Threefold symmetric Hahn-classical multiple orthogonal polynomials
aa r X i v : . [ m a t h . C A ] J u l July 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche
Analysis and Applicationsc (cid:13)
World Scientific Publishing Company
THREEFOLD SYMMETRIC HAHN-CLASSICAL MULTIPLEORTHOGONAL POLYNOMIALS
ANA F. LOUREIRO
School of Mathematics, Statistics and Actuarial Science (SMSAS), Sibson Building, Universityof Kent, Canterbury, Kent CT2 7FS (U.K.)[email protected] (corresponding author)
WALTER VAN ASSCHE
Department of Mathematics, KU Leuven, Celestijnenlaan 200B box 2400, BE-3001 Leuven,[email protected]
We characterize all the multiple orthogonal threefold symmetric polynomial sequenceswhose sequence of derivatives is also multiple orthogonal. Such a property is commonlycalled the Hahn property and it is an extension of the concept of classical polynomialsto the context of multiple orthogonality. The emphasis is on the polynomials whoseindices lie on the step line, also known as 2-orthogonal polynomials. We explain therelation of the asymptotic behavior of the recurrence coefficients to that of the largestzero (in absolute value) of the polynomial set. We provide a full characterization ofthe Hahn-classical orthogonality measures supported on a 3-star in the complex planecontaining all the zeros of the polynomials. There are essentially three distinct families,one of them 2-orthogonal with respect to two confluent functions of the second kind.This paper complements earlier research of Douak and Maroni.
Keywords : orthogonal polynomials, multiple orthogonal polynomials, confluent hyperge-ometric function, Airy function, Hahn classical polynomials, recurrence relation, lineardifferential equationMathematics Subject Classification 2000: 33C45; 42C05
1. Introduction and motivation
In this paper we investigate and characterize all the multiple orthogonal polynomialsof type II that are threefold symmetric and are such that the polynomial sequenceof its derivatives is also a multiple orthogonal sequence of type II. Within the stan-dard orthogonality context there are only four families satisfying this property,commonly referred to as the
Hahn property , and they are the Hermite, Laguerre,Jacobi and Bessel, collectively known as the classical orthogonal polynomials . Thesefour families of polynomials also share a number of analytic and algebraic proper-ties. Several studies are dedicated to extensions of those properties to the context ofmultiple orthogonality. However, those extensions give rise to completely differentsequences of multiple orthogonal polynomials. For the usual orthogonal polynomials uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche A. F. Loureiro and W. Van Assche the Hahn property is equivalent with the Bochner characterization (polynomials sat-isfy a second order differential equation of Sturm-Liouville type) and the existenceof a Rodrigues formula. This is no longer true for multiple orthogonal polynomialssince there are families of multiple orthogonal polynomials with a Rodrigues typeformula [1] that do not satisfy the Hahn property, and various examples of multipleorthogonal polynomials satisfy a higher-order linear recurrence relation, which isnot of Sturm-Liouville type [11]. Hence characterizations for multiple orthogonalpolynomials based on the Hahn property, the Bochner property or the Rodriguesformula give different families of polynomials.A sequence of monic polynomials { P n } n > with deg P n = n is orthogonal withrespect to a Borel measure µ whenever Z P n ( x ) x k d µ ( x ) = 0 if k = 0 , , . . . , n − , n ≥ Z P n ( x ) x n d µ ( x ) = 0 f or n > . Without loss of generality, often we normalize µ so that it is a probability measure.Obviously, an orthogonal polynomial sequence forms a basis of the vector spaceof polynomials P . The measures described above can be represented via a linearfunctional L , defined on P ′ , the dual space of P , and it is understood that the actionof L over a polynomial f corresponds to R f ( x )d µ ( x ). Throughout, we denote thisaction as hL , f ( x ) i := Z f ( x )d µ ( x ) . The derivative of a function f is denoted by f ′ or dd x f ( x ). Properties on P ′ , suchas differentiation or multiplication by a polynomial, can be defined by duality. Adetailed explanation can be found in [29] [31]. In particular, given g ∈ P and afunctional L ∈ P ′ , we define h g ( x ) L , f ( x ) i := hL , g ( x ) f ( x ) i and hL ′ , f ( x ) i := −hL , f ′ ( x ) i (1.1)for any polynomial f .From the definition it is straightforward that an orthogonal polynomial sequencesatisfies a second order recurrence relation P n +1 ( x ) = ( x − β n ) P n ( x ) − γ n P n − ( x ) , n > , (1.2)with P ( x ) = 1, P ( x ) = x − β and γ n = 0 for all integers n >
1. This relation isoften called the three-term recurrence relation, but we will avoid this terminologyas we will be dealing with three-term recurrence relations which are of higher order.There is an important converse of this connection between orthogonal polynomialsand second order recurrence relations, known as the Shohat-Favard theorem orspectral theorem for orthogonal polynomials. It states that any sequence of monicpolynomials { P n } n > with deg P n = n , satisfying the recurrence relation (1.2) withuly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche Threefold symmetric Hahn-classical multiple orthogonal polynomials γ n = 0 for all n > P − ( x ) = 0 and P ( x ) = 1, is always anorthogonal polynomial sequence with respect to some measure µ and, if, in addition, β n ∈ R and γ n +1 > n >
0, then µ is a positive measure on the real line.So, basically, the orthogonality conditions and the second order recurrence relationsare two equivalent ways to characterize an orthogonal polynomial sequence.Multiple orthogonal polynomials of type II correspond to a sequence of polyno-mials of a single variable that satisfy multiple orthogonality conditions with respectto r > ~n = ( n , . . . , n r ) ∈ N r , type II multipleorthogonal polynomials correspond to a (multi-index) sequence of monic polyno-mials P ~n ( x ) of degree | ~n | = n + . . . + n r for which there is a vector of measures( µ , . . . , µ r − ) such that Z x k P ~n ( x )d µ j ( x ) = 0 , k n j − j = 0 , , . . . , r −
1. By setting r = 1, we recover the usual orthogonalpolynomials. Obviously, (1.3) amounts to the same as saying there exists a vectorof r linear functionals ( L , . . . , L r − ) such that hL j , x k P ~n ( x ) i = 0 , k n j − . This polynomial P ~n set may not exist or is not unique unless we impose someextra conditions on the r measures µ , . . . , µ r − , but under appropriate conditionsthe polynomial set satisfies a system of r recurrence relations, relating P ~n withits nearest neighbors P ~n + ~e k and P ~n − ~e j , where ~e k consists of the r -dimensional unitvector with zero entries except for the k th component which is 1 (see [37] for furtherdetails). The focus of the present work is on the polynomials whose multi-indiceslie on the step-line near the diagonal ~n = ( n, n, . . . , n ), and are defined by P rn + j ( x ) = P ~n + ~s j ( x ) , with ~s j = j X ℓ =0 ~e ℓ , j = 0 , , . . . , r − , that is, P rn + j ( x ) = P ( n +1 ,...,n +1 | {z } j ,n,...,n | {z } r − j ) ( x ) , j = 0 , , . . . , r − . This step-line polynomial sequence { P n } n > satisfies a ( r + 1)th order recurrencerelation (involving up to ( r + 2) consecutive terms) and it can be written as xP n ( x ) = P n +1 ( x ) + r X j =0 ζ n,j P n − j ( x ) , n > r,P j ( x ) = x j , j = 0 , . . . , r, (1.4)where ζ n,r = 0 for all n > r . Such a step-line polynomial sequence { P n } n > cor-responds to so-called d -orthogonal polynomials (with d = r ), see [28,15,32] anduly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche A. F. Loureiro and W. Van Assche n n P n +1 ,n ( x )= P n +1 ( x ) P n ,n ( x ) = P n ( x ) Fig. 1. Step-line for the multi-index ( n , n ) when r = 2. Figure 1 for the case of r = 2. So, basically, if there exists a vector of linear func-tionals ( L , . . . , L d − ) and a polynomial sequence { P n } n > satisfying hL j , x k P n ( x ) i = 0 and hL j , x n P dn + j ( x ) i 6 = 0 , k (cid:22) n − jd (cid:23) , n > , then the polynomials { P n } n > are related by (1.4). There is a converse result,which is a natural generalization of the Shohat-Favard theorem, in the sense that ifa polynomial sequence { P n } n > satisfies (1.4) with ζ n,d = 0 for all n > d , then thereis a vector of r linear functionals ( L , . . . , L d − ) with respect to which { P n } n > is d -orthogonal. Such a vector of linear functionals is not unique, but we can considerits components to be the first d elements of the dual sequence { u n } n > associatedto { P n } n > , which always exist and which are defined by h u n , P m i := δ n,m , m, n > , where δ n,m represents the Kronecker symbol . The remaining elements of the dualsequence of a d -orthogonal polynomial sequence can be generated from the firstones: for each pair of integers ( n, j ) there exist polynomials q n,j,ν ( x ) such that [28] u dn + j = d − X ν =0 q n,j,ν ( x ) u ν , for 0 j d − , where deg q n,j,j ( x ) = n , deg q n,j,ν ( x ) n for 0 ν j − j d − q n,j,ν ( x ) n − j + 1 ν d − j d −
2. Unlike the standardorthogonality (case d = 1), this result only provides structural properties for thedual sequence and in practical terms it can be used in that sense.In this paper we mainly deal with 2-orthogonal polynomial sequences, but ourresults and ideas can be extended to the d -orthogonal case. Here, we provide acharacterization of all the 2-orthogonal polynomial sequences that are threefoldsymmetric and possess the Hahn property in the context of 2-orthogonality, i.e., theuly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche
Threefold symmetric Hahn-classical multiple orthogonal polynomials sequence of monic derivatives is also 2-orthogonal. The notion and properties of threefold symmetric Hahn property within the con-text of 2-orthogonality. The orthogonality (1.3) of these polynomials correspondsto an non-hermitian inner product (no conjugation is used) on three-starlike sets.The Airy function, the confluent hypergeometric function and the hypergeometricfunction are analytic functions, so the paths of integration can be deformed awayfrom the three-starlike sets without loosing orthogonality. However, the choice ofthe three-starlike sets is convenient because of the positivity of the measures andthe location of the zeros. It turns out that the orthogonality measures under analy-sis are solutions to a second order differential equation and, as such, the recurrencecoefficients for the whole set of the corresponding multiple orthogonal polynomialsof type II can only be obtained algorithmically via the nearest-neighbor algorithmexplained in [37] [19]. Some of the d -orthogonal polynomials that we found appearin the theory of random matrices, in particular in the investigation of singular val-ues of products of Ginibre matrices, which uses multiple orthogonal polynomialswith weight functions expressed in terms of Meijer G-functions [24]. For r = 2 theseMeijer G-functions are hypergeometric or confluent hypergeometric functions.uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche A. F. Loureiro and W. Van Assche
2. Threefold 2-orthogonal polynomial sequence
A sequence of monic polynomials { P n } n > (with deg P n = n ) is symmetric whenever P n ( − x ) = ( − n P n ( x ) for all n >
0. This means that all even degree polynomialsare even functions while odd degree polynomials are odd functions. Hermite andGegenbauer polynomials are examples of symmetric polynomial sets, which alsohappen to be the only classical orthogonal polynomials that are symmetric. Manyother examples of symmetric orthogonal polynomial sequences are around in theliterature.The notion of symmetry of polynomial sequences has been extended and com-monly referred to as d -symmetric in works by Maroni [28,15] and followed by BenCheikh and his collaborators [6,8,9,25]. The case d = 1 would correspond to theusual symmetric case. We believe the name is misleading, as will soon become ap-parent, and therefore we call it differently as it pictures better the nature of theproblem. Definition 2.1.
A polynomial sequence { P n } n > is m-fold symmetric if P n ( ωx ) = ω n P n ( x ) , for any n > . (2.1)where ω = e iπm . By induction, property (2.1) corresponds to P n ( ω k x ) = ω nk P n ( x ) , for any n > k = 1 , , . . . , m − . So, an m -fold symmetric sequence is a ( m − twofold symmetric polynomials.) In particular, a threefold symmetric se-quence { P n } n > , is such that P n ( ωx ) = ω n P n ( x ) and P n ( ω x ) = ω n P n ( x ) with ω = e iπ , n > , which corresponds to say that there exist three sequences { P [ j ] n } n > with j ∈{ , , } such that P n + j ( x ) = x j P [ j ] n ( x ) , j = 0 , , . (2.2)Throughout, we will refer to { P [ j ] n } n > as the diagonal components of the cubicdecomposition of the threefold symmetric sequence { P n } n > , which is line with theterminology adopted in a more general cubic decomposition framework in [33].In the case of a 2-orthogonal polynomial sequence { P n } n > with respect to avector linear functional u = ( u , u ), we have, as discussed in Section 1, h u , x m P n i = (cid:26) n > m + 1 N ( n ) = 0 for n = 2 m (2.3) h u , x m P n i = (cid:26) n > m + 2 N ( n ) = 0 for n = 2 m + 1 , (2.4)uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche Threefold symmetric Hahn-classical multiple orthogonal polynomials and there exists a set of coefficients { ( β n , α n , γ n ) } n > such that { P n } n > satisfiesa third order recurrence relation (see [28,39]) P n +1 ( x ) = ( x − β n ) P n ( x ) − α n P n − ( x ) − γ n − P n − ( x ) (2.5)with P − ( x ) = P − ( x ) = 0 and P ( x ) = 1. Straightforwardly from the definition,one has γ n +1 = h u , x n +1 P n +2 ih u , x n P n i 6 = 0 , γ n +2 = h u , x n +1 P n +3 ih u , x n P n +1 i 6 = 0 , n > , (2.6)or, equivalently, h u , x n +1 P n +2 i = n Y k =0 γ k +1 and h u , x n +1 P n +3 i = n Y k =0 γ k +2 , for n > . Whenever a 2-orthogonal polynomial sequence is threefold symmetric, the recur-rence relation (2.5) reduces to a three-term relation, where the β - and α -coefficientsall vanish. For this type of 2-orthogonal polynomial sequences more can be said. Proposition 2.1. [14] Let { P n } n > be a 2-orthogonal polynomial sequence withrespect to the linear functional U = ( u , u ) satisfying (2.3) - (2.4) . The followingstatements are equivalent:(a) The sequence { P n } n > is threefold symmetric.(b) The linear functional is threefold symmetric, that is, ( u ν ) n + j = 0 , for ν = 0 , and j = 1 , with j = ν, (2.7) where ( u ) m := h u, x m i for m > .(c) The sequence { P n } n > satisfies the third order recurrence relation P n +1 ( x ) = xP n ( x ) − γ n − P n − ( x ) , n > , (2.8) with P ( x ) = 1 , P ( x ) = x and P ( x ) = x . Each of the components of the cubic decomposition are also 2-orthogonal poly-nomial sequences.
Lemma 2.1. [28] Let { P n } n > be a threefold symmetric -OPS. The three polyno-mial sequences { P [ j ] n } n > (with j = 0 , , ) in the cubic decomposition of { P n } n > described in (2.2) are 2-orthogonal polynomial sequences satisfying: P [ j ] n +1 ( x ) = ( x − β [ j ] n ) P [ j ] n ( x ) − α [ j ] n P [ j ] n − ( x ) − γ [ j ] n − P [ j ] n − ( x ) , (2.9) where β [ j ] n = γ n − j + γ n + j + γ n +1+ j , n > ,α [ j ] n = γ n − j γ n + j + γ n − j γ n − j + γ n − j γ n − j , n > ,γ [ j ] n = γ n − j γ n + j γ n +2+ j = 0 , n > . uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche A. F. Loureiro and W. Van Assche
Moreover, { P [ j ] n } n > is 2-orthogonal with respect to the vector functional U [ j ] =( u [ j ]0 , u [ j ]1 ) with u [ j ] ν = σ ( x j u ν + j ) for each j = 0 , , and ν = 0 , . where { u n } n > represents the dual sequence of { P n } n > and σ : P ′ −→ P ′ is theoperator defined by h σ ( v ) , f ( x ) i := h v, f ( x ) i for any v ∈ P ′ and f ∈ P . The orthogonality measures are supported on a starlike set with three rays.
