Discrete Mathematics 323 (2014) 49–57

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Discrete Mathematics

journal homepage: www.elsevier.com/locate/disc

Counting signed permutations by their alternating runsChak-On Chow a, Shi-Mei Ma b

a P.O. Box 91100, Tsimshatsui Post Office, Hong Kongb School of Mathematics and Statistics, Northeastern University at Qinhuangdao, Hebei 066004, PR China

a r t i c l e i n f o

Article history:Received 25 September 2013Received in revised form 21 January 2014Accepted 24 January 2014Available online 6 February 2014

Keywords:Alternating runsPeaksValleysUp signed permutations

a b s t r a c t

The purpose of this paper is to establish a connection between alternating runs of signedpermutations in the hyperoctahedral group and left peaks of permutations in the symmet-ric group, and then to study some associated enumerative polynomials of signed permuta-tions. Properties of these enumerative polynomials, including combinatorial formulas andrecurrence relations are studied. In particular, we deduce a convolution formula connect-ing the number of alternating runs of permutations in the symmetric group to that in thehyperoctahedral group.

© 2014 Elsevier B.V. All rights reserved.

1. Introduction

Let Sn denote the symmetric group of degree n and let π = π(1)π(2) · · · π(n) ∈ Sn. We say that π changes directionat position i if either π(i− 1)⟨π(i)⟩π(i+ 1), or π(i− 1) > π(i) < π(i+ 1), where i ∈ {2, 3, . . . , n− 1}. We say that π hask alternating runs if there are k − 1 indices i where π changes direction (see [20, A059427]).

André [2] was the first to enumerate permutations by their number of alternating runs. Let R(n, k) denote the number ofpermutations in Sn with k alternating runs. André showed that R(n, k) satisfies the following recurrence relation

R(n, k) = kR(n − 1, k) + 2R(n − 1, k − 1) + (n − k)R(n − 1, k − 2) (1)

for n, k > 1, where R(1, 0) = 1 and R(1, k) = 0 for k > 1. Let Rn(x) =

k>1 R(n, k)xk. It follows from (1) that

Rn+2(x) = x(nx + 2)Rn+1(x) + x1 − x2

R′

n+1(x),

with the initial condition R1(x) = 1. There is a large literature devoted to the polynomials Rn(x). The reader is referred to[6,17,18,21] for recent results on this subject.

The number of peaks of permutations is certainly among the most important combinatorial statistics. See, e.g., [1,3,5,11,14–16] and the references therein. Let π = π(1)π(2) · · · π(n) ∈ Sn. An interior peak of π is an index i ∈ {2, 3, . . . , n − 1}such that π(i − 1) < π(i) > π(i + 1). Let pk (π) denote the number of interior peaks of π . A left peak of π is an indexi ∈ [n− 1] such that π(i− 1) < π(i) > π(i+ 1), where we take π(0) = 0. Let pkl (π) denote the number of left peaks of π .For example, the permutation π = 64713258 ∈ S8 has pk (π) = 2 and pkl (π) = 3. It should be noted that the left peakstatistic first appeared in [1, Definition 3.1].

Define

Wn(x) =

π∈Sn

xpk (π) and Wn(x) =

π∈Sn

xpkl (π).

E-mail addresses: cchow@alum.mit.edu (C.-O. Chow), shimeimapapers@gmail.com (S.-M. Ma).

0012-365X/$ – see front matter© 2014 Elsevier B.V. All rights reserved.http://dx.doi.org/10.1016/j.disc.2014.01.015

50 C.-O. Chow, S.-M. Ma / Discrete Mathematics 323 (2014) 49–57

It is well known (see [20, A008303, A008971]) that the polynomials Wn(x) and Wn(x) satisfy, respectively, the followingrecurrence relations

Wn+1(x) = (nx − x + 2)Wn(x) + 2x(1 − x)W ′

n(x),Wn+1(x) = (nx + 1)Wn(x) + 2x(1 − x)W ′

n(x), (2)

with initial conditionsW1(x) = W1(x) = 1. There is a wealth of enumerative polynomials associated with the polynomialsWn(x) and Wn(x). In particular, there is a close connection between Rn(x) and Wn(x).

