A Toeplitz property of ballot permutations and odd order permutations

We give a new semi-combinatorial proof for the equality of the number of ballot permutations of length $n$ and the number of odd order permutations of length $n$, which is due to Bernardi, Duplantier and Nadeau. Spiro conjectures that the descent number of ballot permutations and certain cyclic weight of odd order permutations of the same length are equi-distributed. We present a bijection to establish a Toeplitz property for ballot permutations with any fixed number of descents, and a Toeplitz property for odd order permutations with any fixed cyclic weight. This allows us to refine Spiro's conjecture by tracking the neighbors of the largest letter in permutations.


Introduction
Let S n be the symmetric group of permutations of the set [n] = {1, 2, . . . , n}. Let π = π 1 π 2 · · · π n ∈ S n . The signature of π is defined to be the sequence (q 1 , q 2 , . . . , q n−1 ) where Niven [8] found a determinantal formula for the number of permutations of length n with a prescribed signature, and showed that this number attains its maximum if and only if the signature is for an André permutation [1]; see de Bruijn [5] for a recursive proof. A pair (π i , π i+1 ) of letters is a descent (resp., an ascent ) if π i < π i+1 (resp., π i > π i+1 ). Denote the number of descents (resp., ascents) of π by des(π) (resp., asc(π)). We call the number h(π) = asc(π 1 π 2 · · · π n ) − des(π 1 π 2 · · · π n ) the height of π. The permutation π is said to be a ballot permutation if the height of any prefix of π is nonnegative, namely, h(π 1 π 2 · · · π i ) ≥ 0 for all i ∈ [n]. The number of ballot permutations of height 0 in S 2n+1 , or Dyck permutations of length n, is the Eulerian-Catalan number; see Bidkhori and Sullivant [3]. A classical coin-tossing game problem concerning the descent-ascent structure in a sequence of independent random variables of values ±1 was considered by Chung and Feller [4].
Bernardi et al. [2] obtained Theorem 1.1 by considering more generalized paths containing horizontal steps, called well-labelled positive paths. They constructed a bijection between well-labelled positive paths of size n with k horizontal steps and matchings on [2n] having k pairs (i, j) with i ∈ [n] and j ∈ [n + 1, . . . , 2n − 1], and thus obtained an explicit formula for the number of well-labelled positive paths of size n having k horizontal steps. Taking k = 0 in their formula yields Theorem 1.1.
Denote by P n the set of odd order permutations of [n], viz., the set of permutations of [n] whose every cycle is of odd length. By considering the neighbours of the letter n, we see that |P n | = |P n−1 | + (n − 1)(n − 2)|P n−2 |.

This recurrence gives
|P n | = p n immediately, where p n is the number defined in Theorem 1.1. In order to find an analogue for the descent statistic in the context of odd order permutations, Spiro [9] introduced the following interesting notion. For a permutation π, Spiro defines where the sum runs over all cycles of π, with the cyclic descent and the cyclic ascent where |c| is the length the cycle c. We call w(c) = min cdes(c), casc(c) the cyclic weight of c, and w(π) = M (π) the cyclic weight of π. In the next section we give a new proof of Theorem 1.1 by using certain combinatorial decomposition of ballot permutations with respect to the neighbors of the letter n. In Section 3, we display a bijection to establish a Toeplitz property for the number of ballot permutations of length n with a fixed number of descents. A slight modification of the bijection gives the Toeplitz property for the number of odd order permutations of length n with a fixed cyclic weight. These Toeplitz properties lead us to a refinement conjecture of Conjecture 1.2; see Conjecture 3.3. In Section 4, we show some easy cases of Conjecture 3.3.

2.
A new semi-bijective proof of Theorem 1.1 We give an overview of notion and notation in combinatorics on words which will be of use; see [6,7]. For any word w of length n, we denote its ith letter by w i , denote its alphabet A(α) = {w 1 , . . . , w n }, denote its length n by ℓ(w), and write w −1 = w n . Denote the reversal of w by w ′ = w n w n−1 · · · w 1 . In particular, the reversal of the empty word ǫ is ǫ itself. We call a word u a factor (resp., prefix ) of w if there exist words x and y such that w = xuy (resp., w = uy). We say that w is ballot if h(u) ≥ 0 for any prefix u of w.
The notions of reversal and factor have a cyclic version. Let π = (c 1 ) · · · (c k ) ∈ S n , where (c i ) are the cycles of π. We say that a word u is a cyclic factor of π if u is a factor of some word v such that (v) is a cycle of π. For example, the permutation π = (145)(26837) has a cyclic factor (372).