Theorem 2.1. [3] If γ n > for n > in (2.8) , then there exists a vector of linearfunctionals ( u , u ) such that the polynomials P n defined by (2.8) satisfy the 2-orthogonal relations (2.3) - (2.4) . Moreover, the vector of linear functionals ( u , u ) satisfies (2.7) and there exist a vector of two measures ( µ , µ ) such that h u , f ( x ) i = Z S f ( x ) d µ ( x ) h u , f ( x ) i = Z S f ( x ) d µ ( x ) where S represents the starlike set S := [ k =0 Γ k with Γ k = [0 , e πik/ ∞ ) , and the measures have a common support which is a subset of S and are invariantunder rotations of π/ . Γ Γ Γ Fig. 2. The three rays Γ , Γ and Γ . Regarding the behavior of the zeros of any threefold symmetric 2-orthogonalpolynomial sequence { P n } n > , it was shown in [9] that between two nonzero con-secutive roots of P n +2 there is exactly one root of P n and one root of P n +1 and allthose roots lie on the starlike set S . Proposition 2.2.
Let { P n } n > be a 2-OPS satisfying (2.8) . If γ n > , then thefollowing statements hold: uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche Threefold symmetric Hahn-classical multiple orthogonal polynomials (a) If x is a zero of P n + j , then ω j x are also zeros of P n + j with ω = e πi/ .(b) is a zero of P n + j of multiplicity j when j = 1 , .(c) P n + j has n distinct positive real zeros { x ( j ) n,k } nk =1 with < x ( j ) n, < . . . < x ( j ) n,n .(d) Between two real zeros of P n + j +3 there exist only one zero of P n + j +2 andonly one zero of P n + j +1 , that is, x ( j +2) n,k < x ( j ) n,k +1 < x ( j +1) n,k +1 < x ( j +2) n,k +1 . Proof.
The result is a consequence of [9, Theorem 2.2] for the case d = 2.So, P n + j with j ∈ { , , } has n zeros ( x ( j ) n, , . . . , x ( j ) n,n ) on the positive real lineand all the other zeros are obtained by rotations of 2 π/ x ( j ) n, , . . . , x ( j ) n,n ) and0 is a single zero for P n +1 and a double zero for P n +2 . The connection betweenthe asymptotic behavior of the γ -recurrence coefficients in (2.8) and the upperbound for the largest zero x ( j ) n,n is discussed in [2], but for bounded recurrencecoefficients. Here, we extend that discussion, by embracing the cases where therecurrence coefficients are unbounded with different asymptotic behavior for evenand odd order indices, which will be instrumental in Section 3. Theorem 2.2. If γ n in (2.8) are positive and, additionally, γ n = c n α + o ( n α ) and γ n +1 = c n α + o ( n α ) for large n , with min { c , c } > , c = max { c , c } > and α > , then largest zero in absolute value | x n,n | behaves as | x n,n | / c / n α/ + o ( n α/ ) , n > . (2.10) Proof.
Consider the Hessenberg matrix H n = · · · · · · γ · · · γ · · · · · · γ n − · · · γ n − · · · γ n − so that the recurrence relation (2.8) can be expressed as H n P ( x ) P ( x )... P n − ( x ) = x P ( x ) P ( x )... P n − ( x ) − P n ( x ) and each zero of P n is an eigenvalue of the matrix H n . The spectral radius of thematrix H n , ρ ( H n ) = max {| λ | : λ is an eigenvalue of H n } , uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche A. F. Loureiro and W. Van Assche is bounded from above by || H n || where || · || denotes a matrix norm (see [23, Section5.6]). We take the matrix norm || H n || S = || S − H n S || ∞ = max i n n X j =1 (cid:12)(cid:12) ( S − H n S ) i,j (cid:12)(cid:12) , where S corresponds to a non-singular matrix and ( S − H n S ) i,j denotes the i throw and j th column entry of the product matrix S − H n S . In particular if S is aninvertible diagonal matrix S = diag( d , . . . , d k , . . . , d n ), then || H n || S = max (cid:26) d d , d d , d + d γ d , . . . , d k + d k − γ k − d k − , . . . , d n + d n − γ n − d n − , d n − γ n − d n (cid:27) . Setting d k = d k ( k !) α/ = 0, for some positive constant d , gives || H n || S α/ (cid:16) d + cd (cid:17) n α/ + o ( n α/ ) as n → + ∞ . The choice of d = (2 c ) / gives a minimum to (cid:0) d + cd (cid:1) , so that || H n || S / ( c n α ) / + o ( n α/ ) as n → + ∞ , which implies the result.To summarize, Proposition 2.2 combined with Lemma 2.1 allow us to concludethat each P [ j ] n has exactly n real zeros { x [ j ] n,k } nk =1 with 0 < x [ j ] n, < . . . < x [ j ] n,n .Moreover, between two consecutive zeros of P [ j +2] n there is exactly one zero of P [ j ] n and another of P [ j +1] n : x [ j +2] n,k < x [ j ] n,k +1 < x [ j +1] n,k +1 < x [ j +2] n,k +1 . Taking into considerationthe asymptotic behavior for the largest zero | x n,n | of P n described in Theorem 2.2,we then conclude that x [ j ] n,k c n α + o ( n α ) . Therefore, for each j ∈ { , , } , all the zeros x ( j ) n,k of P n + j distinct from 0 corre-spond to the cubic roots of x [ j ] n,k >
0, with k ∈ { , . . . , n } , and they all lie on the star-like set S and within the disc centred at the origin with radius b = c n α + o ( n α ).Multiple orthogonal polynomials on starlike sets received attention in recentyears, under different frameworks. This includes the asymptotic behavior of poly-nomial sequences generated by recurrence relations of the type (2.8) when furtherassumptions are taken regarding specific behaviour for the γ -coefficient [2] or forcertain type of the 2-orthogonality measures in [12] and [27]. Faber polynomialsassociated with hypocycloidal domains and stars have also been studied in [21].Here, we describe all the threefold 2-orthogonal polynomial sequences that areclassical in Hahn’s sense, and our study includes the representations for the mea-sures supported on a set containing all the zeros of the polynomial sequence. Fromuly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche Threefold symmetric Hahn-classical multiple orthogonal polynomials Theorem 2.1 and Proposition 2.2, the support lies on a starlike set, that, accord-ing to Theorem 2.2 is bounded if the γ -coefficients are bounded, and unboundedotherwise.
3. Threefold symmetric Hahn-classical 2-orthogonal polynomialsequence
The classical orthogonal polynomial sequences of Hermite, Laguerre, Jacobi andBessel collectively satisfy the so-called
Hahn property : the sequence of its derivativesis again an orthogonal polynomial sequence. In the context of 2-orthogonality, thisalgebraic property is portrayed as follows:
Definition 3.1.
A monic 2-orthogonal polynomial sequence { P n } n > is ”2-Hahn-classical” when the sequence of its derivatives { Q n } n > , with Q n ( x ) := n +1 P ′ n +1 ( x )is also a 2-orthogonal polynomial sequence.The study of this type of 2-orthogonal sequences was initiated in the works byDouak and Maroni [14]-[17]. In those works (as well as in [32]) several properties ofthese polynomials were given, with the main pillars of the study being the structuralproperties, including the recurrence relations, satisfied by the polynomials. All in all,those studies encompassed the analysis of a nonlinear system of equations fulfilledby the recurrence coefficients. Douak and Maroni treated some special solutionsto that system of equations, bringing to light several examples of these threefoldsymmetric ”2-Hahn-classical” polynomials: see [13,16,17]. However, for those casesthe support of the corresponding orthogonality measures that they found consistedof the positive real axis, which does not contain all the zeros. Here, we base ouranalysis on the properties of the orthogonality measures and deduce the propertiesof the recurrence coefficients. We incorporate the works by Douak and Maroni andgo beyond that by fully describing all the threefold ”2-Hahn-classical” polynomialsand bringing up explicitly the orthogonality measures along with the asymptoticbehavior of the largest zero in absolute value as well as a Bochner type result forthe polynomials (i.e., characterizing these polynomials via a third order differentialequation).We start by observing that the threefold symmetry of { P n } n > readily impliesthe threefold symmetry of { Q n ( x ) := n +1 P ′ n +1 ( x ) } n > . This is a straightforwardconsequence of Definition 2.1, as it suffices to take single differentiation of relation(2.1). Such property is valid for any polynomial sequence, regardless any orthogo-nality properties.Regarding threefold symmetric 2-orthogonal polynomial sequences possessingHahn’s property, more can be said. The next result summarizes a characterizationof the orthogonality measures along side with the recurrence relations of the originalsequence { P n ( x ) } n > and the sequence of derivatives { Q n ( x ) := n +1 P ′ n +1 ( x ) } n > .This characterization can be found in the works [14,15,32]. Nonetheless, we revisitthese results for a matter of completion while bringing different approaches to theuly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche A. F. Loureiro and W. Van Assche original proofs, highlighting that all the structural properties for the polynomialscan be derived from the corresponding 2-orthogonality measures.
Theorem 3.1.
Let { P n } n > be a threefold symmetric 2-orthogonal polynomialsequence for ( u , u ) satisfying the recurrence relation (2.8) and let { Q n ( x ) := n +1 P ′ n +1 ( x ) } n > . The following statements are equivalent: (a) { Q n ( x ) } n > is a threefold symmetric 2-orthogonal polynomial sequence, satisfy-ing the third-order recurrence relation Q n +1 ( x ) = xQ n ( x ) − e γ n − Q n − ( x ) , (3.1) with initial conditions Q k ( x ) = x k for k ∈ { , , } . (b) The vector functional ( u , u ) satisfies the matrix differential equation (cid:18) Φ (cid:20) u u (cid:21)(cid:19) ′ + Ψ (cid:20) u u (cid:21) = (cid:20) (cid:21) , (3.2a) where Φ = ϑ (1 − ϑ ) x γ (1 − ϑ ) x ϑ − and Ψ = " γ x (3.2b) for some constants ϑ and ϑ such that ϑ , ϑ = n − n , for all n > . (3.2c)(c) there are coefficients ϑ , ϑ = n − n , such that U = ( u , u ) satisfies (cid:16) φ ( x ) u (cid:17) ′′ + (cid:18) γ ( ϑ + ϑ − x u (cid:19) ′ + 2 γ ( ϑ − xu = 0 (3.3) and ( ϑ −
2) (2 ϑ − u = φ ( x ) u ′ − γ ( ϑ −
1) (2 ϑ − x u , if ϑ = 2 , (3.4a) x u ′ = 2 u ′ , if ϑ = 2 , (3.4b) where φ ( x ) = ϑ (2 ϑ − − γ ( ϑ −
1) ( ϑ − x . (3.5)(d) There exists a sequence of numbers { e γ n +1 } n > such that P n +3 ( x ) = Q n +3 ( x ) + (cid:16) ( n + 1) γ n +2 − ( n + 3) e γ n +1 (cid:17) Q n ( x ) , (3.6) with initial conditions P ( x ) = Q ( x ) = 1 , P ( x ) = Q ( x ) = x and P ( x ) = Q ( x ) = x . uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche Threefold symmetric Hahn-classical multiple orthogonal polynomials Proof.
We prove that (a) ⇒ (d) ⇒ (b) ⇔ (c) ⇒ (a).In order to see (a) implies (d), we differentiate the recurrence relation satisfiedby P n to then replace the differentiated terms by its definition of Q n . A substitutionof the term xQ n ( x ) by the expression provided in (3.1) finally gives (3.6).Now we prove that (d) implies (b). Any linear functional w in P ′ can be writtenas w = ∞ X k =0 h w, P k i u k . Based on this, the relation Q n ( x ) := n +1 P ′ n +1 ( x ) gives a differential relation be-tween the dual sequence { v n } n > of { Q n } n > in terms of { u n } n > (the dual se-quence of { P n } n > ), which is v ′ n = − ( n + 1) u n +1 , n > , (3.7)whilst (3.6) leads to v n = u n + (( n + 1) γ n +2 − ( n + 3) e γ n +1 ) u n +3 , n > . (3.8)The choice of n = 0 and n = 1 in both (3.7) and (3.8) respectively gives (cid:20) v ′ v ′ (cid:21) = − (cid:20) u u (cid:21) and (cid:20) v v (cid:21) = (cid:20) u + ( γ − e γ ) u u + (2 γ − e γ ) u (cid:21) . (3.9)As explained in [32] (and, alternatively, in [14], [15] and [28]), the elements of thedual sequence { u n } n > of the 2-orthogonal polynomial sequence { P n } n > can bewritten as u n = E n ( x ) u + a n − ( x ) u ,u n +1 = b n ( x ) u + F n ( x ) u , where E n and F n are polynomials of degree n , while a n and b n are polynomialsof degree less than or equal to n under the assumption that a − ( x ) = 0, E = 1, F = 1 and b = 0. The recurrence relations (2.8) fulfilled by { P n } n > yield (see[32, Lemma 6.1]) γ n +2 F n +1 ( x ) − xF n ( x ) = − a n − ( x ) , γ n +3 a n +1 ( x ) − xa n ( x ) = − F n ( x ) ,γ n +2 b n +1 ( x ) − xb n ( x ) = − E n ( x ) and γ n +3 E n +2 ( x ) − xE n +1 ( x ) = − b n ( x ) , with initial conditions a ( x ) = 0 and E ( x ) = γ x . In particular, we obtain: b ( x ) = − γ , F ( x ) = 1 γ x, E ( x ) = 1 γ γ x , and a ( x ) = − γ , so that u = 1 γ x u , u = − γ u + 1 γ x u and u = 1 γ γ x u − γ u . (3.10)Consequently, the first identity in (3.9) reads as (cid:20) v ′ v ′ (cid:21) = − Ψ (cid:20) u u (cid:21) with Ψ = " γ x . (3.11a)uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche A. F. Loureiro and W. Van Assche
Using (3.10) in the second identity in (3.9) leads to (cid:20) v v (cid:21) = Φ (cid:20) u u (cid:21) with Φ = (cid:20) φ , φ , φ , φ , (cid:21) , (3.11b)where ( c.f. [32, Eq. (6.17)]) φ ( x ) = 3 e γ γ , φ ( x ) = (cid:18) − e γ γ (cid:19) x,φ ( x ) = 2 (cid:18) γ − e γ γ (cid:19) x and φ ( x ) = − e γ γ . We set ϑ = e γ γ = 0 and ϑ = e γ γ = 0, and now (3.2a)-(3.2b) follows afterdifferentiating both sides of the equation in (3.11b) and then compare with (3.11a),which is a system of two functional equations in ( u , u ): ( ϑ u ′ + (1 − ϑ ) x u ′ + (2 − ϑ ) u = 0 γ (1 − ϑ ) x u ′ + γ (3 − ϑ ) x u − (1 − ϑ ) u ′ = 0 . (3.12)The action of each of these equations over the monomials gives nϑ ( u ) n − + (( n − − nϑ ) ( u ) n = 0 , γ (( n − − nϑ ) ( u ) n +1 + (2 ϑ − n ( u ) n − = 0 , n > , under the assumption that ( u ) − = ( u ) − = 0. All the moments are well definedprovided that (( n − − nϑ ) (( n − − nϑ ) = 0 , n > , which corresponds to condition (3.2c). Under this assumption, the threefold sym-metry implies that ( u ) n = ( u ) n +1 = 0 for n , whilst for n ≡ u ) n ( u ) n +1 = 0, and these are recursively definedby2 γ ( n + 1 − ( n + 2) ϑ ) ( n − ( n + 1) ϑ ) ( u ) n +3 = ϑ (2 ϑ − n + 1)( n + 2)( u ) n , ( n + 1 − ( n + 2) ϑ ) ( n + 3 − ( n + 4) ϑ ) ( u ) n +4 = ϑ (2 ϑ − n + 2)( n + 4)( u ) n +1 , where ( u ) = ( u ) = 1.Concerning the proof of (b) ⇒ (c), we start by noting that the system of equa-tions (3.2a) can be written as in (3.12) which reads as C (cid:20) u ′ u (cid:21) = − D (cid:20) u ′ u (cid:21) , (3.13)with C = (cid:20) (1 − ϑ ) x − ϑ ϑ − (cid:21) and D = " ϑ γ (1 − ϑ ) x γ (3 − ϑ ) x . uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche Threefold symmetric Hahn-classical multiple orthogonal polynomials Observe that det C = − (2 − ϑ )(2 ϑ −
1) is constant. The matrix C is nonsingularif ϑ = 2. For this case, a left multiplication by its adjoint matrix C ∗ gives ( θ −
2) (2 θ − u ′ = − γ ( ϑ −
2) (1 − ϑ ) x u ′ + 2 γ ( ϑ −
2) (2 ϑ − xu , (3.14a)( θ −
2) (2 θ − u = φ ( x ) u ′ − γ ( ϑ −
1) (2 ϑ − x u , (3.14b)where φ ( x ) = det Φ which is given in (3.5). We differentiate (3.14b) once andcompare with (3.14a) to obtain (3.3) and (3.4a).When ϑ = 2, the functional equations (3.12) (or equivalently (3.13)) become ( u ′ = x u ′ , γ (1 − ϑ ) x u ′ + γ (3 − ϑ ) x u − (1 − ϑ ) u ′ = 0 , and this gives (3.3) and (3.4b).The implication (c) ⇒ (b) can be trivially obtained by performing reciprocaloperations to the ones just described.The proof of (b) ⇒ (a) consists of showing that { Q n } n > is 2-orthogonal for( v , v ) given by (3.11b) based on (3.2a)-(3.2b) together with the 2-orthogonality of { P n } n > with respect to ( u , u ) and on account of conditions (3.2c). We will omitthe details which can be followed in [32, Proposition 6.2, p.324], with the obviousadaptations to the threefold symmetric case.The vector functional U = ( u , u ) satisfies a matrix equation of Pearson type(3.2a). This somehow mimics the properties of the classical orthogonal polynomials.However, we refrain ourselves from calling such polynomials 2-orthogonal classical,since other characteristic properties of the classical orthogonal polynomials can beinterpreted in the context of 2-orthogonality (or multiple orthogonality) and giverise to completely different sets of polynomials.As a straightforward consequence of statement (c) in the latter result, we deriveequations for the moments of the Hahn-classical vector linear functional ( u , u ): Corollary 3.1.
