The ascent-plateau statistics on Stirling permutations

In this paper, several variants of the ascent-plateau statistic are introduced, including flag ascent-plateau, double ascent and descent-plateau. We first study the flag ascent-plateau statistic on Stirling permutations by using context-free grammars. We then present a unified refinement of the ascent polynomials and the ascent-plateau polynomials. In particular, by using Foata and Strehl's group action, we prove two bistatistics over the set of Stirling permutations of order n are equidistributed.


Introduction
A Stirling permutation of order n is a permutation of the multiset {1, 1, 2, 2, . . . , n, n} such that for each i, 1 ≤ i ≤ n, all entries between the two occurrences of i are larger than i. Denote by Q n the set of Stirling permutations of order n. Let σ = σ 1 σ 2 · · · σ 2n ∈ Q n . For 1 ≤ i ≤ 2n, we say that an index i is a descent of σ if σ i > σ i+1 or i = 2n, and we say that an index i is an ascent of σ if σ i < σ i+1 or i = 1. Hence the index i = 1 is always an ascent and i = 2n is always a descent. Moreover, a plateau of σ is an index i such that σ i = σ i+1 , where 1 ≤ i ≤ 2n − 1. Let des (σ), asc (σ) and plat (σ) be the numbers of descents, ascents and plateaus of σ, respectively.
Stirling permutations were defined by Gessel and Stanley [8], and they proved that where n k is the Stirling number of the second kind, i.e., the number of ways to partition a set of n objects into k non-empty subsets. A classical result of Bóna [2] says that descents, ascents and plateaus have the same distribution over Q n , i.e., This equidistributed result and associated multivariate polynomials have been extensively studied by Janson, Kuba, Panholzer, Haglund, Chen et al., see [5,9,10,11] and references therein.
Recently, Ma and Toufik [15] introduced the definition of ascent-plateau statistic and presented a combinatorial interpretation of the 1/k-Eulerian polynomials. The purpose of this paper is to explore variants of the ascent-plateau statistic. In the following, we collect some definitions, notation and results that will be needed throughout this paper. 2010 Mathematics Subject Classification. Primary 05A05; Secondary 05A15. Definition 1. An occurrence of an ascent-plateau of σ ∈ Q n is an index i such that σ i−1 < σ i = σ i+1 , where i ∈ {2, 3, . . . , 2n − 1}. An occurrence of a left ascent-plateau is an index i such that σ i−1 < σ i = σ i+1 , where i ∈ {1, 2, . . . , 2n − 1} and σ 0 = 0.
The Eulerian polynomial of type B and the flag descent polynomial are respectively defined by Very recently, we studied the following combinatorial expansions (see [18,Section 4]): It is now well known that F n (x) = (1 + x) n A n (x) (see [1,Theorem 4.4]). This paper is motivated by exploring an expansion of F n (x) in terms of some enumerative polynomials of Stirling permutations. This paper is organized as follows. In Section 2, we present a combinatorial expansion of F n (x). In Section 3, we study a multivariate enumerative polynomials of Stirling permutations. In particular, we consider Foata and Strehl's group action on Stirling permutations.

The flag descent polynomials and flag ascent-plateau polynomials
Context-free grammar is a powerful tool to study exponential structures (see [5,18] for instance). In this section, we first present a grammatical description of the flag descent polynomials by using grammatical labeling introduced by Chen and Fu [5]. And then, we study the flag ascent-plateau statistics over Stirling permutations.