Theorem 1 ([18, Corollary 2]). For n > 2, we have

Rn(x) =x(1 + x)n−2

2n−2Wn

2x

1 + x

. (3)

Since degWn(x) = ⌊n−12 ⌋, Theorem 1 provides an explanation of the fact that Rn(x) is divisible by (1 + x)⌊(n−2)/2⌋ (see

[4, Lemma 2.3]). Furthermore, we can also deduce the following, which was given by Comtet [10, p. 261] as an exercise.

Corollary 2. For n > 2k + 4, we have

1kR(n, 1) + 3kR(n, 3) + 5kR(n, 5) + · · · = 2kR(n, 2) + 4kR(n, 4) + 6kR(n, 6) + · · · . (4)

Proof. Recall that the Stirling number S(k, r) of the second kind satisfies

kr=1

S(k, r)s(s − 1) · · · (s − r + 1) = sk. (5)

Multiplying both sides by R(n, s)ts−k followed by summing over s, we get

s>2

skR(n, s)ts−k=

kr=1

t r−kS(k, r)s>r

s(s − 1) · · · (s − r + 1)R(n, s)ts−r

=

kr=1

t r−kS(k, r)R(r)n (t), (6)

where R(r)n (t) denotes the r-th derivative of Rn(t). Since n > 2k + 4, ⌊ n−2

2 ⌋ > k + 1 so that the exponents of 1 + t in thesummands of (6) are positive. Setting t = −1 then yields

(−1)2−k

s>1

(2s)kR(n, 2s) −

s>0

(2s + 1)kR(n, 2s + 1)

=

kr=1

(−1)r−kS(k, r)R(r)n (−1) = 0,

whence (4). �

Corollary 2 can be interpreted probabilistically. Let X be the random variable taking value of the number of alternatingruns of a randomly selected n-permutation π ∈ Sn. Then the probability generating function is

k>1

P(X = k)tk =1

Rn(1)

k>1

R(n, k)tk =Rn(t)Rn(1)

.

After normalized by Rn(1), the identity (4) states that the sum of contributions from terms with odd indices to the kthmoment of X is equal to that from terms with even indices.

The purpose of this paper is to explore the type B analogue of (3), and then to study some associated enumerativepolynomials of signed permutations. The paper is organized as follows. In Section 2, we give a type B analogue of (3).In Section 3, we show that alternating runs of up signed permutations are closely related to peaks and valleys of thesepermutations. In Section 4, we compute exponential generating functions of up signed permutations by their numbers ofalternating runs, peaks and valleys. We conclude the present work with some remarks in the final section.

C.-O. Chow, S.-M. Ma / Discrete Mathematics 323 (2014) 49–57 51

2. A type B analogue of (3)

In the sequel, we always assume that permutations, signed or not, are prepended by 0. That is, we identify an n-permutation π = π(1) · · · π(n) with the word π(0)π(1) · · · π(n), where π(0) = 0. Denote by Bn the hyperoctahedralgroup of rank n. Elements π of Bn are signed permutations of the set ±[n] such that π(−i) = −π(i) for all i, where±[n] = {±1, ±2, . . . ,±n}. A run of a signed permutation π is defined as a maximal interval of consecutive elementson which the elements of π are monotonic in the order

· · · < 2 < 1 < 0 < 1 < 2 < · · · .

The up signed permutations are signed permutations with π(1) > 0. For example, the up signed permutation 031245 ∈ B5has four runs, i.e., 03, 31, 124 and 45. Let T (n, k) denote the number of up signed permutations in Bn with k alternating runsand let Tn(x) =

nk=1 T (n, k)xk.

The first few members of Tn(x) are given as follows:

T1(x) = x,T2(x) = x + 3x2,T3(x) = x + 12x2 + 11x3,T4(x) = x + 39x2 + 95x3 + 57x4.

Recently, Zhao [22, Theorem 4.2.1] showed that the numbers T (n, k) satisfy the following recurrence relation

T (n, k) = (2k − 1)T (n − 1, k) + 3T (n − 1, k − 1) + (2n − 2k + 2)T (n − 1, k − 2) (7)

for n > 2 and 1 6 k 6 n, where T (1, 1) = 1 and T (1, k) = 0 for k > 1. From (7), we have

Tn+1(x) = (2nx2 + 3x − 1)Tn(x) + 2x1 − x2

T ′

n(x) for n > 1. (8)

Let RZ denote the set of real polynomials having only real zeros. Furthermore, denote by RZ(I) the set of such polynomialsall of whose zeros lie in the real interval I . As pointed out by Zhao, from [19, Theorem 2], we have the following result.