Under the same assumptions of Theorem 3.1, the moments of u and u satisfy γ h (3 n + 2) (cid:16) (3 n + 1) ( ϑ −
1) ( ϑ − − ( ϑ + ϑ − (cid:17) − ( ϑ − i ( u ) n +3 = (3 n + 2)(3 n + 1) ϑ (2 ϑ −
1) ( u ) n , ( u ) n +1 = ( u ) n +2 = 0 and ( u ) = 1 , (3.15)uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche A. F. Loureiro and W. Van Assche while ( u ) n +1 = 2 ( ϑ −
1) ((3 n + 2) ϑ − (3 n + 1)) γ ( ϑ −
2) (2 ϑ −
1) ( u ) n +3 − (3 n + 1) ϑ (2 ϑ −
1) ( u ) n , if ϑ = 2( u ) n +1 = n +1)3 n +2 ( u ) n if ϑ = 2( u ) n = ( u ) n +2 = 0 and ( u ) = 1 . (3.16) Proof.
From (3.3)-(3.4) it follows h (cid:16) φ ( x ) u (cid:17) ′′ + (cid:18) γ ( ϑ + ϑ − x u (cid:19) ′ + 2 γ ( ϑ − xu , x n i = 0 , n > , and ( h ( ϑ −
2) (2 ϑ − u , x n i = h φ ( x ) u ′ − γ ( ϑ −
1) (2 ϑ − x u , x n i , if ϑ = 2 ,x u ′ = 2 h u ′ , x n i , if ϑ = 2 , which, on account of (1.1), leads to (3.15).Similarly, by taking into account the operations defined in (3.43), we deducethat relations (3.4) imply (3.16).The ”2-Hahn-classical” polynomials satisfy a third order recurrence relation(2.5) and the γ -recurrence coefficients have a specific rational structure. Here, weshow that such expression for the γ -coefficients actually characterizes the three-fold symmetric ”2-Hahn-classical” polynomials, by proving the reciprocal conditionfound in the work by Douak and Maroni [14]. For a matter of completion, we ob-tain the γ -coefficients directly from the functional equations (3.2a), rather thanfrom algebraic manipulations on the recurrence relations. Theorem 3.2.
Let { P n } n > be a monic 2-orthogonal polynomial sequence. Then { P n } n > satisfies (2.8) with γ n +2 = n + 3 n + 1 n ( ϑ n −
1) + 1( n + 4)( ϑ n +1 −
1) + 1 γ n +1 , (3.17) where ϑ n = (cid:18) − ( − n (cid:19) − n +12 (1 − ϑ )1 − n − (1 − ϑ ) + (cid:18) − n (cid:19) − n (1 − ϑ )1 − ( n − − ϑ ) , n > , (3.18) with ϑ , ϑ subject to (3.2c) , if and only if the sequence { Q n ( x ) := n +1 P ′ n +1 ( x ) } n > is -orthogonal satisfying the recurrence relation (3.1) with e γ n = nn + 2 ϑ n γ n +1 , for n > . (3.19)uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche Threefold symmetric Hahn-classical multiple orthogonal polynomials Proof.
We start by proving that if { Q n } n > satisfies the recurrence relation (3.1)with the e γ -coefficients given by (3.19), then the γ -coefficients in the recurrencerelation of { P n } n > are given by (3.17)-(3.18). The assumption means that { Q n } n > is 2-orthogonal and therefore { P n } n > is Hahn-classical. According to Theorem 3.1, { P n } n > is 2-orthogonal with respect to a vector functional ( u , u ) satisfying thedifferential equation (3.2a) and { Q n } n > is 2-orthogonal for V = ( v , v ) given by(3.11a)-(3.11b). The relation Q n ( x ) = n +1 P ′ n +1 ( x ) combined with the properties(1.1) implies h v , x n Q n i = − n + 1 h ( x n v ) ′ , P n +1 i and h v , x n Q n +1 i = − n + 2 h ( x n v ) ′ , P n +2 i for any n >
0. In the latter identities we replace ( v ′ , v ′ ) and ( v , v ) by the respectiveexpressions given in (3.11a) and (3.11b), to obtain h v , x n Q n i = (2 n + 1) − (1 − nφ ′ , (0)) h u , x n P n +1 i , for n > , h v , x n Q n +1 i = (2 n + 2) − (cid:18) ψ ′ (0) − n φ ′′ , (0)2 (cid:19) h u , x n +1 P n +2 i , for n > , which are the same as h v , x n Q n i = (2 n + 1) − (1 − (1 − ϑ ) n ) h u , x n P n +1 i , for n > , h v , x n Q n +1 i = (2 n + 2) − γ (1 − (1 − ϑ ) n ) h u , x n +1 P n +2 i , for n > . Taking into account (2.6), from the latter we obtain (3.19) if we set ϑ n as in (3.18).Observe that ϑ n +1 = ( n + 1) ϑ − nnϑ − ( n −
1) and ϑ n +2 = ( n + 1) ϑ − nnϑ − ( n − , n > , and from this, we can clearly see that ϑ n is actually a solution to the Riccatiequation ϑ n +3 + 1 ϑ n +1 = 2 , for n > . (3.20)Conversely, if { P n } n > satisfies (2.8) with the γ -coefficients given by (3.17)where ϑ n is given by (3.18), then we prove that { Q n } n > satisfies a third or-der recurrence relation (3.1) with e γ -coefficients as in (3.19). We differentiatethe recurrence relation (2.8) once and then take into account the definition of Q n ( x ) := n +1 P ′ n +1 ( x ) obtain the structural relation P n ( x ) = ( n + 1) Q n ( x ) − nxQ n − ( x ) + ( n − γ n − Q n − . uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche A. F. Loureiro and W. Van Assche
In (2.8), we replace P n +1 , P n and P n − by the expressions provided in the latteridentity to obtain( n + 2) Q n +1 ( x ) = 2( n + 1) xQ n ( x ) − nx Q n − ( x ) + 2( n − γ n − xQ n − ( x ) − ( n + 1)( γ n − + γ n ) Q n − ( x ) − ( n − γ n − γ n − Q n − ( x ) , (3.21)which is valid for any n >
0, under the assumption that Q − n ( x ) = 0. As { Q n } n > is a basis for P , there are coefficients ξ n +1 ,ν such that xQ n ( x ) = n +1 X ν =0 ξ n +1 ,ν Q ν ( x ) , (3.22)where ξ n +1 ,n +1+ ν = 0 for any ν > ξ n +1 ,n +1 = 1 because Q n is monic. Basedon this, we can also write x Q n ( x ) = n +2 X ν =1 n +1 X σ = ν − ξ n +1 ,σ ξ σ +1 ,ν Q ν ( x ) + n +1 X σ =0 ξ n +1 ,σ ξ σ +1 , Q ( x ) . (3.23)The threefold symmetry of { P n } n > readily implies that of { Q n } n > and this meansthat ξ n,ν = 0 whenever n + ν = 0 mod 3, so that ξ n,n − , ξ n,n − , ξ n,n − , ξ n,n − , . . . are all equal zero. We replace the terms xQ n , xQ n − and x Q n − by the respectiveexpressions given by (3.22) and (3.23) in the relation (3.21) to obtain( n + 2) Q n +1 ( x ) = 2( n + 1) n +1 X ν =0 ξ n +1 ,ν Q ν ( x ) + 2( n − γ n − n − X ν =0 ξ n − ,ν Q ν ( x ) − n n +1 X ν =1 n X σ = ν − ξ n,σ ξ σ +1 ,ν Q ν ( x ) + n X σ =0 ξ n,σ ξ σ +1 , Q ( x ) ! − ( n − γ n − + γ n ) Q n − ( x ) − ( n − γ n − γ n − Q n − ( x ) . (3.24)For n >
2, we equate the coefficients of Q n − , giving0 = 2( n + 1) ξ n +1 ,n − + 2( n − γ n − − nξ n,n − − ( n − γ n − + γ n ) . (3.25)In particular, for n = 2 the latter becomes0 = 4 ξ , − γ − γ , and, because of (3.17), we have γ = (4 ϑ − γ so that ξ , = ϑ γ . If we assume that for some n > ξ n,n − = n − n ϑ n − γ n − , then (3.25) implies( n + 2) ξ n +1 ,n − = (cid:0) ( n − ϑ n − − ( n − (cid:1) γ n − + ( n − γ n , uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche Threefold symmetric Hahn-classical multiple orthogonal polynomials which, after writing γ n − in terms of γ n via (3.17), reads as ξ n +1 ,n − = n − n + 1 ϑ n − γ n , and therefore we conclude that ξ n +3 ,n = n + 1 n + 3 ϑ n +1 γ n +2 , for all n > . (3.26)Equating the coefficients of Q n − in (3.24) leads to0 = ( n + 2) ξ n +1 ,n − + (2( n − γ n − − nξ n,n − ) ξ n − ,n − − nξ n,n − − ( n − γ n − γ n − , (3.27)for n ≥
5. The particular choice of n = 5 in the latter identity becomes0 = 7 ξ , + (2 − ϑ ) ϑ γ γ − γ γ , after replacing ξ , and ξ , by the expressions provided by (3.26). Using the identity(3.20) for n = 4, we then conclude that ξ , = 0. Now, suppose that ξ n,n − = 0 forsome n >
6. Identity (3.27) tells that( n + 2) ξ n +1 ,n − = − (2( n − γ n − − nξ n,n − ) ξ n − ,n − + ( n − γ n − γ n − , which, because of (3.26), becomes( n + 2) ξ n +1 ,n − = − ( n −
4) (2 − ϑ n − ) ϑ n − γ n − γ n − + ( n − γ n − γ n − , and hence, due to (3.20), we conclude ξ n +6 ,n = 0 , for all n > . Now, equating the coefficients of Q n +1 − j for j > n + 2) ξ n +1 ,n +1 − j + (2( n − γ n − − nξ n,n − ) ξ n − ,n +1 − j − n n − X σ = n − j ξ n,σ ξ σ +1 ,n +1 − j , which is also0 = ( n + 2) ξ n +1 ,n +1 − j + (2( n − γ n − − nξ n,n − ) ξ n − ,n +1 − j − n j X σ =2 ξ n,n − σ ξ n +1 − σ,n +1 − j . Suppose that ξ n,n − k = 0 for each k = 2 , , . . . , j − n > k , then the latterbecomes ( n + 2) ξ n +1 ,n +1 − j = nξ n,n − j , which yields ξ n +1 ,n +1 − j = 3 j (3 j + 1)( n + 2)( n + 1) ξ j, . (3.28a)uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche A. F. Loureiro and W. Van Assche
In particular this implies that ξ j +1 , = 3 j (3 j + 2) ξ j, and ξ j +2 , = 3 j + 1(3 j + 3) ξ j, . (3.28b)Finally we compare the coefficients of Q in (3.24) to obtain0 = ( n + 2) ξ n +1 , + 2( n − γ n − ξ n − , − n n − X σ =0 ξ n,σ ξ σ +1 , ! , for n ≥
6. Recall that ξ k, = 0 for k = 0 mod 3, so that the latter identity simplifiesto0 = (3 n + 3) ξ n +3 , + 6 nγ n ξ n, − (3 n + 2) n − X σ =1 ξ n +2 , σ +2 ξ σ +3 , ! − nξ n +2 , ξ , , (3.29)for n ≥
6. We have already seen that ξ , = 0. Proceeding by induction, we promptlyobserve that if ξ j, = 0 for j = 0 , , . . . , n , then identities (3.28a)–(3.28b) allow usto conclude from (3.29) that ξ n +1) , = 0 and this implies ξ n,k = 0 , for 0 k n − . As a result, we conclude that xQ n = Q n +1 + ξ n +1 ,n − Q n − , with ξ n +1 ,n − = n − n + 1 ϑ n − γ n = 0 , for n > , with Q j ( x ) = x j for j = 0 , ,
2. This means that { Q n } n > is 2-orthogonal andthreefold symmetric.In [17] Douak and Maroni have highlighted several properties of threefold-symmetric (therein referred to as ”2-symmetric”) 2-classical polynomials includinga differential equation of third order. Here we show that such differential equationactually characterizes these polynomials, bringing to the theory a Bochner-typecharacterization. Proposition 3.1.
Let { P n } n > be a threefold symmetric 2-orthogonal polynomialsequence satisfying (2.8) . The sequence { P n } n > is Hahn-classical if and only ifeach P n satisfies ( a n x − b n ) P ′′′ n + c n x P ′′ n + d n xP ′ n = e n P n , (3.30) where a n = ( ϑ n − ϑ n +1 −
1) (3.31a) b n = γ n (( n − ϑ n − − n + 2) ( nϑ n − n + 1) (( n + 1) ϑ n +1 − n ) n ( n + 1) (3.31b) c n = ϑ n ϑ n +1 − − ( n − ϑ n − ϑ n +1 −
1) (3.31c) d n = nϑ n +1 − ( n − ϑ n (2 ϑ n +1 −
1) (3.31d) e n = nϑ n +1 , (3.31e)uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche Threefold symmetric Hahn-classical multiple orthogonal polynomials for all n ≥ , and ϑ n given in (3.18) with initial values ϑ = γ + γ )4 γ and ϑ = γ + γ )10 γ + subject to ϑ , ϑ / ∈ { n − n : n > } . Proof.
