be its right and left subtree (with the convention that T (∅) is an empty node).Note that to our purpose we set T (σ L ) to be the 'right' subtree.
Output : A complete increasing binary plane tree with empty leaves.
For example, the tree obtained from 7513426 is shown in Figure 10.From Algorithm 3 we have an easy observation.Lemma 6.1.For 2 ≤ i ≤ n, the node with label σ i has two empty children in T (σ) if and only if σ i is a peak in σ.Proof.For 2 ≤ i ≤ n, if σ i is a peak, then it is not the minimum entry of any consecutive subsequence of σ unless the subsequence only contains σ i itself.This means that when T (σ) is constructed both of the left and right subtrees of σ i are empty nodes.If σ i is not a peak, then at least one of its subtree is nonempty.□ Theorem 6.2.There is a bijection ϕ There must exist a peak |σ p i | in |σ| such that v i < p i < v i+1 for each i = 1, 2, . . ., t−1, and a peak |σ pt | with p t > v t .Now, for 1 ≤ i ≤ t, if σ v i +1 < 0, we remove the two empty children of the node with label |σ p i | in T (σ).Note that these operations are well defined by Lemma 6.1.We result in a tree τ ∈ T • n and set ϕ (B) V (σ) = τ .Furthermore, since σ 1 is in the left subsequence during the recursive construction of T (σ), the node with label σ 1 must be the rightmost labelled node.Hence the map ϕ There must exist a peak |σ p i | in |σ| such that v i < p i < v i+1 for each i = 1, 2, . . ., t−1, and a peak |σ pt | with p t > v t .Now for 1 ≤ i ≤ t, if σ v i +1 < 0, then remove the two empty children of the node with label |σ p i | in T (σ).These operations are well defined by Lemma 6.1.We obtain the resulting tree τ ∈ T * n and set ϕ  For a permutation σ = σ 1 σ 2 . . .σ n , an entry σ i is a left-to-right minimum if σ i = min{σ j : 1 ≤ j ≤ i}.We have a bonus corollary.Proof.Observe that spk([σ]) is equal to the number of labelled leaves of ϕ F ([σ]).Hence the number of non-leaf nodes of ϕ F ([σ]) is n − spk(σ).Moreover, the total number of leaves of a complete binary tree is one plus the number of its non-leaf nodes.Therefore, we have

Concluding remarks
In this paper, we give types B n and type D n definitions of cycle-up-down permutations and flip equivalence classes on signed permutations.We provide new refined Arnold families on them as well as on the valley signed permutations.We also give bijections between these structures with complete increasing binary trees.Together with those bijections in [8] between complete increasing binary trees and other structures (e.g.signed increasing 1-2 trees, signed André permutations, and signed Simsun permutations of type I and type II), we have bijections between all these Arnold families.
However, in the light of the fact that the set of even-signed permutations is just a subset of signed permutations, i.e., D n ⊂ B n , it is not so satisfying that in our structures as well as those presented in [8] realizing the refined Arnold families, the type D n structures are usually not subsets of the corresponding type B n structures.This hints that there could exist totally different constructions of all these structures compatible with D n ⊂ B n .It will be of great interest if one can find them.
Euler numbers are ubiquitous in combinatorics.Except for what we and [8] have investigated, there are still many interesting combinatorial structures counted by them.For examples, the number of sequences (a 1 , . . .a n−1 ) with 0 ≤ a i < i, such that no three terms are equal [4], the number of self-dual rooted edge-labeled trees with n vertices [1], the number of linear extensions of the "zig-zag" poset [16,17], the number of total cyclic orders on {0, 1, . . ., n} under certain conditions [13], and so on.It would also be interesting to find their types B n and D n counterpart definitions and corresponding statistics such that together they realize the refined Arnold families.
Another common generalization of S n and B n is the wreath product S n ≀ C n , the r-colored permutations.The alternating permutations have been generalized along this route [11].It is also a natural problem to see if all the structures considered in [8] and this paper have their colored generalizations.We leave these problems to the interested readers.

n
be the set of valley signed permutations of type D n andVS (D) n,k := {σ ∈ VS (D)n , σ 1 = k} for 1 ≤ k ≤ n.We have VS

Figure 1 .
Figure 1.The complete increasing binary trees with three labelled nodes.

( 1 )
int(C i ) .Algorithm 2: Input : A cycle C i = (a i,1 , a i,2 , . . ., a i,d ) of a signed cycle-up-down permutation σ.First we let A = |a i,1 |, |a i,2 |, . . ., |a i,d | and construct T (A) using Algorithm 1.For 1 ≤ j ≤ d, do the followings.(i) If the node v with label |a i,j | contains two empty children in T (A): -If a i,j < 0, then remove the two empty children.(ii) If the node v with label |a i,j | contains exactly one labelled child in T (A):

Figure 7 .
Figure 7. Map each cycle to a non-plane complete increasing binary tree.

Figure 10 .
Figure 10.The tree obtained from 7513426 by Algorithm 3.

V
(σ) = τ .Furthermore, since the rightmost labelled node of τ is labelled by |σ 1 |, the map ϕ (D) V induces a bijection between VS (D) n,k and T * n,k .□ For example, let σ = 7586341 92 ∈ VS (D) 9,7 .The tree T (|σ|) is shown in the left of Figure 12.First we remove the two empty children of the node 7. The negative entries 9 is corresponding to the peak 9 in |σ|, so we remove its empty children from T (|σ|) and obtain the tree τ ∈ T * 9,7 in the right of Figure 12.

Table 1 .
The Arnold numbers v n,k and Springer numbers.

Table 3 .
Summary of results.