Ordinary Generating Functions of Binary Products of Third-Order Recurrence Relations and 2-Orthogonal Polynomials

Hind Merzouk; Ali Boussayoud; Abdelhamid Abderrezzak

doi:10.1515/ms-2022-0002

Article Open Access

Ordinary Generating Functions of Binary Products of Third-Order Recurrence Relations and 2-Orthogonal Polynomials

Hind Merzouk , Ali Boussayoud and Abdelhamid Abderrezzak

Published/Copyright: February 16, 2022

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From the journal Mathematica Slovaca Volume 72 Issue 1

Abstract

In this paper, we introduce the new generating functions for the product of some numbers and 2-orthogonal polynomials by making use of the symmetrizing endomorphism operators π_e₂e₃π_e₁e₂ to the series ∑n=0+∞Sn(A3)e1nzn.

2020 Mathematics Subject Classification: Primary 06B05, 06D35; Secondary 35R30, 39A10

Keywords: Symmetric functions; generating functions; Tribonacci polynomials; Narayana numbers

1. Introduction

The Tribonacci sequence {Tn}n=0+∞ and Tribonacci Lucas sequence {Kn}n=0+∞ are defined respectively by the third order recurrence relations

Tn=Tn−1+Tn−2+Tn−3andKn=Kn−1+Kn−2+Kn−3

for n ≥ 3, with the initial values T₀ = 0, T₁ = T₂ = 1, K₀ = 3, K₁ = 1, and K₂ = 3, respectively. The Tribonacci numbers T_n were defined in [13] for the first time and some properties for T_n have been investigated in [9, 13]. In [9: Example 3.3], the determinants expression

(1.1)Tn+1=|11−1111−11⋱⋱⋱⋱11−11|n,n≥1

was derived. There are many generalizations of the Tribonacci sequences T_n. One of these generalizations is the generalized Tribonacci sequence V_n(x, y, z; a, b, c) defined for n ≥ 3 by

(1.2)Vn(x,y,z;a,b,c)=xVn−1(x,y,z;a,b,c)+yVn−2(x,y,z;a,b,c)+zVn−3(x,y,z;a,b,c),

where V₀(x, y, z; a, b, c) = a, V₁(x, y, z; a, b, c) = b, V₂(x, y, z; a, b,c) = c are arbitrary integers and x, y, z are real numbers.

The generating function of the generalized Tribonacci sequence V_n(x, y, z; a, b, c) is

(1.3)∑n=0+∞Vn(x,y,z;a,b,c)zn=a+(b−ax)z+(c−xb−ya)z21−xz−yz2−z3.

There have been many studies on the generalized Tribonacci numbers V_n(x, y, z; a, b, c). For more information, please refer to [10,12] and closely related references therein. Some special cases of the generalized Tribonacci sequence V_n(x, y, z; a, b, c) are as follows:

V_n(1, 1, 1; 0, 1, 1) = T_n, the Tribonacci sequence;
V_n(1, 1, 1; 3, 1, 3) = K_n, the Tribonacci–Lucas sequence;
V_n(0, 1, 1; 3, 0, 2) = Q_n, the Perrin sequence;
V_n(0, 1, 1; 1, 1, 1) = P_n, the Padovan (Cordonnier) sequence;
V_n(0, 1, 1; 1, 0, 1) = R_n, the Van Der Laan sequence;
V_n(1, 0, 1; 0, 1, 1) = N_n, the Narayana sequence;
V_n(1, 1, 2; 0, 1, 1) = Jn(3), the third order Jacobsthal sequence;
V_n(1,1, 2; 2,1, 5) = jn(3), the third order Jacobsthal Lucas sequence.

