# Catalan–Qi Numbers, Series Involving the Catalan–Qi Numbers and a Hankel Determinant Evaluation

Authors: Wathek Chammam
Journal: Journal of Mathematics
 (2020)

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## Abstract

In this paper, using a study of the polynomial of Jacobi, we give an evaluation of the Hankel determinants that are associated with the sequence of Catalan–Qi numbers and several sequences of series involving the Catalan–Qi numbers.


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## Body

## 1. Introduction

In [1], Chammam generalizes several formulas and series identities involving the Catalan numbers and establishes several new formulas and series identities. In this paper, we continue our study of analytic and algebraic properties of Catalan–Qi numbers and series involving the Catalan–Qi numbers by giving a complete evaluation of their Hankel determinants. This type of study generalizes the work of Aloui [2] and Chammam [3, 4] and is based on the algebraic analysis of Jacobi orthogonal polynomials. The studies of these determinants are very important in different scientific disciplines, for example, in physics, especially the resolution of Toda’s equation [5].


## 2. Notation and Preliminary Results

Letℙ be the linear space of polynomials in one variable with complex coefficients and ℙ′ its topological dual space. We denote by u,q the action of u∈ℙ′ on q∈ℙ and by un≔u,xn, n≥0, the sequence of moments of u with respect to the polynomial sequence xnn≥0. We define some operations in ℙ′. For linear functionals u and v, any polynomial p and any a,b,c∈ℂ∗×ℂ2, let Du=u′,pu, x−c−1u, τ−bu, and hau be the linear functionals defined by duality [6, 7]:(1)u′,q≔−u,q′,pu,q≔u,pq,x−c−1u,q≔u,θcq=u,qx−qcx−c,τ−bu,q≔u,τbq=u,qx−b,hau,q≔u,haq=u,qax,q∈ℙ.A linear functionalu is called normalized if it satisfies u0=1.With any sequence of complex numberscnn≥0, we can associate a unique linear functional u∈ℙ′ given by un=cn,n≥0.The linear functionalu is said to be regular if the Hankel determinant Hnu=detci+ji,j=0n≠0 for every integer n≥0. In this case, there exists a unique sequence of monic polynomials (SMPs) J^nn≥0; i.e., degJ^n=n in a way that(2)u,xνJ^n=0,0≤ν≤n−1,andu,xnJ^n≠0.The sequenceJ^nn≥0 is said to be the sequence of monic orthogonal polynomials (SMOPs) with respect to u.Notice thatJ^nx can be represented by the following formula:(3)J^0x=1,J^nx=1Hn−1uc0c1⋯cnc1c2⋯cn+1⋮⋮⋱⋮cn−1cn⋯c2n−11x⋯xn,n≥1.The orthogonality ofJ^nn≥0 can be characterized by a three-term recurrence relation (TTRR) [6]:(4)J^n+1x=x−βnj^nx−γnj^n−1x,n≥0.With initial valuesJ^0x=1andJ^−1x=0, where βnn≥0 and γnn≥0 are sequences of complex numbers such as γn≠0,n≥1, and the convention γ0=u0.Furthermore,(5)Hnu=J^n+11xJ^n+1x−J^n1xJ^n+2x=∏ν=0nγν,n≥0.WhenJ^nn≥0 is a SMOP with respect to a linear functional u, then the sequence of monic polynomials J˜nn≥0, where J˜nx=a−nJ^nax+b, is also orthogonal with respect to u˜=ha−1°τ−bu and satisfies the following TTRR [6, 7]:(6)J˜n+1x=x−β˜nJ˜nx−γ˜nJ˜n−1x,n≥0.With initial valuesJ˜0x=1 and J˜−1x=0 and where(7)β˜n=a−1βn−b,n≥0,γ˜n+1=a−2γn+1,n≥0,γ˜0=u0.A linear functionalu is said to be classical when it is quasidefinite and there exist two polynomials ϕ and ψ, ϕ monic, degϕ=t≤2, and degψ=1, such that u satisfies Pearson’s equation (see [6, 7]):(8)ϕu′+ψu=0.In such a case, the corresponding SMOPJ^nn≥0 is said to be classical. Any shift leaves invariant the classical character. Indeed, the shifted linear functional u˜=ha−1°τ−bu fulfils [6]:(9)ϕ˜u˜′+ψ˜u˜=0,where ϕ˜x=a−tϕax+b and ψ˜x=a1−tψax+b.