The necessary condition was proved in [17, Proposition 3.2], where Douakand Maroni have shown that threefold-symmetric (therein referred to as ”2-symmetric”) 2-orthogonal polynomials satisfying Hahn’s property are solutions to(3.30) under the definitions (3.31a) and (3.31c)-(3.31e) with b n = γ n +3 (( n + 3) ϑ n − n + 2) (( n + 4) ϑ n +1 − ( n + 3)) (( n + 5) ϑ n +2 − ( n + 3))( n + 3)( n + 4) , where ϑ n is given in (3.18). It turns out that b n can be written as in (3.31b)because, according to Theorem 3.2, the γ -recurrence coefficients are recursivelygiven by (3.17) which implies γ n +3 = ( n + 4)( n + 3) n ( n + 1) × (cid:16) ( n + 1)( ϑ n +1 −
1) + 1 (cid:17)(cid:16) n ( ϑ n −
1) + 1 (cid:17)(cid:16) ( n − ϑ n − −
1) + 1 (cid:17)(cid:16) ( n + 5)( ϑ n +2 −
1) + 1 (cid:17)(cid:16) ( n + 4)( ϑ n +1 −
1) + 1 (cid:17)(cid:16) ( n + 3)( ϑ n −
1) + 1 (cid:17) γ n . Conversely, suppose that the 2-orthogonal polynomial sequence { P n } n > fulfils(3.30) under the definitions (3.31a)-(3.31e) with ϑ n given in (3.18). Observe that ϑ n satisfies the Riccati equation (3.20) with initial conditions ϑ = γ + γ )4 γ and ϑ = γ + γ )10 γ + , both assumed to be different from k − k for all integers k >
1. Forvalues of n >
3, we consider the expansion P n ( x ) = x n + λ n x n − + . . . and insert itin the differential equation (3.30). We equate the coefficients of x n and this gives λ n = − ( n − n − nb n n − ϑ n − n + 4) (( n − ϑ n +1 − n + 3) , n > . On the other hand, from the recurrence relation (2.8) we deduce γ n = λ n +1 − λ n +2 , n > . After combining the latter two expressions we obtain γ n = − ( n − n ( n + 1) b n +1 n − ϑ n +1 − n + 3) (( n − ϑ n +2 − n + 2)+ n ( n + 1)( n + 2) b n +2 n − ϑ n +2 − n + 2) ( nϑ n +3 − n + 1) , where b n is given in (3.31b). Based on (3.20), we consider the following substitutionsin the latter expression(( n − ϑ n +1 − n + 3) = (( n − ϑ n − − ( n − ϑ n − and ( nϑ n +3 − ( n − n + 1) ϑ n +1 − n ) ϑ n +1 , uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche A. F. Loureiro and W. Van Assche to find γ n = nϑ n (( n + 2) ϑ n +2 − ( n + 1)) (cid:16) − γ n +1 ( n − ϑ n − (( n + 1) ϑ n +1 − n )3( n + 2) (( n − ϑ n − − ( n − γ n +2 ( n + 1) ϑ n +1 (( n + 3) ϑ n +3 − ( n + 2))3( n + 3) ( nϑ n − ( n − (cid:17) , which is the same as γ n n (( n + 3) ϑ n − ( n + 2)) = γ n +2 ( n + 1) (( n + 4) ϑ n +1 − ( n + 3))3( n + 3) ( nϑ n − ( n − − γ n +1 ( n − ϑ n − (( n + 1) ϑ n +1 − n )3( n + 2) (( n − ϑ n − − ( n − ϑ n (( n + 2) ϑ n +2 − ( n + 1)) = (( n + 3) ϑ n − ( n + 2)) , and ϑ n +1 (( n + 3) ϑ n +3 − ( n + 2)) = (( n + 4) ϑ n +1 − ( n + 3)) . We subtract γ n +1 ( n +2)(( n − ϑ n − − ( n − from both sides of (3.32) and this leads to − γ n n (( n + 3) ϑ n − ( n + 2)) γ n +1 (cid:18) − γ n +1 γ n n (( n + 3) ϑ n − ( n + 2))( n + 2) (( n − ϑ n − − ( n − (cid:19) = (cid:18) − γ n +2 γ n +1 ( n + 1) (( n + 4) ϑ n +1 − ( n + 3))( n + 3) ( nϑ n − ( n − (cid:19) , which implies n Y ℓ =1 ( − γ ℓ ℓ (( ℓ + 3) ϑ ℓ − ( ℓ + 2)) γ ℓ +1 ! (cid:18) − γ γ (4 ϑ − (cid:19) = (cid:18) − γ n +2 γ n +1 ( n + 1) (( n + 4) ϑ n +1 − ( n + 3))( n + 3) ( nϑ n − ( n − (cid:19) . The assumption on the initial value for ϑ readily implies the left-hand side of thelatter equality to be zero and therefore we conclude0 = (cid:18) − γ n +2 γ n +1 ( n + 1) (( n + 4) ϑ n +1 − ( n + 3))( n + 3) ( nϑ n − ( n − (cid:19) for all n > . Now, Theorem 3.2 ensures the 2-orthogonality of the sequence { Q n ( x ) := n +1 P ′ n +1 ( x ) } n > , which means that { P n } n > is Hahn-classical. Remark 3.1.
After a single differentiation of (3.30), we obtain a differential equa-tion for the polynomials Q n and these satisfy:( a n x − b n ) Q ′′′ n + ( c n + 3 a n ) x Q ′′ n + ( d n + 2 c n ) xQ ′ n = ( e n − d n ) Q n , n > , with a n , b n , c n , d n and e n given by (3.31a)–(3.31e).uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche Threefold symmetric Hahn-classical multiple orthogonal polynomials Theorem 3.2 shows that a threefold symmetric 2-orthogonal polynomial se-quence is
Hahn-classical if and only if the γ -recurrence coefficients in (2.8) canbe written as (3.17), provided that ϑ , ϑ = nn +1 for all positive integers n .If ϑ , ϑ >
1, then γ n and e γ n are both positive for all integers n >
1. Further-more, from (3.17)-(3.19) we readily see that γ n and e γ n are two rational functionsin n , both having the same asymptotic behavior: γ n = cn α + o ( n α ) as n → + ∞ . Hence, as a result of Theorem 2.1 together with Theorem 2.2 and Theorem 3.1the two linear functionals associated with a Hahn-classical threefold symmetric2-orthogonal polynomial sequence admit the following integral representation.
Theorem 3.3.
Let { P n } n > be a threefold symmetric and 2-orthogonal polynomialsequence with respect to the vector linear functional ( u , u ) fulfilling (3.3) - (3.4) . If ϑ , ϑ > , then { P n } n > satisfies the recurrence relation (2.8) with γ n > and u and u admit the integral representation h u k , f i =13 Z b f ( x ) U k ( x ) d x + ω k − Z bω f ( x ) U k ( ω x ) d x + ω − k Z bω f ( x ) U k ( ωx ) d x ! , (3.33) (with k = 0 , ) for any polynomial f , with ω = e πi/ and b = lim n →∞ (cid:0) γ n (cid:1) ,provided that there exist two twice differentiable functions U and U mapping [0 , b ) to R such that U is solution to (cid:16) φ ( x ) U ( x ) (cid:17) ′′ + (cid:18) ϑ + ϑ − γ x U ( x ) (cid:19) ′ + 2 ( ϑ − γ x U ( x ) = λ g ( x ) , (3.34) and U is given by ( ϑ −
2) (2 ϑ − U ( x )= φ ( x ) U ′ ( x ) − ϑ −
1) (2 ϑ − x γ U ( x ) + λ g ( x ) , if ϑ = 2 , (3.35a) x U ′ ( x ) = 2 U ′ ( x ) + λ g ( x ) , if ϑ = 2 , (3.35b) where φ ( x ) = ϑ (2 ϑ − − ϑ − ϑ − γ x , and satisfying lim x → b f ( x ) d l d x l U ( x ) = 0 , for any l ∈ { , } and f ∈ P , (3.36) Z b U ( x ) d x = 1 , where λ k is a complex constant (possibly zero) and g k a function whose momentsvanish identically on the support of U k . uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche A. F. Loureiro and W. Van Assche
Proof.
According to Theorem 2.1, the threefold symmetry of a 2-orthogonal poly-nomial sequence ensures the existence of two orthogonality measures µ and µ (respectively defined by the vector functional ( u , u )) supported on a starlike set S on the three rays of the complex plane. On the other hand, Theorem 3.1 tellsthat u satisfies (3.3) and u fulfils (3.4a)-(3.4b). Based on these equations, weseek an integral representation for both u and u via a weight function U suchthat (3.33) holds for any polynomial f . Such a representation readily ensures thethreefold symmetry of ( u , u ). Identity (3.3) is the same as * ( φ ( x ) u ) ′′ + (cid:18) ϑ + ϑ − γ x u (cid:19) ′ + 2 ( ϑ − γ xu , f + = 0 , ∀ f ∈ P , which, because of (1.1), reads as (cid:28) u , φ ( x ) f ′′ ( x ) − ϑ + ϑ − γ x f ′ ( x ) + 2 ( ϑ − γ xf ( x ) (cid:29) = 0 , ∀ f ∈ P , (3.37)with φ ( x ) = ϑ (2 ϑ − − ϑ −
1) ( ϑ − γ x . We seek a function W , at least twice differentiable, defined on an open set D containing the piecewise differentiable curve C containing all the zeros of { P n } n > that is a subset of S = Γ ∪ Γ ∪ Γ (represented in Fig.2) and such that h u , f i = Z C f ( x ) W ( x ) d x (3.38)holds for every polynomial f . In the light of Theorem 2.2, the support of the weightfunction is C = e Γ ∪ e Γ ∪ e Γ where e Γ and e Γ are the two straight lines starting atsome point bω and bω (respectively) and ending at the origin, while e Γ correspondsto the straight line on the positive real axis starting at the origin and ending at b . Here b is a positive real number or can represent a point at infinity. Thus, from(3.37), the weight function W we seek must be such that Z C W ( x ) (cid:18) φ ( x ) f ′′ ( x ) − ϑ + ϑ − γ x f ′ ( x ) + 2 ( ϑ − γ xf ( x ) (cid:19) d x = 0 , ∀ f ∈ P . We use (complex) integration by parts to deduce Z C ( φ ( x ) W ( x )) ′′ + (cid:18) ϑ + ϑ − γ x W ( x ) (cid:19) ′ + 2 ( ϑ − γ x W ( x ) ! f ( x ) d x + X j ∈{ , } (cid:18) f ′ ( x ) φ ( x ) W ( x ) + f ( x ) (cid:18) ϑ + ϑ − γ x W ( x ) − ( φ ( x ) W ( x )) ′ (cid:19)(cid:19)(cid:12)(cid:12)(cid:12)(cid:12) bω j b = 0 , which must hold for every polynomial f . This condition is fulfilled if each of thefollowing conditions hold:uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche Threefold symmetric Hahn-classical multiple orthogonal polynomials (a) W is a solution of the second order differential equation( φ ( x ) W ( x )) ′′ + (cid:18) ϑ + ϑ − γ x W ( x ) (cid:19) ′ + 2 ( ϑ − γ x W ( x ) = λ g ( x ) , (3.39)for some constant λ (possibly zero) and a function g representing the nulllinear functional on the vector space of polynomials supported on C , that is Z C g ( x ) f ( x ) d x = 0 , for every polynomial f. (b) The solution W of (3.39) satisfies the boundary conditions (cid:18) f ′ ( x ) φ ( x ) W ( x ) + f ( x ) (cid:18) ϑ + ϑ − γ x W ( x ) − ( φ ( x ) W ( x )) ′ (cid:19)(cid:19)(cid:12)(cid:12)(cid:12)(cid:12) bω j b = 0 , (3.40)with j = 1 ,
2, for every polynomial f .(c) All the moments of W coincide with those of u , that is,( u ) n = Z S W ( x ) x n d x, n > . This corresponds essentially to (3.38), insofar as { x n } n > forms a basis for P .In particular, and in the light of Proposition 2.1, the threefold symmetry of thelinear functional u means that ( u ) n = 0 if n = 0 mod 3.Observe that if y is a solution of (3.39), then so are the functions ω j y ( ω j x ) for j = 1 ,
2. Taking into consideration the threefold symmetry of u , it follows that W ( x ) = U ( x ) if x ∈ e Γ = [0 , b ) , − ω U ( ω x ) if x ∈ e Γ , − ω U ( ωx ) if x ∈ e Γ , x / ∈ C = e Γ ∪ e Γ ∪ e Γ . (3.41)where U : [0 , b ) → R is an at least twice differentiable function that is a solutionof the differential equation (3.39) satisfying the conditions (3.36), so that (3.40)holds. Here, e Γ and e Γ are the two straight lines starting at some point bω and bω (respectively), with ω = e πi/ , and ending at the origin, while e Γ corresponds tothe straight line on the positive real axis starting at the origin and ending at b .Hence (3.33) is proved for k = 0.Similarly, an integral representation for the linear functional u can be obtainedfrom (3.4) by seeking a differentiable function W defined on C such that h u , f ( x ) i = Z C f ( x ) W ( x ) d x, for every polynomial f. (3.42)If ϑ = 2, then u satisfies (3.4a) and this gives h u , f i = ( ϑ − − (2 ϑ − − h φ ( x ) u ′ − ϑ −
1) (2 ϑ − γ x u , f i . uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche A. F. Loureiro and W. Van Assche
Based on the properties (1.1), the latter becomes h u , f i = ( ϑ − − (2 ϑ − − h u , − ( φf ) ′ − ϑ −
1) (2 ϑ − γ x f i . (3.43)Taking into consideration (3.38), we then have h u , f i = − ( ϑ − − (2 ϑ − − × Z C W ( x ) (cid:18) ( φ ( x ) f ( x )) ′ − ϑ −
1) (2 ϑ − γ x f ( x ) (cid:19) d x. We perform integration by parts on the first term of the integral, to obtain( ϑ −
2) (2 ϑ − h u , f i = X j ∈ , (cid:18) φ ( x ) W ( x ) f ( x ) (cid:12)(cid:12)(cid:12) bbω j (cid:19) + Z C f ( x ) (cid:18) φ ( x ) W ′ ( x ) − ϑ −
1) (2 ϑ − γ x W ( x ) (cid:19) d x. The vanishing conditions (3.36) at the end points of the contour readily imply thatthe integrated term vanishes identically for every polynomial f considered. As aconsequence, a weight function associated with u corresponds to W ( x ) = ( ϑ − − (2 ϑ − − (cid:18) φ ( x ) U ′ ( x ) − ϑ −
1) (2 ϑ − γ x W ( x ) (cid:19) + λ g ( x ) , where λ represents a complex constant (possibly zero) and g a function represent-ing the null linear functional. By construction, this choice for the weight functionguarantees that (3.42) is well defined and, in particular, all the moments satisfy thethreefold symmetry property described in Proposition 2.1. So, we conclude that(3.33) also holds for k = 1 with U given by (3.35a) when ϑ = 2.If ϑ = 2, then using an entirely analogous approach we deduce (3.35b) from(3.4b). Remark 3.2.
