Inversion sequences and signed permutations

A signed inversion sequence of length n is a sequence of integers e = e 1 · · · e n , where e i +1 ∈ { 0 , ¯0 , 1 , ¯1 , . . . , i, ¯ i } for every i ∈ { 0 , 1 , . . . , n − 1 } . For a set of signed patterns B , let ¯ I n ( B ) be the set of signed inversion sequences of length n that avoid all the signed patterns from B . We say that two sets of signed patterns B and C are Wilf-equivalent if | ¯ I n ( B ) | = | ¯ I n ( C ) | for every n ≥ 0 . In this paper, by generating trees, we show that the number of Wilf-equivalences among singles of a length-2 signed pattern is 3 and the number of Wilf-equivalences among pairs of a length-2 signed patterns is 30 .


Introduction
The goal of this paper is to give analogies of enumerative results on inversion sequences related to permutations characterized by pattern-avoidance in the inversion sequences related to signed permutations.Let B n be the hyperoctahedral group, which is the natural analogue of the symmetric group S n .We identify classes of restricted inversion sequences of elements of B n with enumerative properties analogous to results in the inversion sequences of elements of S n .
Restricted inversion sequences.Any permutation π = π 1 π 2 • • • π n in S n can be coded as an inversion sequence π 1 π 2 • • • π n of length n, where π j = |{i | π i < π n+1−j , n + 2 − j ≤ i ≤ n}|, for all j = 1, 2, . . ., n.For example, the permutations of S 3 , namely 123, 132, 213, 231, 312, and 321, are coded by the following inversion sequences 000, 010, 001, 011, 002, and 012, respectively.Note that an inversion sequence π = π 1 π 2 • • • π n of length n satisfies 0 ≤ π i < i for all i = 1, 2, . . ., n.We denote the set of all inversion sequences of length n by I n .Let π = π 1 π 2 • • • π n ∈ I n and let τ be any word contains all the letters 0, 1, . . ., k.An occurrence of τ in π is a subsequence is order-isomorphic to τ ; in such a context τ is usually called a pattern.We say that π avoids τ , or is τ -avoiding, if there is no occurrence of τ in π.We denote the set of all τ -avoiding inversion sequences in I n by I n (τ ).For an arbitrary finite collection of patterns T , we say that π avoids T if π avoids any τ ∈ T ; we denote the corresponding subset of I n by I n (T ).
Restricted signed inversion sequences.Let us view the elements of B n as signed permutation s = s 1 s 2 • • • s n in which each of the symbols 1, 2, . . ., n appears once, possibly barred.Clearly, the cardinality of We define the barring operation as the one which changes the symbol θ i to θ i and θ i to θ i .It is thus an involution, that is, θ i = θ i .Furthermore, we define the absolute value |a| = |ā| = a, for all a ≥ 0.
and θ j is barred if and only if π n+1−j is barred, for all j = 1, 2, . . ., n.We denote the set of all signed inversion sequences of length n by Īn .For example, the singed permutation 31 52 4 ∈ B 5 is coded by the following signed inversion sequence 0 02 0 2.
Let θ = θ 1 θ 2 • • • θ n ∈ B n and τ be any word contains all the letters 0, 1, . . ., k possibly barred.We say that θ contains τ , if there is a sequence of k indices and θ ij is barred if and only if τ j is barred, for all 1 ≤ j ≤ k.In such a context τ is usually called a signed pattern.For example, the signed inversion sequence 0 02 0 2 avoids 001 but contains 00 1 and the signed inversion sequence 0 00 20 avoids 000 but contains 0 10 .We denote the set of all τ -avoiding signed inversion sequences in Īn by Īn (τ ).For an arbitrary finite collection of patterns B, we say that π avoids B if π avoids any τ ∈ B; we denote the corresponding subset of Īn by Īn (B).We say that two sets of signed patterns B and C are Wilf-equivalent if | Īn (B)| = | Īn (C)|, for all n ≥ 0.
In the symmetric group S n , for every pattern τ ∈ S 3 , the number of τ -avoiding permutations is given by the nth Catalan number 1 n+1 2n n , see [2].Simion [6] showed that there are similar results for the hyperoctahedral group, for every signed pattern τ ∈ B 2 , the number of τ -avoiding signed permutations in B n is given by n j=0 j! n j 2 (For generalization, see [4]).
As an extension of these works, several researchers studied restricted inversion sequences.In particular, in [1,5], explicit formulas and/or generating functions are derived, which count the inversion sequences of a given length that avoid a length-3 pattern.In this paper, we study the number of the signed inversion sequences in Īn that avoid a length-2 signed pattern.In particular, we show the following result: Theorem 1.1.For signed inversion sequences, we show that (1) the number of Wilf-equivalences among singles of a length-2 signed pattern is 3, and (2) the number of Wilf-equivalences among pairs of length-2 signed patterns is 30.
In the next section, we extend the use of generating functions from the case of inversion sequences to signed inversion sequences.In Section 3, we prove Theorem 1.1(1), and then in Section 4, we prove Theorem 1.1(2).