Proposition 3 ([22, Theorem 4.3.2]). For n > 1, we have Tn(x) ∈ RZ[−1, 0]. More precisely, Tn(x) has ⌈n+12 ⌉ simple zeros

including x = 0, and the zero x = −1 with the multiplicity ⌊n−12 ⌋.

By combining (2) with (8), we have the following result.

Theorem 4. For n > 1, we have

Tn(x) = x(1 + x)n−1Wn

2x

1 + x

.

Proof. For n > 1, let

Fn(x) = x(1 + x)n−1Wn

2x

1 + x

.

It is clear that F1(x) = x = T1(x). Expressing Wn in terms of Fn, we have

Wn(x) =(2 − x)n

2n−1xFn

x

2 − x

.

It follows from (2) that

Fn+1

x

2 − x

=

2nx2 − 4x2 + 10x − 4(2 − x)2

Fn

x

2 − x

+

8x(1 − x)(2 − x)3

F ′

n

x

2 − x

.

Thus, Fn+1(x) = (2nx2 + 3x − 1)Fn(x) + 2x1 − x2

F ′n(x). Since Fn(x) satisfies the same recurrence relation and initial

conditions as those of Tn(x), so they agree. �

Note that deg Wn(x) = ⌊n2⌋. Theorem 4 provides an explanation of the fact that Tn(x) is divisible by (1 + x)⌊(n−1)/2⌋. An

immediate consequence is the following moment-type identity.

Corollary 5. For n > 2k + 3, we have

1kT (n, 1) + 3kT (n, 3) + 5kT (n, 5) + · · · = 2kT (n, 2) + 4kT (n, 4) + 6kT (n, 6) + · · · . (9)

52 C.-O. Chow, S.-M. Ma / Discrete Mathematics 323 (2014) 49–57

Proof. Multiplying both sides of (5) by T (n, s)xs−k followed by summing over s, we gets>1

skT (n, s)xs−k=

kr=1

xr−kS(k, r)s>r

s(s − 1) · · · (s − r + 1)T (n, s)xs−r

=

kr=1

xr−kS(k, r)T (r)n (x), (10)

where T (r)n (x) denotes the r-th derivative of Tn(x). Since n > 2k + 3, ⌊ n−1

2 ⌋ > k + 1 so that the exponents of 1 + x in thesummands of (10) are positive. Setting x = −1 then yields

(−1)1−k

s>0

(2s + 1)kT (n, 2s + 1) −

s>1

(2s)kT (n, 2s)

=

kr=1

(−1)r−kS(k, r)T (r)n (−1) = 0,

whence (9). �

3. Connection between alternating runs and peaks and valleys

Denote by Bn the set of up signed permutations of Bn. Let π = π(0)π(1) · · · π(n) ∈ Bn. A peak (resp., valley) of π is anindex i ∈ {1, 2, 3, . . . , n − 1} such that π(i − 1) < π(i) > π(i + 1) (resp., π(i − 1) > π(i) < π(i + 1)). Let pk (π)(resp., val (π)) denote the number of peaks (resp., valleys) of π . For example, the up signed permutation π = 031245 haspk (π) = 2 and val (π) = 1.

Define

Pn(x) =

π∈Bn

xpk (π)=

k>0

P(n, k)xk,

Pn(x) =

π∈Bn

xval (π)=

k>0

P(n, k)xk.

The first few members of Pn(x) and Pn(x) are given as follows:

P1(x) = 1, P2(x) = 1 + 3x, P3(x) = 1 + 23x, P4(x) = 1 + 134x + 57x2;P1(x) = 1, P2(x) = 4, P3(x) = 13 + 11x, P4(x) = 40 + 152x.

In order to establish the main result of this section, we need the following.