An alternative to the representation (3.33) stated in Theorem 3.3 is h u k , f i = Z C f ( x ) W k ( x ) d x, ( k = 0 , , where W k ( x ) is given in (3.41).The weight functions W and W are threefold symmetric in the sense that theysatisfy the following rotational invariant property: ω j W k ( ω j x ) = W k ( x ) , j = 0 , ± , ± , . . . , for each k = 0 , . We consider the following equivalence relation between two polynomial se-quences { P n } n > and { B n } n > : { P n } n > ∼ { B n } n > iff ∃ a ∈ C \{ } , b ∈ C such that B n ( x ) = a − n P n ( ax + b ) , (3.44)uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche Threefold symmetric Hahn-classical multiple orthogonal polynomials for all n >
0. In this case, if { P n } n > satisfies (2.5) then { B n } n > satisfies B n +1 ( x ) = (cid:18) x − β n − ba (cid:19) B n ( x ) − α n a B n − ( x ) − γ n − a B n − ( x ) , n > , with initial conditions B − ( x ) = B − ( x ) = 0 and B ( x ) = 1.Observe that ϑ n is a solution of a Riccati equation (3.20) for which ϑ n = 1 is atrivial solution. Depending on the initial conditions ϑ and ϑ , there are four setsof independent solutions which in fact give rise to four equivalence classes of thethreefold symmetric 2-Hahn-classical polynomials. In other words, up to a lineartransformation of the variable, there are at most four distinct families of threefoldsymmetric 2-orthogonal Hahn-classical polynomials, which we single out: Case A : ϑ = 1 = ϑ . This implies that ϑ n = 1 for all n > Case B : ϑ = 1 but ϑ = 1 so that by setting ϑ = µ +2 µ +1 it follows that ϑ n − = n + µ + 1 n + µ and ϑ n = 1 , n > . Case B : ϑ = 1 but ϑ = 1 so that by setting ϑ = ρ +2 ρ +1 it follows that ϑ n − = 1 and ϑ n = n + ρ + 1 n + ρ , n > . Case C : ϑ = 1 and ϑ = 1 and hence by setting ϑ = µ +2 µ +1 and ϑ = ρ +2 ρ +1 itfollows that ϑ n − = n + µ + 1 n + µ and ϑ n = n + ρ + 1 n + ρ , n > . All these cases were highlighted in [14], where expressions for the recurrencecoefficients for the three components of the cubic decomposition of each of thesethreefold symmetric Hahn-classical polynomials were deduced. Only some of thesepolynomials were studied in detail: those in case A in [13] and a few subcasesof Case C in [16,17]. The representations for the orthogonality measures that wedescribe in the next sections are new. The support of these orthogonality measurescontain all the zeros of the polynomial sequences. Even for particular cases thatalready appeared in the literature, the integral representations for the orthogonalitymeasures were either not given or given on the positive real line (with oscillatingterms), which is only part of the starlike set S . In the next subsections we fullydescribe all these cases in detail.As observed in [14], the following limiting relations take placecase C −−−−−→ ρ →∞ B −−−−→ µ →∞ case A , and also case C −−−−−→ µ →∞ B −−−−−→ ρ →∞ case A . It turns out that cases B and B are related to each other by differentiation, asexplained in Section 3.2 and Section 3.3.uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche A. F. Loureiro and W. Van Assche
Case A
In the light of Theorem 3.2, for this choice of initial conditions one has ϑ n = 1for all integers n >
1, so that the γ -recurrence coefficients are given by γ n +1 =( n + 1)( n + 2) γ and e γ n +1 = γ n +1 for n >
0. As a consequence, Q n ( x ) = P n ( x ) forall n > { P n } n > is an Appell sequence .Since the 2-orthogonality property is invariant under any linear transformation,we can set γ = 2, and, with this choice, recalling (3.17)-(3.19) it follows that γ n +1 = e γ n +1 = ( n + 2)( n + 1) , n > , so that P n +1 ( x ) = xP n ( x ) − n ( n − P n − ( x ) , n > ,P ( x ) = 1 , P ( x ) = x and P ( x ) = x . (3.45) Fig. 3. Case A: Plot of the largest zero in absolute value of P n against the curve y = / / x / . Figure 3 illustrates the behavior of the largest zero in absolute value (plot gen-erated in
Mathematica ): bounded from above by the curve y = / n / suggestedby Theorem 2.2.Regarding the integral representation of the corresponding orthogonality mea-sures, we have: Proposition 3.2.
The threefold symmetric polynomial sequence { P n } n > definedby the recurrence relation (3.45) is 2-orthogonal with respect to ( u , u ) admittingthe integral representation (3.33) , where U ( x ) = Ai( x ) and U ( x ) = − Ai ′ ( x ) , uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche Threefold symmetric Hahn-classical multiple orthogonal polynomials and b = + ∞ . Moreover, the sequence { Q n ( x ) := n +1 P ′ n +1 ( x ) } n > coincides with { P n } n > (i.e., { P n } n > is an Appell sequence). Here and in what follows Ai and Bi are the
Airy functions of the first and secondkind (see [34, § Proof.
Under the assumptions, by virtue of Theorem 3.2 and Theorem 3.1, thispolynomial sequence is 2-orthogonal with respect to ( u , u ) satisfying (3.3)-(3.4)which reads as ( u ′′ − x u = 0 ,u = − u ′ , from which we conclude that the corresponding sequence of the moments { ( u ) n } n > and { ( u ) n } n > satisfy( u ) n +3 = ( n + 1)( n + 2)( u ) n and ( u ) n = n ( u ) n − , with initial conditions ( u ) = 1 and ( u ) = ( u ) = ( u ) = 0. This implies( u ) n = (3 n )!3 n ( n !) and ( u ) n +1 = ( u ) n +2 = 0 , ( u ) n +1 = (3 n +1)!3 n ( n !) and ( u ) n = ( u ) n +2 = 0 , n > . (3.46)According to Theorem 3.3, u and u admit the representation (3.33) providedthat there exist two twice differentiable functions U and U from R to R that aresolutions to ( U ′′ ( x ) − x U ( x ) = λ g ( x ) , −U ( x ) = U ′ ( x ) + λ g ( x ) , (3.47)and satisfying lim x → ω j ∞ f ( x ) d l d x l U ( x ) = 0 , for j, l ∈ { , , } and f ∈ P , (3.48) Z ∞ U ( x ) d x = 1 , where λ k ∈ C (possibly zero) and g k ( x ) = 0 are rapidly decreasing functions, locallyintegrable, representing the null functional ( k = 0 , λ = 0, the generalsolution of the first equation in (3.47) can be written as y ( x ) = c Ai( x ) + c Bi( x ) , for arbitrary constants c , c . Observe that (3.48) is realized if we take c = 0 (see[34, (9.7.5)-(9.7.8)]). Since [34, (9.10.17)] Z + ∞ x n Ai( x ) d x = Γ( n + 1)3 n +1 Γ( n + 1) < + ∞ , for all n > , the result now follows if we take c = 1 and λ = 0.uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche A. F. Loureiro and W. Van Assche
The differential equation
Under these assumptions (3.30) becomes − P ′′′ n ( x ) + xP ′ n ( x ) = nP n ( x ) , n > , whose general solution is given by P n ( x ) = c F − n , ; x ! + c x F − n − , ; x ! + c x F − n − , ; x ! , with integration constants c , c and c . Observe that for each n there is onepolynomial solution and we have P n ( x ) = P [0] n ( x ) with P [0] n ( x ) = ( − n (1 / n (2 / n F (cid:18) − n , ; x (cid:19) ,P n +1 ( x ) = x P [1] n ( x ) with P [1] n ( x ) = ( − n (2 / n (4 / n F (cid:18) − n , ; x (cid:19) ,P n +2 ( x ) = x P [2] n ( x ) with P [2] n ( x ) = ( − n (4 / n (5 / n F (cid:18) − n , ; x (cid:19) . The cubic decomposition
In [7, §
5] the cubic decomposition of an Appell 2-orthogonal sequence has been high-lighted. This also happens to be threefold symmetric, and the corresponding poly-nomials were called Hermite-type 2-orthogonal polynomials. The three 2-orthogonalpolynomial sequences { P [ j ] n } n > ( j = 0 , ,
2) in the cubic decomposition of { P n } n > are 2-orthogonal with respect to weights involving modified Bessel functions of thesecond kind, studied in [7,38].Following Lemma 2.1, each of the three polynomial sequences { P [ j ] n } n > , with j ∈ { , , } , is 2-orthogonal and satisfies the third order recurrence relation (2.9),where β [ j ] n = 3 (cid:0) j + 6 jn + j + 9 n (cid:1) + 9 n + 2 , n > ,α [ j ] n = 3( j + 3 n − j + 3 n − ( j + 3 n ) , n > ,γ [ j ] n = ( j + 3 n − j + 3 n − j + 3 n )( j + 3 n + 1)( j + 3 n + 2)( j + 3 n + 3) , n > . It should be noted that the Appell polynomials { P n } n > and the components inthe cubic decomposition were treated in [7] under a different normalisation, namelyby considering 2 /γ = 9. However the integral representation provided in [7] issupported on the positive real axis, and therefore different from the one given here.In [25, Cor. 5,7] the authors have also studied these polynomials, where the focuswas put on algebraic properties, including generating functions. Remark 3.3.
This Appell sequence { P n } n > has already appeared in the literaturein other contexts, often not recognized as 2-orthogonal polynomial sequences. Forinstance, it is linked to the Vorob’ev-Yablonski polynomials associated with rationaluly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche Threefold symmetric Hahn-classical multiple orthogonal polynomials solutions of the second Painlev´e equations [10]. Furthermore, in [40] Widder studiedthe so-called Airy transform defined as follows u ( x, t ) = Z + ∞−∞ f ( y ) 1(3 t ) / Ai (cid:18) y − x t / (cid:19) d y. Up to a scaling, it maps the sequence of monomials to the Appell 2-orthogonalpolynomial sequence { P n } n > [40, § § π/ Fig. 4. Case A: Zeros of P (empty circles), P (stars) and P (solid squares). uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche A. F. Loureiro and W. Van Assche
Case B In this case, ϑ n = 1 and ϑ n − = n + µ +1 n + µ , under the constraint that µ = − n , forall n >
1. Therefore, the threefold symmetric sequence { P n ( · ; µ ; γ ) } n > satisfiesthe recurrence relation (2.8), where γ n +1 = ( n + 1)(2 n + 1)( µ + 2)(3 n + µ + 2) γ ,γ n +2 = ( n + 1)(2 n + 3)( n + µ + 1)( µ + 2)(3 n + µ + 2)(3 n + µ + 5) γ , n > . Observe that γ is a mere scaling factor and, from this point forth, we set γ = µ +2) . The discussion is therefore about on { P n ( · ; µ ) := P n ( · ; µ, µ +2) ) } n > . Thegeneral case { P n ( · ; µ ; γ ) } n > can be deduced after a linear transformation of thevariable: P n ( x ; µ ; γ ) = a − n P n ( ax ; µ ) , with a = (cid:16) µ +2) γ (cid:17) / , n > . The two sequences are equivalent, under (3.44).Hence, { P n ( · ; µ ) } n > satisfies (2.8) with γ n +1 := γ n +1 ( µ ) given by γ n +1 = 23 ( n + 1)(2 n + 1)(3 n + µ + 2) , γ n +2 = 23 ( n + 1)(2 n + 3)( n + µ + 1)(3 n + µ + 2)(3 n + µ + 5) , n > , (3.49)and it is 2-orthogonal for ( u , u ) for which13 u ′′ + x u ′ − ( µ − xu = 0 and u = − ( µ +2) µ (cid:16) u ′ + 3 x u (cid:17) if µ = 0 ,xu ′ = 2 u ′ if µ = 0 . (3.50)Consequently, ( u ) n = ( ) n ( ) n (cid:0) µ +23 (cid:1) n and ( u ) n +1 = ( u ) n +2 = 0 , (3.51)and ( u ) n +1 = (cid:0) (cid:1) n (cid:0) (cid:1) n (cid:0) µ +53 (cid:1) n and ( u ) n = ( u ) n +2 = 0 . (3.52)Observe that in order to have γ n +1 > n >
0, the constraint µ > − µ make the γ -coefficients linear functions in n , namely:for µ = − / γ n ( − /
2) = 4 n , γ n +1 ( − /
2) = 4( n + 1)9 , whilst for µ = 1 : γ n (1) = 2(2 n + 1)27 , γ n +1 (1) = 2(2 n + 1)9 . uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche Threefold symmetric Hahn-classical multiple orthogonal polynomials For each µ > −
1, we have γ n +1 ( µ ) = 4 n − µ )27 O (1) whilst γ n +2 ( µ ) = 4 n
27 + 2(7 + 2 µ )81 O (1) n → + ∞ . In the light of Theorem 2.2, an upper bound for the absolute value of the largestzero | x n,n | of P n is given by (2.10) with c = and α = 1 so that we obtain | x n,n | / n / + o ( n / ) , n > . The accuracy of the result is illustrated in Figure 5, where we have only plottedthe positive zeros of P n ( x ), but similar results hold for the zeros of P n +1 ( x ) and P n +2 ( x ), which are just rotations of the positive zeros. Fig. 5. Plot of of the curve y = 3 / x / (grey solid line) and the largest zeros in absolute value(black dots) of P n ( x ; 3) with n ranging from 0 to 100. The zeros of three consecutive polynomials interlace and lie on the three-starlikeset S : Figure 6 illustrates this.In the light of Theorem 2.1, the two orthogonality weights can be expressed viathe confluent hypergeometric function of the second kind U ( a, b ; x ), which admitsthe following integral representation [34, (13.4.4)] U ( a, b ; x ) = 1Γ( a ) Z ∞ t a − ( t + 1) − a + b − e − tx d t, provided that ℜ ( a ) > | arg( x ) | < π/
2, whilst U (0 , b ; x ) = 1 , and one has the identity U ( a, b ; x ) = x − b U ( a − b + 1 , − b ; x ). Proposition 3.3.
Let µ > − . The threefold symmetric polynomial sequence { P n ( · : µ ) } n > defined by the recurrence relation (2.8) , with γ n given by (3.49) , uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche A. F. Loureiro and W. Van Assche
Fig. 6. Zeros of P ( x ; µ ) (circle), P ( x ; µ ) (star) and P ( x ; µ ) (square) with µ = 3, where P n ( x ; µ ) is the 2-orthogonal polynomial sequence studied in case B1. is 2-orthogonal with respect to ( u , u ) admitting the integral representation (3.33) where U ( x ) := U ( x ; µ ) = 3Γ( µ +23 )Γ( )Γ( ) e − x U ( µ , ; x ) , (3.53) U ( x ) := U ( x ; µ ) = 9Γ (cid:0) µ +53 (cid:1) Γ (cid:0) (cid:1) Γ (cid:0) (cid:1) x e − x U (cid:0) µ + 1 , , x (cid:1) , (3.54) and b = + ∞ , where U represents the Kummer (confluent hypergeometric) functionof second kind. Proof.