In [9: Example 3.4], the determinants expression

(1.4)Qn+1=|21−3010−1011−10⋱⋱⋱⋱11−10|n,n≥1

was derived. In [16], the Tribonacci polynomials T_n(x) were defined for n ≥ 3 by

Tn(x)=x2Tn−1(x)+xTn−2(x)+Tn−3(x),

where T₀(x) = 0, T₁(x) = 1, and T₂(x) = x². In [9: Theorem 3.10], the determinant

(1.5)Tn+1(x)=|x21−xx211−xx2⋱⋱⋱⋱11−xx2|n,n≥1

was established. Now, we recall the notion of d-orthogonal polynomials. A remarkable property of the d-monic orthogonal polynomials sequence is that those sequences satisfy a (d + 1)-order recurrence relation, written in the form

Pm+d+1(x)=(x−βm+d)Pm+d(x)−∑ν=0d−1γm+d−νd−1−νPm+d−1−ν(x),m≥0,

with the initial conditions P₀(x) = 1, P_–1(x) = 0. So, if d ≥ 2:

Pn(x)=(x−βn−1)Pn−1(x)−∑ν=0n−2γd−1−νn−1−νPn−2−ν(x),2≤n≤d,

and the regularity conditions γm+10≠0. m ≥ 0. The 2-orthogonal monic Chebyshev polynomials (2-classical) of the first kind {T^n}n≥0 studied in [11], and defined by the following relations, where α and γ are constants (see also [19]):

T^0(x)=1,T^1(x)=x,T^2(x)=x2−αT^n+3(x)=xT^n+2(x)−αT^n+1(x)−γT^n(x),n≥0,γ≠0.

DEFINITION 1.

A d-orthogonal polynomials sequence {P_n}_n≥0 is called d-classical d-orthogonal polynomials sequence if both {P_n}_n≥0 and its derivative {Pn′}n≥0 are d-orthogonal. For more details, see [8].

Note that the 2-classical 2-orthogonal polynomials sequence is analogous to the Chebyshev orthogonal polynomials sequence of the first kind {T^n}n≥0 (see [11]). In this contribution, we shall define a new useful operator denoted by π_e₂e₃π_e₁e₂ for which we can formulate, extend and prove new results based on our previous ones, (see [10], [16], [17]). In order to determine generating functions for third-order recurrence relations and 2-classical 2-orthogonal polynomials sequence, we combine between our indicated past techniques and these presented polishing approaches. The further contents of this paper are as follows: Section 1 gives introduction. In Section 2, we introduce a new symmetric function and give some properties of this symmetric function. We also give some more useful definitions which are used in the subsequent sections. In Section 3, we prove our main result which relates the symmetric function defined in the previous section with the symmetrizing operator. This main theorem unifies several previously known results about the generating functions. It is then used to find the new generating functions of products for some know numbers and polynomials, in Section 4.

2. Definitions and some properties

We need some preliminaries on symmetric functions and divided differences. We will follow the conventions of [15] rather than those of [18], so that you can conveniently use multiple alphabets at once.

DEFINITION 2 ([1]).

Let A and E be any sets of given variables. Then we give S_n (A – E) by the following form:

(2.1)∏e∈E(1−ez)∏a∈A(1−az)=∑n=0∞Sn(A−E)zn,

with the condition S_n(A – E) = 0 for n < 0.

Equation (2.1) can be rewritten in the following form:

∑n=0∞Sn(A−E)zn=(∑n=0∞Sn(A)zn)×(∑n=0∞Sn(−E)zn),

where

Sn(A−E)=∑i=0∞Si(A)Sn−i(−E).

Remark 1.

Taking A = {0} in (2.1) gives

∑n=0∞Sn(−E)zn=∏e∈E(1−ez).

The summation is in fact limited to a finite number of non-zero terms, in particular:

(2.2)∏e∈E(x−e)=Sn(x−E)=xnS0(−E)+xn−1S1(−E)+xn−2S1(−E)+⋯

The S_j(–E) are the coefficients of the polynomials S_n(x – E) for 0 ≤ j ≤ n. These are therefore at the sign near the elementary symmetric functions of the alphabet E, who are void for j ≥ n. In particular, when E = {e,e,e,…, e} (we notice E = ne), we have S_n(x – ne) = (x – e)ⁿ. The specialization e = 1, i.e., E = {1,1,1,… } (we notice E = n), gives the binomial coefficients:

Sj(−n)=(−1)j(nj),Sj(n)=(n+j+1j),andSj(−ne)=(−e)j(nj).

