A Unified Approach to Combinatorial Triangles: a Generalized Eulerian Polynomial

arXiv:2007.12602v1 [math.CO] 24 Jul 2020 S Let numbers Eulerian 1.1 Introduction 1 n nelement An . ∗ mi address: Email upre yteNtoa aua cec onaino hn (N China of Foundation Science Natural National the by Supported S colo ahmtc n ttsis ins omlUniversity, Normal Jiangsu Statistics, and Mathematics of School n nfidapoc ocmiaoiltinls a triangles: combinatorial to approach unified A T esuyatinua ra [ array stair triangular from a array study triangular we an groups, hyperoctahedral the tn ucin;Ra eo;Lgcnaiy Strong Log-concavity; zeros; Real functions; ating Keywords: MSC: alternating-runs the and approach. tableaux unified staircase from array lar oml o h lsia ueinplnma otoeo g Eule generalized a the of for polynomial. those identity derivative an to general get the polynomial also Eulerian We classical polynomial. the for formula with o-ocvt,ra otdes xlctfruaads on so the and formula, formula Frobenius explicit famous rootedness, real log-concavity, o t o-eeaigfunction row-generating its for rnfrain ecngtpoete of properties get can we transformation, nte ra [ array another eoetesmercgopon group symmetric the denote n,k oiae ytecasclElra ubr ecn n exc and descent number, Eulerian classical the by Motivated ial,w pl u eut oa ra rmteLmetfun Lambert the from array an to results our apply we Finally, = T 0 λ , 0 51;0A9 52;26C10 05A20; 05A19; 05A15; ( a and 1 = 0 i n xh@sueuc B-.Zhu) (B.-X. [email protected] ∈ ueinnmes ueinplnmas eurnerelat Recurrence polynomials; Eulerian numbers; Eulerian + eeaie ueinpolynomial Eulerian generalized [ a n A 1 − n,k k T n,k ]i aldadsetof descent a called is 1] + ] n,k a nes0 unless 0 = 2 ) aifigatotr eurnerlto.Bsdo this on Based relation. recurrence two-term a satisfying T n − 1 T ,k n,k ( + a-unZhu Bao-Xuan γ T ] n,k n n b oiiiydcmoiinadteDavid-Barton the and decomposition positivity ( ≤ 0 x eeet [ -elements ≥ Abstract n rmterwgnrtn function row-generating the from ) 0 k + aifigtercrec relation: recurrence the satisfying ≤ b 1 1 n T k edrv ucinltransformation functional a derive We . n,k π + if b n and 2 π = ] ) T ( i q n ∗ ) -log-convexity; − T { π > 1 n 1 ,k ( , x − 2 nldn nonnegativity, including ) 1 ( n , . . . , i + aetbeu n oon, so and tableaux case hnw xedthe extend we Then . ) h lsia Eulerian classical The 1). + uhu211,P China PR 221116, Xuzhou rageo type of triangle .11971206). o. cd λ inplnma with polynomial rian ( nrlzdEulerian eneralized dnenmesin numbers edance n } Let . − to,atriangu- a ction, γ -positivity; k 1) + os Gener- ions; π = A T n n π B − ( (1) x 1 na in ,k of ) − · · · 2 π ( n ) ∈ polynomial is defined by n E (x)= x1+des (π) = xk, n k π Sn k X∈ X n where des (π) denotes the number of descents of π and is called the Eulerian k number. Eulerian numbers and polynomials have many nice properties. For example: n (i) The Eulerian number satisfies the recurrence relation k n n 1 n 1 = k − +(n k + 1) − k k − k 1 − 0 0 with initial conditions = 1 and = 0 for k 1 or k < 0, see [13]. 0 k ≥ n +1 Clearly, if let = −→E for n, k 1, then shift Eulerian numbers −→E k +1 n,k ≥ n,k satisfy the next recurrence

−→E n,k =(k + 1)−→E n 1,k +(n k + 1)−→E n 1,k 1 (1.1) − − − − with initial conditions −→E = 1 and −→E = 0 for k 1 or k < 0. 0,0 0,k ≥ n (ii) It is well-known that E (x) is related to Stirling numbers of the second kind n k by the famous Frobenius formula n n i n i E (x)= i! x (1 x) − . (1.2) n i − i=1 X This implies an explicit formula for Eulerian numbers: n n n n i k i = i! − ( 1) − , (1.3) k i k i − i=1 X − where i n i j i n i!= ( 1) − j (1.4) i − j j=1 X for n, i 0, see B´ona [5] for instance. ≥ (iii) It is also well known that En(x) has only real zeros and therefore is log-concave. In addition, Foata and Sch¨utzenberger [22] and Foata and Strehl [23] proved for En(x) that there exists nonnegative numbers γn,k such that

n/2 k n 1 2k En(x)= γn,kx (1 + x) − − (1.5) Xk=0 for n 0. This is an important property (called the γ-positivity [2]) because it ≥ implies symmetry and unimodality of En(x).

2 (iv) The Eulerian polynomial En(x) also has a closely connection with the famous alternating-runs polynomial Rn(x) by

n 1 1+ x − 1 w 1 x R (x)= (1 + w)n+1E − ,w = − , for n 2, (1.6) n 2 n 1+ w 1+ x ≥ r see David and Barton [20] and Knuth [28].

The Eulerian number and polynomial have a rich history and appear in many contexts in combinatorics; see [34] for a detailed exposition.

