The peak statistics on simsun permutations

In this paper, we study the relationship among left peaks, interior peaks and up-down runs of simsun permutations. Properties of the generating polynomials, including the recurrence relation, generating function and real-rootedness are studied. Moreover, we introduce and study simsun permutations of the second kind.

Let a i (n) be the number of distinct S n -orbits such that the stabiliser of a maximal chain in the orbit is conjugate to the Young subgroup S i 2 × S n−2i 1 . Following Sundaram [24,Theorem 3.2], the numbers a i (n) satisfy the recurrence relation a i (n + 1) = ia i (n) + (n − 2i + 2)a i−1 (n), with initial conditions a 0 (1) = 1 = a 1 (2), a 0 (n) = 0 for n > 1 and a i (n) = 0 if 2i > n. Let RS n be the set of simsun permutations of length n. Simion and Sundaram [24, p. 267] discovered that a i (n) is the number of permutations in RS n−2 with i − 1 descents and #RS n = E n+1 , where E n is the nth Euler number, which also is the number alternating permutations in S n .
The descent number of π ∈ S n is defined by des (π) = #{i ∈ [n − 1] : π(i) > π(i + 1)}. Let S(n, k) = #{π ∈ RS n : des (π) = k}. We define S n (x) = ⌊n/2⌋ k=0 S(n, k)x k . Then the numbers S(n, k) satisfy the recurrence relation S(n, k) = (k + 1)S(n − 1, k) + (n − 2k + 1)S(n − 1, k − 1), (1) with the initial conditions S(0, 0) = 1 and S(0, k) = 0 for k ≥ 1, which is equivalent to with S 0 (x) = 1. Let S(x, z) = n≥0 S n (x) z n n! . Chow and Shiu [5,Theorem 1] obtained that For convenience, here we list the first few terms of S n (x): The number of peaks of permutations is certainly among the most important combinatorial statistics. See, e.g., [2,16,18] and the references therein. A left peak in π is an index i ∈ [n − 1] such that π(i − 1) < π(i) > π(i + 1), where we take π(0) = 0. Let lpk (π) denote the number of left peaks in π. For example, lpk (21435) = 2. Sundaram discovered that a i (n) is also the number of André permutations in S n−1 with i left peaks (see [26, p. 175]). In fact, since any descent of a simsun permutation is a left peak, we have Let W (n, k) = #{π ∈ S n : lpk (π) = k}. Let W n (x) = k≥0 W (n, k)x k . The polynomials W n (x) satisfy the recurrence relation By comparing (3) with (4), we observe that S(x, z) = W (2x, z/2) 2 , which leads to the following formula: Denote by B n the hyperoctahedral group of rank n. Elements π of B n are signed permutations of the set ±[n] such that π(−i) = −π(i) for all i, where ±[n] = {±1, ±2, . . . , ±n}. A snake of type B n is a signed permutation π(1)π(2) · · · π(n) ∈ B n such that 0 < π(1) > π(2) < · · · π(n). The nth Springer number S n is the number of snakes of of type B n . Springer [22] derived the following generating function: which equals W (2, z). As a special case of (5), we get We refer the reader to [4] for various structures related to Springer numbers. Motivated by (5), it is natural to study peak statistics on simsun permutations. This paper is organized as follows. In Section 2, we give a constructive proof of a connection between S(n, k) and the number of permutations in S n+1 with k interior peaks. In Section 3, we count simsun permutations by their interior peaks. In Section 4, we count simsun permutations by their up-down runs. In Section 5, we introduce simsun permutations of the second kind.
Moreover, we have In the rest of this section, we give a constructive proof of (7). Let It should be noted that if we get a permutation π ′ ∈ RS n+1 from a permutation π ∈ RS n by inserting the entry n + 1 into π, then the entry n + 1 can not be inserted right after π(j), where j ∈ ← − D (π). In this section, we always assume that permutations in RS n are prepended by 0. That is, we identify a permutation π(1)π(2) · · · π(n) ∈ RS n with the word π(0)π(1)π(2) · · · π(n), where π(0) = 0. Define RS n,k = {π ∈ RS n | des (π) = k}. We can now introduce a definition of labeled simsun permutations.
