Eulerian polynomials , Stirling permutations of the second kind and perfect matchings

In this paper, we introduce Stirling permutations of the second kind. In particular, we count Stirling permutations of the second kind by their cycle ascent plateaus, fixed points and cycles. Moreover, we get an expansion of the ordinary derangement polynomials in terms of the Stirling derangement polynomials. Finally, we present constructive proofs of a kind of combinatorial expansions of the Eulerian polynomials of types A and B.

Let M n (x) be a sequence of polynomials defined by −1) . ( Combining (1) and an explicit formula of the Ehrhart polynomial of the s-lecture hall polytope, Savage and Viswanathan [27] proved that M n (x) = E for n, k 1, where N (1, 1) = 1 and N (1, k) = 0 for k 2 or k 0 (see [24, Proposition 1]).Let N n (x) = n k=1 N (n, k)x k .The first few of the polynomials N n (x) are The exponential generating function for N n (x) is given as follows (see [21, Eq. ( 25)]): Combining ( 1) and (2), we get M n (x) = x n N n ( 1 x ) for n 0. Let ol (M) be the number of blocks of M ∈ M 2n with odd larger entries.Then x ol (M) .
the electronic journal of combinatorics 24(4) (2017), #P4.27 Context-free grammar was introduced by Chen [6] and it is a powerful tool for studying exponential structures in combinatorics.We refer the reader to [8,11,22] for further information.Using [22,Theorem 10], one can present a grammatical proof of the following result.
Proposition 1.For n 0, we have Recall that the exponential generating function for xA n (x) is An equivalent formula of ( 3) is given as follows: Motivated by Proposition 1, the main purpose of this paper is to introduce some cycle structure related to N (x, z) or M (x, z).This paper is organized as follows.In Section 2, we introduce the Stirling permutations of the second kind, which is a disjoint union of Stirling permutations.In Section 3, we count Stirling permutations of the second kind by their cycle ascent plateaus, fixed points and cycles.In Section 4, we present a constructive proof of Proposition 1.