1.2 Descent and excedance numbers in the hyperoctahedral groups n Let B be the hyperoctahedral group of rank n. Let be the Eulerian number n k B n of type B counting the elements of B with k B-descents. Then satisﬁes the n k recurrence B n n 1 n 1 = (2k + 1) − + (2n 2k + 1) − (1.7) k k − k 1 B B − B 0 n with initial conditions = 1 and = 0 unless 0 k n, see Brenti [8]. Let 0 k ≤ ≤ B B n B (x)= n xk be the Eulerian polynomials of type B. In fact, Eulerian num- n k=0 k B bers (resp.P polynomials) of type B share many properties of classical Eulerian numbers (resp. polynomials), see [37, A060187] for details. The excedance number in Sn is deﬁned by:

exc(π)= i 1, 2,...,n : π(i) > i, π S . |{ ∈{ } ∈ n}| It is well-known that the excedance number and the descent number have the same dis- n tribution and they both are the classical Eulerian number . k Bagno et al. gave a deﬁnition of the excedance number in Bn, see [4] for details. Let X be the number of permutations π B having exc (π)= k. It was proved that n,k ∈ n A

Xn,k =(n k)Xn 1,k 1 +(n + 1)Xn 1,k +(k + 1)Xn 1,k+1 − − − − − for n 1 with X = 1 [4]. In addition, in [4], they showed that the sequence (X )n ≥ 0,0 n,k k=0 is log-concave by induction. Let Xn,k∗ = Xn,n k. Then the array [Xn,k∗ ]n,k 0 satisﬁes the − ≥ recurrence relation

Xn,k∗ = kXn∗ 1,k +(n + 1)Xn∗ 1,k 1 +(n k + 1)Xn,k∗ 2 (1.8) − − − − − for n 1 with X∗ = 1. Using the recurrence relation (1.8), we proved many properties ≥ 0,0 n of X similar to those of the Eulerian number , see [48, Propostion 3.3] for details. n,k∗ k 3 1.3 Staircase tableaus The Partially Asymmetric Simple Exclusion Process (PASEP) is a physical model in which n sites on a one-dimensional lattice are either empty or occupied by a single particle. This model is also a Markov chain. Staircase tableaux are combinatorial objects which appear as key tools in the study of the PASEP physical model, see Corteel, Mandelshtam and Williams [15], Corteel, Stanley, Stanton, and Williams [16], Corteel and Williams [17, 18, 19] for instance. Let us recall the deﬁnitions as follows. A staircase tableau T of size n is a Ferrers diagram of staircase shape (n, n 1, , 2, 1) such that boxes are either empty or labeled with α, β, γ or δ, and satisfying− the following··· conditions: (1) no box along the diagonal of T is empty; (2) all boxes in the same row and to the left of a β or a δ are empty; (3) all boxes in the same column and above an α or a γ are empty. For staircase tableaux without any γ label, we denote by Tn,k the number of such tableaux T of size n with k labels α or δ in the diagonal. Aval, Boussicault and Dasse- Hartaut [3] proved for n 0 and all k Z that ≥ ∈

Tn,k = (k + 1)Tn 1,k +(n + 1)Tn 1,k 1 +(n k + 1)Tn 1,k 2, (1.9) − − − − − − T T k T where n(x) = k 0 n,kq for n 0. In [3], it was proved that the polynomial n(x) ≥ ≥ has only real zeros and is log-concave. P 1.4 An array from the Lambert function The Lambert W function was deﬁned in [14] by W eW = x. The n-th derivative of W is given implicitly by

n nW (x) d W (x) e− pn(W (x)) n = 2n 1 dx (1 + W (x)) −

n 1 n 1 k for n 1, where p (x)=( 1) − − β x are polynomials satisfying the recurrence ≥ n − k=0 n,k relation P p (x)= (nx +3n 1)p (x)+(1+ x)p′ (x) n+1 − − n n

for n 1, where the array [βn,k]n 1,k 0 satisﬁes the recurrence relation ≥ ≥ ≥

βn+1,k = (3n k 1)βn,k + nβn,k 1 (k + 1)βn,k+1 (1.10) − − − −

for n, k 0 with β1,0 = 1. Kalugin and Jeffrey [27] proved that if each polynomial n 1 ≥ dW (x) ( 1) − p (w) has all positive coefficients, then it follows that is a completely mono- − n dx tonic function. In order to get the positivity, by nontrivial computations, they proved that the coefficients are unimodal and log-concave. If let βn,k◦ = βn+1,n k for n, k 0, − ≥ then the array [βn,k◦ ]n,k 0 exactly satisfies the recurrence relation ≥

βn,k◦ = nβn◦ 1,k + (2n + k 1)βn◦ 1,k 1 (n k + 1)βn◦ 1,k 2 (1.11) − − − − − − − − for n, k 0 with β◦ = 1, which easily implies the positivity of β◦ , see Section 4. ≥ 0,0 n,k 4 1.5 Structure of this paper As we know that Eulerian polynomials and Eulerian numbers have many nice properties. In addition, different generalizations of Eulerian polynomials and Eulerian numbers were considered, see Haglund-Zhang [24], Han-Mao-Zeng [25], Rz¸adkowski-Urlińska [35], Zhu [48], Zhuang [52]. But recently in many new combinatorial enumerations, there bring out more and more combinatorial triangles satisfying certain three-term recurrence similar to recurrence relations (1.8), (1.9) and (1.11). Motivated by these, the aim of this paper is to consider their generalization and extend those properties of Eulerian polynomials and Eulerian numbers to those of the generalized case. Then we can apply to some combinatorial triangles in a unified approach. + Let R (resp. R , R≥) be the set of all (resp., positive, nonnegative) real numbers. Let + b ,c R, λ R and a , a , a , b , b ,d R≥. Define a generalized triangular array { 1 } ∈ ∈ { 0 1 2 0 2 } ⊆ [ n,k]n,k 0 by the recurrence relation: T ≥ cd n,k = λ(a0n + a1k + a2) n 1,k +(b0n + b1k + b2) n 1,k 1 + (n k +1) n 1,k 2 (1.12) T T − T − − λ − T − − with 0,0 = 1 and n,k = 0 unless 0 k n. Also denote its row-generating function by T Tk ≤ ≤ n(x)= k 0 n,kx for n 0. T ≥ T ≥ In Section 2, we derive a functional transformation of n(x) from the row-generating P T function of another array [An,k]n,k satisfying a two-term recurrence relationship, see The- orem 2.1. Based on the transformation, we can immediately apply nonnegativity, log- concavity and zeros of An(x) to those of n(q). In Section 3, we first give a generalizationT of the formula (1.4)(Theorem 3.1). Then we extend the identity (1.2) and formula (1.3) to the generalized Frobenius formula (Theorem 3.2). We also present a γ positivity result for a generalized Eulerian polynomial (Theorem 3.3), which gives a unified proof for γ positivity of Eulerian polynomials of two kinds. Finally, for a generalized Eulerian polynomial, we present a David-Barton type formula, which can be looked as a generalization of (1.6). In Section 4, we apply our results to some important combinatorial triangles in a unified manner, including the array [βn,k]n,k from the Lambert function, the triangular array [Tn,k]n,k from staircase tableaux, and alternating-runs triangle of type B. Throughout this paper, for an array [Mn,k]n,k 0, we let its row-generating function n k ≥ Mn(x)= Mn,kx for n 0 and its reciprocal array [Mn,k∗ ]n,k 0 by Mn,k∗ = Mn,n k. k=0 ≥ ≥ − P 2 An important functional transformation of (x) Tn In this section, we will first present a functional transformation of n(x) from a triangular array satisfying a two-term recurrence. T