Lemma 6. For n > 2, we have

P(n, k) = (4k + 1)P(n − 1, k) + (2n − 4k + 2)P(n − 1, k − 1) + P(n − 1, k − 1), (11)

P(n, k) = (4k + 3)P(n − 1, k) + (2n − 4k)P(n − 1, k − 1) + P(n − 1, k). (12)

Proof. We first prove (11). There are three ways in which a permutation π̃ ∈ Bn with k peaks can be obtained from apermutation π ∈ Bn−1.

(a) If π has k peaks, then we can insert n or −n into π without increasing the number of peaks. We can achieve thisby inserting n immediately before or right after a peak of π , or by appending n to π . We can always insert −nimmediately before each ascending run except the first (since the first run is always ascending), or append −n rightafter each descending run, or insert −n immediate before π(n − 1) if π ends with a descending run. This accounts for(2k + 1 + k + k − 1 + 1)P(n − 1, k) = (4k + 1)P(n − 1, k) possibilities.

(b) If π has k − 1 peaks, then we can insert n into any of the remaining n − (2k − 1) = n − 2k + 1 positions, or we caninsert −n into any of the remaining n − 1 − 2(k − 1) = n − 2k + 1 positions. This gives (2n − 4k + 2)P(n − 1, k − 1)possibilities.

(c) Let π = 0π(1)π(2) · · · π(n − 1) ∈ Bn−1. Define a bijection ϕ as follows:

ϕ(π) = 0nπ(1)π(2) · · · π(n − 1),

where π(i) = −π(i), 1 6 i 6 n − 1. Clearly, ϕ(π) ∈ Bn and if π has k − 1 valleys, then ϕ(π) has k peaks. This givesP(n − 1, k − 1) possibilities.

This completes the proof of (11). Along the same lines, one can obtain (12). �

C.-O. Chow, S.-M. Ma / Discrete Mathematics 323 (2014) 49–57 53

Corollary 7. For n > 2, we have

Pn(x) = (2nx − 2x + 1)Pn−1(x) + 4x(1 − x)P ′

n−1(x) + xPn−1(x),

Pn(x) = (2nx − 4x + 3)Pn−1(x) + 4x(1 − x)P′

n−1(x) + Pn−1(x).

We can now conclude the following.

Theorem 8. For n > 1, we have

(1 + x)Tn(x) = xPn(x2) + x2Pn(x2).

Proof. We proceed by induction on n. Since

(1 + x)T1(x) = x(1 + x) = xP1(x2) + x2P1(x2),

the result holds for n = 1. Assume that the result holds for n > 1, i.e.,

(1 + x)Tn(x) = xPn(x2) + x2Pn(x2). (13)

Differentiating (13) with respect to x, we have

Tn(x) + (1 + x)T ′

n(x) = Pn(x2) + 2x2P ′

n(x2) + 2xPn(x2) + 2x3Pn(x2). (14)

From (8), we have

(1 + x)Tn+1(x) = (2nx2 + 3x − 1)(1 + x)Tn(x) + 2x(1 − x2)(1 + x)T ′

n(x). (15)

It remains to show that the right side of (15) is equal to xPn+1(x2) + x2Pn+1(x2). Substituting (14) into the right side of (15)and using Corollary 7, we have

(2nx2 + 3x − 1)(1 + x)Tn(x) + 2x(1 − x2)(1 + x)T ′

n(x)

= ((2n + 2)x2 + x − 1)(1 + x)Tn(x) + 2x(1 − x2)(Pn(x2) + 2x2P ′

n(x2) + 2xPn(x2) + 2x3P

′

n(x2))

= ((2n + 2)x2 + x − 1)(xPn(x2) + x2Pn(x2)) + 2x(1 − x2)(Pn(x2) + 2x2P ′

n(x2) + 2xPn(x2) + 2x3P

′

n(x2))

= x(2nx2 + 1)Pn(x2) + 4x3(1 − x2)P ′

n(x2) + x3Pn(x2)

+ x2(2nx2 − 2x2 + 3)Pn(x2) + 4x4(1 − x2)P′

n(x2) + x2Pn(x2)

= xPn+1(x2) + x2Pn+1(x2).