## 3. Hankel Determinants of Catalan–Qi Numbers

In 2019, Chammam [3] introduced a sequence of complex numbers that are called the Jacobi numbers, which is defined as follows.Letξ,α,β∈ℂ3, with ξ≠0, α≠−n, β≠−n, and α+β≠−n−1, n≥1:(10)jnξ,α,β=ξnΓn+α+1Γα+1Γα+β+2Γn+α+β+2,n≥0.With any sequence of complex numberscnn≥0, we can associate a unique linear functional u∈ℙ′ given by un=cn,n≥0. If we consider the form J˜ξ,α,β, moments with respect to the sequence xnn≥0 are jξ,α,β=jnξ,α,βn≥0. Equivalently, J˜ξ,α,β is monic i.e.,J˜ξ,α,β0=1 and satisfies the following functional equation:(11)xx−ξJ˜ξ,α,β′−α+β+2x−ξα+1J˜ξ,α,β=0.While (9) and [6], J˜ξ,α,β is a shifted Jacobi; more precisely,(12)J˜ξ,α,β=hξ/2°τ1Jα,β.Under the assumptionsξ≠0, α≠−n, β≠−n et α+β≠−n−1, n≥1, the linear functionals Jα,β and their shifted J˜ξ,α,β are regular. Denoting by J^nx;α,βn≥0, (resp., J˜nx;ξ,α,βn≥0) be the SMOP with respect to Jα,β (resp., J˜ξ,α,β).In view of (12), the following connection relation is immediate [6, 7]:(13)J˜nx;ξ,α,β=ξ2nJ^n2x−ξξ;α,β,n≥0.From the TTRR satisfied by the Jacobi SMOPJ˜nx;ξ,α,βn≥0, we can obtain the TTRR satisfied by J^nx;ξ,α,βn≥0:(14)J^−1x;α,β=0,J^0x;α,β=1,J^n+1x;α,β=x−βnα,βJ^nx;α,β−γnα,βJ^n−1x;α,β,n≥0,where(15)β0α,β=α−βα+β+2,βn+1α,β=α2−β22n+α+β+22n+α+β+4,n≥0,γnα,β=4nn+α+βn+αn+β2n+α+β−12n+α+β22n+α+β+1,n≥1.So,(16)J˜−1x;ξ,α,β=0,J˜0x;ξ,α,β=1,J˜n+1x;ξ,α,β=x−β˜nξ,α,βJ˜nx;ξ,α,β−γ˜nξ,α,βJ˜n−1x;ξ,α,β,n≥0,where(17)β˜0ξ,α,β=ξα+1α+β+2,β˜n+1ξ,α,β=ξα2−β222n+α+β+22n+α+β+4+ξ2,n≥0,γ˜nξ,α,β=ξ2nn+α+βn+αn+β2n+α+β−12n+α+β22n+α+β+1,n≥1.In the same article, Chammam has determined a complete evaluation of the Hankel determinant associated with the sequencejξ,α,β. It proves the following result.Theorem 1 (see [3], Theorem 1).
Ifξ, α, and β are complex numbers such that ξ≠0, α≠−n, β≠−n, and α+β≠−n−1, n≥1, then(18)Hnjξ,α,β=ξnβ+1α+β+1αn+1n+1!∏ν=0n2ν+α+βν+α2ν+α+β+1ν+α+β,n≥0.
Recall a binomial coefficient is indexed by a pair of integers0≤k≤n and is written as nk=n!/n−k!k!=Γn+1/Γk+1Γn−k+1, and this coefficient can be generalized to all complex s≠−n,t≠−n, and s−t≠−n by the following formula:(19)st=Γs+1Γt+1Γs−t+1.