For any letters i and j, denote by B n,d (i, j) the set of permutations in B n,d that contain the factor inj, by P n,d (i, j) the set of permutations in P n,d that contain the cyclic factor inj. For example, we have We use lower-case letters b and p, in replace of B and P respectively, to denote the corresponding set cardinality, such as p n,d (i, j) = |P n,d (i, j)|. It is clear that Proof. Fix a letter i. Suppose that i < j. Then j = i + 1. For π ∈ B n,d (i, i + 1), define φ(π) to be the permutation obtained from π by removing the letters i + 1 and n and arranging the remaining letters in the order-preserving manner so that φ(π) ∈ S n−2 . It is clear that φ is a bijection between the sets B n,d (i, i + 1) and B n−2, d−1 . The other case i > j can be handled similarly. For odd order permutations, one obtains the desired equality by using the same operation of φ and keeping the cycle structure.
For any permutation π on a set of n positive integers, we define the standard form of π to be the permutation σ ∈ S n such that π i < π j if and only if σ i < σ j for all pairs (i, j). Denote by B the set of finite permutations on positive integers whose standard forms are ballot permutations. Let ω ∈ B \ {ǫ}. We say that a ballot permutation π ∈ B n is ω-decomposable if π is the concatenation αωγδ of factors α, ω, γ, δ such that where α, γ and δ are allowed to be the empty word ǫ. Let X n (ω) be the set of ω-decomposable ballot permutations π ∈ S n . Define the ω-decomposition of a permutation π ∈ X n (ω) to be the 4-tuple (α, ω, γ, δ) such that π = αωγδ and that γ is the longest word satisfying From definition we see that the ω-decomposition of any permutation in X n (ω) uniquely exists. In this case, we write π = (α, ω, γ, δ). In addition, we will use the convenience Proof. Denote by χ the characteristic function defined by χ(P ) = 1 if a proposition P is true, and Then the factor η = γδ 1 δ 2 · · · δ j , which is longer than γ, satisfies On the other hand, note that h(δ j−1 · · · δ 2 δ 1 ) = h(δ 1 δ 2 · · · δ j−1 ) = 0 and δ j−1 · · · δ 2 δ 1 ∈ B.
contradicting the premise π ∈ B. This completes the proof.
As will be seen, Theorem 2.3 is the key in the new proof of Theorem 1.1.
Note that (1) to (3) are independent to each other. We show (2) first for it will be of use in showing (1).
Together with Eq. (2.3), we find On the other hand, since ρ ′ αµ −1 is ballot, so is its prefix ρ ′ . Since δ is ballot, so is its prefix ρ. Therefore, the word ρ must be of height 0, and h It remains to show that f : X n (λ) → X n (µ) is a bijection. For σ = (α, µ, γ, δ) ∈ X n (µ), define g(σ) = γ ′ λα ′ δ. Same to the above, one may show that g(σ) ∈ X n (λ) and the λ-decomposition of g(σ) is (γ ′ , λ, α ′ , δ). By definition, we can derive that Thus the composition gf is the identity on X n (λ). In the same fashion, one may show that f g is the identity on X n (µ). Hence f is a bijection, with the inverse g. Now we are in a position to give the new proof of Theorem 1.1.
Proof of Theorem 1.1. Let n ≥ 4, i ≥ 1 and i + 2 ≤ j ≤ n − 1. Let b n (i, j) be the number of ballot permutations in B n containing the factor inj. Recall that λ = in(j − 1)j and µ = (j − 1)jni. We claim that In fact, consider the involution Φ of exchanging the letters j − 1 and j on the set We shall show that Φ(Y n ) = B n (i, j), which implies Eq. (2.5) immediately. Let π ∈ Y n . If the letters j − 1 and j are not adjacent in π, then Φ(π) ∈ B n (i, j) and the letters j − 1 and j are not adjacent in Φ(π). Suppose that j − 1 and j are adjacent in π. Since π ∈ X n (λ), there is no factor γ satisfying Eq. (2.1). In other words, the height of any prefix of π that is longer than αλ is at least 2, where αλ is the prefix of π ending at λ. Therefore Φ(π) ∈ B n (i, j). It is clear that the preimage of every permutation σ ∈ B n (i, j) lies in Y n . This proves Eq. (2.5).
Similarly, one may show that By Theorem 2.3, Eqs. (2.5) and (2.6), and Lemma 2.1, we infer that Since the number of ballot permutations in B n ending with the letter n is b n−1 , we obtain It is trivial to check that b 1 = b 2 = 1. Since the sequence p n admits the same recurrence and initial values, we conclude that b n = p n .

