Stirling permutations, cycle structures of permutations and perfect matchings

In this paper we provide a unified combinatorial approach to establish a connection between Stirling permutations, cycle structures of permutations and perfect matchings. The main tool of our investigations is MY-sequences. In particular, we discover that the Eulerian polynomials have a simple combinatorial interpretation in terms of some statistics on MY-sequences.


Introduction
Stirling permutations were defined by Gessel and Stanley [6]. Let j 2 := {j, j} for j ≥ 1. A Stirling permutation of order n is a permutation of the multiset {1 2 , 2 2 , . . . , n 2 } such that every element between the two occurrences of i are greater than i for each i ∈ [n], where [n] = {1, 2, . . . , n}. Denote by Q n the set of Stirling permutations of order n. For σ = σ 1 σ 2 · · · σ 2n ∈ Q n , an occurrence of an ascent (resp. a plateau) is an index i such that σ i < σ i+1 (resp. σ i = σ i+1 ), Recently, there is a large literature devoted to Stirling permutations and their generalizations. The reader is referred to [1,8,9,13,14] for recent progress on the study of statistics on Stirling permutations.
We define Analyzing the placement of 2 copies of (n + 1), it is easy to deduce that the polynomials N n (x) satisfy the recurrence relation N n+1 (x) = (2n + 1)xN n (x) + 2x(1 − x)N ′ n (x) with the initial value N 0 (x) = 1. The exponential generating function for N n (x) is given as follows (see [11,Section 5]): The first few of N n (x) are Let S n denote the permutation group on the set [n] and π = π(1)π(2) · · · π(n) ∈ S n . An excedance in π is an index i such that π(i) > i. Let exc (π) denote the number of excedances in π. The classical Eulerian polynomials A n (x) are defined by x exc (π) for n ≥ 1, and have been extensively investigated (see [4,5,7] for instance). In [5], Foata and Schützenberger introduced a q-analog of the Eulerian polynomials defined by where cyc (π) is the number of cycles in π. Brenti [2,3] further studied q-Eulerian polynomials and established the link with q-symmetric functions arising from plethysm. In particular, Brenti [3,Proposition 7.3] obtained the exponential generating function for A n (x; q): Let e = (e 1 , e 2 , . . . , e n ) ∈ Z n . Let I n,k = {e|0 ≤ e i ≤ (i − 1)k}, which known as the set of n-dimensional k-inversion sequences (see [15]). The number of ascents of e is defined by Recently, Savage and Viswanathan [16] discovered that x asc (e) = k n A n (x; 1/k).
From (2) and (3), we get Hence Let N n (x) = x n N n 1 x . It follows from (3) and (4) that A perfect matching of [2n] is a set partition of [2n] with blocks (disjoint nonempty subsets) of size exactly 2. Let M 2n be the set of matchings of [2n], and let M ∈ M 2n . The standard form of M is a list of blocks (i 1 , j 1 )/(i 2 , j 2 )/ . . . /(i n , j n ) such that i r < j r for all 1 ≤ r ≤ n and 1 = i 1 < i 2 < · · · < i n . Throughout this paper we always write M in standard form. It is well known that M can be regarded as a fixed-point-free involution on [2n] and also as a Brauer diagram on [2n] (see [10] for instance). Let so (M) be the number of blocks of M with odd smaller entries. It is well known that which has been studied in [11,12]. In particular, from [11, Theorem 9], we have Combining (1), (5) and (6), we get σ∈Qn x ap (σ) = π∈Sn x n−exc (π) 2 n−cyc (π) = M∈M 2n x so (M) .
It is natural to consider the following question: Is there existing a unified combinatorial approach to prove (7)? The main object of this paper is to provide a solution to this problem. The paper is organized as follows. In Section 2, we present the main results, and recall some definitions that will be used throughout the rest of this work. In other Sections, we give combinatorial proofs of these main results.