Stirling permutations of the second kind
Stirling permutations were introduced by Gessel and Stanley [14].A Stirling permutation of order n is a permutation of the multiset [n] 2 such that every element between the two occurrences of i is greater than i for each i ∈ ) be the number of Stirling permutations of [n] 2 with k descents.Following [14, Eq. ( 6)], the numbers C(n, k) satisfy the recurrence relation for n 2, with the initial conditions C(1, 1) = 1 and C(1, 0) = 0.The second-order Eulerian polynomial is defined by the electronic journal of combinatorics 24(4) (2017), #P4.27 In recent years, there has been much work on Stirling permutations (see [2,12,15,16,23]).In particular, Bóna [2] introduced the plateau statistic on Stirling permutations, and proved that descents and plateaus have the same distribution over Q n .Given σ ∈ Q n , the index i is called a plateau if σ i = σ i+1 .We say that an index i ∈ [2n − 1] is an ascent plateau if σ i−1 < σ i = σ i+1 , where σ 0 = 0. Let ap (σ) be the number of the ascent plateaus of σ.For example, ap (221133) = 2. Very recently, we present a combinatorial proof of the following identity (see [24,Theorem 3]): Motivated by Proposition 1 and (6), we shall introduce Stirling permutations of the second kind.
Let [k] n denote the set of words of length n in the alphabet [k].
the reduction of ω, denoted by red (ω), is the unique word of length n obtained by replacing the ith smallest entry by i.For example, red (33224547) = 22113435.Definition 2. A permutation σ of the multiset [n] 2 is a Stirling permutation of the second kind of order n whenever σ can be written as a nonempty disjoint union of its distinct cycles and σ has a standard cycle form satisfying the following conditions: (i) For each i ∈ [n], the two copies of i appear in exactly one cycle; (ii) Each cycle is written with one of its smallest entry first and the cycles are written in increasing order of their smallest entry; (iii) The reduction of the word formed by all entries of each cycle is a Stirling permutation.In other words, if (c 1 , c 2 , . . ., c 2k ) is a cycle of σ, then red (c Let Q 2 n denote the set of Stirling permutations of the second kind of order n.In the following discussion, we always write σ ∈ Q 2 n in standard cycle form.Example 3. the electronic journal of combinatorics 24(4) (2017), #P4.27 Proof.There are two ways in which a permutation σ ∈ Q 2 n with k cycle plateaus can be obtained from a permutation σ ∈ Q 2 n−1 .If cplat (σ) = k, then we can put the two copies of n right after a cycle plateau of σ.This gives k possibilities.If cplat (σ) = k − 1, then we can append a new cycle (nn) right after σ or insert the two copies of n into any of the remaining 2n − 2 − (k − 1) = 2n − k − 1 positions.This gives 2n − k possibilities.Comparing with (5), this completes the proof of C n (x) = π∈Q 2 n x cplat (π) .Along the same lines, one can easily prove the assertion for cycle ascents.This completes the proof.
Theorem 5.The polynomials Q n (x, q) satisfy the recurrence relation for n 0, with the initial condition Q 0 (x) = 1.Moreover, Proof.Given σ ∈ Q 2 n .Let σ i be an element of Q 2 n+1 obtained from σ by inserting the two copies of n + 1, in the standard cycle decomposition of σ, right after i ∈ Therefore, we have the electronic journal of combinatorics 24(4) (2017), #P4.27 and ( 7) follows.By rewriting (7) in terms of the exponential generating function Q(x, q; z), we have It is routine to check that the generating function q satisfies (9).Also, this generating function gives Q(x, q; 0) = 1, Q(x, 0; z) = 1 and Q(0, q; z) = e qz .Hence Q(x, q; z) = Q(x, q; z).
Combining ( 1) and ( 8), we get Q(x, q; z) = M q (x, z).Thus Q n (x, 1) = M n (x).Moreover, it follows from (7) that So the following corollary is immediate.Corollary 6.For n 1, we have k are called Eulerian numbers of type A and satisfy the recurrence relation with the initial conditions 0 0 = 1 and 0 k = 0 for k 1 (see [3,5] for instance).Hence Let CQ n denote the set of Stirling permutations of Q 2 n with only one cycle, which can be named as the set of cyclic Stirling permutations.Define Comparing (7) with (10), we get the following corollary.