### 3.1. Application: Hankel Determinant of Catalan–Qi Numbers

For the past few years, the Professor Feng Qi and his coauthors, in defining the Catalan–Qi numbers, have published out important results of this nice number [8–12].This generalization of the Catalan numbersCn was introduced according to Euler–Gamma function as follows:(20)Ca,b;n=ΓbΓabanΓn+aΓn+b=bananbn,Ra,Rb>0,where the rising factorials xn are defined by(21)xn=xx+1x+2,…,x+n−1,n≥1,1,n=0,−xn=−1nx+n−1n.Ifa and b are complex numbers such that Rb,Ra>0,Rb>Ra+1, and for ξ,α,β=b/a,a−1,b−a−1, by substitution in the last theorem, we obtain a completed evaluation of the Hankel determinants that are associated with the sequence of Catalan–Qi numbers:(22)HnCa,b;n=b/ab−ab−1a−1n+1n+1!∏ν=0n2ν+b−2ν+a−12ν+b−1ν+b−2,n≥0.Fora=1/2 and b=2 and using the Gamma duplication formula, we obtained(23)HnCn=HnC12,2;n=1,n≥0.


## 3.1. Application: Hankel Determinant of Catalan–Qi Numbers

For the past few years, the Professor Feng Qi and his coauthors, in defining the Catalan–Qi numbers, have published out important results of this nice number [8–12].This generalization of the Catalan numbersCn was introduced according to Euler–Gamma function as follows:(20)Ca,b;n=ΓbΓabanΓn+aΓn+b=bananbn,Ra,Rb>0,where the rising factorials xn are defined by(21)xn=xx+1x+2,…,x+n−1,n≥1,1,n=0,−xn=−1nx+n−1n.Ifa and b are complex numbers such that Rb,Ra>0,Rb>Ra+1, and for ξ,α,β=b/a,a−1,b−a−1, by substitution in the last theorem, we obtain a completed evaluation of the Hankel determinants that are associated with the sequence of Catalan–Qi numbers:(22)HnCa,b;n=b/ab−ab−1a−1n+1n+1!∏ν=0n2ν+b−2ν+a−12ν+b−1ν+b−2,n≥0.Fora=1/2 and b=2 and using the Gamma duplication formula, we obtained(23)HnCn=HnC12,2;n=1,n≥0.