DEFINITION 3 ([2])

Given a function f on ℝⁿ, the divided difference operator is defined as follows:

∂xixi+1f(x1,…,xi,xi+1,…,xn)=f(x1,…,xi,xi+1,…,xn)−f(x1,…,xi−1,xi+1,xi,…,xn)xi−xi+1.

DEFINITION 4 ([3,6])

Let n be positive integer and A₂ = {a₁, a₂} are set of given variables. Then, the n -th symmetric function S_n(a₁ + a₂) is defined by

Sn(A2)=Sn(a1+a2)=a1n+1−a2n+1a1−a2,

with

S0(A2)=S0(a1+a2)=1,S1(A2)=S1(a1+a2)=a1+a2,S2(A2)=S2(a1+a2)=a12+a1a2+a22,...

DEFINITION 5 ([4,5])

The symmetrizing operator δa1a2k is defined by

(2.3)δa1a2kf(a1)=a1kf(a1)−a2kf(a2)a1−a2forallk∈ℕ0.

If f (a₁) = a₁, the operator (2.3) gives us

δa1a2kf(a1)=a1k+1−a2k+1a1−a2Sk(a1+a2).

3. Lemmas and main results

In this section, we state and prove our main results.

DEFINITION 6

Given an alphabet E₃ = {e₁, e₂, e₃}, the symmetrizing operator is defined by

πe2e3πe1e2f(e1)=e12f(e1)e2e3πe2e3f(e2)−e1πe2e3(e2f(e2))(e1−e2)(e1−e3).

PROPOSITION 1

Let E₃ = {e₁, e₂, e₃} be an alphabet. Then we have

Sn(E3)=e2Sn(e1+e2)−e3Sn(e1+e3)e2−e3.

Proof. We have (see [7])

1∑n=0+∞Sn(e1+e2)zn=1(1−e1z)(1−e2z).

Then

2∑n=0+∞Sn(e1+e3)zn=1(1−e1z)(1−e3z).

Multiplying the equation (1) by e₂ and subtracting it from (2) multiplying by e₃, we have

∑n=0+∞[e2Sn(e1+e2)−e3Sn(e1+e3)]zn=e2−e3(1−e1z)(1−e2z)(1−e3z).

Thus

∑n=0+∞e2Sn(e1+e2)−e3Sn(e1+e3)e2−e3zn=1(1−e1z)(1−e2z)(1−e3z)=∑n=0+∞Sn(e1+e2+e3)zn.

Therefore

e2Sn(e1+e2)−e3Sn(e1+e3)e2−e3=Sn(e1+e2+e3).

This completes the proof. □

PROPOSITION 2

Let E₃ = {e₁, e₂, e₃} be an alphabet. Then we have

Sn(E3)=e1n+1(e1−e2)(e1−e3)+e2n+2(e2−e1)(e2−e3)+e3n+3(e3−e1)(e3−e3).

If E₃ = {1, 2, 3} in Proposition 2 gives

Sn(E3)=3n+2+12−2n+2.

COROLLARY 1

The following identity holds true:

πe2e3Sn(e1+e2)=Sn(E3).

In order to prove our main results, we recall several lemmas below.

LEMMA 1.

Let A₃ = {a₁, a₂, a₃} and E₃ = {e₁, e₂, e₃} be two alphabets. Then action operatorπ_e₂e₃π_e₁e₂on the series∑n=0+∞Sn(A3)e1nznis given by

πe2e3πe1e2∑n=0+∞Sn(A3)e1nzn=∑n=0+∞Sn(A3)Sn(E3)zn.