This finishes the induction and the proof of theorem. �

In the rest of this section, we will present a connection between the zeros of the polynomials Pn(x) and Pn(x). We needsome notations from [13] concerning the zeros of polynomials. A polynomial p ∈ R[x] is (Hurwitz or asymptotically) stable ifevery zero of p is in the open left half plane LHP = {z: Re z < 0}; p is standard if its leading coefficient is positive. Supposethat p, q ∈ R[x] both have only real zeros, that those of p are ξ1 6 · · · 6 ξn, and that those of q are θ1 6 · · · 6 θm. We saythat p interlaces q if deg q = 1 + deg p and the zeros of p and q satisfy

θ1 6 ξ1 6 θ2 6 · · · 6 ξn 6 θn+1.

We also say that p alternates left of q if deg p = deg q and the zeros of p and q satisfy

ξ1 6 θ1 6 ξ2 6 · · · 6 ξn 6 θn.

More generally, we say that p interlaces (resp., alternates left of) q in I ⊆ R if the zeros of p and q in I satisfy the correspondingproperty. This refined notion of interlacing and alternating properties will make statements of results handy. We use thenotation p Ď q for ‘‘p interlaces q’’, p ≪ q for ‘‘p alternates left of q’’, and p ≺ q for either p Ď q or p ≪ q.

A classical result concerning the location of zeros of polynomials is the Hermite–Biehler theorem [12, p. 228], of whichan extended version is given in [13, Theorem 3]. We shall need the following version in what follows.

Theorem 9 (Hermite–Biehler). Let F(x) = f (x2) + xg(x2) ∈ R[x] be standard. Then F is stable if and only if both f and g arestandard, have only nonpositive zeros, and g ≺ f .

Since x = 0 is a simple zero of Tn(x), (1 + x)Tn(x)/x ∈ RZ[−1, 0). From Theorem 8, we then have

(1 + x)Tn(x)/x = Pn(x2) + xPn(x2).

Since (1 + x)Tn(x)/x is stable and both Pn(x) and Pn(x) are standard, Theorem 9 then yields the following.

54 C.-O. Chow, S.-M. Ma / Discrete Mathematics 323 (2014) 49–57

Theorem 10. For n > 1, Pn(x) and Pn(x) are (−∞, 0)-rooted and Pn(x) ≺ Pn(x).

Define

Mn(x) = xWn(x2) + Wn(x2).

Ma [16, Theorem 6] showed thatMn(x) ∈ RZ[−1, 0) andMn(x) interlacesMn+1(x). By Theorem 9, we obtainWn(x) ≺ Wn(x).Therefore, as an application of Theorems 1 and 4, we get the following result.

Theorem 11. For n > 1, the polynomial Rn(x) interlaces Tn(x).

4. Generating functions

In a series of papers [7–9], Carlitz studied the generating function for the numbers R(n, k). In particular, Carlitz [7] provedthat

H(x, z) =

∞n=0

zn

n!

nk=0

R(n + 1, k)xn−k=

1 − x1 + x

√1 − x2 + sin(z

√1 − x2)

x − cos(z√1 − x2)

2

. (16)

In recent years, Deutsch and Gessel [20, A059427], Stanley [21] also studied the generating function for these numbers. Aspointed out by Canfield andWilf [6, Section 6], the generating function for the numbers R(n, k) can be elusive. In this section,we compute exponential generating functions of Tn(x), Pn(x) and Pn(x), and deduce a convolution formula (Theorem 13)relating the polynomials Rn(x) and Tn(x).

Let T (x, z) :=

n>0 Tn(x)zn/n! be the exponential generating function of {Tn(x)}, where T0(x) :≡ 1.

Theorem 12. The formal power series T (x, z) ∈ Q[x][[z]] satisfies the following partial differential equation

(1 − 2x2z)∂T∂z

+ 2x(x2 − 1)∂T∂x

= (3x − 1)T − 2x + 1,

which together with the initial condition T (x, 0) = 1 admit the following solution

T (x, z) =1

1 + x+

x√x − 1

(1 + x)x − cos(2z

√x2 − 1) −

√x2 − 1 sin(2z

√x2 − 1)

.