A Toeplitz property
Computer calculus gives that the matrices b n (i, j) i,j for 3 ≤ n ≤ 8 are respectively A square matrix (a ij ) is said to be Toeplitz if a i+1, j+1 = a i,j for all well defined entries a i+1, j+1 and a i,j . In this section, we manage to show that the matrices and P (n, d) = p n,d (i, j) are Toeplitz for any number d. Proof. Suppose that n ≥ 4, 1 ≤ d ≤ ⌊(n − 1)/2⌋, and i, j ∈ [n − 2]. This proof is organized as follows. First we give necessary notion and notation to define a map T between the sets B n,d (i, j) and B n,d (i + 1, j + 1). Second we interpret some facts implicitly contained in the definition. We then make some efforts to show that T is well defined. Finally we prove that T is bijective, which will complete the proof.
Let m = min(i, j) and M = max(i, j). We introduce the notation and define l π = 0 if M and M + 1 are adjacent. Note that l π ≤ M − m + 1 since the letter M + 1 occurs twice in the word (m; M − m + 2). Define the lower core of π to be the word Then κ π is a factor of π of length l π + 2. For example, κ 382549671 = 96 and κ 134875962 = 7596.
We call the first operation the core replacement, and the second the straightening. We will show that T is well defined, i.e., T (π) ∈ B n,d (i + 1, j + 1) for any π ∈ B n,d (i, j), and that T is bijective. From definition, we observe that the map T operates in the following 3 steps: (1) The core replacement is length-preserving and position-preserving, namely the words T | κπ and κ π have the same length and the same position. Every letter in the set A(κ π ) = {m, m + 1, . . . , m + l π − 1, M, n} is replaced by a letter in the set In particular, the letter n is contained in both words κ π and T | κπ . Now, we show that T (π) ∈ B n,d (i + 1, j + 1). More precisely, we need to show (i) the word T (π) contains the factor (i + 1)n(j + 1); (ii) des T (π) = des(π); and (iii) T (π) ∈ B.
Suppose that l π = 0. Then the κ π ∈ {nM, M n}, and π contains the factor ι = mnM (M + 1) or its reversal. From definition of the core replacement, T | κπ is the factor (m + 1)n or its reversal. Since the letter m (resp., M + 1) is the smallest (resp., largest) one in the interval [m, M + 1], it is invariant in the order-preserving straightening. Therefore, the image T (π) contains the factor T | ι = m(m + 1)n(M + 1) or its reversal. This proves (i). Note that each of the words ι and T | ι has a unique descent, and the reversal operation exchanges descents and ascents. Since the straightening is order-preserving, the map T preserves all descents and ascents that has empty intersection with the cores. As a result, we obtain (ii). Moreover, the unique descent in T | ι appears later than the unique descent appears in ι, and the unique ascent in T | ′ ι appears earlier than that in ι. Since π is ballot, so is T (π). This proves (iii) and T (π) ∈ B n,d (i + 1, j + 1) for l π = 0.
Below we can suppose that l π ≥ 1. Then (i) is clear from the definition of the core replacement. In order to show (ii), we claim the equivalence for any adjacent letters u and v such that {u, v} ⊆ A(κ π ). Since the straightening is order-preserving, Equivalence (3.4) is true if {u, v} ∩ A(κ π ) = ∅. Below we can suppose that v ∈ {M, m + l π − 1} ⊆ A(κ π ) and u ∈ A(κ π ).
If v = m + l π − 1, then T π (v) = m + 1 by the core replacement. Since u ∈ A(κ π ), the letter u is not mapped in the core replacement. If T π (u) = m, then the letter u is mapped in the straightening operation. Since m is the smallest image in the straightening, its preimage u must be the smallest element in the straightening, that is, u = T −1 π (m) = m + l π . Now, the elements m + l π = u and m + l π − 1 = v are adjacent, contradicting the definition of the lower width l π . Therefore, T π (u) = m. We then obtain the equivalence On the other hand, by the premise u ∈ A(κ π ) and v ∈ A(κ π ), we deduce the equivalence Since all letters in the set [m − 1] are fixed under T π , we obtain the desired Equivalence (3.4) by combining the above two equivalences.
By Equivalence (3.4), the map T preserves all descents and ascents that are not entirely contributed by the lower core. For those entirely contained in the lower core, we note both the numbers of descents of the words κ π and T | κπ equal the lower width l π . Therefore, T preserves the total number of descents. This proves (ii).
For (iii), we observe that in the core replacement, either • both the preimage κ π and its image T | κπ contain a unique ascent, and the ascent in T | κπ appears earlier than the ascent in κ π appears, or • both of them contain a unique descent, and the decent in T | κπ appears later than the descent in κ π appears.