## 4. Hankel Determinants of Series Involving the Catalan–Qi Numbers

Our second objective, essential although complex, concerns the evaluation of the Hankel determinant of ordern, n∈ℕ associated with the sequence Sξ,α,β≔Snξ,α,βn≥0 given by(24)Snξ,α,β=∑ν=0nξνjn−νξ,α,β,n≥0.The series involving the Catalan–Qi numbers are a special case, and it is clear [1] that(25)Snba,a−1,b−a−1=∑ν=0nbaνCa,b;n−ν,n≥0.The sequenceSξ,α,β=Snξ,α,βn≥0 satisfies(26)S−1ξ,α,β=0,Snξ,α,β=ξSn−1ξ,α,β+jnξ,α,β,n≥0.Using (26), elementary row operations give(27)HnSξ,α,β=S0ξ,α,βS1ξ,α,β…Snξ,α,βj1ξ,α,βj2ξ,α,β…jn+1ξ,α,βj2ξ,α,βj3ξ,α,β…jn+2ξ,α,β⋮⋮⋮jnξ,α,βjn+1ξ,α,β…j2nξ,α,β,n≥1.From (24), we can write(28)Snξ,α,β=J˜ξ,α,βxnξ,n≥0,where J˜ξ,α,βxn is the right multiplication of J˜ξ,α,β by xn.Taking into account (28), we can write(29)HnSξ,α,β=J˜ξ,α,βTnx;ξ,α,βξ,n≥0,where T0x;ξ,α,β=1 and(30)Tnx;ξ,α,β=1x…xnj1ξ,α,βj2ξ,α,β…jn+1ξ,α,βj2ξ,α,βj3ξ,α,β…jn+2ξ,α,β⋮⋮⋮jnξ,α,βjn+1ξ,α,β…j2nξ,α,β,n≥1.Observe that(31)α+β+2jn+1ξ,α,β=ξα+1jnξ,α+1,β,n≥0.It follows that(32)Tnx;ξ,α,β=ξα+1α+β+2n1x…xnj0ξ,α+1,βj1ξ,α+1,β…jnξ,α+1,βj1ξ,α+1,βj2ξ,α+1,β…jn+1ξ,α+1,β⋮⋮⋮jn−1ξ,α+1,βjnξ,α+1,β…j2n−1ξ,α+1,β.From the regularity of the linear functionalJ˜ξ,α,β of which its corresponding SMOP is J˜nx;ξ,α+1,βn≥0, we find(33)Tnx;ξ,α,β=Θnξ,α,βJ˜nx;ξ,α+1,β,n≥0,where(34)Θnξ,α,β=−1nHn−1jξ,α+1,βξα+1α+β+2n.Substituting (29) into (33), we find(35)HnSξ,α,β=Θnξ,α,βJ˜ξ,α,βJ˜n;ξ,α+1,βξ,n≥0.Lemma 1.
Ifξ, α, and β are complex numbers such that ξ≠0, α≠−n, β≠−n, and α+β≠−n−1, n≥1, then(36)HnSξ,α,β=Θnξ,α,βJ˜nξ;ξ,α+1,β+ξα+1α+β+2Jn−11ξ;ξ,α+1,β,where(37)Θnξ,α,β=−1nHn−1jξ,α+1,βξα+1α+β+2n,n≥0.
In the above,J˜n1x;ξ,α+1,βn≥0 is the sequence of associated polynomials of the first kind of the sequence J˜nx;ξ,α+1,βn≥0.Proof.
Set(38)Rnx≔J˜ξ,α,βJ˜n.;ξ,α+1,β,n≥0.
SinceJ˜nx;ξ,α+1,β are normalized with degJ˜nx;ξ,α+1,β=n and J˜ξ,α,β is monic, then Rnn≥0 is a sequence of monic polynomials with degRn=n.
In order to establish a recurrence relation verified byRnn≥0, we need the following formula [6]:(39)uxpx=xupx+yuy,py,p∈ℙ,p∈ℙ′.
Using (39) with u=J˜ξ,α,β and px=J˜nx;ξ,α+1,β, we get(40)Rn+1x=x−β˜nξ,α+1,βRnx−γ˜nξ,α+1,βRn−1x+yJ˜ξ,α,βy,J˜ny;ξ,α+1,β,n≥0.
But, in view of (31),(41)α+β+2xJ˜ξ,α,β=ξα+1J˜ξ,α+1,β.
Then,(42)Rn+1x=x−β˜nξ,α+1,βRnx−γ˜nξ,α+1,βRn−1x+ξα+1α+β+2J˜ξ,α+1,βy,J˜ny;ξ,α+1,β,n≥0.
SinceJ˜nx;ξ,α+1,βn≥0 is a SMOP with respect to J˜ξ,α+1,β, it comes that J˜ξ,α+1,βy,J˜ny;ξ,α+1,β=δn,0 (δn,0 is the “Kronecker delta”). Obviously, the SMP Rnn≥0 is orthogonal because it satisfies the following TTRR:(43)R0x=1,R1x=x−β˜0ξ,α+1,β+ξα+1α+β+2,Rn+1x=x−β˜nξ,α+1,βRnx−γ˜nξ,α+1,βRn−1x,n≥1.
While referring to [6], the SMOP Rnn≥0 is the corecursive of the SMOP J˜nx;ξ,α+1,βn≥0 since it is generated by the TTRR (16), where α is replaced by α+1 and in which β˜0ξ,α+1,β is replaced by β˜0ξ,α+1,β−ξα+1α+β+2−1. To this topic, let us recall that the following connection relation holds:(44)Rnx=J˜nx;ξ,α+1,β+ξα+1α+β+2J˜n−11x;ξ,α+1,β,n≥0.