Proof. Multiplying both sides of (8) by zn/n! followed by summing over n, we haven>1

Tn+1(x)zn

n!= 2x2z

n>1

Tn(x)zn−1

(n − 1)!+ (3x − 1)

n>1

Tn(x)zn

n!+ 2x(1 − x2)

n>1

T ′

n(x)zn

n!∂T∂z

− x = 2x2z∂T∂z

+ (3x − 1)(T − 1) + 2x(1 − x2)∂T∂x

,

hence the PDE. Next we solve for the characteristic, which satisfies the following ordinary differential equation:

dzdx

=1 − 2x2z2x(x2 − 1)

which integrates to yield

12tan−1

x2 − 1 − z

x2 − 1 = c1,

where c1 is the constant parametrizing the characteristic. Along the characteristic, the PDE becomes the following ODE:

2x(x2 − 1)dTdx

= (3x − 1)T − 2x + 1,

whose first integral is

(1 + x)T√x(x − 1)

=1

√x(x − 1)

+ f12tan−1

x2 − 1 − z

x2 − 1

,

C.-O. Chow, S.-M. Ma / Discrete Mathematics 323 (2014) 49–57 55

where f is an arbitrary function to be determined. When z = 0, T = 1 so that

(1 + x)√x(x − 1)

=1

√x(x − 1)

+ f12tan−1

x2 − 1

.

Let u =12 tan−1

√x2 − 1 and express x in terms of u, we obtain

f (u) =1

√1 − cos(2u)

.

Reporting f into the first integral, we get

(1 + x)T√x(x − 1)

=1

√x(x − 1)

+1

1 − cos(tan−1√x2 − 1 − 2z

√x2 − 1)

=1

√x(x − 1)

+

√x

x − cos(2z√x2 − 1) −

√x2 − 1 sin(2z

√x2 − 1)

.

Multiplying both sides by√x(x − 1)/(1 + x), the expression of T = T (x, z) follows. �

The exponential generating functions of {Pn(x)} and {Pn(x)} can be approached via Theorem 8. Note that xPn(x2) andx2Pn(x2) are the odd and even parts of (1 + x)Tn(x), respectively. Since

n>1

(1 + x)Tn(x)zn

n!= (1 + x)

1

1 + x+

x√x − 1

(1 + x)x − cos(2z

√x2 − 1) −

√x2 − 1 sin(2z

√x2 − 1)

− 1

=x√x − 1

x − cos(2z√x2 − 1) −

√x2 − 1 sin(2z

√x2 − 1)

− x, (17)

the odd part of which is given byn>1

xPn(x2)zn

n!=

12

n>1

[(1 + x)Tn(x) − (1 − x)Tn(−x)]zn

n!

= −x +x2

(x − 1)(x + cos u +

√x2 − 1 sin u) +

(x + 1)(x − cos u −

√x2 − 1 sin u)

x2 − (cos u +√x2 − 1 sin u)2

,

where u = 2z√x2 − 1. Replacing x by

√x, we get

n>1

Pn(x)zn

n!= −1 +

12

(√x − 1)(

√x + cos v +

√x − 1 sin v) +

(√x + 1)(

√x − cos v −

√x − 1 sin v)

x − (cos v +√x − 1 sin v)2

,

where v = 2z√x − 1. Similar consideration yields that

n>1

Pn(x)zn

n!=

12√x

(√x − 1)(

√x + cos v +

√x − 1 sin v) −

(√x + 1)(

√x − cos v −

√x − 1 sin v)

x − (cos v +√x − 1 sin v)2

.

We can now conclude the following result.

Theorem 13. We havex + (1 + x)

n>1

Tn(x)zn

n!

2

= x2 + x(1 + x)n>1

2n−1Rn+1(x)zn

n!, (18)

or equivalently,

2n−1Rn+1(x) = 2Tn(x) +1 + xx

n−1k=1

nk

Tk(x)Tn−k(x) for n > 2. (19)

56 C.-O. Chow, S.-M. Ma / Discrete Mathematics 323 (2014) 49–57

Proof. Define

R(x, z) =

n>0

Rn+1(x)zn

n!.

Comparing with (16), it is clear that R(x, z) = H( 1x , xz). Hence

R(x, z) =

x − 1x + 1

√x2 − 1 + x sin(z

√x2 − 1)

1 − x cos(z√x2 − 1)

2

.