Corollary 7.
For n 1, we have the electronic journal of combinatorics 24(4) (2017), #P4.27 3 The joint distribution of cycle ascent plateaus and fixed points on The number of fixed points of σ is defined by fix (σ) = #{k ∈ [n] : (kk) is a cycle of σ}.For example, fix ((1133)( 22)) = 1.Define Now we present the main result of this section.
In the following, we shall prove Theorem 8 by using 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 [6], a context-free grammar over A is a function ] that replace a letter in A by a formal function over A. The formal derivative D is a linear operator defined with respect to a context-free grammar G.More precisely, the derivative n (i, j, k).We now introduce a labeling scheme for σ: (i) Put a superscript label a at the end of σ and a superscript q before each cycle of σ; the electronic journal of combinatorics 24(4) (2017), #P4.27 (ii) If k is a fixed point of σ, then we put a superscript label b right after each k; (iii) Put superscript labels c immediately before and right after each cycle ascent plateau; (iv) In each of the remaining positions except the first position of each cycle, we put a superscript label d.
When n = 1, we have Suppose we get all labeled permutations in Q 2 m (i, j, k) for all i, j, k, where m 2. We consider the case n = m + 1.Let σ ∈ Q 2 m+1 be obtained from σ ∈ Q 2 m (i, j, k) by inserting two copies of the entry m + 1 into σ.Now we construct a correspondence, denoted by ϑ, between σ and σ .Consider the following cases: (c 1 ) If the two copies of m + 1 are put at the end of σ as a new cycle ((m + 1)(m + 1)), then we leave all labels of σ unchanged except the last cycle.In this case, the correspondence ϑ is defined by In this case, σ ∈ Q 2 m+1 (i, j, k + 1).By induction, we see that ϑ is the desired correspondence between permutations in Q 2 m and Q 2 m+1 , which also gives a constructive proof of ( 14).Lemma 10.
the electronic journal of combinatorics 24(4) (2017), #P4.27 there follows We see that the coefficients F (n, k) satisfy the same recurrence relation and initial conditions as n k , so they agree.Proof of Theorem 8: By Lemma 9 and Lemma 10, we get Taking c 2 = x, b 2 = y and d 2 = 1 in both sides of the above identity, we immediately get (11).Set S n (i, j, k) = #Q 2 n (i, j, k).The following recurrence relation follows easily from the proof of Lemma 9: Multiplying both sides of the last recurrence relation by q i y j x k and summing for all i, j, k, we immediately get (12).Note that Thus P (x, y, q; z) = e qz(y−1) Q(x, q; z).This completes the proof of Theorem 8.
It should be noted that there exists a straightforward proof of (13).Note that each object of Q 2 n is a disjoint union of one object counted by P (x, 0, q; z) and some fixed points.Since each fixed point contributes no cycle ascent plateau but one cycle, by rules of exponential generating function one has Q(x, q; z) = e qz P (x, 0, q; z) and P (x, y, q; z) = e yqz P (x, 0, q; z).
Given σ ∈ Q 2 n .We say that σ is a Stirling derangement if σ has no fixed points.Let DQ n be the set of Stirling derangements of [n] 2 .Let R n,k (x, q) be the coefficient of y k in P n (x, y, q).Note that R n,0 (x, q) is the corresponding enumerative polynomials on DQ n .Set R n (x, q) = R n,0 (x, q).Note that Comparing the coefficients of both sides of ( 12), we get the following result.
Let q n = #DQ n .Then the following corollary is immediate.
Corollary 12.For n 1, the numbers q n satisfy the recurrence relation with the initial conditions q 0 = 1, q 1 = 0 and q 2 = 2.
which can be easily proved by using the exponential formula (see [3,Theorem 3.50]).It should be noted that q n+1 is also the number of minimal number of 1-factors in a 2nconnected graph having at least one 1-factor (see [1]).It would be interesting to study the relationship between Stirling permutations of the second kind and 2n-connected graphs.
Proposition 14.For n 2, the polynomial R n (x) is symmetric and has only simple real zeros.
Let ı 2 = √ −1.From ( 17), putting x = −1, we deduce the expression Note that sec(z) is an even function.Therefore, for n 1, we have where the number h n is defined by the following series expansion: The first few of the numbers h n are h 0 = 1, h 1 = 2, h 2 = 28, h 3 = 1112, h 4 = 87568.It should be noted that the numbers h n also count permutations of S 4n having the following properties: (a) The permutation can be written as a product of disjoint cycles with only two elements; (b) For i ∈ [2n], indices 2i−1 and 2i are either both excedances or both anti-excedances.
4 Perfect matchings and a constructive proof of Proposition 1 Given M ∈ M 2n .The standard form of the perfect matching 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 In the following discussion we always write M in standard form.For convenience, we call (i, j) a marked block (resp.an unmarked block ) if j is even (resp.odd) and large than i.