Theorem 2.1. Let [ n,k]n,k 0 be deﬁned in (1.12). If b1 = da1 c, then there exists an T ≥ − array [An,k]n,k 0 satisfying the recurrence relation ≥ An,k =(a0n + a1k + a2)An 1,k + [[b0 + d(a1 a0)]n (c + da1)k + b2 + d(a1 a2)]An 1,k 1(2.1) − − − − − − with A0,0 = 1 and An,k = 0 unless 0 k n such that their row-generating functions satisfy ≤ ≤ x (x)=(λ + dx)nA ( ) (2.2) Tn n λ + dx

5 for n 0. In addition, ≥ n i n k k i = A − λ − d − Tn,k n,i k i i 0 X≥ − for n 0 and k 0. ≥ ≥ Proof. Because the array [An,k]n,k 0 satisﬁes the recurrence relation ≥

An,k =(a0n + a1k + a2)An 1,k + [[b0 + d(a1 a0)]n (c + da1)k + b2 + d(a1 a2)]An 1,k 1 − − − − − −

with A0,0 = 1 and An,k = 0 unless 0 k n, its reciprocal array [An,k∗ ]n,k 0 satisﬁes the ≥ recurrence relation ≤ ≤

An,k∗ = [(b0 da0 c)n +(c + da1)k + b2 + d(a1 a2)]An∗ 1,k + [(a0 + a1)n a1k + a2]An∗ 1,k 1 − − − − − − −

with A0∗,0 = 1 and An,k∗ = 0 unless 0 k n. It follows for n 1 that its row-generating functions satisfy the next recurrence≤ ≤ ≥

An∗ (x) = [(b0 da0 c)n + b2 + d(a1 a2)+ x[(a0 + a1)n a1 + a2]]An∗ 1(x)+ − − ′ − − − x(c + da1 a1x)(An∗ 1(x)) . (2.3) − − This implies that

An∗ (λx + d) = [(b0 da0 c)n + b2 + d(a1 a2)+(λx + d)[(a0 + a1)n a1 + a2]] An∗ 1(λx + d) − − ′ − − − +(λx + d)(c a1λx)An∗ 1(λx + d). − −

If let B (x)= A∗ (λx + d) for n 0, then for n 1, we have n n ≥ ≥ ′ [Bn 1(x)] Bn(x) = (b0 + da1 c)n + b2 + λx[(a0 + a1)n a1 + a2] Bn 1(x)+(λx + d)(c λa1x) − , { − − } − − λ

which implies that coeﬃcients array [Bn,k]n,k 0 of Bn(x) satisﬁes the next recurrence ≥ relation

Bn,k = [(b0 + da1 c)n +(c da1)k + b2]Bn 1,k + λ[(a0 + a1)n a1k + a2]Bn 1,k 1 − − − − − − cd + (k + 1)Bn 1,k+1 λ −

for n 1. Thus its reciprocal array [Bn,k∗ ]n,k 0 satisﬁes the next recurrence relation ≥ ≥

Bn,k∗ = [b0n +(da1 c)k + b2]Bn∗ 1,k 1 + λ(a0n + a1k + a2)Bn∗ 1,k − − − − cd + (n k + 1)Bn∗ 1,k 2 λ − − −

with B∗ = 1 and B∗ = 0 unless 0 k n. 0,0 n,k ≤ ≤ Obviously, if b1 = da1 c, then arrays [Bn,k∗ ]n,k 0 and [ n,k]n,k 0 satisfy the same − ≥ T ≥ recurrence relations and the same initial conditions. Thus B∗ = . n,k Tn,k In addition, it follows from

n 1 ∗(x)= B (x)= A∗ (λx + d)=(λx + d) A ( ) (2.4) Tn n n n λx + d

6 that

n n n λ 1 n x n(x)= x T ∗(1/x)= x + d An =(λ + dx) An . T n x λ + d λ + dx x !

Finally, using the identity ∗(x)= A∗ (λx + d), we have Tn n

i k i k n,k∗ = An,n i λ d − , T − k i 0 X≥ n i n k k i which implies n,k = i 0 An,i n−k λ − d − . This completes the proof. T ≥ − Remark 2.2. The formulaP in (2.2) plays an important role. If we know certain property of A (x), then by (2.2) we immediately get that of (x). n Tn n 2 Let ak k=0 be a sequence of positive numbers. It is called log-concave if akak+2 ak+1 { } n k ≤ for all k 0. It is well-known that if the polynomial k=0 akx has only real zeros, then n ≥ ak k=0 is log-concave. In fact, log-concave sequences arise often in combinatorics and {have} been extensively investigated. We refer the readerP to Br¨and´en [7], Brenti [9], Liu and Wang [31], Stanely [39], and Wang and Yeh [41] for the log-concavity and real rootedness.

Proposition 2.3. Let [ n,k]n,k 0 and [An,k]n,k 0 be deﬁned in (1.12) and (2.1), respec- T ≥ ≥ tively. Assume that b = da c and the triangle [A ] is positive, then we have 1 1 − n,k n,k (i) each row-sequence n is positive and log-concave; {Tn,k}k=0 (ii) if A (x) has only real zeros, then so does (x) and zeros of (x) locate in [ λ , 0]. n Tn Tn − d Proof. (i) By (2.1), it follows from Kurtz [29, Theorem 2] that each row A n is { n,k}k=0 log-concave. Thus the reciprocal sequence An,n,An,n 1, ...,An,1, An,0 is log-concave. This − obviously implies that An∗ (dx) is log-concave. Let g(x) := An∗ (dx) and f(x) := g(x + 1). Then by [9, Corollary 8.3](it says that if coeﬃcients of a polynomial P (x) is positive and λ log-concave, then so is P (x + 1)), we have f(x) = g(x + 1) is log-concave. So is f( d x). λ Hence we get that ∗(x) is log-concave since ∗(x)= A∗ (λx + d)= f( x) in (2.4). This Tn Tn n d implies that , ,..., is log-concave. Tn,0 Tn,1 Tn,n (ii) If An(x) has only real zeros, then all zeros of An(x) is non-positive. Thus by (2.2), all zeros of A (x) is non-positive and locate in [ λ , 0]. The proof is complete. n − d

Since An∗ (x) satisﬁes the recurrence relation (2.3), for real rootedness of An(x), we can use the following result a special case of Corollary 2.4 in Liu and Wang [31].

Lemma 2.4. [31] Let Pn(x) n 0 be a sequence of polynomials with nonnegative coeﬃ- { } ≥ cients and 0 deg Pn deg Pn 1 1. Suppose that ≤ − − ≤

Pn(x)=(anx + bn)Pn 1(x)+ x(cnx + dn)Pn′ 1(x) − − where a , b R and c 0,d 0. Then P (x) has only real zeros and the zeros of n n ∈ n ≤ n ≥ n Pn(x) interlace those of Pn 1(x) for n 1. − ≥ By Remark 2.2, we will explore other properties of A and apply to those of . n,k Tn,k 7 3 Extensions of properties of Eulerian numbers

In this section, we will generalize some properties of Eulerian numbers.

3.1 A generalized Frobenius formula In what follows we will generalize the famous Frobenius formula (1.2) and (1.3). Let n Fn,i = i! for n, i 0. It is known that the triangle [Fn,k]n,k 0 satisﬁes the recurrence i ≥ ≥ relation

Fn,k = kFn 1,k + kFn 1,k 1 (3.1) − − − with F = 1. In convenience, we deﬁne a generalized Frobenius triangle [ ] by 0,0 Fn,k n,k

n,k = (a1k + a2) n 1,k +(b1k + b2) n 1,k 1, F F − F − −

with 0,0 = 1 and n,k = 0 unless 0 k n. AsF a generalizationF of (1.4), we have≤ ≤ the next result.

Theorem 3.1. Let a , b R and a , b R+. Then we have the explicit formula { 2 2} ⊆ { 1 1} ⊆

b2 k k b1 + k b1 k k j n = ( 1) − (a + a j) Fn,k k a j − 2 1 1 j=0 X for n, k 0. ≥ Proof. Let the exponential generating function tn F (q, t)= qk . Fn,k n! n,k 0 X≥ Then by the recurrence relation of , we have the partial diﬀerential equation Fn,k F (b q + a )qF = [a +(b + b )q]F t − 1 1 q 2 1 2 with the initial condition F (q, 0) = 1. Solve this equation and get

b (1+ 2 ) b q(1 ea1t) − b1 F (q, t)= ea2t 1+ 1 − (3.2) a 1 (It is also routine to check that F (q, t) is a solution of the above equation with the initial

8 condition). Then we have

b (1+ 2 ) n a1t − b1 k t a2t b1q(1 e ) n,kq = e 1+ − F n! a1 n,k 0 ≥ X b (1+ 2 ) b q(ea1t 1) − b1 = ea2t 1 1 − − a 1 b2 + i b q(ea1t 1) i = ea2t b1 1 − i a i 0 1 X≥ b2 i i a2t b1 + i b1q i i j a1jt = e ( 1) − e i a j − i 0 1 j=0 X≥ X b2 i i b1 + i b1q i i j (a2+a1j)t = ( 1) − e i a j − i 0 1 j=0 X≥ X b2 i i n n b1 + i b1q i i j (a2 + a1j) t = ( 1) − i a j − n! i 0 1 j=0 n 0 X≥ X X≥ k b2 k n b1 + k b1 k k j n k t = ( 1) − (a2 + a1j) q , k a1 j − n! n,k 0 j=0 X≥ X which implies

b2 k k b1 + k b1 k k j n = ( 1) − (a + a j) . Fn,k k a j − 2 1 1 j=0 X This completes the proof. By Theorem 2.1 for c = 0 and Theorem 3.1, we get a generalized Frobenius formula as follows. Obviously, it generalizes the identity (1.2) and formula (1.3).

+ Theorem 3.2. Let a1, b1 R and a2, b2 R≥. If an array [ n,k]n,k satisﬁes the recurrence relation: { } ⊆ { } ⊆ D

n,k = (a1k + a2) n 1,k +(b1n b1k + b2) n 1,k 1, (3.3) D D − − D − −

where 0,0 = 1 and n,k = 0 unless 0 k n, then there exists an array [ n,k]n,k 0 ≥ satisfyingD the recurrenceD relation ≤ ≤ F

a2 n,k =(a1k + a2) n 1,k +(b1k + b2 b1 + b1) n 1,k 1 F F − − a1 F − − for n 1 with =1 such that the row generating functions ≥ F0,0

b1 n q n(q)=(1 q) n b1 . D − a1 F 1 q − a1 !

9 In addition, we have the explicit formula

n i b1 k i = − ( ) − , Dn,k Fn,i k i −a i 0 1 X≥ − a2 b2 i a1 + b1 1+ i i b1 i j n = − ( 1) − (a + a j) . Fn,i i j a − 2 1 j 0 1 X≥ 3.2 A generalization of γ positivity of Eulerian polynomials In this subsection, we will generalize (1.5). In order to do so, we deﬁne a generalized Eulerian triangle [ ] by En,k n,k

n,k = (a1k + a2) n 1,k +(a1n a1k + a2) n 1,k 1, (3.4) E E − − E − − with = 1 and = 0 unless 0 k n. Its row-generating function has the following E0,0 En,k ≤ ≤ γ-positivity decomposition.

Theorem 3.3. Let a , a R+ and the generalized Eulerian triangle [ ] is deﬁned { 1 2} ⊆ En,k n,k in (3.4). Then there exists a nonnegative triangle [ ] satisfying the recurrence relation Sn,k n,k

n,k =(a1k + a2) n 1,k + c(n 2k + 1) n 1,k 1 (3.5) S S − − S − − with k n+1 and =1 such that their row-generating functions ≤ 2 S0,0 2a1 x (x)=(1+ x)n c En Sn (1 + x)2 for n 1, where c> 0. ≥

Proof. We will prove this result by induction on n. Let λ =2a1/c. It is obvious for n = 0. For n = 1, (x)= a + a x and (x)= a . Therefore we have E1 2 2 S1 2 λx (x)=(1+ x) . E1 S1 (1 + x)2 In the following we assume n 2. By the inductive hypothesis, we have ≥ n 1 n 2 λx (1 + x) − λ(1 x) λx n′ 1(x)=(n 1)(1 + x) − n 1 + − n′ 1 (3.6). E − − S − (1 + x)2 (1 + x)3 S − (1 + x)2 On the other hand, by the recurrence relations (3.5) and (3.4), we have

2a1(n 1)x 4a1 n(x) = [a2 + − ] n 1(x)+ x(a1 x) n′ 1(x) S λ S − − λ S − and

n(x) = [a2 + x(a1n a1 + a2)] n 1(x)+ a1x(x 1) n′ 1(x). E − E − − E −

10 Then we derive that

n 1 λx n(x) = [a2(x +1)+ a1x(n 1)](1 + x) − n 1 + a1x(x 1) E − S − (1 + x)2 − n 1 n 2 λx (1 + x) − λ(1 x) λx (n 1)(1 + x) − n 1 + − n′ 1 × − S − (1 + x)2 (1 + x)3 S − (1 + x)2 n a1(n 1)x λx n = (1+ x) a2 + − n 1 + a1x(1 x)(1 + x) 1+ x S − (1 + x)2 − (n 1) λx λ(1 x) λx − n 1 + − Sn′ 1 × (1 + x)2 S − (1 + x)2 (1 + x)4 − (1 + x)2 n 2a1(n 1) λx λx n a1 λx = (1+ x) a2 + − n 1 +(1+ x) λ (1 + x)2 S − (1 + x)2 × λ × (1 + x)2 × λx λx λ 4 n′ 1 − (1 + x)2 × S − (1 + x)2 λx = (1+ x)n . Sn (1 + x)2 Finally, obviously, the nonnegativity of [ ] follows from its recurrence relation (3.5). Sn,k n,k The proof is complete. Remark 3.4. By recurrence relations (1.1)and(1.7), respectively, it is clear that Theorem 3.3 gives a uniﬁed proof for γ positivity of Eulerian polynomials of two kinds.

3.3 An identity with the general derivative polynomial In this subsection, we will prove a formula of the generalized Eulerian polynomial (x) En in terms of the general derivative polynomial. Let x = tan t and y = sec t. Let Dt denote the diﬀerential operator. Obviously, D (x)= y2 and D (y)= xy. For δ 1, deﬁne t t ≥ dn yδ = Q (δ, t)yδ. (3.7) dtn n

The Qn(δ, t) is called the general derivative polynomial in t in [42] and satisﬁes the recurrence relation 2 Qn+1(δ, t)=(1+ t )DtQn(δ, t)+ δtQn(δ, t)

with Q0(δ, t) = 1. Obviously, if deg(Qn(δ, x)) is even (resp., odd) then so is the degree of each term of Qn(δ, t). Thus we can assume that

n ⌊ 2 ⌋ n 2k Qn(δ, t) := Qn,k(δ)t − . Xk=0 It is also known in [42] that the exponential generating function zn 1 Q (δ, x) = . n n! (cos z x sin z)δ n 0 X≥ − 11 In addition, Qn(1, 1) = Sn, where Sn is the Springer number which is a type B analog of the Euler number for the group of signed permutation and counts also the number of Weyl chambers in the principal Springer cone of the Coxeter group Bn of symmetries of an n dimensional cube, see [37, A001586]. In order to get a connection between the generalized Eulerian polynomial (x) and the En general derivative polynomial Qn(δ, x), we prove another identity for Qn(δ, x) as follows.

+ Theorem 3.5. Let a, b, c R . Assume that [ n,k]n,k 0 is an array satisfying the { } ⊆ S ≥ recurrence relation:

n,k =(ak + b) n 1,k + c(n 2k + 1) n 1,k 1 (3.8) S S − − S − − with 0,0 = 1 and n,k = 0 unless 0 k (n + 1)/2. Then for n 0, we have their row-generatingS functionsS ≤ ≤ ≥

a n 2c n/2 2b 1 n(q)= q 1 Qn , . S 2 a −  a 2c  q 1 a −  q  Proof. Suppose that an array [Sn,k◦ ]n,k 0 satisﬁes the recurrence relation ≥ 2b Sn,k◦ = (2k + )Sn◦ 1,k +(n 2k + 1)Sn◦ 1,k 1 a − − − −

n+1 with S◦ = 1 and S◦ = 0 unless 0 k . It is clear for n, k 0 that 0,0 n,k ≤ ≤ 2 ≥ n k a − k = c S◦ , Sn,k 2 n,k which implies a n 2cq (q)= S◦( ). Sn 2 n a a n 2c n/2 2b 1 Thus, in order to show n(q) = 2 ( a q 1) Qn( a , 2c ) for n 0, it suﬃces to S − √ a q 1 ≥ prove − n/2 1 S◦(q)=(q 1) Q (δ, ) n − n √q 1 − 2b with δ = a for n 0. Let x = tan t and≥ y = sec t. Then we have d yδ = δxyδ, dt d2 yδ = (δ2x2 + δy2)yδ, dt2 d3 yδ = [δ3x3 + (3δ2 +2δ)xy2]yδ, dt3 . . .

12 Thus we can also assume that

n/2 n ⌊ ⌋ d δ δ ⋆ n 2k 2k y = y S x − y . (3.9) dtn n,k Xk=0 ⋆ By induction on n, we get that the array [Sn,k]n,k 0 satisﬁes the recurrence relation ≥ ⋆ ⋆ ⋆ Sn,k = (2k + δ)Sn 1,k +(n 2k + 1)Sn 1,k 1 − − − − ⋆ n+1 ⋆ ⋆ with S = 0 unless 0 k and S = 1. Clearly, S = S◦ . Then combining n,k ≤ ≤ 2 0,0 n,k n,k n ⋆ 2 2 x Sn(y /x )= Qn(δ, x)

and the identity 1 + x2 = y2, we have 1 S⋆(q)=(q 1)n/2Q (δ, ) n − n √q 1 − for q > 1 and n 0. The general derivative polynomial Qn(δ, t) satisﬁes the recurrence relation ≥ 2 Qn+1(δ, t)=(1+ t )DtQn(δ, t)+ δtQn(δ, t)

with Q0(δ, t) = 1. It is obvious that if deg(Qn(δ, t)) is even (resp., odd) then so is the degree of each term of Q (δ, t). This implies that (q 1)n/2Q (δ, 1 ) is a polynomial in n − n √q 1 q. Thus we have polynomials − 1 S⋆(q)=(q 1)n/2Q (δ, ) n − n √q 1 − for n 0. The proof is complete. ≥

Theorem 3.6. Assume that the generalized Eulerian polynomial n(x) is deﬁned in (3.4) and Q (δ, x) is the general derivative polynomial. If 0 < a 2a E, then we have n 1 ≤ 2 n a √ 1 2a √ 1 (x)= 1 − (1 x)nQ 2 , − , (3.10) En 2 − n a w2 1

1 x where w = 1+−x . q Proof. For [ n,k]n,k 0, by Theorem 3.3, there exists a nonnegative triangle [ n,k]n,k satis- ≥ fying the recurrenceE relation S

n,k =(a1k + a2) n 1,k + c(n 2k + 1) n 1,k 1, (3.11) S S − − S − − where = 1 and = 0 unless 0 k (n + 1)/2 such that for n 1 their S0,0 Sn,k ≤ ≤ ≥ row-generating functions

2a1 x (x)=(1+ x)n c . En Sn (1 + x)2

13 Then by Theorem 3.5, we have its row-generating function

n n 2 a1 2cx 2a2 1 n(x)= 1 Qn , (3.12) S 2 a1 −  a1 2cx 1 a1 −  q  for n 0. ≥ Thus we have n a1√ 1 n 2a2 1+ x n(x)= − (1 x) Qn , . (3.13) E 2 − a1 √ 1(1 x) − − This proof is complete. Remark 3.7. For Eulerian polynomials of two kinds in Section 1, by recurrence relations (1.1) and (1.7), respectively, using Theorem 3.6 we immediately get n √ 1 √ 1 −→E (x)= − (1 x)nQ 2, − , (3.14) n 2 − n w2 n √ 1 B (x)= √ 1 (1 x)nQ 1, − , (3.15) n − − n w2 1 x where w = 1+−x . q 3.4 A generalization of formula of David and Barton In what follows we will give a generalization of the formula (1.6) of David and Barton. Let us recall alternating-runs polynomials Rn(x). For π = π(1)π(2) π(n) n, we say that π changes direction at position i if either π(i 1) < π(i) > π(···i +1), or∈π S(i 1) > − − π(i) < π(i +1). We say that π has k alternating runs if there are k 1 indices i such that π changes direction at these positions. Denote by R(n, k) the number− of permutations in Sn having k alternating runs. Then we have R(n, k)= k R(n 1,k)+2R(n 1,k 1)+(n k)R(n 1,k 2) (3.16) − − − − − − for n 2 and k 1 where R(2, 1) = 2 and R(n, k) = 0 unless for 1 k n 1, see André[≥1]. For n ≥ 2, we have the alternating-runs polynomial R (x)=≤ n≤ R(−n, k)xk. ≥ n k=0 Beginning with André[1], there are many interesting results related to alternating runs, see Bóna and Ehrenborg [6], Canfield and Wilf [10], David and Barton [20P], Stanley [38], Zhu [46], Zhu-Yeh-Lu [49], Zhuang [51] for instance. If let R(n, k)= R(n +2,k +1)/2 for n, k 0, then the array [R(n, k)]n,k satisfies the recurrence relation ≥ e R(n, k)=(k + 1)R(n 1,k)+2R(n 1,k 1)+(n k + 1)R(n 1,k 2) (3.17) e− − − − − − for ne k 0, where Re (0, 0)= 1 andeR(n, k) = 0 unless for 0 k e n. Let≥ a≥, a ,c,d R+ and b R. Let a triangle [T ] ≤satisfy≤ the next recurrence { 1 2 } ⊆ { 2} ⊆ n,k n,k relation: e e Tn,k = d(a1k + a2)Tn 1,k + b2Tn 1,k 1 + c(n k + 1)Tn 1,k 2, (3.18) − − − − − − where T0,0 = 1 and Tn,k = 0 unless 0 k n. Denote its row-generating function by n k ≤ ≤ Tn(x)= k=0 Tn,kx . We present a generalization of (1.6) as follows. P 14 Theorem 3.8. Let the generalized Eulerian polynomial (x) and the polynomial T (x) En n be defined by (3.4) and (3.18), respectively. Let b2 = d(a2 + σa1). If c = da1 and σ 0, 1, 1 , then ∈{ − } n+σ σ n 1+ w 1 w T (x) = a− (d + dx) − n 2 2 En+σ 1+ w 1 x for n 1, where w = − . ≥ 1+x Proof. For [ n,k]n,k 0,q by Theorem 3.3, there exists a nonnegative triangle [ n,k]n,k satis- E ≥ S fying the recurrence relation

n,k =(a1k + a2) n 1,k + da1(n 2k + 1) n 1,k 1, (3.19) S S − − S − − where = 1 and = 0 unless 0 k (n + 1)/2 such that for n 1 their S0,0 Sn,k ≤ ≤ ≥ row-generating functions 2 x (x)=(1+ x)n d , En Sn (1 + x)2 which implies 1+ w n 1 w 1+ w n 2 n 1 w2 − = − 2 En 1+ w 2 1+ w Sn 2d x = . (3.20) Sn d(1 + x) On the other hand, for the array [Tn,k]n,k, it follows from Theorem 2.1 with a0 = b0 = b1 = 0 and λ = d, and c = da1 that there exists a triangle [An,k]n,k 0 satisfying the ≥ recurrence relation

An,k =(a1k + a2)An 1,k + da1(n 2k +1+ σ)An 1,k 1 (3.21) − − − − with A0,0 = 1 and An,k = 0 unless 0 k (n +1+ σ)/2 such that row-generating functions ≤ ≤ x T (x)=(d + dx)nA . (3.22) n n d + dx (i) Assume σ = 0. Obviously, = A and (x)= A (x) for n, k 0. Combining Sn,k n,k Sn n ≥ identities (3.20) and (3.22), we have for n 0 that ≥ 1+ w n 1 w T (x) = (d + dx)n − . n 2 En 1+ w (ii) Assume σ = 1. Obviously, = A /a and A (x)= a (x) for n, k 0. − Sn,k n+1,k 2 n+1 2Sn ≥ Thus, combining identities (3.20) and (3.22), we have for n 1 that ≥ n 1 n 1+ w − 1 w Tn(x) = a2(d + dx) n 1 − . 2 E − 1+ w (iii) Assume σ = 1. Obviously, An,k = n+1,k/a2 and n+1(x)= a2An(x) for n, k 0. Thus, combining identities (3.20) and (3.22S), we have forSn 0 that ≥ ≥ n+1 1 n 1+ w 1 w T (x) = a− (d + dx) − . n 2 2 En+1 1+ w This proof is complete.

15 4 Applications

In this section, we will give some applications of our results.

4.1 The array from the Lambert function

Our results can immediately be applied to the array [βn,k]n,k 0 in Subsection 1.4. ≥ Proposition 4.1. (i) Each row β n is positive; { n,k}k=0 (ii) Each row β n is log-concave; { n,k}k=0

(iii) Its row generating functions βn(q) form a strongly q-log-convex sequence. Proof. Let T = β for n, k 0. Then by the recurrence relation (1.10), we have the n,k n+1,k ≥ array [Tn,k]n,k 0 satisﬁes the recurrence relation ≥

Tn,k = (3n k 1)Tn 1,k + nTn 1,k 1 (k + 1)Tn 1,k+1 − − − − − − −

for n, k 0 with T0,0 = 1. For its reciprocal array [Tn,k∗ ]n,k 0, it satisﬁes the recurrence ≥ relation ≥

Tn,k∗ = nTn∗ 1,k + (2n + k 1)Tn∗ 1,k 1 (n k + 1)Tn∗ 1,k 2 − − − − − − − − for n, k 0 with T ∗ = 1. By Theorem 2.1 with c = 1 and d = λ, there exists an array ≥ 0,0 − [An,k]n,k 0 satisfying the recurrence relation ≥ n An,k = An 1,k +(n + k 1)An 1,k 1 d − − − − for n, k 0 with A = 1 such that ≥ 0,0 n n x T ∗(x)= d (1 + x) A ( ) n n d(1 + x)

n n for n 1. It is obvious that A is positive for d > 0. Thus, T ∗ is positive ≥ { n,k}k=0 { n,k}k=0 and log-concave for 0 k n by Proposition 2.3 (In fact, for d = 1, An(x) is exactly the Ramanujan polynomial≤ [30≤]). In addition, it follows from [12, 46] that the row-generating functions An(q) form a strongly q-log-convex sequence for n 0. Thus An∗ (q) n 0 is a strongly q-log-convex ≥ { } ≥ sequence. Noting for the generating function βn(q) of the array [βn,k]n,k 0 that ≥

βn+1(q)= Tn(q)= An∗ (d(q + 1))

for n 0, we immediately get that βn(q) n 0 is a strongly q-log-convex sequence. ≥ { } ≥ Remark 4.2. For a polynomial sequence fn(q) n 0, it is called q-log-convex if { } ≥ 2 fn+1(q)fn 1(q) fn(q) − − is a polynomial with nonnegative coeﬃcients for n 1, see Liu and Wang [32]. It is called ≥ strongly q-log-convex deﬁned by Chen et al. [11], if

fn+1(q)fm 1(q) fn(q)fm(q) − − 16 is a polynomial with nonnegative coeﬃcients for any n m 1. Many famous polynomials were proved to be q-log-convex or strongly q-log-convex,≥ ≥ e.g., the Bell polynomials, the classical Eulerian polynomials, the Narayana polynomials of type A and B, Jacobi- Stirling polynomials, and so on, see Liu and Wang [32], Chen et al. [11], Zhu [45, 46, 47], Zhu et al. [49], Zhu and Sun [50] for instance.

4.2 Staircase tableaus

We will apply our results to the array Tn,k from staircase tableaus and it generating function Tn(q) in subsection 1.3.

Proposition 4.3. Let Tn,k and Tn(q) be deﬁned in subsection 1.3. Then

(i) The polynomial Tn(q) has only real zeros and is log-concave.

(ii) An explicit formula for Tn,m can be written as

n k n i 1 + i i T = − − 2 ( 2)k( 1)j(1 + j)n. n,m m k k i i j − − k,i,j 0 X≥ − −

(iii) The exponential generating function of Tn(q) can be expressed as

(q 1)t/3 3/2 T (q) (q 1)e − n tn = − . n! 2q (q + 1)e(q 1)t n>0 − X −

Proof. (i) By Theorem 2.1, there exists an array [An,k]n,k satisfying the recurrence relation

An,k = (1+ k)An 1,k + (2n 2k + 1)An 1,k 1 − − − −

with A0,0 = 1 such that

q T (q)=(1+ q)nA (4.1) n n 1+ q k for n 0, where An(q)= k 0 An,kq . In addition, An(x) satisﬁes the recurrence relation ≥ ≥ P An(q) = [1 + (2n 1)q]An 1(q)+ x(1 2x)An′ 1(q) − − − −

with A0(q) = 1. By Lemma 2.4, An(q) has only real zeros. It follows from (4.1) that T (q) has only real zeros. Therefore T (q) is log-concave for n 0. (In fact, the triangle n n ≥ [An,k]n,k 0 is called the ﬂower triangle, see [37, A156920].) ≥ (ii) For array [An,k]n,k, it follows from Theorem 3.1 (i), we have

n i k i A = − ( 2) − , n,k Fn,i k i − i 0 ≥ − X 1 2 + i i i i j n = 2 ( 1) − (1 + j) . Fn,i i j − j 0 X≥ 17 Thus by Theorem 2.1, we get for n, m 0 that ≥ n k n k n i 1 + i i T = A − = − − 2 ( 2)k( 1)j(1 + j)n. n,m n,k m k m k k i i j − − k 0 k,i,j 0 X≥ − X≥ − − (iii) Moreover, by (3.2) and Theorem 3.2, we deduce the expression

(2q 1)t/3 3/2 A (q) (2q 1)e − n tn = − , n! 2q e(2q 1)t n>0 − X − which by (4.1) implies the desired result in (iii).

4.3 Alternating runs of type B A run of a signed permutation π is deﬁned as a maximal interval of consecutive elements on 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. Let Z(n, k) denote··· the number of up signed permutations in Bn with k alternating runs. In [44, Theorem 4.2.1], Zhao demonstrated that the numbers Z(n, k) satisfy the recurrence relation

Z(n, k)=(2k 1)Z(n 1,k)+3Z(n 1,k 1)+(2n 2k + 2)Z(n 1,k 2) (4.2) − − − − − − − for n > 2 and 1 6 k 6 n, where Z(1, 1) = 1 and Z(1,k)= 0 for k > 1. The alternating n k runs polynomials of type B are deﬁned by Zn(x)= k=1 Z(n, k)x . The ﬁrst few of the polynomials Zn(x) are given as follows: P 2 2 3 Z1(x)= x, Z2(x)= x +3x , Z3(x)= x + 12x + 11x .

Zhao proved that Zn(x) has only real zeros and is log-concave. Let Z(n, k)= Z(n+1,k+1) for n, k 0. Then it follows from (4.2) that ≥ e Z(n, k)=(2k + 1)Z(n 1,k)+3Z(n 1,k 1)+2(n k + 1)Z(n 1,k 2) (4.3) − − − − − − foren 1 with Z(0, 0)e = 1. Using oure results, we immediately get thee following properties. ≥ Proposition 4.4.b Let Zn(x) be deﬁned above. We have the following results.

(i) The polynomial Zne(x) has only real zeros and is log-concave. (ii) For n 0, we have ≥ e 2x Z (x)=(1+ x)nW l , n n+1 1+ x l el k l where Wn(x)= k 0 Wn,kx and Wn,k satisﬁes the recurrence relation ≥ P l l l Wn,k = (2k + 1)Wn 1,k +(n 2k + 1)Wn 1,k 1 (4.4) − − − − l with initial condition W0,0 =1, see Petersen [33].

18 (iii) Let Bn(x) be the Eulerian polynomials of type B in subsection 1.2. Then we have a David-Barton type formula

1+ w n+1 1 w Z (x)=(1+ x)n B − . n 2 n+1 1+ w e (iv) Let Q (δ, t) be the general derivative polynomial. Then for n 0 we have n ≥ n+1 n n 1 Zn( x) = 2 (1 x) Qn+1 1, √x 1 . − − √x 1 − − e Proof. (i) and (ii) Applying Theorem 2.1 with c = 2 and λ = d = 1 to Z(n, k), there exists an array [An,k]n,k 0 satisfying the recurrence relation ≥ e An,k = (2k + 1)An 1,k + 2(n 2k + 2)An 1,k 1 − − − − for n 1 with A = 1 such that ≥ 0,0 x Z (x)=(1+ x)nA n n (1 + x) e for n 0. Note that An(x)=(1+2x)An 1(x)+2x(1 2x)An′ 1(x) for n 1. By Lemma ≥ − − − ≥ 2.4, An(x) has only real zeros. Thus Zn(x) has only real zeros and is log-concave. Then by the recurrence relation (4.4) we have A = W l 2k for n, k 0. Thus we n,k n+1,k ≥ get e 2x Z (x)=(1+ x)nW l n n+1 1+ x for n 0 and d = 2. e For≥ (iii), it follows from Theorem 3.8 with σ = 1. For (iv), it follows from (i) and Theorem 3.5.

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