Let u = sec−1 x, i.e., sec u = x. Then√x2 − 1 + x sin(z

√x2 − 1)

1 − x cos(z√x2 − 1)

2

=

sin u + sin(z

√x2 − 1)

cos u − cos(z√x2 − 1)

2

=

cos2

u−z√

x2−12

sin2

u−z

√x2−1

2

=

1 + cos(u − z√x2 − 1)

1 − cos(u − z√x2 − 1)

=x + cos(z

√x2 − 1) +

√x2 − 1 sin(z

√x2 − 1)

x − cos(z√x2 − 1) −

√x2 − 1 sin(z

√x2 − 1)

,

where in the second, third and fourth equality, we have applied the following trigonometric identities:

sin u + sin(zx2 − 1) = 2 sin

u + z

√x2 − 1

2

cosu − z

√x2 − 1

2

,

cos u − cos(zx2 − 1) = −2 sin

u + z

√x2 − 1

2

sinu − z

√x2 − 1

2

,

cos(u − zx2 − 1) = 2 cos2

u − z

√x2 − 1

2

− 1 = 1 − 2 sin2

u − z

√x2 − 1

2

,

cos(u − zx2 − 1) =

cos(z√x2 − 1) +

√x2 − 1 sin(z

√x2 − 1)

x,

respectively. Thus, we have

R(x, z) =

x − 1x + 1

x + cos(z

√x2 − 1) +

√x2 − 1 sin(z

√x2 − 1)

x − cos(z√x2 − 1) −

√x2 − 1 sin(z

√x2 − 1)

. (20)

It follows from (20) that

x2 + x(1 + x)n>1

2n−1Rn+1(x)zn

n!= x2 +

x + x2

2

n>1

Rn+1(x)(2z)n

n!

= x2 +x + x2

2(R(x, 2z) − 1)

=x2(x − 1)

x − cos(2z√x2 − 1) −

√x2 − 1 sin(2z

√x2 − 1)

.

Comparing with (17), we immediately obtain (18). Comparing the coefficients of zn/n! on both sides of (18), we get (19).This completes the proof. �

It would be interesting to have a combinatorial proof of (19).

5. Concluding remarks

Since x(1 + x)⌊(n−1)/2⌋ divides Tn(x), a natural expansion is the following:

Tn(x) = (1 + x)ntn

x

1 + x

,

C.-O. Chow, S.-M. Ma / Discrete Mathematics 323 (2014) 49–57 57

or equivalently,

tn(x) =

nk=1

T (n, k)xk(1 − x)n−k.

The first few members of the polynomials tn(x) are given as follows:

t1(x) = x, t2(x) = x + 2x2, t3(x) = x + 10x2, t4(x) = x + 36x2 + 20x3.

Note that tn(1) = Sn, where Sn is the well known Springer number [20, A001586]. It would be interesting to have acombinatorial interpretation of the coefficients of the polynomial tn(x).

Denote by Dn the group of even-signed permutations of rank n, which is also known as a Coxeter group of type D. It iswell known that Dn forms a subgroup of Bn of index 2. Denote by Dn the set of up signed permutations of Dn. Let G(n, k)denote the number of permutations in Dn with k alternating runs and let Gn(x) =

nk=1 G(n, k)xk.

Define

En(x) =

π∈Dn

xpk (π)=

k>0

E(n, k)xk,

En(x) =

π∈Dn

xval (π)=

k>0

E(n, k)xk.

The first few members of the polynomials En(x), En(x) and Gn(x) are respectively given as follows:

E1(x) = 1, E2(x) = 1 + x, E3(x) = 1 + 11x, E4(x) = 1 + 66x + 29x2,E1(x) = 1, E2(x) = 2, E3(x) = 7 + 5x, E4(x) = 20 + 76x,G1(x) = x, G2(x) = x + x2, G3(x) = x + 6x2 + 5x3, G4(x) = x + 19x2 + 47x3 + 29x4.

Based on empirical evidence, we propose the following.

Conjecture 14. For n > 1, we have

(1 + x)Gn(x) = xEn(x2) + x2En(x2).

Moreover, we have Gn(x) ∈ RZ[−1, 0] and En(x) ≺ En(x).

Acknowledgments

This research was supported by NSFC (11126217) and Natural Science Foundation of Hebei Province (A2013501070) forthe second author. The authors thank the anonymous referee for careful reading and helpful suggestions.

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