, then we put the superscript label y s right after Let S n,k = {π ∈ S n | pk (π) = k}. We introduce a definition of labeled permutations.
Definition 3. Let π ∈ S n,k . Suppose i 1 < i 2 < · · · < i k are indices of the interior peaks of π. Then we put the superscript labels p r immediately before and right after π(i r ), where 1 ≤ r ≤ k.
In the following discussion, we always add labels to permutations in RS n,k and S n,k . As an example, for π = 34125, if we say that π ∈ RS 5,1 , then the labels of π is given by y 1 34 x 1 1 y 2 2 y 3 5; if we say that π ∈ S 5,1 , then the labels of π is given by 3 p 1 4 p 1 1 q 1 2 q 2 5. Now we construct a correspondence, denoted by Φ, between RS n,k and S n+1,k . When n = 1, the correspondence between RS 1,0 and S 2,0 is given by When n = 2, the correspondence between RS 2,k and S 3,k is given by Let n = m. Suppose Φ is a correspondence between RS m,k and S m+1,k for all k. More precisely, given an element π ∈ RS m,k . Suppose we have the correspondence Consider the case n = m + 1. Suppose π ∈ RS m+1 is obtained from π by inserting the entry m + 1 into π. We distinguish three cases: (i) If π(m + 1) = m + 1, then we insert the entry m + 2 at the front or at the end of each σ i . In this case, the obtained elements in Φ( π) all have k interior peaks. Therefore, we get 2 · 2 m−k = 2 m+1−k elements in S m+2,k . (ii) If the entry m + 1 is inserted to the position of π with label x r , then we insert the entry m + 2 to one of the positions of each σ i with label p r . In this case, des ( π) = k and we get 2 · 2 m−k = 2 m+1−k elements in S m+2,k . (iii) If the entry m + 1 is inserted to the position of π with label y s , then we insert the entry m + 2 to the position of each σ i with label q s . In this case, des ( π) = k + 1 and we get It is straightforward to show that each labeled permutation in Φ( π) will be obtained exactly once in this way. Conversely, given an element τ of S m+2,k . Removing the entry m + 2 of τ , we can find the position of the largest entry of the corresponding simsun permutation in RS m+1 . As illustrated in example 4, we can get an unique element of RS m+1 by repeatedly removing the largest entry. By induction, we see that Φ is the desired correspondence between RS m,k and S m+1,k , which also gives a constructive proof of (7).
Equivalently, the polynomials P + n (x) and P − n (x) satisfy the following recurrence relations Proof. We now prove (8). There are three ways we can get a permutation π ′ ∈ RS + n+1 with k interior peaks from a permutation π ∈ RS n by inserting the entry n + 1 into π: (a) If π ∈ RS + n and pk (π) = k, then we can insert the entry n + 1 right after an interior peak of π or put the entry n + 1 at the end of π. As we have P + (n, k) choices for π, this accounts for (k + 1)P + (n, k) possibilities. (b) If π ∈ RS + n and pk (π) = k−1, then there are n−2k positions could be inserted the entry n + 1, since we cann't insert the entry n + 1 immediately before or right after each left peak of π. As we have P + (n, k − 1) choices for π, this accounts for (n − 2k)P + (n, k − 1) possibilities. (c) If π ∈ RS − n and pk (π) = k, then we have to put the entry n + 1 at the front of π. This completes the proof of (8). In the same way, one can get (9).
The first few terms of the P n (x), P + n (x) and P − n (x) are respectively given as follows: By Lemma 5, it is easy to deduce that
It is clear that P (n, k) = P + (n, k) + P − (n, k) for n ≥ 2. We can now conclude the following result from the discussion above.
It should be noted that (14) follows immediately from (1) and (12). We now recall some notations from [14] concerning the zeros of polynomials. Let RZ denote the set of real polynomials with only real zeros. Furthermore, denote by RZ(I) the set of such polynomials all whose zeros are in the interval I. Suppose that p, q ∈ RZ, that those of p are ξ 1 · · · ξ n , and that those of q are θ 1 · · · θ m . We say that p interlaces q if deg q = 1+deg p and the zeros of p and q satisfy θ 1 ξ 1 θ 2 · · · ξ n θ n+1 .