Under the assumptions, it follows that u and u satisfy (3.50) and { P n ( · : µ ) } n > satisfies the recurrence relation (2.8) with γ n > n >
1. In the lightof Theorem 2.1, there exists a function W and a domain C containing all the zerosuly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche Threefold symmetric Hahn-classical multiple orthogonal polynomials of { P n } n > so that h u , f ( x ) i = Z C f ( x ) W ( x ) d x, is valid for every polynomial f . By virtue of Theorem 2.2 and the asymptotic behav-ior of γ n for large n , the curve that contains all the zeros of { P n } n > correspondsto the starlike set S in Fig. 2. According to Theorem 3.3, u and u admit therepresentation (3.33), provided that there exist two twice differentiable functions U and U from R to R that are solutions to U ′′ ( x ) + x U ′ ( x ) − ( µ − x U ( x ) = λ g ( x ) if µ > − , U ( x ) = U ′ ( x ) + 2 x U ( x ) + λ g ( x ) if µ = 0 and µ > − ,x U ′ ( x ) = 2 U ′ ( x ) + λ g ( x ) if µ = 0 , (3.55)and satisfying lim x → ω j ∞ f ( x ) d l d x l U ( x ) = 0 , for j, l ∈ { , , } and f ∈ P , (3.56) Z ∞ U ( x )d x = 1 , (3.57)where λ k ∈ C (possibly zero) and g k = 0 are rapidly decreasing functions, locallyintegrable, representing the null functional ( k = 0 , λ = 0, the generalsolution of the first equation in (3.55) can be written as y ( x ) = c F − µ ; − x ! + c x F − µ ; − x ! , (3.58)with c , c two arbitrary constants. Based on [34, Eqs. (13.2.39) and (13.2.42)] wehavee − x U (cid:0) µ , , x (cid:1) = Γ( )Γ( µ +13 ) F − µ ; − x ! + µ Γ( − )3Γ( µ + 1) x F − µ ; − x ! , which is valid when b is not an integer. Observe that (see [34, (13.7.3)]) G ( x ; µ ) = e − x U (cid:0) µ , , x (cid:1) ∼ e − x x − µ ∞ X s =0 (cid:0) µ (cid:1) s (cid:0) µ +13 (cid:1) s s ! ( − z ) − s , and also that [34, (13.3.27)]dd x G ( x ; µ ) = − x e − x U ( µ , , x ) , so that, for both l = 0 ,
1, we obtain lim x →∞ f ( x ) d l G ( x ; µ )d x l = 0 for every polynomial f . According to [20, Eq. (7.621.6)], the following identity Z + ∞ t b − U ( a, c ; t ) e − t d t = Γ( b )Γ( b − c + 1)Γ( a + b − c + 1) , (3.59)uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche A. F. Loureiro and W. Van Assche holds, provided that ℜ ( b ) > max(0 , ℜ ( c ) − Z + ∞ x n e − x U ( µ , ; x ) d x = 13 Z + ∞ t n +13 − e − t U ( µ , ; t ) d t = 13 Γ( n +13 )Γ( n +23 )Γ( n + µ +23 ) , for n >
0, which, upon the substitution n → n , becomes Z + ∞ x n e − x U ( µ , ; x ) d x = Γ( )Γ( )3Γ( µ +23 ) ( ) n ( ) n ( µ +23 ) n , n > . Consequently, the particular choices of c = 3Γ( µ +23 )Γ( )Γ( µ +13 ) and c = − µ Γ( µ +23 )Γ( µ + 1)Γ( )in the general solution (3.58) gives the function U ( x ; µ ) in (3.53), which meets therequirements (3.56)-(3.57). Furthermore, we have( u ) n = (cid:18)Z Γ d x − ω Z Γ d x − ω Z Γ d x (cid:19) x n Γ( µ +23 )Γ( )Γ( ) e − x U ( µ , ; x ) ! , which matches with (3.51).Now, for the integral representation of the linear functional u we consider (3.55)with λ = 0 and this gives U ( x ; µ ) = − ( µ + 2) µ (cid:16) U ′ ( x ) + 3 x U ( x ) (cid:17) = 9Γ (cid:0) µ +53 (cid:1) Γ (cid:0) (cid:1) Γ (cid:0) (cid:1) x e − x U (cid:0) µ + 1 , , x (cid:1) if µ = 0 , U ( x ; µ ) = ( ) Γ ( ) Γ ( ) Γ (cid:0) , x (cid:1) if µ = 0 , where, for the first line of the latter identity we have used [34, (13.3.27)] as wellas the contiguous relation [34, (13.3.10)]. The second line of the latter identitycorresponds to the first line after taking µ = 0 because of [34, Eq. (13.6.6)], sothat we obtain (3.54). Here, Γ( α, z ) is the incomplete Gamma function: Γ( α, z ) = Z + ∞ z t α − e − t d t provided that α >
0. Moreover, based on (3.59), we have Z + ∞ x n +1 U ( x ; µ ) d x = (2 / n (4 / n (cid:0) µ +53 (cid:1) n , µ > − , which coincides with the moments of u given in (3.52).3.2.1. Particular cases
For the following choices of the parameter µ the expressions (3.53)-(3.54) relate toother functions. Namely we have:uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche Threefold symmetric Hahn-classical multiple orthogonal polynomials • µ = − /
2, then U ( x ; − /
2) = 32 r π e − x U (cid:18) − , , x (cid:19) and U ( x ; − /
2) = 9 √ e − x xK (cid:16) x (cid:17) π , where K ν is the modified Bessel function of the second kind (see [34, Ch. 10.25]). • µ = 0, then U ( x ; 0) = 3Γ( ) e − x and U ( x ; 0) = 9Γ (cid:0) (cid:1) Γ (cid:0) (cid:1) Γ (cid:0) (cid:1) Γ (cid:18)
23 ; x (cid:19) . • µ = 1, then U ( x ; 1) = 3 √ e − x √ xK (cid:16) x (cid:17) π / and U ( x ; 1) = 9 √ e − x U (cid:0) , , x (cid:1) π . • µ = 2, then U ( x ; 2) = √ (cid:0) (cid:1) π Γ (cid:0) , x (cid:1) and U ( x ; 2) = 2 √ (cid:0) (cid:1) π Γ (cid:0) , x (cid:1) . The differential equation
Following Lemma 3.1, the polynomial P n ( x ; µ ) is a solution of the differential equa-tion − y ′′′ ( x )+2 x y ′′ ( x )+2 x (cid:18) µ + 34 (( − n + 3) − n (cid:19) y ′ ( x ) = 2 n (cid:18) µ + n − n (cid:19) y ( x ) , whose general solution can be written as y ( x ) = c F − n , n + ( − n + µ + , ; x ! + c x F − n , n + ( − n + µ + , ; x ! + c x F − n , n + ( − n + µ + , ; x ! . For each positive integer n , there is only one (monic) polynomial solution, butit is not always the same: it depends on whether n equals 0 , A. F. Loureiro and W. Van Assche precise, we have P n ( x ; µ ) := P [0] n ( x ; µ )= ( − n (cid:0) (cid:1) n (cid:0) (cid:1) n (cid:0) n + ( − n + µ + (cid:1) n F − n, n + ( − n + µ + , ; x ! ,P n +1 ( x ; µ ) := xP [1] n ( x ; µ )= x ( − n (cid:0) (cid:1) n (cid:0) (cid:1) n (cid:0) n + ( − n +1 + µ + (cid:1) n F − n, µ + n + ( − n +1 + , ; x ! ,P n +2 ( x ; µ ) := x P [2] n ( x ; µ )= x ( − n (cid:0) (cid:1) n (cid:0) (cid:1) n (cid:0) n + ( − n +2 + µ + (cid:1) n F − n, µ + n + ( − n +2 + , ; x ! . These polynomial sequences { P [ k ] n ( · ; µ ) } n > , with k ∈ { , , } , are precisely the2-orthogonal polynomial sequences in the cubic decomposition of { P n ( · ; µ ) } n > .In fact, from Lemma 2.1, these three 2-orthogonal polynomial sequences are notthreefold symmetric and satisfy the recurrence relation (2.9), whose recurrence co-efficients are given in the Appendix for completeness. These coefficients have beencomputed in [14, Tableau 4, 8 and 12 - Case C], for a different choice of the ”free”parameter γ . We have included them here for a matter of completeness.3.2.3. The sequence of derivatives
Following Theorem 3.1 along with Theorem 3.2, { Q n ( · ; µ ) := n +1 P ′ n +1 ( x ; µ ) } n > is2-orthogonal for the vector functional ( v , v ), which admits an integral representa-tion via two weight functions ( f W ( · ; µ ) , f W ( · ; µ )) with support on the three-starlikeset S which are given by " f W ( x ; µ ) f W ( x ; µ ) = " µ +2 µ +1 − xµ +1 W ( x ; µ ) W ( x ; µ ) (cid:21) . Proposition 3.3 allows us to conclude h v k , f i = (cid:18)Z Γ d x − ω Z Γ d x − ω Z Γ d x (cid:19) ( f ( x ) V k ( x ; µ )) , ∀ f ∈ P , k = 0 , , where V ( x ; µ ) = µ + 2 µ + 1 U ( x ; µ ) − xµ + 1 U ( x ; µ ) = U ( x ; µ + 3) , V ( x ; µ ) = U ( x ; µ ) . uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche Threefold symmetric Hahn-classical multiple orthogonal polynomials Furthermore, { Q n ( · ; µ ) } n > satisfies the second order recurrence relation (3.1)where e γ n := e γ n ( µ ) = 23 n (2 n + 1)(3 n + 2 + µ ) , n > , e γ n +1 := e γ n +1 ( µ ) = 23 ( n + 1)(2 n + 1)( n + 2 + µ )(3 n + 2 + µ )(3 n + 5 + µ ) , n > . (3.60)For each integer n >
0, the polynomial Q n ( x ; µ ) is a solution to the differentialequation − y (3) ( x )+2 x y ′′ ( x ) − x ( − µ + 3 (( − n −
5) + 2 n ) y ′ ( x ) = 12 n (4 µ + 2 n − − n + 11) y ( x ) , from which we deduce Q n ( x ; µ ) = ( − n ( ) n ( ) n ( n − ( − n + µ + ) n F − n, n − ( − n + µ + , ; x ! ,Q n +1 ( x ; µ ) = x ( − n ( ) n ( ) n ( µ + n + ( − n + ) n F − n, µ + n + ( − n + , ; x ! ,Q n +2 ( x ; µ ) = x ( − n ( ) n ( ) n ( µ + n − ( − n + ) n F − n, µ + n − ( − n + , ; x ! , n > . It turns out that the forthcoming case B is closely connected to this case B :the sequence of derivatives in case B belongs to case B . Case B For this family ϑ n +1 = 1 and we can write ϑ n = n + ρ +1 n + ρ , under the relation ϑ = ρ +2 ρ +1 with ρ = − n . As a result, the threefold symmetric Hahn-classical sequence { P n ( · ; ρ ; γ ) } n > satisfies the recurrence (2.8) with γ n = n (2 n + 1)( ρ + 3)(3 n + ρ ) γ , n > ,γ n +1 = ( n + 1)(2 n + 1)( n + ρ )( ρ + 3)(3 n + ρ )(3 n + ρ + 3) γ , n > . Analogously to the former cases, the parameter γ is redundant for the study, andtherefore we can choose a representative value for γ . Here we set γ = ρ +3) . Hencethe 2-orthogonal polynomial sequence { P n ( · ; ρ ) := P n ( · ; ρ ; ρ +3) ) } n > satisfies (2.8)with γ n := γ n ( ρ ) = 2 n (2 n + 1)3(3 n + ρ ) , n > ,γ n +1 := γ n +1 ( ρ ) = 2( n + 1)(2 n + 1)( n + ρ )3(3 n + ρ )(3 n + ρ + 3) , n > . (3.61)uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche A. F. Loureiro and W. Van Assche
Furthermore, the2-orthogonal polynomial sequence { Q n ( x ; ρ ) := n +1 P ′ n +1 ( x ; ρ ) } n > satisfies (3.1)with e γ n := e γ n ( ρ ) = 2 n (2 n + 1)( n + ρ + 1)3(3 n + ρ )(3 n + ρ + 3) , n > , e γ n +1 := e γ n ( ρ ) = 2( n + 1)(2 n + 1)3(3 n + ρ + 3) , n > . (3.62)Observe that the recurrence coefficients uniquely determine a 2-orthogonal poly-nomial sequence. A comparison between (3.49) and (3.62) and between (3.60) and(3.61) shows that Q case B n ( x ; µ ) = P case B n ( x ; µ + 1) , for all n > , while Q case B n ( x ; µ ) = P case B n ( x ; µ + 2) , for all n > , where the notation P case B n and P case B n is used for the threefold 2-orthogonalpolynomial sequences defined by the recurrence coefficients (3.49), in case B , and(3.61), in case B , respectively. Likewise, Q case B n and Q case B n denote the monicderivatives of P case B n and P case B n , respectively.From this, we conclude that the the polynomial sequences { Q case B n ( x ; µ ) := n +1 dd x P case B n +1 ( x ; µ ) } n > and { Q case B n ( x ; ρ ) := n +1 dd x P case B n +1 ( x ; ρ ) } n > are alsoHahn-classical. Furthermore, we have1( n + 2)( n + 1) d d x P n +2 ( x ; µ ) = P n ( x ; µ + 3) , in both cases B and B . These observations also serve as an alternative to theproof of the following result: Proposition 3.4.
Let ρ > . The threefold symmetric polynomial sequence { P n ( · : ρ ) } n > defined by the recurrence relation (2.8) with γ n given by (3.61) is 2-orthogonal with respect to ( u , u ) admitting the integral representation (3.33) where U ( x ) := U ( x ; ρ ) = 3Γ (cid:0) ρ +33 (cid:1) Γ (cid:0) (cid:1) Γ (cid:0) (cid:1) e − x U (cid:18) ρ + 13 , , x (cid:19) , U ( x ) := U ( x ; ρ ) = 9Γ (cid:0) ρ +33 (cid:1) Γ (cid:0) (cid:1) Γ (cid:0) (cid:1) x e − x U (cid:18) ρ + 13 , , x (cid:19) , and b = + ∞ , where U represents the Kummer (confluent hypergeometric) functionof second kind. Proof.
The proof is entirely analogous to the proof of Proposition 3.3.In fact, we have U case B ( x ; ρ ) = U case B ( x ; ρ + 1) and U case B ( x ; ρ ) = U case B ( x ; ρ − , provided that ρ > Threefold symmetric Hahn-classical multiple orthogonal polynomials Case C
Here we have ϑ n − = n + µ + 1 n + µ and ϑ n = n + ρ + 1 n + ρ , n > . so that (3.17) reads as γ n = 2 n (2 n + 1)( n + µ )(3 n + µ − n + µ + 2)(3 n + ρ ) ( µ + 2)( ρ + 3) γ , n > ,γ n +1 = 2( n + 1)(2 n + 1)( n + ρ )(3 n + µ + 2)(3 n + ρ )(3 n + ρ + 3) ( µ + 2)( ρ + 3) γ , n > . The Hahn-classical polynomial sequence obtained under these assumptionsclearly depends on the pair of parameters ( µ, ρ ) and on γ , so it makes sense to in-corporate this information, and therefore we shall refer to it as { P n ( · ; µ, ρ ; γ ) } n > .Similar to the precedent cases, without loss of generality, it suffices to study the se-quence for a particular choice of γ . A scaling of the variable would then reproduceall the other sequences { P n ( · ; µ, ρ ; γ ) } n > within this equivalence class. Hence, weset γ = 2( µ + 2)( ρ + 3) , and the analysis is on n P n ( · ; µ, ρ ) := P n (cid:16) · ; µ, ρ ; µ +2)( ρ +3) (cid:17)o n > , which satisfies the recurrence relation (2.8) where the γ -coefficients are given by γ n := γ n ( µ, ρ ) = 2 n (2 n + 1)( n + µ )(3 n + µ − n + µ + 2)(3 n + ρ ) , n > ,γ n +1 := γ n ( µ, ρ ) = 2( n + 1)(2 n + 1)( n + ρ )(3 n + µ + 2)(3 n + ρ )(3 n + ρ + 3) , n > . (3.63)Observe that γ n +1 > n > µ + 1 , ρ > γ n = 427 + o (1) , as n → + ∞ . Theorem 2.2 ensures that the largest zero of P n ( x ; µ, ρ ) in absolute value is alwaysless than or equal to 1. Figure 7 illustrates the curve defined by these zeros.Regarding the orthogonality measures for the polynomial sequence { P n ( · ; µ, ρ ) } n > we have: Proposition 3.5.