Generating trees and signed inversion sequences
To establish a meaningful link between generating trees and the problem of avoiding patterns in signed inversion sequences, let us extend the generating trees [7] that discussed in [3] to pattern avoidance on signed inversion sequences.Given a set of patterns B, we define Ī(B) = ∪ n≥0 Īn (B).We proceed to build a pattern-avoidance tree, denoted as T (B), for the set Ī(B).If does not exists an nonempty signed inversion sequence that avoids the set B, then the tree T (B) has only a root labeled by empty word .Starting from this initial root that remains at level 0, the nodes at level n + 1 within the tree T (B) can be generated from the nodes at level n in a manner that the descendants of e = e 1 • • • e n ∈ Īn (B) are e = e 1 • • • e n j ∈ Īn+1 (B), where j = n, . . ., 1, 0, 0, 1, . . ., n.Now, we proceed to relabel the vertices of the tree T (B) as follows.Let T (B; e) be the subtree comprises the signed inversion sequence e as its root along with its subsequent descendants in T (B).Let e, e be any two nodes in T (B), e is said to be equivalent to e , denoted by e ∼ e if and only if T (B; e) ∼ = T (B; e ) as plane tree isomorphism.We define the set of all equivalent classes in the quotient set T (B)/ ∼ by E(B).We denote each equivalence class in E(B) by the label of the singular node found on the tree T (B) as the first node (from top to bottom, left to right).Let T [B] be the identical tree T (B), where each node of T (B) is replaced with its corresponding equivalence class label.
• the only child of 0 is 0 0, so 0 0 0. Note that there are no children for 0 0.
• After we guessed and proved (if possible) the rules of the generating tree T [B], we translate these rules into a system of equations and we solve for F B (x) = n≥0 | Īn (B)|x n .Note that the rule e v (1) , . . ., v (s) can be translated to where I w (x) = n≥0 (#the nodes at level n in T (B; w))x n is the generating function for the number of nodes at level n ≥ 0 in the subtree of T (B; w), where its root stays at level 0. Clearly, F B (x) = I (x).

Example 2.2. As continuation of Example 2.1, we have
where A m (x) = I am (x), B m (x) = I bm (x), and C m (x) = I cm (x).
Note that I 0 0(x) = 1 and I 0 (x) = 1 + x.So, by induction on m, we have that B m (x) = (1 + x) m .This implies that C m (x) = (1 + x) m+2 , and By iterating this equation infinity number of times (here we assumed that |x| < 1), we obtain Hence, by solving for I (x), we obtain

Avoiding a length-2 signed pattern
In this section, we deal with avoiding a length-2 signed pattern.To shorten the notation, we define S m to be either the word SS  (1).By symmetric operation baring, we see that 00 ∼ 00 and 0 0 ∼ 00.Hence, it is enough to show that 00 ∼ 0 0. Note that the generating tree T [00] is given by a root and the following rules: where Also, the generating tree T [0 0] is given by a root and the following rules: We leave to the reader that the rules of the generating trees T [00] and T [0 0] are holding (similar to the proofs of the Case ( 2)).Hence, Hence, 00 ∼ 0 0.
(3).By symmetric operation baring, we see that 10 ∼ 10 and 1 0 ∼ 10.Hence, it is enough to show that 10 ∼ 1 0. Note that the generating trees T [10] and T [1 0] are given by a root and the following rules where a m = 0m and b m,j = a m j.We leave to the reader that the rules of the generating trees T [10] and T [1 0] are holding (similar to the proofs of the Case (2)).Hence, T [00] ∼ = T [0 0], which implies that 00 ∼ 0 0

Avoiding a pair of length-2 signed patterns
In this section, we deal with avoiding a pair of length-2 signed patterns.By finding the number of signed inversion sequences of length n that avoid a length-2 signed pattern, for all n = 1, 2, . . ., 7, we can guess that there are 30 Wilf classes, see Table 2.Note that by baring operation, we see that Based on Table 2, in order to find the number of Wilf classes when signed inversion sequences of length n avoid a pair of length-2 signed pattern, we have to consider the Classes 6, 11, 12, 19, 26 in Table 2.In the next 5 propositions, we prove that each class of these classes create exactly one Wilf class.Proposition 4.1.We have {00, 00 } ∼ {01, 01}.
By iterating, we have (from now, we assume that |x| < 1) Note that I (x) = 1 + xI 0 (x) + xI0(x), which is equivalent to I (x) = 1 + xB 0,0 (x) + xB 1 (x).Thus, Hence, the number of signed inversion sequences of length n that avoid B is given by

Table 1 :
• • • S with m letters or the sequence S, S, ..., S with m terms, for any sequence S and a nonnegative integer m.By finding the number of signed inversion sequences of length n that avoid a length-2 signed pattern, for all n = 1, 2, ..., 7, we can guess that there are three Wilf classes, see Table1.Number signed inversion sequences in Īn (τ ), where n = 1, 2, . . ., 7 and τ is any length-2 signed pattern.