It remains to show that T is bijective. We define the upper width l ′ σ to be the length l of the longest factor of the form (M + 1; l) or (M + 1; l) ′ , if the letters m and m + 1 are not adjacent in σ; and define l ′ σ = 0 otherwise. Define the upper core of σ to be the word Define a map T ′ : B n,d (i + 1, j + 1) → B n,d (i, j) as follows. For σ ∈ B n,d (i + 1, j + 1), define T ′ (σ) to be the permutation obtained from σ by firstly replacing its upper core by in(j; l ′ σ ), if i > j, and then replacing each letter in the set [m, M + 1]\A(ρ σ ) to a letter in the set [m, M + 1]\A(T ′ | ρσ ) in the order-preserving manner.
We shall show that In fact, if l π = 0, then π contains the factor ι or its reversal. It follows that σ contains the factor T | ι or its reversal. Thus l ′ σ = 0 = l π . Suppose that l π ≥ 1. Then m + 1 and m are not adjacent in σ, since otherwise the letters m + l π − 1 = T −1 π (m + 1) and m + l π = T −1 π (m) would be adjacent in π, which contradicts the definition of l π . Thus l ′ σ ≥ l π from the definition of the core replacement. Furthermore, the letters M − l π + 2 and M − l π + 1 are not adjacent in σ, since otherwise the letters would be adjacent in π, which contradicts l π ≥ 1. This proves Eq. (3.5).
In the same fashion one may prove that l ′ σ = l T ′ (σ) . As a consequence, both compositions T T ′ and T ′ T are identities. This completes the whole proof.
We remark that the map T π can be represented letter-wise using a piecewise function, with the aid of the number l π .
Note that the key notion of lower and upper core widths, and that of lower and upper cores, are defined in a local structure of a permutation. In fact, the map T looks for the longest factor consisting of discretely continuous numbers starting from m. This localness perspective inspires us to translate the idea application from permutations to certain cyclic words, especially to each cycle of a permutation in its cycle representation, and obtain Theorem 3.2.
Theorem 3.2. The matrix P (n, d) is symmetric and Toeplitz for all n and d.
Proof. In this proof, we keep in mind that permutations are considered to be unions of cycles, and forget their nature as maps on the set [n]. Since taking reversal cycle-wise is an involution between the sets P n,d (i, j) and P n,d (j, i), the matrix P (n, d) is symmetric. Let n ≥ 4 and 1 ≤ d ≤ ⌊(n − 1)/2⌋. In order to show the Toeplitz property, namely P (n, d) i,j = P (n, d) i+1, j+1 , it suffices to consider 1 ≤ i < j ≤ n − 2 by virtue of the symmetry.
For π ∈ P n,d (i, j), let C π be the cycle of π containing the maximum letter n. Paralleling the notion of lower width by considering cyclic factors instead of factors, we define the lower width l π to be the length l of the longest cyclic factor of C π that is of the form (m; l) or (m; l) ′ , if the letters M and M + 1 are not adjacent in C π ; and define l π = 0 otherwise. The subsequent notion of the lower core and core replacement are exactly the same as those in the proof of Theorem 3.1, using the cyclic version of the lower width. Define the map T : P n,d (i, j) → P n,d (i + 1, j + 1) by running firstly the core replacement and then the straightening.
While the core replacement affects letters only inside the cycle C π , the straightening works on letters that are both inside and outside C π . Same to the proof of Theorem 3.1, we can deduce the following one by one: (i) the width l π ≤ M − m + 1, where M = max(i, j) and m = min(i, j); (ii) the image T (π) is a permutation of length n, and contains the cyclic factor (i + 1)n(j + 1); (iii) the map T preserves the length of each cycle of π, and the permutation T (π) is of odd order; (iv) the map T preserves the cyclic descents outside κ π , and thus preserves the cyclic descent number of each cycle except C π ; (v) the map T preserves the cyclic descent number inside the core κ π , and the cyclic descents formed by one letter inside κ π and the other letter outside κ π ; thus T preserves the cyclic descent number of C π if C π = (κ π ); (vi) the map T preserves the cyclic descent number of C π if C π = (κ π ).
In conclusion, the map T preserves the cyclic descent number of each cycle, thus preserves the cyclic weight d of π.
This proves (vi) and completes the proof.
We propose the following refinement conjecture of Conjecture 1.2.  Proof. Fix n, d, i, j. Since b n,d (1, j) + b n,d (j, 1) = 2p n,d (1, j) for all n, d and j, we can infer by Theorems 3.1 and 3.2 that Since the number of permutations in B n,d with the letter n appearing at the last position is b n−1, d , we obtain Since the number of permutations in P n,d with the letter n forming a singleton is p n−1, d , we obtain (3.7) p n,d = p n−1, d + i =j p n,d (i, j).
It follows that c and u have the same weight, and so do the permutations π and φ(π). This completes the proof. This completes the proof.