Let µ > − and ρ > . The threefold symmetric polynomialsequence { P n ( · ; µ, ρ ) } n > defined by the recurrence relation (2.8) with γ n given by (3.63) is -orthogonal with respect to ( u , u ) , admitting the integral representation uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche A. F. Loureiro and W. Van Assche
Fig. 7. Plot of the largest zero in absolute value of P n ( x ; µ, ρ ) for each n = 1 , . . . ,
100 when µ = 2 , ρ = 3, against the curve y = 1. (3.33) with b = 1 and U ( x ) := U ( x ; µ, ρ ) = 3Γ (cid:0) µ +23 (cid:1) Γ (cid:0) ρ + 1 (cid:1) Γ (cid:0) (cid:1) Γ (cid:0) (cid:1) Γ (cid:0) µ + ρ +23 (cid:1) (1 − x ) µ + ρ − F µ , ρ +13 µ + ρ +23 ; 1 − x ! , (3.64) U ( x ) := U ( x ; µ, ρ ) = 3Γ (cid:0) µ +53 (cid:1) Γ (cid:0) ρ + 1 (cid:1) Γ (cid:0) (cid:1) Γ (cid:0) (cid:1) Γ (cid:0) µ + ρ +23 (cid:1) x (1 − x ) µ + ρ − F µ + 1 , ρ +13 µ + ρ +23 ; 1 − x ! . (3.65) Proof.
According to Theorem 3.1, { P n ( · ; µ, ρ ) } n > is 2-orthogonal with respect tothe vector of linear functionals ( u , u ) fulfilling the distributional equations (cid:0) − x (cid:1) u ′′ + x ( µ + ρ − u ′ − ( µ − ρ − xu = 0 , ( µ ( µ +2) u = (cid:0) x − (cid:1) u ′ − ( ρ − x u if µ = 0 ,xu ′ = 2 u ′ if µ = 0 , which yields( u ) n = (cid:0) (cid:1) n (cid:0) (cid:1) n (cid:0) µ +23 (cid:1) n (cid:0) ρ +33 (cid:1) n , ( u ) n +1 = ( u ) n +2 = 0 , n > , as well as( u ) n +1 = (cid:0) (cid:1) n (cid:0) (cid:1) n (cid:0) µ +53 (cid:1) n (cid:0) ρ +33 (cid:1) n , ( u ) n = ( u ) n +2 = 0 , n > , because ( u ) = ( u ) = 1 and ( u ) = ( u ) = ( u ) = ( u ) = 0.Since lim n →∞ γ n = , then by virtue of Theorem 3.3, u and u admit the inte-gral representation (3.33) with b = 1, provided that there exists a pair of functionsuly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche Threefold symmetric Hahn-classical multiple orthogonal polynomials U and U with support on [0 ,
1) satisfying (3.34)-(3.35) and subject to the bound-ary condition (3.36). To be precise, under the assumptions, we seek U ( x ; µ, ρ ) and U ( x ; µ, ρ ) defined on defined on [0 ,
1) such that (cid:0) − x (cid:1) U ′′ ( x ; µ, ρ ) + x ( µ + ρ − U ′ ( x ; µ, ρ ) − ( µ − ρ − x U ( x ; µ, ρ ) = λ g ( x ) , (3.66) µ ( µ + 2) U ( x ; µ, ρ ) = (cid:0) x − (cid:1) U ′ ( x ; µ, ρ ) − ( ρ − x U ( x ; µ, ρ ) + λ g ( x ) if µ = 0 ,x U ′ ( x ; µ, ρ ) = 2 U ′ ( x ; µ, ρ ) + λ g ( x ) if µ = 0 , (3.67)where λ j , for j ∈ { , } , are constants (possibly zero) and g j are functions repre-senting the null linear functional, i.e., Z C g j ( x ) x n d x = 0 , n > , and, in addition,lim x → f ( x ) d l d x l U ( x ; µ, ρ ) = 0 , for any l ∈ { , } and f ∈ P , (3.68) Z x n U ( x ; µ, ρ ) d x = (cid:0) (cid:1) n (cid:0) (cid:1) n (cid:0) µ +23 (cid:1) n (cid:0) ρ +33 (cid:1) n , n > . (3.69)For λ = 0, the differential equation (3.66) becomes (cid:0) − x (cid:1) U ′′ ( x ; µ, ρ ) + x ( µ + ρ − U ′ ( x ; µ, ρ ) − ( µ − ρ − x U ( x ; µ, ρ ) = 0 , whose general solution can be written as U ( x ; µ, ρ ) = c F − µ , − ρ ; x ! + c x F − µ , − ρ ; x ! , with integration constants c and c which must be chosen such that (3.68)-(3.69)hold. By taking c = − Γ (cid:0) (cid:1) Γ (cid:0) µ +13 (cid:1) Γ (cid:0) ρ +23 (cid:1) Γ (cid:0) (cid:1) Γ (cid:0) µ (cid:1) Γ (cid:0) ρ +13 (cid:1) c , we have U ( x ; µ, ρ ) = c Γ (cid:0) µ +13 (cid:1) Γ (cid:0) ρ +23 (cid:1) Γ (cid:0) (cid:1) Γ (cid:0) (cid:1) Γ (cid:0) µ +13 (cid:1) Γ (cid:0) ρ +23 (cid:1) F − µ , − ρ ; x ! − Γ (cid:0) (cid:1) x Γ (cid:0) µ (cid:1) Γ (cid:0) ρ +13 (cid:1) F − µ , − ρ ; x !! = c Γ (cid:0) µ +13 (cid:1) Γ (cid:0) ρ +23 (cid:1) Γ (cid:0) (cid:1) (1 − x ) µ + ρ − F µ , ρ +13 µ + ρ +23 ; 1 − x ! , uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche A. F. Loureiro and W. Van Assche where, for the latter identity, we have used [34, (15.10.18)]. In fact, for this choiceof c , it follows that lim x → − U ( x ; µ, ρ ) = 0, because F (cid:18) a, bc ; 1 (cid:19) = Γ( c )Γ( c − a − b )Γ( c − a )Γ( c − b )is valid whenever ℜ ( c − a − b ) >
0. Moreover, condition (3.68) is fulfilled, and, inaddition, we successively have Z x n U ( x ; µ, ρ ) d x = c Γ (cid:0) µ +13 (cid:1) Γ (cid:0) ρ +23 (cid:1) Γ (cid:0) (cid:1) Z x n (1 − x ) µ + ρ − F µ , ρ +13 µ + ρ +23 ; 1 − x ! d x = c Γ (cid:0) µ +13 (cid:1) Γ (cid:0) ρ +23 (cid:1) (cid:0) (cid:1) Z (1 − x ) n − x µ + ρ − F µ , ρ +13 µ + ρ +23 ; x ! , d x = c Γ (cid:0) (cid:1) Γ (cid:0) µ +13 (cid:1) Γ (cid:0) ρ +23 (cid:1) Γ (cid:0) µ + ρ +23 (cid:1) (cid:0) µ +23 (cid:1) Γ (cid:0) ρ + 1 (cid:1) (cid:0) (cid:1) n (cid:0) (cid:1) n (cid:0) µ +23 (cid:1) n (cid:0) ρ + 1 (cid:1) n , where, for the last identity, we have used [20, (7.512.4)]. If we take c = 3Γ (cid:0) µ +23 (cid:1) Γ (cid:0) ρ + 1 (cid:1) Γ (cid:0) (cid:1) Γ (cid:0) µ +13 (cid:1) Γ (cid:0) ρ +23 (cid:1) Γ (cid:0) µ + ρ +23 (cid:1) then (3.69) is fulfilled. As a result, the first orthogonality measure can be representedas in (3.33) (case k = 0) where U is given by (3.64).Now, in order to obtain the second orthogonality measure, we take λ = 0 in(3.67), which involves a derivative of U ( x ; µ, ρ ), that, according to [34, (15.5.4)],corresponds to U ′ ( x ; µ, ρ ) = − (cid:0) µ +23 (cid:1) Γ (cid:0) ρ + 1 (cid:1) Γ (cid:0) (cid:1) Γ (cid:0) (cid:1) Γ (cid:0) µ + ρ +23 (cid:1) ( µ + ρ − x (1 − x ) µ + ρ +23 − × F µ , ρ +13 µ + ρ +23 − − x ! . For µ = 0, it follows from (3.67) that µ ( µ + 2) U ( x ; µ, ρ ) = 9Γ (cid:0) µ +23 (cid:1) Γ (cid:0) ρ + 1 (cid:1) Γ (cid:0) (cid:1) Γ (cid:0) (cid:1) Γ (cid:0) µ + ρ +23 (cid:1) x (1 − x ) µ + ρ − (cid:0) µ + ρ − (cid:1) F µ , ρ +13 µ + ρ − ; 1 − x ! − (cid:0) ρ − (cid:1) F µ , ρ +13 µ + ρ +23 ; 1 − x ! ! = 3 µ Γ (cid:0) µ +23 (cid:1) Γ (cid:0) ρ + 1 (cid:1) Γ (cid:0) (cid:1) Γ (cid:0) (cid:1) Γ (cid:0) µ + ρ +23 (cid:1) x (1 − x ) µ + ρ − F µ + 1 , ρ +13 µ + ρ +23 ; 1 − x ! , where, for the last identity we have used [34, (15.5.15)]( c − F (cid:18) a, bc − − z (cid:19) − ( − a + c − F (cid:18) a, bc ; 1 − z (cid:19) = a F (cid:18) a + 1 , bc ; 1 − z (cid:19) . uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche Threefold symmetric Hahn-classical multiple orthogonal polynomials As a result, we obtain the expression (3.65) for U ( x ; µ, ρ ) and we have Z x n +1 U ( x ; µ, ρ ) d x = (cid:0) (cid:1) n (cid:0) (cid:1) n (cid:0) µ +53 (cid:1) n (cid:0) ρ +33 (cid:1) n = ( u ) n +1 , n > . When µ = 0, then (3.67) reads as U ′ ( x ; 0 , ρ ) = − (cid:0) ρ + 1 (cid:1) Γ (cid:0) (cid:1) Γ (cid:0) ρ − (cid:1) x (cid:0) − x (cid:1) ρ − , so that U ( x ; 0 , ρ ) = K − x Γ (cid:0) ρ + 1 (cid:1) Γ (cid:0) (cid:1) Γ (cid:0) ρ − (cid:1) F , − ρ ; x ! , for some integration constant K . We set K = ( ρ +1 ) Γ ( ) Γ ( ) Γ ( ρ +13 ) , use identity [34,(15.10.18)] to obtain U ( x ; 0 , ρ ) = 6Γ (cid:0) ρ + 1 (cid:1) Γ (cid:0) (cid:1) Γ (cid:0) ρ +23 (cid:1) x (1 − x ) ρ − F , ρ +13 ρ +23 ; 1 − x ! , which is the same as (3.65) with µ = 0.3.4.1. Differential equation and the cubic decomposition of { P n ( x ; µ, ρ ) } n > Let us assume µ > − ρ >
0. For each positive integer n , the polynomial P n ( x ; µ, ρ ) satisfies the differential equation (cid:0) x − (cid:1) y (3) ( x ) + x ( µ + ρ + 5) y ′′ ( x )+ x (cid:0) µρ + 14 µ − n − n ( µ + ρ + 2) − − n (2 µ − ρ + 1) + 18 ρ + 27 (cid:1) y ′ ( x )= n (4 µ + 2 n + 3( − n + 5) (2 n − − n + 4 ρ + 3) y ( x ) . The general solution of the latter equation can be written as P n ( x ; µ, ρ ) = c F (cid:18) − n , n + ( − n + µ + , n − ( − n + ρ + , ; x (cid:19) + c x F (cid:18) − n , n + ( − n + µ + , n − ( − n + ρ + , ; x (cid:19) + c x F (cid:18) − n , n + ( − n + µ + , n − ( − n + ρ + , ; x (cid:19) . which has only one polynomial solution for each n >
0. Similar to the precedentcases, for each positive integer n , there is only one (monic) polynomial solution anduly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche A. F. Loureiro and W. Van Assche it differs depending on whether n equals 0 , P n ( x ; µ, ρ ) := P [0] n ( x ; µ, ρ )= ( − n (cid:0) (cid:1) n (cid:0) (cid:1) n (cid:0) n + ( − n + µ + (cid:1) n (cid:0) n − ( − n + ρ + (cid:1) n · F − n, n + ( − n + µ + , n − ( − n + ρ + , ; x ! ,P n +1 ( x ; µ, ρ ) := xP [1] n ( x ; µ, ρ )= x ( − n (cid:0) (cid:1) n (cid:0) (cid:1) n (cid:0) n − ( − n + µ + (cid:1) n (cid:0) n + ( − n + ρ + (cid:1) n · F − n, n − ( − n + µ + , n + ( − n + ρ + , ; x ! ,P n +2 ( x ; µ, ρ ) := x P [2] n ( x ; µ, ρ )= x ( − n (cid:0) (cid:1) n (cid:0) (cid:1) n (cid:0) n + ( − n + µ + (cid:1) n (cid:0) n − ( − n + ρ + (cid:1) n · F − n, n + ( − n + µ + , n − ( − n + ρ + , ; x ! . Here, the sequences { P [ k ] n ( · ; µ, ρ ) } n > , with k ∈ { , , } , are precisely the 2-orthogonal polynomial sequences in the cubic decomposition of { P n ( · ; µ, ρ ) } n > .From Lemma 2.1, these three 2-orthogonal polynomial sequences are not threefoldsymmetric and satisfy the recurrence relation (2.9). These coefficients have beencomputed in [14, Tableau 1, 5 and 9 - Case A], for a different choice of the ”free”parameter γ . We have included them in the Appendix for completeness, where wehave used the software Mathematica .3.4.2.