This leads to the desired expression ofHnsξ,α,β.Lemma 2.
Ifξ, α, and β are complex numbers such that ξ≠0, α≠−n, β≠−n, and α+β≠−n−1, n≥1, then the following holds:(i)
J˜nξ;ξ,α+1,β=ξnΓn+α+β+2Γn+β+1/Γ2n+α+β+2Γβ+1,n≥0,.(ii)
J˜n1ξ;ξ,α+1,β=α+β+2ξnXnα,β/βΓα+2Γβ+1Γ2n+α+β+4,n≥0.(iii)
Ifβ≠0 and where Xnα,β=Γα+2Γn+α+β+3Γn+β+2−n+1!Γβ+1Γα+β+2Γn+α+3.Proof.
(i)
By (13), it is clear that J˜nξ;ξ,α+1,β=ξ/2nJ^n1;α+1,β. It is well known that J^nx;α+1,βn≥0 satisfies a first structure relation [6]:(45)x2−1J^n+1′x;α+1,β=n+1x+α+1−β2n+α+β+3J^n+1x;α+1,β−2n+α+β+4γn+1α+1,βJ^nx;α+1,β,n≥0,where(46)γn+1α+1,β=4n+1n+α+β+2n+α+2n+β+12n+α+β+22n+α+β+322n+α+β+4.
Letx=1 in (45) to get(47)J^n+11;α+1,β=2n+α+β+2n+β+12n+α+β+22n+α+β+3J^n1;α+1,β,n≥0.
Then,(48)J^n1;α+1,β=2nΓn+α+β+2Γn+β+1Γ2n+α+β+2Γβ+1,n≥0.
And so that(49)J˜nξ;ξ,α+1,β=ξnΓn+α+β+2Γn+β+1Γ2n+α+β+2Γβ+1,n≥0.
Using (5) with Pnx=J˜nx;ξ,α+1,β and x=ξ, we get for n≥0:(50)J˜n+11ξ;ξ,α+1,βJ˜n+2ξ;ξ,α+1,β−J˜n1ξ;ξ,α+1,βJ˜n+1ξ;ξ,α+1,β=∏ν=0nγ˜ν+1ξ,α+1,βJ˜n+1ξ;ξ,α+1,βJ˜n+2ξ;ξ,α+1,β.
But(51)∏ν=0nγ˜ν+1ξ,α+1,β=ξ2n+1n+1!Γn+α+β+3Γn+α+3Γn+β+2Γα+β+3Γ2n+α+β+4Γ2n+α+β+5Γα+2Γβ+1.
And so that(52)∏ν=0nγ˜ν+1α+1,βP˜n+1ξ;ξ,α+1,βP˜n+2ξ;ξ,α+1,β=κξ,α,βn+1!2n+α+β+5Γn+α+3Γn+α+β+4Γn+β+3,where(53)κξ,α,β=Γα+β+3Γβ+1ξΓα+2.
Then, (6) becomes(54)J˜n+11ξ;ξ,α+1,βJ˜n+2ξ;ξ,α+1,β−J˜n1ξ;ξ,α+1,βJ˜n+1ξ;ξ,α+1,β=κξ,α,βn+1!2n+α+β+5Γn+α+3Γn+α+β+4Γn+β+3,n≥0.
Forβ≠0, we find this result.
Observe that(55)n+1!2n+α+β+5Γn+α+3Γn+α+β+4Γn+β+3=n+1!Γn+α+3βΓn+α+β+3Γn+β+2−n+2!Γn+α+4βΓn+α+β+4Γn+β+3.
Then,(56)J˜n1ξ;ξ,α+1,βJ˜n+1ξ;ξ,α+1,β=−κξ,α,βn+1!Γn+α+3βΓn+α+β+3Γn+β+2+α+β+2ξβ,n≥0.
And so that(57)J˜n1ξ;ξ,α+1,β=α+β+2ξnXnα,ββΓα+2Γβ+1Γ2n+α+β+4,n≥0,where Xnα,β=Γα+2Γn+α+β+3Γn+β+2−n+1!Γβ+1Γα+β+2Γn+α+3.Using the previous lemmas, we will be able to give the following results.Theorem 2.
Forξ,α,β∈ℂ3, where ξ≠0, α≠−n, β≠−n+1, and α+β≠−n−1, n≥1, we have(58)HnSξ,α,β=Δnξ,α,β2n+α+β+2n+α+β+1∏ν=0n2ν+α+β+1ν+α+12ν+α+β+2ν+α+β+1,n≥0,
with(59)Δnξ,α,β=−1nΥα,βξnβ+1α+β+1αn+1n!βΓα+β+2Γn+α+2Γn+β+1,n≥0,where(60)Υ0α,β=βΓα+1Γβ+1Γα+β+2,Υα,β=α+β+1Γn+α+β+2Γn+β+1−n!Γβ+1Γα+β+2Γn+α+2,n≥1.Proof.
From Lemma2 where β≠0, we have(61)Hnsξ,α,β=Δnξ,α,βJ˜nξ;ξ,α+1,β+ξα+1α+β+2J˜n−11ξ;ξ,α+1,β,where(62)Δnξ,α,β=−1nHn−1jξ,α+1,βξα+1α+β+2n.
But, on the one hand, by Lemma1, we have(63)Hn−1jξ,α,β=ξnn−1Γα+β+3Γα+2Γβ+1n×∏ν=0n−1ν!Γν+α+β+2Γν+α+2Γν+β+1Γ2ν+α+β+2Γ2ν+α+β+3,n≥1.
Then,(64)Δnξ,α,β=−1nξn2Γα+β+2Γα+1Γβ+1n×∏ν=0n−1ν!Γν+α+β+2Γν+α+2Γν+β+1Γ2ν+α+β+2Γ2ν+α+β+3,n≥1.
On the other hand, by Lemma2, we have(65)J˜nξ;ξ,α+1,β+ξα+1α+β+2J˜n−11ξ;ξ,α+1,β=ξnΥnα,ββΓα+1Γβ+1Γ2n+α+β+2,n≥0,where(66)Υ0α,β=βΓα+1Γβ+1Γα+β+2,Υnα,β=Γα+1α+β+1Γn+α+β+2Γn+β+1−n!Γβ+1Γα+β+2Γn+α+2,n≥1.
Consequently,(67)HnSξ,α,β=−1nξnn+1βΓα+β+2Γα+β+2Γα+1Γβ+1n+1×Γ2n+α+β+3Υnα,βn!Γn+α+β+2Γn+α+2Γn+β+1×∏ν=0nν!Γν+α+β+2Γν+α+2Γν+β+1Γ2ν+α+β+2Γ2ν+α+β+3,n≥0.
Using the binomial coefficient, we obtained the result.