Context-free grammars.
For an alphabet A, let Q[[A]] be the rational commutative ring of formal power series in monomials formed from letters in A. Following Let us now recall a result on context-free grammars.
For n ≥ 0, we have D n (xy) = xy π∈Bn y fdes (π) z 2n−fdes (π) . Moreover, The grammatical labeling is illustrated in the following proof of (4). Let π ∈ B n . As usual, denote by i the negative element −i. We define an ascent (resp. a descent) of π to be a position i ∈ {0, 1, 2 . . . , n − 1} such that π(i) < π(i + 1) (resp. π(i) > π(i + 1)). Now we give a labeling of π ∈ B n as follows: is an ascent, then put a superscript label z and a subscript label z right after π(i); (L 2 ) If i ∈ [n − 1] is a descent, then put a superscript label y and a subscript label y right after π(i); (L 3 ) If π(1) > 0, then put a superscript label z and a subscript label x right after π(0); (L 4 ) If π(1) < 0, then put a superscript label x and a subscript label y right after π(0); (L 5 ) Put a superscript label y and a subscript label z at the end of π.
Note that D(xy) = xyz 2 + xy 2 z. Thus the sum of weights of the elements of B 1 is given by D(xy).
In general, the insertion of n + 1 (resp. n + 1) into π corresponds to the action of the formal derivative D on a superscript label (resp. subscript label). By induction, we get a grammatical proof of (4).

Example 3.
For example, let π = 043152. Then π can be generated as follows:
The number of flag ascent-plateau of σ is defined by We can now present the first main result of this paper.
Theorem 5. Let D be the formal derivative with respect to the grammar (3). For n ≥ 1, we have Therefore, Proof. We first introduce a grammatical labeling of σ ∈ Q n as follows: is an ascent-plateau, then put a superscript label y immediately before σ i and a superscript label y right after σ i ; (L 2 ) If σ 1 = σ 2 , then put a superscript label y immediately before σ 1 and a superscript x right after σ 1 ; (L 3 ) If σ 1 < σ 2 , then put a superscript label x immediately before σ 1 ; (L 4 ) The rest of positions in σ are labeled by a superscript label z.
Note that the weight of σ is given by For example, The labeling of 1223314554 and 661223314554 are respectively given as follows: x 1 y 2 y 2 y 3 y 3 z 1 z 4 y 5 y 5 z 4 z , y 6 x 6 z 1 y 2 y 2 y 3 y 3 z 1 z 4 y 5 y 5 z 4 z .
In this case, the insertion corresponds to the rule z → y 2 z and σ ′ ∈ S n+1 (i + 2).
It is routine to check that each element of Q n+1 can be obtained exactly once. By induction, we present a constructive proof of (5). Using the Leibniz's formula, we have D n (xy) = n k=0 D k (x)D n−1 (y). Combining (5) and Proposition 2, we get the desired formula (6). Let From the proof of (5), we see that the numbers T (n, k) satisfy the recurrence relation with the initial conditions T (0, 0) = 1, T ( 1, 1) = 1 and T (1, k) = 0 for k = 1. It should be noted that T (n, k) is also the number of dual Stirling permutations of order n with k alternating runs (see [16]). Recall that (see [20, A008292]): Write the formula (6) as follows: Combining (2) and (7), we have Therefore, a dual formula of (6) is given as follows:
Proof. Now we give a labeling of σ ∈ Q n as follows: (L 1 ) If i is a left ascent-plateau, then put a superscript label y immediately before σ i and a superscript label x right after σ i ; (L 2 ) If i is a double ascent, then put a superscript label q immediately before σ i ; (L 3 ) If i is a descent-plateau, then put a superscript label p right after σ i ; (L 4 ) The rest positions in σ are labeled by a superscript label z.
For example, the labeling of 552442998813316776 is as follows: y 5 x 5 z 2 y 4 x 4 z 2 y 9 x 9 z 8 p 8 z 1 y 3 x 3 z 1 q 6 y 7 x 7 z 6 z .
We proceed by induction on n. Note that Q 1 = { y 1 x 1 z } and Thus the weight of y 1 x 1 z is given by D(z) and the sum of weights of elements in Q 2 is given by D 2 (z), since D(z) = xyz and D 2 (x) = z(xyqz + xypz + x 2 y 2 ).
Assume that the result holds for n = m − 1, where m ≥ 3. Let σ be an element counted by P m−1 (i, j, k), and let σ ′ be an element of Q m obtained by inserting the pair mm into σ. We distinguish the following five cases: (c 1 ) If the pair mm is inserted at a position with label x, then the change of labeling is illustrated as follows: In this case, the insertion corresponds to the rule x → xzq and produces i permutations in Q m with i left ascent-plateaus, j + 1 double ascents and k descent-plateaus; (c 2 ) If the pair mm is inserted at a position with label y, then the change of labeling is illustrated as follows: In this case, the insertion corresponds to the rule y → yzp and produces i permutations in Q m with i left ascent-plateaus, j double ascents and k + 1 descent-plateaus; (c 3 ) If the pair mm is inserted at a position with label z, then the change of labeling is illustrated as follows: · · · σ z i σ i+1 · · · → · · · σ y i m x m z σ i+1 · · · .
In this case, the insertion corresponds to the rule z → xyz and produces 2m−2−2i−j −k permutations in Q m with i + 1 left ascent-plateaus, j double ascents and k descentplateaus; (c 4 ) If the pair mm is inserted at a position with label q, then the change of labeling is illustrated as follows: · · · σ q i σ i+1 · · · → · · · σ y i m x m z σ i+1 · · · .
In this case, the insertion corresponds to the rule q → xyz and produces j permutations in Q m with i + 1 left ascent-plateaus, j − 1 double ascents and k descent-plateaus; (c 5 ) If the pair mm is inserted at a position with label p, then the change of labeling is illustrated as follows: In this case, the insertion corresponds to the rule p → xyz and produces k permutations in Q m with i + 1 left ascent-plateaus, j double ascents and k − 1 descent-plateaus.
By induction, we see that grammar (9) generates all of the permutations in Q m . Combining the above five cases, we see that Multiplying both sides of the above recurrence relation by x i y j z k for all i, j, k, we get (10)

Equidistributed statistics.
Let i ∈ [2n] and let σ = σ 1 σ 2 . . . σ 2n ∈ Q n . We define the action ϕ i as follows: • If i is a double ascent, then ϕ i (σ) is obtained by moving σ i to the right of the second σ i , which forms a new pleateau σ i σ i ; • If i is a descent-plateau, then ϕ i (σ) is obtained by moving σ i to the right of σ k , where k = max{j ∈ {0, 1, 2, . . . , i − 1} : σ j < σ i }.
It is clear that the ϕ ′ i 's are involutions and that they commute. Hence, for any subset S ⊆ [2n], we may define the function ϕ ′ S : Q n → Q n by ϕ ′ S (σ) = i∈S ϕ ′ i (σ). Hence the group Z 2n 2 acts on Q n via the function ϕ ′ S , where S ⊆ [2n]. The third main result of this paper is given as follows, which is implied by (10).

Connection with Eulerian numbers.
Recall that the Eulerian numbers are defined by n k = #{π ∈ S n : des (π) = k}.
Since the numbers a(n, k) and n k satisfy the same recurrence relation and initial conditions, so they agree. This completes the proof.
A bijective proof of Theorem 11: Proof. Let σ ∈ Q n . Note that every element of [n] appears exactly two times in σ. Let α(σ) be the permutation of S n obtained from σ by deleting all of the first i from left to right, where i ∈ [n]. Then α is a map from Q n to S n . For example, α(344355661221) = 435621. Let D n = {σ ∈ Q n : lap (σ) = i, dasc (σ) = n − i, dp (σ) = 0}. Let x be a given element of [n]. For any σ ∈ Q n , we define the action β x on Q n as follows: • Read σ from left to right and let i be the first index such that σ i = x; • Move σ i to the right of σ k , where k = max{j ∈ {0, 1, 2, . . . , i − 1} : σ j < σ i }, where σ 0 = 0.

Concluding remarks
In this paper, we introduce several variants of the ascent-plateau statistic on Stirling permutations. Recall that Park [19] studied the (p, q)-analogue of the descent polynomials of Stirling permutations: C n (x, p, q) = σ∈Qn x des (σ) p inv (σ) q maj (σ) .
It would be interesting to study the relationship between C n (x, p, q) and the following polynomials: σ∈Qn x ap (σ) y lap (σ) p inv (σ) q maj (σ) .
In [6], Egge introduced the definition of Legendre-Stirling permutation, which shares similar properties with Stirling permutation. One may study the ascent-plateau statistic on Legendre-Stirling permutations.