The sequence of derivatives
The 2-orthogonal polynomial sequence { Q n ( x ; µ, ρ ) := n +1 P ′ n +1 ( x ; µ, ρ ) } n > satis-fies the relation (3.1) with e γ n := e γ n ( µ, ρ ) = 2 n (2 n + 1)( n + ρ + 1)( µ + 3 n + 2)(3 n + ρ )(3 n + ρ + 3) , n > , e γ n +1 := e γ n +1 ( µ, ρ ) = 2( n + 1)(2 n + 1)( µ + n + 2)( µ + 3 n + 2)( µ + 3 n + 5)(3 n + ρ + 3) , n > , (3.70)whose expressions are derived from (3.17) and (3.63). A straightforward comparisonbetween (3.63) and (3.70) readily shows that e γ n +1 ( µ, ρ ) = γ n +1 ( ρ + 1 , µ + 2) , n > . uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche Threefold symmetric Hahn-classical multiple orthogonal polynomials This, combined with the uniqueness of a polynomial sequence defined by the recur-rence relation (3.1) implies Q n ( x ; µ, ρ ) = P n ( x ; ρ + 1 , µ + 2) which means1 n + 1 P ′ n +1 ( x ; µ, ρ ) = P n ( x ; ρ + 1 , µ + 2) , n > . (3.71)3.4.3. Particular cases
The so called
Humbert polynomials introduced in [22] correspond to { P n ( x ; ν − , ν ) } n > , up to a scaling of the variable. In fact, by setting µ = ν − and ρ = ν , this 2-orthogonal polynomial sequence satisfies P n +2 ( x ; ν − , ν )= xP n +1 ( x ; ν − , ν ) − n ( n + 1)(3 ν + n − ν + n − ν + n )( ν + n + 1) P n − ( x ; ν − , ν ) , with initial conditions P = 1 , P ( x ) = x and P ( x ) = x . Baker [5, p.60] gaveexplicit expressions for this particular case via F hypergeometric series and thesecoincide with those obtained above. Pincherle polynomials , introduced in [35], area particular case obtained with ν = 1 /
2, which received special attention. For in-stance, they were discussed in [30] and several properties were further analyzed in[26] with the focus of the study on the algebraic properties, including generatingfunctions, as well as the analysis of the components arising in the cubic decomposi-tion, but integral representations for the orthogonality measures were not included.Prior to this work, Douak and Maroni [16] analyzed in detail the case where ν = 1(and therefore µ = 1 and ν = 3 / γ -coefficients, namely P n +2 ( x ; 1 , ) = xP n +1 ( x ; 1 , ) − P n − ( x ; 1 , ) , with the same initial conditions. In other words, this means that the associatedsequence coincides with the original one, and, regarding the nature of the problem,they referred to this polynomial sequence as Chebyshev type polynomials . Therein,an integral representation for the orthogonality functionals was given, with supporton an interval on the real line (which obviously does not contain all the zeros ofthe polynomial sequence [16, Theorem 4.1]), which is, from our point of view, a bitartificial. Following Proposition 3.5, the polynomial sequence { P n ( x ; 1 , ) } n > is 2-orthogonal with respect to ( u , u ) which admit the integral representation (3.33)with b = 1 and U ( x ; 1 , ) = 9 √ π (cid:16) p − x (cid:17) / − (cid:16) − p − x (cid:17) / ! , U ( x ; 1 , ) = 27 √ π (cid:18)(cid:16)p − x + 1 (cid:17) / − (cid:16) − p − x (cid:17) / (cid:19) , after using [34, Eq. (15.4.9)].Another particular case that made its appearance in the literature correspondsto the case where µ = − / ρ = 0, and therefore γ n +2 = e γ n = 4 /
27 and γ =uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche A. F. Loureiro and W. Van Assche /
9. In fact, { P n ( x ; − , } n > is a particular case of Faber polynomials , named afterthe author of [18], and this polynomial sequence with almost constant coefficientshas been studied in [17] and [21]. The latter paper was essentially devoted to thestudy of the zeros, while the former was mainly dedicated to the algebraic propertiesas well as a representation of the measure on an interval on the real line. Here, wecombine the two approaches in a more general setting. So, { P n ( x ; − , } n > is 2-orthogonal with respect to ( u , u ) which admit the integral representation (3.33)with b = 1 and U ( x ; − ,
0) = 3 √ (cid:16)(cid:0) − √ − x (cid:1) / + (cid:0) √ − x (cid:1) / (cid:17) π √ − x , U ( x ; − ,
0) = 9 √ (cid:16)(cid:0) − √ − x (cid:1) / + (cid:0) √ − x (cid:1) / (cid:17) π √ − x . The penultimate and the latter particular cases are related to each other: if werecall (3.71) we readily see that1 n + 1 P ′ n +1 ( x ; − ,
0) = P n ( x ; 1 , ) , n > . It turns out that, the sequence { R n ( x ; − , } n > of the associated polynomials of { P n ( x ; − , } n > , defined by R n ( x ; − ,
0) := (cid:28) u , P n ( x ; − , − P n ( t ; − , x − t (cid:29) actually coincides with { P n ( x ; 1 , ) } n > , as discussed in [17]. Acknowledgments
We thank the anonymous referees for their careful reading and for the sugges-tions and corrections, which led to an improvement of the paper. The research ofWVA was supported by FWO research grant G.0864.16N and EOS project PRIMA30889451.
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Multiple orthogonal polynomials associated withMacDonald functions , Integral Transform Special Funct. (2000), 229–244.[39] J. Van Iseghem, Vector orthogonal relations, vector QD-algorithm , J. Comput. Appl.Math. (1987), 141–150.[40] D. V. Widder, The Airy Transform , Amer. Math. Monthly (1979), 271–277. uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche Threefold symmetric Hahn-classical multiple orthogonal polynomials Appendix
As discussed in Lemma 2.1, the polynomial sequences { P [ j ] n ( x ) } n > , with j = 0 , , P n ( x ) are 2-orthogonal polynomials. The expressions for the correspondingrecurrence coefficients can be obtained from the expressions of the γ -coefficients.So, in the light of Lemma 2.1, we computed the expressions of the recurrence co-efficients for each of the cubic components of the polynomial sequences discussedin Cases B and C, and this is detailed below. We have included them here fora matter of completion, as these polynomial sequences { P [ j ] n ( x ) } n > were alreadydescribed in [14] up to a linear change of variable. Case B β [0]2 n ( µ ) = 2( µ + 3 n (3 µ + 9 n (2 µ + 14 n + 3) − − µ + 9 n − µ + 9 n + 2) β [0]2 n +1 ( µ ) = 2(19 µ + 3 n (21 µ + 9 n (2 µ + 10 n + 17) + 77) + 32)3( µ + 9 n + 2)( µ + 9 n + 8) α [0]2 n ( µ ) = 4 n (3 n − n − (cid:16) − µ − µ + 702 n + 27(8 µ − n + 3(6( µ − µ − n + 8 (cid:17) µ + 9 n − µ + 9 n − ( µ + 9 n + 2) α [0]2 n +1 ( µ ) = 4(2 n + 1)(3 n + 1)(6 n + 1) (cid:16) µ + 3 (cid:16) µ + 117 n + 9(4 µ + 9) n + (3 µ ( µ + 6) + 5) n (cid:17) − (cid:17) µ + 9 n − µ + 9 n + 2) ( µ + 9 n + 5) γ [0]2 n ( µ ) = 8 n (2 n + 1)(3 n − n + 1)(6 n − n + 1)( µ + 3 n − µ + 3 n )( µ + 3 n + 1)3( µ + 9 n − µ + 9 n − ( µ + 9 n + 2) ( µ + 9 n + 5) γ [0]2 n +1 ( µ ) = 8( n + 1)(2 n + 1)(3 n + 1)(3 n + 2)(6 n + 1)(6 n + 5)3( µ + 9 n + 2)( µ + 9 n + 5)( µ + 9 n + 8) β [1]2 n ( µ ) = 8( µ −
1) + 6 n (9 µ + 9 n (2 µ + 10 n + 7) + 5)3( µ + 9 n − µ + 9 n + 5) β [1]2 n +1 ( µ ) = 2(31 µ + 3 n (27 µ + 9 n (2 µ + 14 n + 31) + 202) + 143)3( µ + 9 n + 5)( µ + 9 n + 8) α [1]2 n ( µ ) = 4 n (cid:16) n − (cid:17) (cid:16) − µ + 3 n (cid:16) µ − µ + 117 n + 36( µ − n − (cid:17) + 4 (cid:17) µ + 9 n − µ + 9 n − ( µ + 9 n + 2) α [1]2 n +1 ( µ ) = 4(2 n + 1)(3 n + 1)(3 n + 2)3( µ + 9 n + 2)( µ + 9 n + 5) ( µ + 9 n + 8) (cid:16) ( µ + 2)(9 µ + 37) + 702 n + 27(8 µ + 43) n + 3(6 µ ( µ + 13) + 187) n (cid:17) γ [1]2 n ( µ ) = 8 n (2 n + 1)(3 n + 1)(3 n + 2)(6 n − n + 1)3( µ + 9 n − µ + 9 n + 2)( µ + 9 n + 5) γ [1]2 n +1 ( µ ) = 8( n + 1)(2 n + 1)(3 n + 1)(3 n + 2)(6 n + 5)(6 n + 7)( µ + 3 n + 1)( µ + 3 n + 2)( µ + 3 n + 3)3( µ + 9 n + 2)( µ + 9 n + 5) ( µ + 9 n + 8) ( µ + 9 n + 11) uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche A. F. Loureiro and W. Van Assche β [2]2 n ( µ ) = 20( µ + 2) + 6 n (15 µ + 9 n (2 µ + 14 n + 17) + 58)3( µ + 9 n + 2)( µ + 9 n + 5) β [2]2 n +1 ( µ ) = 2(46 µ + 3 n (33 µ + 9 n (2 µ + 10 n + 27) + 209) + 170)3( µ + 9 n + 5)( µ + 9 n + 11) α [2]2 n ( µ ) = 4 n (3 n + 1)(6 n + 1) (cid:16) µ + 3 (cid:16) µ + 234 n + 9(8 µ + 17) n + (6 µ ( µ + 5) + 7) n (cid:17) − (cid:17) µ + 9 n − µ + 9 n + 2) ( µ + 9 n + 5) α [2]2 n +1 ( µ ) = 4(2 n + 1)(3 n + 2)(6 n + 5)3( µ + 9 n + 2)( µ + 9 n + 5) ( µ + 9 n + 8) (cid:16) µ + 2)(3 µ + 10) + 351 n + 54(2 µ + 11) n + 3(3 µ ( µ + 14) + 98) n (cid:17) γ [2]2 n ( µ ) = 8 n (2 n + 1)(3 n + 1)(3 n + 2)(6 n + 1)(6 n + 5)( µ + 3 n )( µ + 3 n + 1)( µ + 3 n + 2)3( µ + 9 n − µ + 9 n + 2) ( µ + 9 n + 5) ( µ + 9 n + 8) γ [2]2 n +1 ( µ ) = 8( n + 1)(2 n + 1)(3 n + 2)(3 n + 4)(6 n + 5)(6 n + 7)3( µ + 9 n + 5)( µ + 9 n + 8)( µ + 9 n + 11) Case C β [0]2 n ( µ, ρ ) = 2(2 n + 1)(3 n + 1)(6 n + 1)(9 n + µ + 2)(9 n + ρ + 3) − n (3 n − n − n + µ − n + ρ − β [0]2 n +1 ( µ, ρ ) = 4( n + 1)(3 n + 2)(6 n + 5)(9 n + µ + 8)(9 n + ρ + 6) − n + 1)(3 n + 1)(6 n + 1)(9 n + µ + 2)(9 n + ρ + 3) α [0]2 n ( µ, ρ ) = 6 n (6 n − n − n + µ − n + µ − (9 n + ρ − (9 n + ρ ) (6 n − n + µ − n + ρ − n − n + µ − n + ρ − n + ρ − n + ρ − n + 1)(3 n + µ − n + µ )(9 n + ρ − n + µ + 2 ! α [0]2 n +1 ( µ, ρ ) = 6(2 n + 1)(6 n + 1)(9 n + µ + 2) (9 n + µ + 5)(9 n + ρ )(9 n + ρ + 3) n + µ + 1)(3 n + ρ )(3 n + 1) + 6 n (3 n + µ )(3 n + µ + 1)(9 n + ρ + 3)(3 n + 1)9 n + µ − n + 2)(6 n + 2)(9 n + µ + 2)(3 n + ρ )(3 n + ρ + 1)9 n + ρ + 6 ! γ [0]2 n ( µ, ρ ) = 6 n (6 n − n − n + 1)(6 n + 2)(6 n + 3)( µ + 3 n − µ + 3 n )( µ + 3 n + 1)( µ + 9 n − µ + 9 n − ( µ + 9 n + 2) ( µ + 9 n + 5)(9 n + ρ − n + ρ )(9 n + ρ + 3) γ [0]2 n +1 ( µ, ρ ) = 6 n (6 n − n − n + 1)(6 n + 2)(6 n + 3)( µ + 3 n − µ + 3 n )( µ + 3 n + 1)( µ + 9 n − µ + 9 n − ( µ + 9 n + 2) ( µ + 9 n + 5)(9 n + ρ − n + ρ )(9 n + ρ + 3) uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche Threefold symmetric Hahn-classical multiple orthogonal polynomials β [1]2 n ( µ, ρ ) = 4(2 n + 1)(3 n + 1)(3 n + 2)(9 n + µ + 5)(9 n + ρ + 3) + 2 n − n (9 n + µ − n + ρ ) β [1]2 n +1 ( µ, ρ ) = 2( n + 1)(6 n + 5)(6 n + 7)(9 n + µ + 8)(9 n + ρ + 9) − n + 1)(3 n + 1)(3 n + 2)(9 n + µ + 5)(9 n + ρ + 3) α [1]2 n ( µ, ρ ) = 6 n (6 n − n + 1)(9 n + µ − (9 n + µ + 2)(9 n + ρ − n + ρ ) n (3 n + µ )(3 n + ρ − n + 2)(9 n + µ − n + ρ )(3 n + ρ − n + ρ + 3 + (6 n − n + µ − n + µ )(9 n + ρ )9 n + µ − ! α [1]2 n +1 ( µ, ρ ) = 6(2 n + 1)(9 n + µ + 2)(9 n + µ + 5) (9 n + ρ + 3) (9 n + ρ + 6) × n + 1)(3 n + 1)(3 n + 2)(3 n + µ + 1)(3 n + ρ + 1)+ (3 n + 2)(6 n + 1)(6 n + 2)(9 n + µ + 5)(3 n + ρ )(3 n + ρ + 1)9 n + ρ + (3 n + 1)(6 n + 4)(6 n + 5)(3 n + µ + 1)(3 n + µ + 2)(9 n + ρ + 3)9 n + µ + 8 ! γ [1]2 n ( µ, ρ ) = 6 n (6 n − n + 1)(6 n + 2)(6 n + 3)(6 n + 4)(3 n + ρ − n + ρ )(3 n + ρ + 1)( µ + 9 n − µ + 9 n + 2)( µ + 9 n + 5)(9 n + ρ − n + ρ ) (9 n + ρ + 3) (9 n + ρ + 6) γ [1]2 n +1 ( µ, ρ ) = 6 n (6 n − n + 1)(6 n + 2)(6 n + 3)(6 n + 4)(3 n + ρ − n + ρ )(3 n + ρ + 1)( µ + 9 n − µ + 9 n + 2)( µ + 9 n + 5)(9 n + ρ − n + ρ ) (9 n + ρ + 3) (9 n + ρ + 6) uly 16, 2019 0:50 WSPC/INSTRUCTION FILEaa˙style˙Loureiro˙VanAssche A. F. Loureiro and W. Van Assche β [2]2 n ( µ, ρ ) = 2(2 n + 1)(3 n + 2)(6 n + 5)(9 n + µ + 5)(9 n + ρ + 6) − n (3 n + 1)(6 n + 1)(9 n + µ + 2)(9 n + ρ ) β [2]2 n +1 ( µ, ρ ) = 4( n + 1)(3 n + 4)(6 n + 7)(9 n + µ + 11)(9 n + ρ + 9) − n + 1)(3 n + 2)(6 n + 5)(9 n + µ + 5)(9 n + ρ + 6) α [2]2 n ( µ, ρ ) = 6 n (6 n + 1)(6 n + 2)(9 n + µ − n + µ + 2) (9 n + ρ ) (9 n + ρ + 3) (6 n + 1)(3 n + µ )(3 n + ρ )+ (6 n − n + µ + 2)(3 n + ρ − n + ρ )9 n + ρ −
3+ (6 n + 3)(3 n + µ )(3 n + µ + 1)(9 n + ρ )9 n + µ + 5 ! α [2]2 n +1 ( µ, ρ ) = 6(2 n + 1)(6 n + 5)(9 n + µ + 5) (9 n + µ + 8)(9 n + ρ + 3)(9 n + ρ + 6) n + µ + 2)(3 n + ρ + 1)(3 n + 2) + 6( n + 1)(9 n + µ + 5)(3 n + ρ + 1)(3 n + ρ + 2)(3 n + 2)9 n + ρ + 9+ (3 n + 1)(6 n + 4)(3 n + µ + 1)(3 n + µ + 2)(9 n + ρ + 6)9 n + µ + 2 ! γ [2]2 n ( µ, ρ ) = 6 n (6 n + 1)(6 n + 2)(6 n + 3)(6 n + 4)(6 n + 5)( µ + 3 n )( µ + 3 n + 1)( µ + 3 n + 2)( µ + 9 n − µ + 9 n + 2) ( µ + 9 n + 5) ( µ + 9 n + 8)(9 n + ρ )(9 n + ρ + 3)(9 n + ρ + 6) γ [2]2 n +1 ( µ, ρ ) = 6 n (6 n + 1)(6 n + 2)(6 n + 3)(6 n + 4)(6 n + 5)( µ + 3 n )( µ + 3 n + 1)( µ + 3 n + 2)( µ + 9 n − µ + 9 n + 2) ( µ + 9 n + 5) ( µ + 9 n + 8)(9 n + ρ )(9 n + ρ + 3)(9 n + ρρ