Permutations and pairs of perfect matchings
Let the entry π(i) be called a descent (resp.an ascent) of π if π(i) > π(i + 1) (resp.π(i) < π(i + 1)).By using the reverse map, it is evident that ascent and descent are equidistributed.Let asc (π) be the number of ascents of π.Hence Throughout this subsection, we shall always use (20) as the definition of the Eulerian polynomial A n (x).
We now constructively define a set of decorated permutations on [n] with some entries of permutations decorated with hats and circles, denoted by P n .Let w = w 1 w 2 • • • w n ∈ P n .We say that w i with a hat (resp.circle) if We now construct entries of P n by inserting n, n, n or n into v according the following rules: (r 1 ) We can only put n or n at the end of v; (r 2 ) For 1 i n − 1, if v i with no hat, then we can only put n or n immediately before v i ; if v i with a hat, then we can only put n or n immediately before v i .
It is clear that there are 2n elements in P n that can be generated from any v ∈ P n−1 .By induction, we obtain be a permutation of S n obtained from w ∈ P n by deleting the hats and circles of all w i .For example, ϕ( 3 1 4 2) = 3142.Let P n (π) = {w ∈ P n : ϕ(w) = π}.Let k be a consecutive subword of π ∈ S n .We see that if k < , then k can be decorated as k , k , k , k .If k > , then k can be decorated as k , k , k or k .Therefore, |P n (π)| = 2 n for any π ∈ S n .It should be noted that k or k is a consecutive subword of w ∈ P n if and only if k < .Let the entry w i be called an ascent (resp.a descent) of w if ϕ(w i ) < ϕ(w i+1 ) (ϕ(w i ) > ϕ(w i+1 )).Also a conventional ascent is counted at the beginning of w.That is, we identify a decorated permutation w = w 1 • • • w n with the word w 0 w 1 • • • w n , where w 0 = 0. Let asc (w) be the number of ascents of w.Therefore, we obtain 2 n xA n (x) = w∈Pn x asc (w) .
the electronic journal of combinatorics 24(4) (2017), #P4.27Secondly, suppose that 2 with a hat.We say that 2 is a hat-ascent-top (resp.hatdescent-bottom) if ϕ( 1 ) < ϕ( 2 ) (resp.ϕ( 1 ) > ϕ( 2)).Consider the following two cases:  After the above step, we write the obtained perfect matchings in standard form.Suppose that w ∈ P n,k and Φ(w) = (S 1 , S 2 , I n,k ).Then asc (w) = i + j if and only if el (S 1 ) + el (S 2 ) = i + j.By induction, we see that Φ is the desired bijection between P n,k and PM n,k for all k, which also gives a constructive proof of (3).

Concluding remarks
Given σ = σ 1 σ 2 • • • σ 2n ∈ Q n .We say that σ i is a left-to-right minimum of σ if σ i < σ j for all 1 j < i.We now present a bijection between Q n and Q 2 n .Define σ to be the Stirling permutation of the second kind obtained from σ by inserting a left parenthesis the electronic journal of combinatorics 24(4) (2017), #P4.27 in σ preceding every left-to-right minimum.Then insert a right parenthesis before every internal left parenthesis and at the end.For example, if σ = 331221, then σ = (33)(1221).It is easy to verify that we can uniquely recover σ from σ by requiring that (a) each cycle is written with its smallest element first, (b) the cycles are written in decreasing order of their smallest element, and (c) we then erase all parentheses.
A natural generalization of Stirling permutations is k-Stirling permutations.Let j i denote the i copies of j, where i, j 1.We call a permutation of the multiset {1 k , 2 k , . . ., n k } a k-Stirling permutation of order n if for each i, 1 i n, all entries between the two occurrences of i are at least i.One can introduce k-Stirling permutations of the second kind along the same line as in Definition 2.Moreover, it would be interesting to investigate an analog of (18) on Coxeter groups of types B and D. Furthermore, one may find some multivariate extensions of Proposition 1.

2 )
If the two copies of m + 1 are inserted to a position of σ with label b, then ϑ corresponds to the operation b → b −1 c 2 d 2 .In this case, σ ∈ Q 2 m+1 (i, j − 1, k + 1).(c 3 ) If the two copies of m + 1 are inserted to a position of σ with label c, then ϑ corresponds to the operation c → cd 2 .In this case, σ ∈ Q 2 m+1 (i, j, k).(c 4 ) If the two copies of m + 1 are inserted to a position of σ with label d, then ϑ corresponds to the operation d → c 2 d.