Definitions and main results
Let N n (x) = n k=1 N (n, k)x k . Hence N (n, k) is the number of perfect matchings in M 2n with the restriction that only k matching pairs have odd smaller entries. We partition the blocks of M into four subsets: We now prove (8). Let (a, b) be a given subset of M. Let ϕ be the construction of M ′ ∈ M 2n+2 that replacing (a, b) by (a, 2n + 1)/(b, 2n + 2) or (a, 2n + 2)/(b, 2n + 1). We distinguish two cases.
(c 1 ) If (a, b) ∈ OE or (a, b) ∈ EE, then the construction ϕ does not increasing the number odd smallers. Combining (9), this accounts for 2kN (n, k) possibilities. (c 2 ) If (a, b) ∈ OO or (a, b) ∈ EO, then the construction ϕ does form a new odd smaller.
Moreover, we can also append (2n + 1, 2n + 2) to M. This gives (2n − 2(k − 1) For k ≥ 1 and ℓ ≥ 0, we define be the number of positive (resp. negative) entries of Y n . Denote by star (Y n ) the number of ⋆ of Y n . Careful consideration of Proposition 1 yields the following definition.
We can now present the first main result of this paper.
In this paper we always write π ∈ S n by its standard cycle decomposition, in which each cycle is written with its smallest entry first and the cycles are written in ascending order of their smallest entry.  (c 1 , c 2 , . . . , c i ) be one cycle of π. We say that c j is a cycle descent if c j > c j+1 for 1 ≤ j < i.
Note that neg (Y n ) + star (Y n ) + pos (Y n ) = n. The following corollary is immediate.
In the following sections, we give bijective proofs of the main results. As a consequence, we get a desired proof of (7). (10) In the following discussion, we always put any 2-elements block of M into exactly a parenthesis or a square bracket. For (S 1 ) Note that M(P 1 N 0 ) = [1,2]. We take [1,2] as the starting point, which corresponds to the first term of a MY-sequence. Since P 1 N 0 = {1, 2, ⋆}, we can perform one of the following three operations: (c 1 ) Using the element 1 of P 1 N 0 to break the square bracket [1,2] such that [1,2] is replaced by (1, 3)/ [2,4]. (c 2 ) Using the element 2 of P 1 N 0 to break the square bracket [1,2] such that [1,2] is replaced by [1,4]/(2, 3). (c 3 ) As for ⋆, we append [3,4] right after [1,2]. (S 2 ) For k ≥ 2, given a M ∈ M(P 1+s k N t k ). We use entries of P 1+s k N t k to break some square brackets or parentheses of M. Assume that c, d, e and f are positive integers. For the entries of P 1+s k , we distinguish two cases: (c 1 ) Using the element 2i − 1 to break the i-th square bracket with the restriction that (i) if the i-th square bracket with elements 2c − 1 and 2d, then we replace Given a MY-sequence Y n with neg (Y n ) + star (Y n ) = i. Repeat the BB algorithm n-times, we can get a unique perfect matching M of [2n] with oe (M) + ee (M) = i. Conversely, given a perfect matching M ∈ M 2n with oe (M) + ee (M) = i. If we delete 2n − 1 and 2n, then we can find the n-th element of the corresponding MY-sequence. Along the same lines, we can get a unique MY-sequence Y n with neg (Y n ) + star (Y n ) = i. Thus the BB algorithm gives a bijective proof of the left equality of (10). In fact, using the BB algorithm, we give a bijective proof of the following result: 4. Proof of the right equality of (10) Set r = 1 + s k + t k . Denote by Q (P 1+s k N t k ) the set of Stirling permutations of order r with exactly 1 + s k ascent plateaus. In particular, Q (P 1 N 0 ) = {11}, Q (P 1 N 1 ) = {2211, 1221} and Q (P 2 N 0 ) = {1122}. Let us first give a definition of labeled Stirling permutations. Definition 8. Let σ ∈ Q (P 1+s k N t k ). If i 1 < i 2 < . . . < i 1+s k are the ascent plateaus of σ, then we put the superscript labels 2ℓ − 1 before i ℓ and 2ℓ after it, where 1 ≤ ℓ ≤ 1 + s k . In the remaining positions, we put the superscript labels −1, −2, . . . , −2t k and ⋆ from left to right.

Proof of the left equality of
For example, the labels of 13324421 are given as follows: Given a MY-sequence Y n = (y 1 , y 2 , . . . , y n ). Now we present a labeled Stirling permutations algorithm (LSP algorithm for short): (S 1 ) Since Q (P 1 N 0 ) = {11}, we take 11 as the start point, which corresponds to y 1 = ⋆.
It is straightforward to show that each such labeled Stirling permutation will be obtained exactly once. Indeed, given a labeled Stirling permutation of order n with i ascent plateaus, we can just read the indices of ascent plateaus. Deleting the pair (2n)(2n) gives the entry y n . Along the same lines, we can get a unique MY-sequence Y n with neg (Y n ) + star (Y n ) = i. In conclusion, LSP algorithm gives the desired proof of the right equality of (10).

Proof of (11)
Let us first give a definition of labeled permutations.
Definition 10. Let π ∈ S n with p excedances. If i 1 < i 2 < · · · < i p are the excedances, then we put superscript labels −k between i k and π(i k ), where 1 ≤ k ≤ p. In the remaining positions except the first position, we put the superscript labels 1, 2, . . . , n − p from left to right.
A labeled permutations algorithm (LP algorithm for short) for generating MY-sequences is given as follows: Take (1 1 ) as the start point, which corresponds to the first term ⋆ of a MY-sequence. Let π be a labeled permutation in S n . There are n + 1 permutations of S n+1 can be obtained from π by inserting n + 1 to the positions with superscript labels or as a new cycle (n + 1). We distinguish three cases: (C 1 ) If we insert the entry n + 1 to the position with superscript label k, then the n-th term of the corresponding MY-sequence is y n ∈ {2k − 1, 2k}. (C 2 ) If we insert the entry n + 1 to the position with superscript label −ℓ, then the n-th term of the corresponding MY-sequence is y n ∈ {−2ℓ + 1, −2ℓ}. (C 3 ) If we insert the entry n + 1 at the end of π to form a new cycle (n + 1), then the n-th term of the corresponding MY-sequence is ⋆.
In particular, consider π = (1 1 ). Note that (1, 2) is obtained from (1 1 ) by inserting 2 to the position with superscript label 1. Hence y 2 ∈ {1, 2} is the MY-sequence that corresponds to (1,2). Note that (1)(2) is obtained from (1 1 ) by inserting 2 as a new cycle (2). Hence y 2 = ⋆ is the MY-sequence that corresponds to (1)(2). Take (1 1 ) as a start point. Repeat the LP algorithm n times, it is easy to verify that each MY-sequence of length n will be obtained exactly once. Note that 2 cdes (π)+exc π is the number of MY-sequences that corresponds to π. Therefore, the LP algorithm gives a combinatorial proof of (11).
Therefore, the corresponding MY-sequences of π are given as follows: