Linear Algebra and its Applications 451 (2014) 260–289

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Linear Algebra and its Applications

www.elsevier.com/locate/laa

On symmetric polynomials with only real zeros andnonnegative γ-vectors

José AgapitoCentro de Estruturas Lineares e Combinatórias, Dep. Matemática, Fac. Ciências,Univ. de Lisboa, Campo Grande, 1749-016 Lisboa, Portugal

a r t i c l e i n f o a b s t r a c t

Article history:Received 18 December 2013Accepted 13 March 2014Available online 13 April 2014Submitted by R. Brualdi

MSC:05A11B3711B8311C11Y5526C33F34A25

Keywords:Log-concave symmetric unimodalpolynomialPolynomial with only real zerosEulerian polynomialNarayana polynomialf-Eulerian polynomialγ-vectorGeneralized Stirling number ofsecond kind

We present a family of polynomials with nonnegative coef-ficients that are symmetric and have only real nonpositivezeros. Therefore, they are also log-concave, unimodal and havenonnegative γ-vectors. This family of polynomials arises byapplying the operator (ZrDs)n to the rational function 1

1−z.

Two distinguished examples are the classical Eulerian polyno-mials and the Narayana polynomials. We use two special linearbases, besides the standard basis, to provide various character-izations for these polynomials and their coefficients, in termsof generalized Stirling numbers of the second kind. The typeof generalized Stirling numbers we deal with, emerge from thenormal ordering of (ZrDs)n. We generalize several formulasknown for Eulerian and Narayana polynomials. Likewise, weprovide an explicit formula to compute the so-called γ-vectorof any symmetric polynomial of the type consider in this pa-per. Further related polynomials are obtained by applying theoperator (ZrDs)n to a general Möbius transformation. Alongthe way, associated families of novel nonnegative integer se-quences are put in evidence as a byproduct of our polynomialcharacterizations. We provide explicit formulas for some ofthese integral sequences.

© 2014 Elsevier Inc. All rights reserved.

E-mail address: jagruiz@cii.fc.ul.pt.

http://dx.doi.org/10.1016/j.laa.2014.03.0180024-3795/© 2014 Elsevier Inc. All rights reserved.

J. Agapito / Linear Algebra and its Applications 451 (2014) 260–289 261

1. Introduction

Let D and Z denote respectively the usual derivative and multiplication by z operators.Implicit in Euler’s work [11] is the use of the operator (ZD)n to obtain the classicalEulerian polynomial An(z) =

∑nk=1 A(n, k)zk, as the numerator of the rational function

(ZD)n( 11−z ); that is,

(ZD)n(

11 − z

)= An(z)

(1 − z)n+1 . (1)

The Narayana polynomial Nn(z) =∑n

k=1 N(n, k)zk (another well-known polynomialwith many combinatorial interpretations) can be obtained in a similar way; namely,

(ZD2)n( 1

1 − z

)= n!(n + 1)!Nn(z)

(1 − z)2n+1 . (2)

The symbol D2 above stands for the usual second order derivative operator. Althoughthe identity (1) is known, the identity (2) is seemingly new. The polynomials An(z)and Nn(z) (see Table 1) share many similar features. For instance, they both havesymmetric positive integral coefficients, and their zeros are all nonpositive real numbers.Furthermore, they are also log-concave, unimodal, and their γ-vectors are nonnegative.

Recall that a sequence (a1, . . . , aN ) is called log-concave if a2k+1 akak+2 for 1

k N − 2. It is said to be symmetric if ak = aN+1−k for 1 k N+12 , and

is called unimodal if there exists an index 1 N c N such that ak ak+1 for1 k N c − 1 and ak ak+1 for N c k N − 1. The sequence (a1, . . . , aN ) issaid to have no internal zeros if there are no three indices 1 i < j < k N suchthat ai, ak = 0 and aj = 0. Equivalently, the polynomial h(z) =

∑Ni=1 hiz

i is calledlog-concave (respectively symmetric, unimodal, with no internal zeros) if the sequence(h1, . . . , hN ) has the corresponding property. For any polynomial h(z), symmetric or not,there is a unique vector (γ1, . . . , γN ), called the γ-vector of h(z), such that

h(z) =Nc∑k=1

γkzk(1 + z)N+1−2k +

N∑k=Nc+1

γkzk, (3)

where N c = N+12 . If h(z) is symmetric then γk = 0 for all N c + 1 k N . However,

for 1 k N c, the numbers γk may be positive or negative, even if the coefficients ofh(z) are all positive (see Example 4.9).

The operator (ZD)n used in the identity (1), has an expansion related to the classicalStirling numbers S(n, k) of the second kind [28]; that is,

(ZD)n =n∑

S(n, k)ZkDk. (4)

k=1

262 J. Agapito / Linear Algebra and its Applications 451 (2014) 260–289

Table 1The first six Eulerian and Narayana polynomials.

An(z): Eulerian polynomials Nn(z): Narayana polynomials

A1(z) = z. N1(z) = z.A2(z) = z + z2. N2(z) = z + z2.A3(z) = z + 4z2 + z3. N3(z) = z + 3z2 + z3.A4(z) = z + 11z2 + 11z3 + z4. N4(z) = z + 6z2 + 6z3 + z4.A5(z) = z + 26z2 + 66z3 + 26z4 + z5. N5(z) = z + 10z2 + 20z3 + 10z4 + z5.A6(z) = z + 57z2 + 302z3 + 302z4 + 57z5 + z6. N6(z) = z + 15z2 + 50z3 + 50z4 + 15z5 + z6.

The aim of this work is to show that, for a pair of positive integers (r, s) such that1 r s, the operator (ZrDs)n applied on the rational function 1

1−z yields a plethoraof (generally shifted) f -Eulerian polynomials [31] that have the same list of propertiesthe Eulerian and Narayana polynomials have. An extension of the expansion (4), calledthe normal ordering of (ZrDs)n, is used to give a generalization of the following classicalformula for Eulerian polynomials.

Theorem 1.1 (Frobenius). Given any integer n 1, the Eulerian polynomial An(z) isgiven by1

An(z) =n∑

k=1

k!S(n, k)zk(1 − z)n−k.

The paper is organized as follows. Section 2 reviews the generalized Stirling numbersof the second kind Sr,s(n, k) that arise in the normal ordering of the operator (ZrDs)n.A useful code in Mathematica is given to compute these numbers when the integral pa-rameters r and s are in the first quadrant. Known recursive and explicit formulas forthese numbers are clearly stated for later use, and restatements of the explicit formula insome particular interesting cases are also considered. Along the way, some comparisonswith other type of generalized Stirling numbers are made. Section 3 introduces the poly-nomials Ar,s,Nr(n)(z), which are the object of our study. The computational definitionof Ar,s,Nr(n)(z) (Definition 3.4) embraces already the symmetry property. It also putsin evidence a number factor which is important in the characterization of Ar,s,Nr(n)(z)(see Remark 3.5). A general recursive differential equation formula (Theorem 3.7) followsstraightforwardly from Definition 3.4. Such a formula is known in the case of Eulerianpolynomials (Corollary 3.8), but is new in the case of the Narayana polynomials (Corol-lary 3.9). The polynomials Ar,s,Nr(n)(z) are indeed f -Eulerian polynomials, in the senseof Stanley [31], for a suitable polynomial f whose zeros are integral and nonnegative.We show that the polynomials Ar,s,Nr(n)(z) possess the list of highlighted propertiesthe Eulerian and Narayana polynomials have (Theorem 3.13). In Section 4.1, we give

1 Since An(z) = zn+1An( 1z ), we equivalently have An(z) = z

∑nk=1 k!S(n, k)(z − 1)n−k [9].

J. Agapito / Linear Algebra and its Applications 451 (2014) 260–289 263

explicit formulas for Ar,s,Nr(n)(z) (Theorem 4.1) and their standard coefficients Ar,s(n, k)(Theorem 4.3), in terms of the generalized Stirling numbers Sr,s(n, k). We introduce inSection 4.2, the γ-basis for univariate polynomials (with no constant term). Havingonly real zeros guarantees that a symmetric polynomial with nonnegative coefficientshave a nonnegative γ-vector (Lemma 4.10). The converse is clearly false (take for in-stance h(z) = z + 5z2 + 9z3 + 5z4 + z5. The γ-vector of h(z) is γ(h) = (1, 1, 1, 0, 0)and 0 is its only real zero). The main result in this section is an explicit formulathat expresses the γ-vector of any symmetric polynomial (Theorem 4.11), in terms ofa Catalan triangle and the inverse of a Pascal triangle. This yields a useful formulafor the γ-vector of Ar,s,Nr(n)(z) (Theorem 4.13). Section 5 extends, in a natural way,the two-parameter family of polynomials Ar,s,Nr(n)(z) to a four-parameter family ofpolynomials Ac,d

r,s,Nr(n)(z). These polynomials are no longer symmetric nor unimodal ingeneral. A characterization for such polynomials is given in Theorem 5.2. Finally, in Sec-tion 6, we speak about some of the combinatorial interpretations known for the Eulerianand Narayana polynomials, and their corresponding γ-vectors, and give some generalremarks.

2. Generalized Stirling numbers arising from the normal ordering of (ZrDs)n

There are several generalizations of the classical Stirling numbers in the mathe-matical literature, see for example [16,18,21,22,28] and references therein. We elab-orate in this section on a generalization coming from the normal ordering of bosonstrings (see e.g., Blasiak [2]). Note that, when acting over a nice space of functions,for example the algebra of polynomials in one variable C[z], the operators Z and D

do not commute. Nevertheless, they satisfy the relation [D,Z] := DZ − ZD = 1.For r and s positive integers, we substitute DZ = 1 + ZD into (ZrDs)n as manytimes as needed until a linear combination of monomials of the form ZaDb is reached.This standard expansion is called the normal ordering of (ZrDs)n and the coefficientsthat arise in such expansion are, by the type of substitution performed, positive inte-gers.

Example 2.1. The normal ordering of (ZD2)2 is

(ZD2)2 = ZD2ZD2 = ZDDZD2 = ZD(1 + ZD)D2

= ZD3 + ZDZD3 = ZD3 + Z(1 + ZD)D3 = 2ZD3 + Z2D4.

With a bit of patience or the help of a mathematics software, one can systematicallycompute these coefficients. For instance, here is a code to obtain the normal ordering of(ZrDs)n applying the substitution DZ = 1 + ZD, using Mathematica and the packageNCAlgebra [15]:

264 J. Agapito / Linear Algebra and its Applications 451 (2014) 260–289

rules = D ** Z -> 1 + Z ** D;Z[r_] := Nest[Z ** # &, Z, r-1];D[s_] := Nest[D ** # &, D, s-1];simp[expr_] := NCE[Substitute[expr,rules]];ZrDsn[r_,s_,n_] := NCUnMonomial[FixedPoint[simp[#] &,

Nest[Z[r] ** D[s] ** # &, Z[r] ** D[s], n-1]]];

Example 2.2. The normal ordering of (Z2D2)4 is

ZrDsn[2,2,4] = 8 Z^2 ** D^2 + 208 Z^3 ** D^3 + 652 Z^4 ** D^4+ 576 Z^5 ** D^5 + 188 Z^6 ** D^6 + 24 Z^7 ** D^7+ Z^8 ** D^8

We can as well consider the normal ordering of (ZrDs)n for general integers r, s notnecessarily both positive. For instance, let r 0 and s > 0. Working over the algebraC[z, z−1] of Laurent polynomials, it holds the relation [D,Z−1] = DZ−1 − Z−1D =−Z−2. This time positive and negative integer coefficients arise in the normal ordering of(ZrDs)n, when the substitution DZ−1 = −Z−2+Z−1D is performed. Similarly, denotingby D−1 the usual integration operator, we have [D−1, Z] = D−1Z−ZD−1 = −D−2. Thisrelation in turn gives rise to positive and negative integer coefficients when is used toobtain the normal ordering of (ZrDs)n, for any r > 0 and s 0. Observe that a nice dualpattern is emerging. One might guess that [D−1, Z−1] = −1. However, this is not true.Working on any reasonable space, it is easy to check that [D−1, Z−1] = D−1Z−2D−1.Nevertheless, changing the point of view, we could respectively lift D and Z to formalvariables U and V satisfying the first three aforementioned commutation relations anddecree that [U−1, V −1] = −1. It can be checked that the normal ordering of (V rUs)n isthen given by

(V rUs

)n =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

∑rnk=r Sr,s(n, k)V kUn(s−r)+k if s 0 < r or 0 < r s∑snk=s Sr,s(n, k)V n(r−s)+kUk if r 0 < s or 0 < s r∑|r|nk=|r| Sr,s(n, k)V −kUn(s−r)−k if s r < 0 or r < 0 = s∑|s|nk=|s| Sr,s(n, k)V n(r−s)−kU−k if r s < 0 or s < 0 = r

(5)

Definition 2.3. The integral numbers Sr,s(n, k) that emerge in the normal ordering of(V rUs)n are called generalized Stirling numbers of the second kind.

Recall that the generalized Stirling numbers Sr,s(n, k) of the second kind are positivewhen r, s > 0. Formula (5) extends the expansion considered in (4), which explains thename given to Sr,s(n, k). As an illustration, we have S1,2(2, 1) = 2 and S1,2(2, 2) = 1in Example 2.1 and S2,2(4, 2) = 8, S2,2(4, 3) = 208, S2,2(4, 4) = 652, S2,2(4, 5) = 576,S2,2(4, 6) = 188, S2,2(4, 7) = 24 and S2,2(4, 8) = 1 in Example 2.2.

We set Sr,s(0, 0) = 1 for all r, s ∈ Z and S0,0(n, 0) = 1 for any n > 0. It is alsoreasonable to assume that Sr,s(n, k) = 0, whenever (k < |r| with n > 0 or |r|n < k) or

J. Agapito / Linear Algebra and its Applications 451 (2014) 260–289 265

(k < |s| with n > 0 or |s|n < k). Note that S0,1 = S1,0 is the identity matrix. In general,for any |s| = 0, the matrix S0,s = Ss,0 is obtained from the identity matrix by inserting|s| − 1 zero columns between two consecutive columns.

One can computationally verify that the numbers Sr,s(n, k) are symmetric in (r, s);that is,

Sr,s(n, k) = Ss,r(n, k), (6)

for any n, k, r, s ∈ Z, with n, k > 0. (Blasiak and Flajolet [3, Note 2] provides a nicejustification of this property.) Furthermore, for r, s > 0, it holds that

S−r,−s(n, k) = (−1)n−kSr,s(n, k). (7)

We shall use, from now on, the representation U → D and V → Z whenever it makessense. We thus obtain the following useful identity.

Lemma 2.4. Let l and s be any integers such that s 0. It holds

DsZl =s∑

p=0

(s

p

)lpZl−pDs−p. (8)

Proof. Recall that, for p 1, the notation lp stands for the falling factorial of l (oforder p) defined by lp = l(l − 1) · · · (l − p + 1). By convention, l0 = 1. We proceed byinduction on s. The case s = 0 is trivial. In addition, the case s = 1 follows from theLeibnitz’s rule, DZl = lZl−1 + ZlD. Assuming that Lemma 2.4 is valid for any s > 0,we have

Ds+1Zl = D

s∑p=0

(s

p

)lpZl−pDs−p =

s∑p=0

(s

p

)lpDZl−pDs−p

=s∑

p=0

(s

p

)lp((l − p)Zl−p−1Ds−p + Zl−pDs−p+1)

= ZlDs+1 +s∑

p=1

[(s

p− 1

)+

(s

p

)]lpZl−pDs−(p−1)

+(s

s

)ls+1Zl−(s+1)Ds−s =

s+1∑p=0

(s + 1p

)lpZl−pDs+1−p.

By Lemma 2.4 and after some algebraic manipulations, similar to those performed byBlasiak [2], which involve applying the normal expansion of (ZrDs)n to the exponentialfunction, we obtain the next recursive and explicit formulas.

266 J. Agapito / Linear Algebra and its Applications 451 (2014) 260–289

Proposition 2.5. Let r and s be any integers such that r 0 < s or 0 < s r. We have

Sr,s(n, k) =s∑

p=0

(s

p

)(n(r − s) + k − r + p

)pSr,s(n− 1, k − s + p), (9)

Sr,s(n, k) = 1k!

k∑p=s

(−1)k−p

(k

p

) n∏j=1

(p + (j − 1)(r − s)

)s. (10)

By symmetry, we also get

Proposition 2.6. For any integers s 0 < r or 0 < r s, we have

Sr,s(n, k) =r∑

p=0

(r

p

)(n(s− r) + k − s + p

)pSr,s(n− 1, k − r + p), (11)

Sr,s(n, k) = 1k!

k∑p=r

(−1)k−p

(k

p

) n∏j=1

(p + (j − 1)(s− r)

)r. (12)

Remark 2.7. Blasiak [2, p. 216, formula (92)] states the recurrence relation (9) of Propo-sition 2.5 in a slightly different way. He is primarily concerned with the case 0 < s r.However, his and our recursive formulas are equivalent. Note that both recursive andexplicit formulas for the generalized Stirling numbers Sr,s(n, k) given in Propositions 2.5and 2.6 are valid for all integral points (r, s) in the union of the half planes s > 0 andr > 0. Moreover, such formulas remain valid for any r ∈ C and any integer s > 0(correspondingly, for any s ∈ C and any integer r > 0). In particular, the case s = 1yields the generalized Stirling numbers S(r;n, k) of the second kind studied by Lang[18]. A different but related parametrization is considered by Mansour et al. [21]. Theirtype of generalized Stirling numbers are denoted by S(s,h)(n, k). It can be checked; forinstance, that S(1/2,2)(n, k) = S2,1(n, k) and S(2,1)(n, k) = S−1,1(n, k).

We will focus on the instance 0 < r s along this paper (see Remark 3.3 for ajustification of this choice). Recall that in this case the generalized Stirling numbersSr,s(n, k) of the second kind are positive integers when r k rn and zero otherwise.By straightforward algebraic manipulations, formula (12) is equivalent to

Sr,s(n, k) = (r!)n

k!

k∑p=r

(−1)k−p

(k

p

) n∏j=1

(p + (j − 1)(s− r)

r

). (13)

In particular, for r = s, we have

Sr,r(n, k) = (r!)n

k!

k∑(−1)k−p

(k

p

)(p

r

)n

. (14)

p=0

J. Agapito / Linear Algebra and its Applications 451 (2014) 260–289 267

Note also that

Sr,s(n, rn) = (r!)n

(rn)!

rn∑p=r

(−1)rn−p

(rn

p

) n∏j=1

(p + (j − 1)(s− r)

r

).

It follows that Sr,s(1, r) = 1. By induction on n and using the recursive formula (11), weget Sr,s(n, rn) = 1. Observe that for 0 p r − 1 and 1 n, we have

∏nj=1(p + (j −

1)(s − r))r = 0. Hence, summing over all integers 0 p k, we can express formula(12) as

Sr,s(n, k) = 1k!

k∑p=0

(−1)k−p

(k

p

) n∏j=1

(p + (j − 1)(s− r)

)r. (15)

It makes sense to replace∏n

j=1(p+(j− 1)(s− r))r by 1 when n = 0, so that Sr,s(0, k) =δ0,k. Reversing the order of summation and using the identity

(k

k−p

)=

(kp

), formula (15)

can equivalently be written as

Sr,s(n, k) = 1k!

k∑p=0

(−1)p(k

p

) n∏j=1

(k − p + (j − 1)(s− r)

)r. (16)

It is convenient to obtain additional formulas for Sr,s(n, k) that be equivalent to theones seen so far. To this purpose, note that each factor in the product of the right handside of formula (15) is in turn a subproduct; that is,

(p + (j − 1)(s− r)

)r =r∏

l=1

(p + (j − 1)(s− r) − (l − 1)

).

Interchanging the order of the double product in (15) yields

n∏j=1

(p + (j − 1)(s− r)

)r =r∏

i=1

(p− (i− 1)

∣∣(s− r))n

,

where the notation (z|α)n stands for the generalized rising factorial of z with increment α;namely,

(z|α)n =n∏

l=1

(z + (l − 1)α

).

Similarly, one can define the generalized falling factorial of z with reduction α by

(z|α)n =n∏(

z − (l − 1)α).

l=1

268 J. Agapito / Linear Algebra and its Applications 451 (2014) 260–289

Clearly, we have (z|α)n = (z| − α)n. Observe that (z|1)n = zn and (z|1)n = zn arerespectively the usual rising and falling factorials of z. Note also that (z|0)n = (z|0)n =zn. The generalized factorial (z|α)n (also denoted by (z|α)n ) has been used by Hsu andShiue [16] to construct generalized Stirling numbers S(n, k, α, β, r) that englobe variouswell-known generalizations of Stirling numbers of the first and second kinds. It is worthobserving that the generalized Stirling numbers S(n, k, α, β, r) do not contain all thegeneralized Stirling numbers Sr,s(n, k) considered in this paper. For instance, it holdsthat S(1, 0, α, β, r) = r and S(n, k, α, β, r) = 0 for k > n, regardless the values of α

and β; whereas Sr,s(1, 0) = 0 and Sr,s(n, rn) = 1 for any integer r 1.Using the notation introduced above, we can write (15) as

Sr,s(n, k) = 1k!

k∑p=0

(−1)k−p

(k

p

) r∏i=1

(p− (i− 1)

∣∣(s− r))n

. (17)

Let us consider the two distinguishable cases separately.Case r = s. Identity (17) yields

Sr,r(n, k) = 1k!

k∑p=0

(−1)k−p

(k

p

) r∏i=1

(p− (i− 1)

)n = 1k!

k∑p=0

(−1)k−p

(k

p

)(pr

)n. (18)

Example 2.8 (Stirling numbers of the second kind). Setting r = s = 1 in (18) and (16)we obtain the next well-known formulas for the Stirling numbers of the second kind [9,p. 204],

S1,1(n, k) = 1k!

k∑p=1

(−1)k−p

(k

p

)pn = 1

k!

k∑p=0

(−1)p(k

p

)(k − p)n.

Case r < s. It is useful to define, for any integer α 0, the polynomial

F (z, α, n) =n∏

l=1l ≡0 mod α

(z + l),

assuming by convention that F (z, 1, n) = 1. Specializations of F (z, α, n) give well-knownsequences of numbers. For instance, F (0, 0, n) = n! and F (0, 2, n) = n!! (the doublefactorial of n) when n is odd. Now, for any integers z 1 and α 1, we can write

(z|α)n = (z + (n− 1)α)!(z − 1)!F (z, α, n) = (1 + (n− 1)α)!

F (z, α, n)

(z + (n− 1)α

z − 1

). (19)

It is important to take α 1 in (19) because, when α = 0, the identity above clearlydoes not hold. Replacing z with p− (i− 1) and α with s− r in (19) yields the following

J. Agapito / Linear Algebra and its Applications 451 (2014) 260–289 269

formula for the generalized Stirling numbers Sr,s(n, k); that is,

Sr,s(n, k) = (((n− 1)(s− r) + 1)!)r

k!

k∑p=0

(−1)k−p

(k

p

) r∏i=1

(p−(i−1)+(n−1)(s−r)

p−i

)F (p− (i− 1), s− r, n) . (20)

Example 2.9 (Unsigned Lah numbers). Let r = 1, s = 2 (see [9, p. 156]). It follows from(20) that

S1,2(n, k) = n!k!

k∑p=0

(−1)k−p

(k

p

)(p + n− 1p− 1

)

= n!k!

(n− 1k − 1

).

The last identity is a particular case of the more general signed sum (see e.g., Riordan[27]):

(n + l

k + l

)=

k∑p=0

(−1)k−p

(k

p

)(p + n + l

p + l

)(l ∈ Z). (21)

(If l 0, identity (21) holds for any n, k 0, and if l < 0, it holds for n k −l.)

Example 2.10 (s = r + 1). It follows from (13) and (20) that

Sr,r+1(n, k) = (r!)n

k!

k∑p=0

(−1)k−p

(k

p

) n∏j=1

(p + j − 1

r

)

= (n!)r

k!

k∑p=0

(−1)k−p

(k

p

) r∏i=1

(p + n− i

p− i

)

= (n!)r

k!

k∑p=0

(−1)k−p

(k

p

) r∏i=1

(p + n− i

n

).

3. A family of shifted symmetric f -Eulerian polynomials

Here is a useful code in Mathematica that helps us to compute the rational expression(zrDs)k( 1

1−z ) for every integer 0 k n, once the integer parameters r, s and n arefixed.

ZrDsn[r_,s_,n_]:=NestList[Simplify[z^r D[#,z,s]] &, 1/(1-z),n];

270 J. Agapito / Linear Algebra and its Applications 451 (2014) 260–289

Example 3.1.

ZrDsn[1, 1, 6] =

11− z

,z

(1− z)2 ,z + z2

(1− z)3 ,z + 4z2 + z3

(1− z)4 ,z + 11z2 + 11z3 + z4

(1− z)5 ,

z + 26z2 + 66z3 + 26z4 + z5

(1− z)6 ,z + 57z2 + 302z3 + 302z4 + 57z5 + z6

(1− z)7

.

ZrDsn[1, 2, 6] =

11− z

,2(z)

(1− z)3 ,12(z + z2)(1− z)5 ,

144(z + 3z2 + z3)(1− z)7 ,

2880(z + 6z2 + 6z3 + z4)(1− z)9 ,

86 400(z + 10z2 + 20z3 + 10z4 + z5)(1− z)11 ,

3 628 800(z + 15z2 + 50z3 + 50z4 + 15z5 + z6)(1− z)13

.

ZrDsn[2, 2, 5]

=

11− z

,2z(z)

(1− z)3 ,4z(z + 4z2 + z3)

(1− z)5 ,8z(z + 20z2 + 48z3 + 20z4 + z5)

(1− z)7 ,

16z(z + 72z2 + 603z3 + 1168z4 + 603z5 + 72z6 + z7)(1− z)9 ,

32z(z + 232z2 + 5158z3 + 27664z4 + 47290z5 + 27664z6 + 5158z7 + 232z8 + z9)(1− z)11

.

ZrDsn[1, 3, 6] =

11− z

,6(z)

(1− z)4 ,360(z + z2)

(1− z)7 ,15120(5z + 14z2 + 5z3)

(1− z)10 ,

5 443 200(7z + 37z2 + 37z3 + 7z4)(1− z)13 ,

1 796 256 000(21z+ 176z2 + 334z3 + 176z4 + 21z5)(1− z)16 ,

1 961 511 552 000(33z+ 397z2 + 1202z3 + 1202z4 + 397z5 + 33z6)(1− z)19

.

ZrDsn[2, 3, 5]

=

11− z

,6z(z)

(1− z)4 ,144z(z + 3z2 + z3)

(1− z)7 ,8640z(z + 10z2 + 20z3 + 10z4 + z5)

(1− z)10 ,

1036800z(z + 22z2 + 113z3 + 190z4 + 113z5 + 22z6 + z7)(1− z)13 ,

217728000z(z + 40z2 + 400z3 + 1456z4 + 2212z5 + 1456z6 + 400z7 + 40z8 + z9)(1− z)16

.

J. Agapito / Linear Algebra and its Applications 451 (2014) 260–289 271

Example 3.2.

ZrDsn[2, 1, 6] =

11− z

,z2

(1− z)2 ,2z3

(1− z)3 ,6z4

(1− z)4 ,24z5

(1− z)5 ,120z6

(1− z)6 ,720z7

(1− z)7

.

ZrDsn[3, 1, 6] =

11− z

,z3

(1− z)2 ,z5(3− z)(1− z)3 ,

3z7(5− 4z + z2)(1− z)4 ,

3z9(35− 47z + 25z2 − 5z3)(1− z)5 ,

15z11(63− 122z + 102z2 − 42z3 + 7z4)(1− z)6 ,

45z13(231− 593z + 686z2 − 434z3 + 147z4 − 21z5)(1− z)7

.

ZrDsn[3, 2, 6] =

11− z

,2z3

(1− z)3 ,12z4(1 + z)

(1− z)5 ,144z5(1 + 3z + z2)

(1− z)7 ,

2880z6(1 + 6z + 6z2 + z3)(1− z)9 ,

86400z7(1 + 10z + 20z2 + 10z3 + z4)(1− z)11 ,

3628800z8(1 + 15z + 50z2 + 50z3 + 15z4 + z5)(1− z)13

.

Remark 3.3. We can see in Example 3.1 that a striking pattern arises when regardingintegers 0 < r s. Families of symmetric polynomials with positive coefficients are put inevidence in the numerator of the rational expression (ZrDs)n( 1

1−z ), some of them beingwell-known classes of polynomials. For instance, for r = s = 1, we have the classicalEulerian polynomials. For r = 1, s = 2, up to a positive scalar, we have the well-studiedNarayana polynomials. More generally, the polynomials corresponding to the case r 1and s = r + 1 are, up to a positive scalar and a factor of z, the (r + 1)-Narayanapolynomials studied by Sulanke [34]. The case r > s > 0, although interesting too, willnot be studied in this paper because the polynomials obtained by this process are notalways symmetric nor are their coefficients positive integers in general. See Example 3.2.Note, in particular, that (Z3D2)n( 1

1−z ) = z1+n(Z1D2)n( 11−z ).

Definition 3.4. Let r, s and n be any integers such that 1 r s and n 1. The formula

(ZrDs

)n( 11 − z

)=

Kr,s,nzr−1Ar,s,Nr(n)(z)

(1 − z)ns+1 , (22)

defines a zero constant term polynomial Ar,s,Nr(n)(z) =∑Nr(n)

k=1 Ar,s(n, k)zk of degreeNr(n) = r(n − 1) + 1, such that Ar,s(n, k) ∈ Z for every 1 k Nr(n). Moreover,Ar,s,Nr(n)(z) is symmetric; namely, it holds

Ar,s(n, k) = Ar,s

(n,Nr(n) + 1 − k

), for each 1 k

⌊Nr(n)+1

2

⌋.

The number Kr,s,n is a positive integer that in general depends on r, s and n.

272 J. Agapito / Linear Algebra and its Applications 451 (2014) 260–289

It is convenient, for all pairs (r, s) and all n, k 1, to set Ar,s(0, 0) = 1, Ar,s(n, 0) = 0,and Ar,s(0, k) = 0.

Remark 3.5. By carrying out the calculation proposed in Eq. (22), one can identify, caseby case, the integer number Kr,s,n for any given values of r and s and every n 1. Forexample, we obtain, by direct observation, the formula Kr,r,n = (r!)n, for any integerr 1. A formula for K1,3,n, although not as immediate as the previous case, is given by

K1,3,n = (n + 1)!(2n + 1)!2a(n) . (23)

The power a(n) above is the highest integer such that 2a(n) divides (n+ 1)!. The on-lineencyclopedia of integer sequences (OEIS) [17] is in this regard, a useful tool that helpus find explicit formulas for Kr,s,n, in many cases. For instance, we determine K1,2,n =n!(n + 1)! A010790, K2,3,n = n!(n+1)!(n+2)!

2 A090443, and K3,4,n = n!(n+1)!(n+2)!(n+3)!12

A090444. Based on this evidence, we conjecture that Kr,r+1,n =∏r

i=0(n+i)!

i! . Comparedwith the actual values for Kr,r+1,n, obtained from Eq. (22), such a conjecture is validated.It is worth mentioning that the reciprocal sequence 1/Kr,r+1,n is a key factor associatedwith the (r + 1)-dimensional Catalan numbers (see the specialization of Corollary 4.2below for s = r + 1). It would be interesting to find explicit formulas for Kr,s,n, as wellas a combinatorial significance for these sequences, for all integers 1 r s. It is worthpointing out that neither the sequence Kr,r+1,n, for r 4, nor the sequence K1,s,n, forany s 3, are currently catalogued in the OEIS.

Proposition 3.6. Set f(z) =∏n

j=1(z − (j − 1)(s− r))s. We can write (22) as

∑i0

f(i)zi =Kr,s,nz

n(s−r)+r−1Ar,s,Nr(n)(z)(1 − z)ns+1 . (24)

Proof. Observe that the left hand side of (22) is also equivalent to

(ZrDs

)n( 11 − z

)=

(ZrDs

)n(∑i0

zi)

=∑i0

n∏j=1

(i− (j − 1)(s− r)

)szi−n(s−r)

= 1zn(s−r)

∑i0

n∏j=1

(i− (j − 1)(s− r)

)szi. (25)

Comparing (22) with (25), we get the identity (24). We can derive from identity (24) a useful recursive formula to compute the polynomial

Ar,s,Nr(n)(z). In this sense, it is convenient to emphasize the role played by n in thedefinition of f(z); thus, let us rewrite f(z) as fn(z) =

∏nj=1(z − (j − 1)(s − r))s. Note

J. Agapito / Linear Algebra and its Applications 451 (2014) 260–289 273

that (z − n(s− r))sfn(z) = fn+1(z). Multiplying out both sides of (24) by z−n(s−r), weobtain

∑i0

fn(i)zi−n(s−r) =Kr,s,nz

r−1Ar,s,Nr(n)(z)(1 − z)ns+1 . (26)

Theorem 3.7. The polynomial Ar,s,Nr(n)(z) satisfies the following recursive formula.

Ar,s,Nr(n+1)(z) = Kr,s,n

Kr,s,n+1

s∑j=0

j∑k=0

k∏l=1

(l + ns)(s

j

)(j

k

)(r − 1)s−j

× zr−s+j(1 − z)s−kA(j−k)r,s,Nr(n)(z). (27)

Proof. On the one hand, differentiating s times the LHS of (26) with respect to z, weget

Ds

(∑i0

fn(i)zi−n(s−r))

=∑i0

fn+1(i)zi−n(s−r)−s =Kr,s,n+1Ar,s,Nr(n+1)(z)

z(1 − z)ns+s+1 . (28)

On the other hand, differentiating s times the RHS of (26) with respect to z, we get

Ds

(Kr,s,nz

r−1Ar,s,Nr(n)(z)(1 − z)ns+1

)= Kr,s,n

s∑j=0

j∑k=0

k∏l=1

(l + ns)(s

j

)(j

k

)(r − 1)s−j

× zr−s+j−1(1 − z)−ns−k−1A(j−k)r,s,Nr(n)(z), (29)

where A(j−k)r,s,Nr(n)(z) = Dj−kAr,s,Nr(n)(z). We assume that

∏0l=1(l + ns) = 1 and 00 = 1.

Identity (27) follows from equating (28) with (29). Corollary 3.8 (r = 1, s = 1: Eulerian polynomials). Write A1,1,N1(n)(z) = An(z) andrecall that K1,1,n = 1. The polynomials An(z) satisfy the following recursive formula

An+1(z) = z(1 − z)A′n(z) + (1 + n)zAn(z), (30)

with initial condition A0(z) = 1.

Corollary 3.9 (r = 1, s = 2: Narayana polynomials). Set A1,2,N1(n)(z) = Nn(z) andrecall that K1,2,n = n!(n + 1)!. The polynomials Nn(z) satisfy the following recursiveformula

Nn+1(z) = z((1 − z)2N ′′n (z) + 2(1 + 2n)(1 − z)N ′

n(z) + (1 + 2n)(2 + 2n)Nn(z))(n + 2)(n + 1) , (31)

with initial conditions N0(z) = 1 and N1(z) = z.

274 J. Agapito / Linear Algebra and its Applications 451 (2014) 260–289

Formula (30) is a well-known identity satisfied by Eulerian polynomials (actually, it isa slight variation of a classical formula satisfied by the polynomials An(z)

z , see for instance[4]). Formula (31) is new, to the best of our knowledge (compare with the alternativeknown recurrence (2) in [33]). Note also that, for r = 1 and any s 1, we have 0s−j = 0in (27), unless j = s (since 00 = 1). Therefore, formula (27) yields in this case thefollowing simple identity.

Corollary 3.10. Let r = 1 and s 1 and write A1,s,N1(n)(z) = A1,s,n(z). We have

A1,s,n+1(z) = K1,s,n

K1,s,n+1z

s∑k=0

k∏l=1

(l + ns)(s

k

)(1 − z)s−kA

(s−k)1,s,n (z). (32)

The recursive formula (27), or any of its specializations, can be used to find recurrencerelations for the coefficients of Ar,s,Nr(n). For instance, writing A1,1(n, k) = A(n, k)and A1,2(n, k) = N(n, k), the following recurrence relations are easy consequences ofCorollaries 3.8 and 3.9.

Corollary 3.11. The Eulerian numbers satisfy

A(n, k) = (n− k + 1)A(n− 1, k − 1) + kA(n− 1, k). (33)

Corollary 3.12. The Narayana numbers satisfy

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

+ 2k(2n− k)(n + 1)n N(n− 1, k) + (k + 1)k

(n + 1)nN(n− 1, k + 1). (34)

Formula (33) is a classical result (see for instance [9]); while formula (34) is new,although implicitly equivalent to several other known recurrence relations that hold forthe Narayana numbers (compare for instance with [33,35]).

Theorem 3.13. Let 1 r s and n 1 be integers. The coefficients in the standard ex-pansion of the polynomial Ar,s,Nr(n)(z) are nonnegative integers. Moreover, Ar,s,Nr(n)(z)has only nonpositive real zeros, and it is log-concave and unimodal, with no internal zeros.

Proof. Let us consider again the polynomial f(z) =∏n

j=1(z− (j−1)(s− r))s. The poly-nomial Wr,s,n(z) = Kr,s,nz

n(s−r)+r−1Ar,s,Nr(n)(z) is called the f -Eulerian polynomialassociated with f by (24), in the sense of Stanley [31, p. 209]. It is easy to see thatWr,s,n(z) has degree ns. Note also that f(z) has only integral zeros, the smallest onebeing λ(f) = 0 and the largest one being Λ(f) = (n − 1)(s − r) + s − 1. A result ofBrenti (see [6, Theorem 3.7]) implies that the polynomial Ar,s,Nr(n)(z) has nonnegativecoefficients and only real zeros. Moreover, if a polynomial has nonnegative coefficients

J. Agapito / Linear Algebra and its Applications 451 (2014) 260–289 275

and only real zeros then the zeros must be nonpositive (see Edrei [10]). Furthermore, itis well-known that a polynomial with nonnegative coefficients and only real zeros is log-concave and unimodal, with no internal zeros (see for instance Comtet [9, Theorem B,p. 270]). 4. Some convenient expansions of the polynomial Ar,s,Nr(n)(z)

Besides the ordinary presentation for the polynomial Ar,s,Nr(n)(z) in terms of thestandard basis Br,n

1 = zkNr(n)k=1 , as used in Definition 3.4, we shall also consider two

other worthy expansions for Ar,s,Nr(n)(z). We shall expand Ar,s,Nr(n)(z), on the one hand,with respect to the unnormalized Bernstein basis (without the binomial coefficients)Br,n

2 = zk(1−z)Nr(n)−kNr(n)k=1 , and, on the other hand, with respect to the basis Br,n

3 =zk(1+z)Nr(n)+1−2kN

cr (n)

k=1 ∪zkNr(n)k=Nc

r (n)+1, where N cr (n) = (Nr(n)+1)/2. We refer to

Br,n3 as the γ-basis. We derive in this section some general formulas for the polynomials

Ar,s,Nr(n)(z) and the numbers Ar,s(n, k), that extend several known formulas that holdfor the Eulerian and Narayana polynomials.

4.1. Regarding the unnormalized Bernstein basis

Let us recall that the normal ordering of (ZrDs)n is given by

(ZrDs

)n =rn∑k=r

Sr,s(n, k)ZkDn(s−r)+k.

Theorem 4.1. The polynomial Ar,s,Nr(n)(z) satisfies the following formula

Ar,s,Nr(n)(z) = 1Kr,s,n

rn∑k=r

(n(s− r) + k

)!Sr,s(n, k)zk+1−r(1 − z)rn−k

= 1Kr,s,n

Nr(n)∑k=1

(n(s− r) + k + r − 1

)!Sr,s(n, k + r − 1)zk(1 − z)Nr(n)−k.

Proof. This identity follows straightforwardly from Definition 3.4 and by applying theformula Dm( 1

1−z ) = m!(1−z)m+1 , for any integer m > 0.

Theorem 4.1 extends Frobenius’ characterization for the classical Eulerian polynomials(Theorem 1.1) to all polynomials Ar,s,Nr(n)(z). In particular, for s = r + 1, we have

Ar,r+1,Nr(n)(z) =r∏

i=0

i!(n + i)!

Nr(n)∑k=1

(n + k + r − 1)!Sr,r+1(n, k + r − 1)

× zk(1 − z)Nr(n)−k. (35)

276 J. Agapito / Linear Algebra and its Applications 451 (2014) 260–289

The polynomials Ar,r+1,Nr(n)(z)z are the (r+1)-Narayana polynomials studied by Sulanke

[34]. Thus, formula (35) provides a novel characterization for these polynomials in termsof generalized Stirling numbers.

Corollary 4.2. The sum of coefficients of the polynomial Ar,s,Nr(n)(z) is given by

Ar,s,Nr(n)(1) = (sn)!Kr,s,n

.

Thus, we have for example, A1,1,n(1) = n! and A1,2,n(1) = (2n)!n!(n+1)! (the n-th Catalan

number); two well-known properties of the sum of Eulerian and Narayana numbers.Moreover, we have Ar,r+1,Nr(n)(1) = ((r + 1)n)!

∏ri=0

i!(n+i)! . This number is known as

the (r + 1)-dimensional Catalan number [34, p. 2]. In general, since Ar,s,Nr(n)(1) is aninteger, we conclude that Kr,s,n is an integral divisor of (sn)!.

Let us denote the coefficients of Ar,s,Nr(n)(z) with respect to the unnormalized Bern-stein basis Br,n

2 = zk(1 − z)Nr(n)−kNr(n)k=1 by

br,s,nk = 1Kr,s,n

(n(s− r) + k + r − 1

)!Sr,s(n, k + r − 1), for 1 k Nr(n).

These numbers are clearly nonnegative integers since they can be obtained by a straight-forward change of basis; that is, by

(br,s,n1 , . . . , br,s,nNr(n)

)T = CBr,n2 Br,n

1

(A(r,s)(n, 1), . . . , A(r,s)

(n,Nr(n)

))T, (36)

where the transition matrix CBr,n2 Br,n

1= (CBr,n

1 Br,n2

)−1 is a Pascal triangle arranged ina convenient way; namely, we have (CBr,n

2 Br,n1

)k,j =(Nr(n)−j

k−j

). Similarly, by (36), the

numbers Ar,s(n, k) can be expressed in terms of the br,s,nk .

Theorem 4.3. For 1 j k Nr(n), the numbers Ar,s(n, k) are given by

Ar,s(n, k) = 1Kr,s,n

k∑j=1

(−1)k−j

(Nr(n) − j

k − j

)

×(n(s− r) + j + r − 1

)!Sr,s(n, j + r − 1). (37)

Example 4.4 (s = r = 1: Eulerian numbers). Recall that K1,1,n = 1 and N1(n) = n;hence

A1,1(n, k) =k∑

j=1(−1)k−j

(n− j

k − j

)j!S1,1(n, j)

=k∑

(−1)k−j

(n− j

k − j

) j∑(−1)j−p

(j

j − p

)pn. (38)

j=1 p=1

J. Agapito / Linear Algebra and its Applications 451 (2014) 260–289 277

Identity (38) is a known formula that relates Eulerian and Stirling numbers of the secondkind (see e.g., Bóna [4, Corollary 1.18]). Now, interchanging the order of summation, weget

A1,1(n, k) =k∑

p=1(−1)k−ppn

k∑j=p

(n− j

k − j

)(j

j − p

)

=k∑

p=1(−1)k−ppn

k−p∑i=0

(n− p− i

k − p− i

)(i + p

i

)

=k∑

p=1(−1)k−ppn

(n + 1k − p

)=

k∑p=0

(−1)p(n + 1p

)(k − p)n. (39)

We have used in (39) the following variation of Vandermonde’s convolution formula [27,p. 10]:

(n + p

m

)=

m∑i=0

(n− 1 − i

m− i

)(i + p

i

). (40)

Example 4.5 (s = r). Recall that Kr,r,n = (r!)n. We then have

Ar,r(n, k) = 1(r!)n

k∑j=1

(−1)k−j

(Nr(n) − j

k − j

)(j + r − 1)!Sr,r(n, j + r − 1)

= 1(r!)n

k∑j=1

(−1)k−j

(Nr(n) − j

k − j

)(j + r − 1)!(r!)n

(j + r − 1)!

×j+r−1∑p=r

(−1)j+r−1−p

(j + r − 1

p

)(p

r

)n

=k+r−1∑p=r

(−1)k+r−1−p

(p

r

)n k+r−1∑j+r−1=p

(Nr(n) + r − 1 − (j + r − 1)

k + r − 1 − (j + r − 1)

)(j + r − 1

p

)

=k+r−1∑p=r

(−1)k+r−1−p

(p

r

)n k+r−1−p∑i=0

(Nr(n) + r − p− 1 − i

k + r − 1 − p− i

)(i + p

i

)

=k+r−1∑p=0

(−1)k+r−1−p

(p

r

)n(Nr(n) + r

k + r − 1 − p

)

=k+r−1∑p=0

(−1)p(Nr(n) + r

p

)(k + r − 1 − p

r

)n

.

278 J. Agapito / Linear Algebra and its Applications 451 (2014) 260–289

Example 4.6 (s = 2, r = 1: Narayana numbers). We have K1,2,n = n!(n + 1)!; hence

A1,2(n, k) = 1n!(n + 1)!

k∑j=1

(−1)k−j

(n− j

k − j

)(n + j)!S1,2(n, j).

Recall that S1,2(n, j) = n!j!(n−1j−1

). Using the following well known identities

(n

j

)= n

j

(n− 1j − 1

),

(n− j

k − j

)(n

j

)=

(n

k

)(k

j

)and

(k − 1j − 1

)=

(k

j

)−

(k − 1j

),

together with identity (21) (for l = 0), we obtain

A1,2(n, k) = 1n!(n + 1)!

k∑j=1

(−1)k−j

(n− j

k − j

)(n + j)!n!

j!

(n− 1j − 1

)

= 1n + 1

k∑j=1

(−1)k−j

(n− j

k − j

)(n + j

j

)(n− 1j − 1

)

= 1n + 1

k∑j=1

(−1)k−j j

n

(n

k

)(k

j

)(n + j

j

)

= 1n(n + 1)

(n

k

) k∑j=1

(−1)k−jjk

j

(k − 1j − 1

)(n + j

j

)

= k

n(n + 1)

(n

k

) k∑j=1

(−1)k−j

(n + j

j

)[(k

j

)−

(k − 1j

)]

= k

n(n + 1)

(n

k

)(n + 1k

)= 1

n

(n

k − 1

)(n

k

).

There is another useful polynomial basis, that we shall call the γ-basis, worth beingstudied separately. The chosen name for this basis comes from its relation with importantenumerative invariants in geometric and algebraic combinatorics (see Section 6 for furtherdetails).

4.2. The γ-basis

Let h(z) be any polynomial in one variable of degree n 1 and without constantterm; that is,

h(z) = a1z + · · · + anzn, a1, . . . , an ∈ R. (41)

J. Agapito / Linear Algebra and its Applications 451 (2014) 260–289 279

Setting nc = n+12 , a basis for the space of polynomials of type (41) is

Bn3 =

zk(1 + z)n+1−2knc

k=1 ∪zk

n

k=nc+1. (42)

We say that Bn3 is the γ-basis (of order n).

Definition 4.7. The γ-vector of h(z) is given by

γ(h) =(γ1(h), . . . , γn(h)

)=

(CBn

3 Bn1 (a1, . . . , an)T

)T, (43)

where CBn3 Bn

1 is the transition matrix from the standard basis Bn1 = zknk=1 to the

basis Bn3 . Sometimes we shall write γ(n) instead of γ(h), for the sake of brevity. The

components γk(h) of γ(h) are called the γ-numbers of h(z).

Example 4.8. The γ-vectors γ(n), for n = 5, 6, 7, 8, are given below.

γ(5) = (a1,−4a1 + a2, 2a1 − 2a2 + a3,−a2 + a4,−a1 + a5),

γ(6) = (a1,−5a1 + a2, 5a1 − 3a2 + a3,−a3 + a4,−a2 + a5,−a1 + a6),

γ(7) = (a1,−6a1 + a2, 9a1 − 4a2 + a3,−2a1 + 2a2 − 2a3 + a4,−a3 + a5,−a2 + a6,

− a1 + a7),

γ(8) = (a1,−7a1 + a2, 14a1 − 5a2 + a3,−7a1 + 5a2 − 3a3 + a4,−a4 + a5,−a3 + a6,

− a2 + a7,−a1 + a8).

Note in particular that, if h(z) is symmetric, the last n − nc components of γ(n) inExample 4.8 are zero. This holds in general for the γ-vector of any symmetric polynomialh(z) of degree n 1. The other components of γ(n) may be positive, negative or zero,even if h(z) is symmetric with nonnegative coefficients.

Example 4.9. The γ-vector of h(z) = z + 3z2 + 7z3 + 3z4 + z5 is γ(h) = (1,−1, 3, 0, 0).

We have seen (in the proof of Theorem 3.13) that symmetric polynomials with non-negative coefficients and only real zeros have many interesting properties. Here is anotherimportant property regarding their γ-vectors.

Lemma 4.10. If h(z) = a1z + · · · + anzn is a symmetric polynomial with nonnegativecoefficients and only real zeros then the γ-vector of h is nonnegative.

Proof. Since h(z) is symmetric then zn+1h( 1z ) = h(z) when z = 0. Hence, the zeros of

h(z), except 0 and −1, are negative and come in pairs, say a and 1/a. The polynomialh(z) can be then written as a product of factors of the form

(z − a)(z − 1/a) = (1 + z)2 − (2 + a + 1/a)z,

280 J. Agapito / Linear Algebra and its Applications 451 (2014) 260–289

together with a positive scalar and factors of z and (1+ z). Thus, the coefficients of h(z)in its expansion with respect to the basis Bn

3 are necessarily nonnegative. The property stated in Lemma 4.10 was already observed by Brändén [5, Lemma 4.1]

(see also Stembridge [32, Section 1.4]).Let us now further expand on the product matrix (43), in order to determine a general

explicit formula for the components of γ(h) in terms of the coefficients a1, . . . , an. To findsuch a direct formula, we first look closely at some particular instances of the transitionmatrices CBn

1 Bn3 and CBn

3 Bn1 = (CBn

1 Bn3 )−1. The transition matrices for n = 7 and n = 8

are shown below.

CB71B7

3=

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

1 0 0 0 0 0 06 1 0 0 0 0 015 4 1 0 0 0 020 6 2 1 0 0 015 4 1 0 1 0 06 1 0 0 0 1 01 0 0 0 0 0 1

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

, CB73B7

1=

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

1 0 0 0 0 0 0−6 1 0 0 0 0 09 −4 1 0 0 0 0−2 2 −2 1 0 0 00 0 −1 0 1 0 00 −1 0 0 0 1 0−1 0 0 0 0 0 1

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

,

CB81B8

3=

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

1 0 0 0 0 0 0 07 1 0 0 0 0 0 021 5 1 0 0 0 0 035 10 3 1 0 0 0 035 10 3 1 1 0 0 021 5 1 0 0 1 0 07 1 0 0 0 0 1 01 0 0 0 0 0 0 1

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

,

CB83B8

1=

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

1 0 0 0 0 0 0 0−7 1 0 0 0 0 0 014 −5 1 0 0 0 0 0−7 5 −3 1 0 0 0 00 0 0 −1 1 0 0 00 0 −1 0 0 1 0 00 −1 0 0 0 0 1 0−1 0 0 0 0 0 0 1

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

.

In general, for any integer n 1, we have

(CBn1 Bn

3 )k,j =(

n−2j+1k−j

), 1 j k n

0, (n < 2j − 1 and k = j) or k < j n.

Although it is easy to come up with a formula for the entries of CBn1 Bn

3 by simple in-spection, a handy formula for the entries of CBnBn is not that immediate. However, the

3 1

J. Agapito / Linear Algebra and its Applications 451 (2014) 260–289 281

transition matrix CBn3 Bn

1 is closely related to a certain product matrix. Note in this re-gard that the first nc columns of the transition matrix CBn

1 Bn3 , and the odd columns of

the Pascal triangle P[n]; its entries given by the formula P[n]k,j =(n−jk−j

), look alike. One

factor matrix of the referred product matrix is the inverse P[n]−1 of P[n], whose entriesare given by P[n]−1

k,j = (−1)k−j(n−jk−j

). We have, for instance,

P[7]−1 =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

1 0 0 0 0 0 0−6 1 0 0 0 0 015 −5 1 0 0 0 0−20 10 −4 1 0 0 015 −10 6 −3 1 0 0−6 5 −4 3 −2 1 01 −1 1 −1 1 −1 1

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

,

P[8]−1 =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

1 0 0 0 0 0 0 0−7 1 0 0 0 0 0 021 −6 1 0 0 0 0 0−35 15 −5 1 0 0 0 035 −20 10 −4 1 0 0 0−21 15 −10 6 −3 1 0 07 −6 5 −4 3 −2 1 0−1 1 −1 1 −1 1 −1 1

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

.

On the other hand, let c(x) = 1−√

1−4x2x be the generating function of the classical

Catalan numbers Ck = 1k+1

(2kk

). There is in the literature on Riordan arrays, an infinite

lower triangular matrix C generated by (1, xc(x)), whose nonzero entries are given byCk,j = [xk](xc(x))j = [xk−j ](c(x))j ; the first column and row being indexed by 0. (Seefor instance Barry [1, where C is denoted by Cat].) The first entries of C are shown next,

C =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

1 0 0 0 0 0 0 · · ·0 1 0 0 0 0 0 · · ·0 1 1 0 0 0 0 · · ·0 2 2 1 0 0 0 · · ·0 5 5 3 1 0 0 · · ·0 14 14 9 4 1 0 · · ·0 42 42 28 14 5 1 · · ·...

......

......

......

. . .

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

We index the columns and rows of C, for convenience, from 1 to infinity. The numbersappearing on the submatrix of C formed by erasing the first column and row, are knownas ballot numbers (the two leftmost columns of that submatrix corresponding to thesequences of Catalan numbers Ck and Ck+1, for k 1). Now, define C[n] to be the

282 J. Agapito / Linear Algebra and its Applications 451 (2014) 260–289

submatrix formed by the first n columns and rows of C. The entries of C[n] are explicitlygiven by

C[n]k,j =

⎧⎪⎨⎪⎩

j−1k−1

(2k−j−2k−2

), 2 j k n,

0, 1 k < j n,

δk,1, j = 1, 1 k n.

Thus, for instance, we have

C[7] =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

1 0 0 0 0 0 00 1 0 0 0 0 00 1 1 0 0 0 00 2 2 1 0 0 00 5 5 3 1 0 00 14 14 9 4 1 00 42 42 28 15 5 1

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

,

C[8] =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

1 0 0 0 0 0 0 00 1 0 0 0 0 0 00 1 1 0 0 0 0 00 2 2 1 0 0 0 00 5 5 3 1 0 0 00 14 14 9 4 1 0 00 42 42 28 14 5 1 00 132 132 90 48 20 6 1

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

.

The product matrix that we previously referred to is C[n]P[n]−1. Both the transitionmatrix CBn

3 Bn1 and the product matrix C[n]P[n]−1 have two important similarities. To

illustrate these features, the cases n = 7 and n = 8 are shown below.

CB73B7

1=

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

1 0 0 0 0 0 0−6 1 0 0 0 0 09 −4 1 0 0 0 0−2 2 −2 1 0 0 00 0 −1 0 1 0 00 −1 0 0 0 1 0−1 0 0 0 0 0 1

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

,

C[7]P[7]−1 =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

1 0 0 0 0 0 0−6 1 0 0 0 0 09 −4 1 0 0 0 0−2 2 −2 1 0 0 00 0 −1 0 1 0 00 −1 −2 0 2 1 0

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

,

−1 −4 −5 0 5 4 1

J. Agapito / Linear Algebra and its Applications 451 (2014) 260–289 283

CB83B8

1=

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

1 0 0 0 0 0 0 0−7 1 0 0 0 0 0 014 −5 1 0 0 0 0 0−7 5 −3 1 0 0 0 00 0 0 −1 1 0 0 00 0 −1 0 0 1 0 00 −1 0 0 0 0 1 0−1 0 0 0 0 0 0 1

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

,

C[8]P[8]−1 =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

1 0 0 0 0 0 0 0−7 1 0 0 0 0 0 014 −5 1 0 0 0 0 0−7 5 −3 1 0 0 0 00 0 0 −1 1 0 0 00 0 −1 −1 1 1 0 00 −1 −3 −2 2 3 1 0−1 −5 −9 −5 5 9 5 1

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

.

By comparison, we observe that the submatrix formed by the first nc columns and rowsof CBn

3 Bn1 and C[n]P[n]−1, for n = 7 and n = 8, coincide (such submatrices are shown

in red2 above). We also notice that there is a signed mirror symmetry in the lower partof CBn

3 Bn1 and C[n]P[n]−1 (look at the entries in green, for the rows indexed from nc + 1

to n above).

Theorem 4.11. The γ-vector of any symmetric polynomial h(z) of type (41) is given by

γ(h) =(γ1(h), . . . , γnc(h), 0, . . . , 0

)=

(C[n]P[n]−1(a1, . . . , an)T

)T. (44)

Proof. A computational verification shows that the similarities observed between thematrices CBn

3 Bn1 and C[n]P[n]−1, for n = 7 and n = 8, hold in general for any integer

n 1. Since we are interested on symmetric polynomials, we readily obtain the identity(44).

It is easy to obtain a handy formula for the entries of C[n]P[n]−1. It is given by

(C[n]P[n]−1)

k,j=

⎧⎪⎨⎪⎩

∑ki=2(−1)i−j i−1

k−1(2k−i−2

k−2)(

n−ji−j

), 1 j k n, k = 1,

0, 1 k < j n,

δ1,j , 1 j n.

(45)

Corollary 4.12. The γ-numbers of any symmetric polynomial h(z) of type (41) are givenby

2 For interpretation of the references to colors in this paper, the reader is referred to the web version ofthis article.

284 J. Agapito / Linear Algebra and its Applications 451 (2014) 260–289

γ1(h) = a1 and γk(h) =k∑

j=1

[k∑

i=2(−1)i−j i− 1

k − 1

(2k − i− 2

k − 2

)(n− j

i− j

)]aj

for 2 k nc.

Proof. This characterization for the γ-numbers of h(z) is a direct consequence of usingformula (45) in Theorem 4.11.

Note, in particular, that γ2(h) = −(n− 1)a1 + a2. In addition, if n is odd, we have

γnc(h) =nc−1∑j=1

(−1)nc−j2aj + anc ,

and, if n is even, we have

γnc(h) =nc−1∑j=1

(−1)nc−j

(2(nc − j

)+ 1

)aj + anc .

Now, let us set N cr (n) = Nr(n)+1

2 , and consider the polynomial Ar,s,Nr(n)(z) and theγ-basis Br,n

3 given by

Br,n3 =

zk(1 + z)Nr(n)+1−2kNc

r (n)k=1 ∪

zk

Nr(n)k=Nc

r (n)+1.

Theorem 4.13. The polynomial A(r,s),Nr(n)(z) belongs to the cone span+zk(1 +z)Nr(n)+1−2kN

cr (n)

k=1 and its γ-vector is given by

γ(Ar,s,Nr(n)(z)

)=

(C[Nr(n)

]P[Nr(n)

]−1P[Nr(n)

]−1(br,s,n1 , . . . , br,s,nNr(n)

)T )T,

where br,s,nk = 1Kr,s,n

(n(s− r) + k + r − 1)!Sr,s(n, k + r − 1) for 1 k Nr(n).

Proof. Theorem 4.13 is a direct consequence of Lemma 4.10, and the use of formula (36)in Theorem 4.11, noting that P[Nr(n)] = CBr,n

2 Br,n1

. 5. On the rational function (ZrDs)n(az+b

cz+d)

The rational function (ZrDs)n( 11−z ) is a particular case of the more general rational

function (ZrDs)n(az+bcz+d ), where a, b, c and d are any complex numbers, with c, d = 0. We

give in this section a brief discussion about a family of polynomials Ac,dr,s,Nr(n)(z) that

extend the polynomials Ar,s,Nr(n)(z) regarded in Section 3. In general, these polynomi-als are no longer symmetric nor unimodal and their coefficients may not be integers.However, they have similar expansions as those considered in the previous section.

J. Agapito / Linear Algebra and its Applications 451 (2014) 260–289 285

By way of example, let Δ = ad− bc. For any integer m 1 and c = 0, we have

Dm

(az + b

cz + d

)= Δ(−c)m−1m!

(cz + d)m+1 . (46)

On the one hand, using the normal ordering of (ZrDs)n, we obtain

(ZrDs

)n(az + b

cz + d

)=

rn∑k=r

Sr,s(n, k)(ZkDn(s−r)+k

)(az + b

cz + d

)

=rn∑k=r

Sr,s(n, k)Δ(−c)n(s−r)+k−1(n(s− r) + k)!(cz + d)n(s−r)+k+1 zk. (47)

On the other hand, computational evidence similar to the one that gave rise to thepolynomials Ar,s,Nr(n)(z) render the following description.

Definition 5.1. For each integer n 1 and c = 0, the expression (ZrDs)n(az+bcz+d ) is given

by

(ZrDs

)n(az + b

cz + d

)=

Kr,s,nΔ(−c)n(s−r)+r−1zr−1Ac,dr,s,Nr(n)(z)

(cz + d)ns+1 , (48)

where Kr,s,n is the same positive integer of Definition 3.4, and Ac,dr,s,Nr(n)(z) is a polyno-

mial in z with no constant term and of degree Nr(n) = r(n− 1) + 1, given by

Ac,dr,s,Nr(n)(z) =

Nr(n)∑k=1

dN−k(−c)k−1Ar,s(n, k)zk. (49)

The numbers Ar,s(n, k) can be computed by formula (37).

Theorem 5.2. The polynomial Ac,dr,s,Nr(n)(z) satisfies the following formula

Ac,dr,s,Nr(n)(z) = 1

Kr,s,n

Nr(n)∑k=1

(n(s− r) + k + r − 1

)!(−c)k−1Sr,s(n, k + r − 1)

× zk(cz + d)Nr(n)−k. (50)

Proof. Formula (50) follows from equating (47) with (48). 6. Final remarks

The most widely studied polynomials among the polynomials Ar,s,Nr(n)(z) are un-doubtedly the Eulerian and Narayana polynomials (see for instance [7,12,13,19,23,33]

286 J. Agapito / Linear Algebra and its Applications 451 (2014) 260–289

and references therein). Several authors have proved the common properties they share,separately or together, by using geometric, algebraic, combinatorial, and analytic tech-niques. This paper provides a unified and simple explanation of these properties, byobserving that the Ar,s,Nr(n)(z) are f -Eulerian polynomials, for a suitable polynomial f .Of course, this observation is redundant for the Eulerian polynomials, but it is not inall the other cases, including the Narayana polynomials. The γ-vectors of the Eulerianpolynomials have been studied by Foata and Schützenberger [13, Théorème 5.6], andalso by Shapiro [29], in connection with permutations of a given set with the samenumber of runs and slides. Similarly, the γ-vectors of the Narayana polynomials havebeen studied by Simion and Ullman [30], in connection with enumerative aspects ofthe lattice of noncrossing partitions, and also by Coker [8] in the context of latticepaths. Several nonnegative γ-vectors have been studied related to a large class of gen-eralized permutahedra which are flag simple polytopes; with Coxeter complexes, andmore generally, with flag homology spheres (see for instance [14,20,24–26,32]). Since theγ-vector of Ar,s,Nr(n)(z) is integral and nonnegative, the question is, of course: whatdo they count in the cases which are not well-known? Actually, we can ask a similarquestion for the polynomials Ar,s,Nr(n)(z) themselves. It is known, for instance, thatthe Eulerian polynomials A1,1,n(z)

z are the h-polynomials of the usual permutahedra,and the Narayana polynomials A1,2,n(z)

z are the h-polynomials of the usual Stasheff as-sociahedra. We have also pointed out that, for the cases s = r + 1, with r 1, thepolynomials Ar,r+1,Nr(n)(z)

z are the higher dimensional Narayana polynomials studiedby Sulanke [34]. A satisfactory significance for all the other cases seems to be un-known.

Acknowledgements

The author was partially supported by the Portuguese Science and Technology Foun-dation (FCT) through the program Ciência 2008.

Appendix A

Tables A.1–A.7 below show some of the numbers Ar,s(n, k) for the reader’s conve-nience.

Table A.1A1,4(n, k).

n\k 0 1 2 3 4 5 60 11 0 12 0 1 13 0 7 19 74 0 1 5 5 15 0 91 707 1311 707 916 0 52 570 1655 1655 570 52

J. Agapito / Linear Algebra and its Applications 451 (2014) 260–289 287

Table A.2A1,5(n, k).

n\k 0 1 2 3 4 5 60 11 0 12 0 1 13 0 3 8 34 0 13 63 63 135 0 663 4938 9038 4938 6636 0 17 177 502 502 177 17

Table A.3A1,6(n, k).

n\k 0 1 2 3 4 5 60 11 0 12 0 1 13 0 11 29 114 0 4 19 19 45 0 44 319 579 319 446 0 13 131 366 366 131 13

Table A.4A1,7(n, k).

n\k 0 1 2 3 4 5 60 11 0 12 0 1 13 0 13 34 134 0 19 89 89 195 0 1235 8788 15 858 8788 12356 0 155 1527 4222 4222 1527 155

Table A.5A2,4(n, k).

n\k 0 1 2 3 4 5 6 7 8 9 10 110 11 0 12 0 3 8 33 0 1 8 15 8 14 0 1 16 70 112 70 16 15 0 3 80 630 2016 2940 2016 630 80 36 0 1 40 495 2640 6930 9504 6930 2640 495 40 1

Table A.6A2,5(n, k).

n\k 0 1 2 3 4 5 6 7 8 9 10 110 11 0 12 0 2 5 23 0 28 200 363 200 284 0 77 1056 4263 6650 4263 1056 775 0 1001 22 088 154 739 464 126 662 927 464 126 154 739 22 088 10016 0 1768 56 729 605 607 2 933 823 7 292 331 9 825 279 7 292 331 2 933 823 605 607 56 729 1768

288 J. Agapito / Linear Algebra and its Applications 451 (2014) 260–289

Table A.7A3,4(n, k).

n\k 0 1 2 3 4 5 6 7 8 9 10 11 12 130 11 0 12 0 1 6 6 13 0 1 22 113 190 113 22 14 0 1 53 710 3548 7700 7700 3548 710 53 15 0 1 105 2856 30 422 151 389 385 029 523 200 385 029 151 389 30 422 2856 105 1

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