We also say that p alternates left of q if deg p = deg q and the zeros of p and q satisfy ξ 1 θ 1 ξ 2 · · · ξ n θ n .
We use the notation p † q for "p interlaces q," p ≪ q for "p alternates left of q," and p ≺ q for either p † q or p ≪ q. For notational convenience, let a ≺ bx + c for any real constants a, b, c.
We now recall a result on the real-rootedness of S n (x). Let sgn denote the sign function defined on R.
Proof. There are three ways in which a permutation π ′ ∈ RS n with uprun (π ′ ) = k can be obtained from a permutation π ∈ RS n−1 by inserting the entry n into π.
(a) If uprun (π) = k, then we can insert the entry n right after the end of each ascending run. This accounts for ⌈k/2⌉T (n − 1, k) possibilities. (b) If uprun (π) = k − 1, then we distinguish two cases: when π ends with an ascending run, we insert the entry n to the front of the last entry of π; when π ends with descending run, we insert the entry n at the end of π. This gives T (n − 1, k − 1) possibilities. (c) If uprun (π) = k −2, then we can insert the entry n into the remaining n−k +1 positions.
This completes the proof of (15).
We call the simsun permutations discussed above to be the simsun permutations of the first kind. In the next section, we shall introduce the simsun permutations of the second kind.

Simsun permutations of the second kind
In this section, we always write π ∈ S n in standard cycle decomposition, where each cycle is written with its smallest entry first and the cycles are written in increasing order of their smallest entry. For each π ∈ S n , we say that π has an excedance at i if π(i) > i. The excedance number of π is defined by exc (π) = #{i ∈ [n − 1] : π(i) > i}. Following [20], for π ∈ S n , a value x = π(i) is called a double excedance if i = π −1 (x) < x < π(x), and we say that x = π(i) is a cyclic peak if i = π −1 (x) < x > π(x). Let cpk (π) denote the number of cyclic peaks of π.
Definition 12. We say that π ∈ S n is a simsun permutation of the second kind if for all k ∈ [n], after removing the k largest letters of π, the resulting permutation has no double excedances.
For example, (1,5,3,4) (2) is not a simsun permutation of the second kind since when we remove the letter 5, the resulting permutation (1,3,4)(2) contains a double excedance. Let SS n be the set of the simsun permutations of the second kind of length n. It is clear that exc (π) = cpk (π) for π ∈ SS n .
In the following, we first present a constructive proof of the following identity: Let SS n,k = {π ∈ SS n | exc (π) = k}. As a variant of Definition 2, we introduce a definition of labeled simsun permutations of the the second kind.
Definition 13. Let σ ∈ SS n,k . Suppose i 1 < i 2 < · · · < i k are the excedances of σ. Then we put the superscript labels u r right after i r , where 1 ≤ r ≤ k. In the remaining positions except the first position of each cycle and the positions right after σ(i r ), we put the superscript labels v 1 , v 2 , . . . , v n−2k from left to right.
By induction, we see that Ψ is the desired bijection between RS m,k and SS m,k for all k, which also gives a constructive proof of (17).
Example 14. Given π = 3412 ∈ RS 4,1 . The correspondence between π and Ψ(π) is built up as follows: We now consider the following enumerative polynomials where cyc (π) is the number of cycles of π. Let S = S(x, q; z) = n≥0 S n (x, q) z n n! .
Corollary 16. If q > 0, then S n (x, q) has nonpositive and simple zeros for n ≥ 2.

Concluding remarks
In this paper we study the peak statistics on simsum permutations. It is well known that the descent statistic is equidistributed over n-simsun permutations and n-André permutations (see [5]), and there are bijections between simsun permutations and increasing 1-2 trees (see [6] for instance). Therefore, one can find corresponding results on André permutations and increasing 1-2 trees. For example, S(1, q; z) also is the (shifted) exponential generating function that counts André permutations with respect to the size and the number of right-to-left minima (see [9, Proposition 1]) Furthermore, it would be interesting to derive similar results on signed simsum permutations introduced by Ehrenborg and Readdy [10].