### 4.1. Application: Hankel Determinant of Series Involving the Catalan–Qi Numbers

Forξ,α,β=b/a,a−1,b−a−1, by substitution in the last theorem, we obtain a completed evaluation of the Hankel determinants that are associated with the sequence of series involving the Catalan–Qi numbers.Proposition 1.
Ifa and b are complex numbers such that Rb,Ra>0,Rb>Ra+1, we have(68)Hn∑ν=0nbaνCa,b;n−ν=Ωa,b;n2n+bn+b−1∏ν=0n2ν+b−1ν+a2ν+bν+b−1,n≥0,where(69)Ωa,b;0=b−a−1ΓaΓb−aΓb,Ωa,b;n=Γab−1Γn+bΓn+b−a−n!Γb−aΓbΓn+a+1,n≥1.
Fora=1/2 and b=2 and using the Gamma duplication formula, we obtained(70)Hn∑ν=0n4νCn−ν=−1n2n+1,n≥0.


## 4.1. Application: Hankel Determinant of Series Involving the Catalan–Qi Numbers

Forξ,α,β=b/a,a−1,b−a−1, by substitution in the last theorem, we obtain a completed evaluation of the Hankel determinants that are associated with the sequence of series involving the Catalan–Qi numbers.Proposition 1.
Ifa and b are complex numbers such that Rb,Ra>0,Rb>Ra+1, we have(68)Hn∑ν=0nbaνCa,b;n−ν=Ωa,b;n2n+bn+b−1∏ν=0n2ν+b−1ν+a2ν+bν+b−1,n≥0,where(69)Ωa,b;0=b−a−1ΓaΓb−aΓb,Ωa,b;n=Γab−1Γn+bΓn+b−a−n!Γb−aΓbΓn+a+1,n≥1.
Fora=1/2 and b=2 and using the Gamma duplication formula, we obtained(70)Hn∑ν=0n4νCn−ν=−1n2n+1,n≥0.



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