Proof. We have

πe2e3πe1e2∑n=0+∞Sn(A3)e1nzn=πe2e3(e1∑n=0+∞Sn(A3)e1nzn−e2∑n=0+∞Sn(A3)e2nzn(e1−e2))=πe2e3(∑n=0+∞Sn(A3)e1n+1−e2n+1(e1−e2)zn)=πe2e3(∑n=0+∞Sn(A3)Sn(e1+e2)zn)=e2∑n=0+∞Sn(A3)(e1+e2)zn−e3∑n=0+∞Sn(A3)Sn(e1+e3)zn(e2−e3)=∑n=0+∞Sn(A3)e2Sn(e1+e2)−e3Sn(e1+e3)(e2−e3)zn

according to Proposition 1, we have then

πe2e3πe1e2∑n=0+∞Sn(A3)e1nzn=∑n=0+∞Sn(A3)Sn(E3)zn.

This completes the proof. □

LEMMA 2

Let A₃ = {a₁, a₂, a₃} and E₂ = {e₁, e₂} be two alphabets. Then action operator π_e₁e₂on the series1∑n=0+∞Sn(−A3)e1nznis given by

πe1e21∑n=0+∞Sn(−A3)e1nzn=1−e1e2S2(−A3)z2−e12e2S3(−A3)z3−e1e22S3(−A3)z3(∑n=0+∞Sn(−A3)e1nzn)(∑n=0+∞Sn(−A3)e2nzn).

Proof. We have

πe1e21∑n=0+∞Sn(−A3)e1nzn=1(e1−e2)(e1∑n=0+∞Sn(−A3)e1nzn−e2∑n=0+∞Sn(−A3)e2nzn)=e1(∑n=0+∞Sn(−A3)e2nzn)−e2(∑n=0+∞Sn(−A3e1nzn)(e1−e2)(∑n=0+∞Sn(−A3)e1nzn)(∑n=0+∞Sn(−A3)e2nzn)=∑n=0+∞Sn(−A3)e1e2n−e2e1n(e1−e2)zn(∑n=0+∞Sn(−A3)e1nzn)(∑n=0+∞Sn(−A3)e2nzn)=1−e1e2∑n=2+∞Sn(−A3)Sn−2(e1+e2)zn(∑n=0+∞Sn(−A3)e1nzn)(∑n=0+∞Sn(−A3)e2nzn)=1−e1e2(S2(−A3)z2+S3(−A3)S1(e1+e2)z3)(∑n=0+∞Sn(−A3)e1nzn)(∑n=0+∞Sn(−A3)e2nzn)=1−e1e2S2(−A3)z2−e12e2S3(−A3)z3−e1e22S3(−A3)z3(∑n=0+∞Sn(−A3)e1nzn)(∑n=0+∞Sn(−A3)e2nzn)

This completes the proof. □

LEMMA 3

Given an alphabet E₂ = {e₂, e₃}, we have

πe2e3(1∑n=0+∞Sn(−A3)e2nzn)=1−e2e3(S2(−A3)z2−S3(−A3)S1(e2+e3)z3)(∑n=0+∞Sn(−A3)e2nzn)(∑n=0+∞Sn(−A3)e3nzn).

Proof. We have

πe2e3(1∑n=0+∞Sn(−A3)e2nzn)=1e2−e3(e2∑n=0+∞Sn(−A3)e2nzn−e3∑n=0+∞Sn(−A3)e3nzn)=(∑n=0+∞Sn(−A3)e2e3n−e3e2ne2−e3zn)(∑n=0+∞Sn(−A3)e2nzn)(∑n=0+∞Sn(−A3)e3nzn)=1−e2e3∑n=2+∞Sn(−A3)Sn−2(e2+e3)zn(∑n=0+∞Sn(−A3)e2nzn)(∑n=0+∞Sn(−A3)e3nzn)=1−e2e3S2(−A3)z2−e2e3S3(−A3)S1(e2+e3)z3(∑n=0+∞Sn(−A3)e2nzn)(∑n=0+∞Sn(−A3)e3nzn).

This completes the proof. □

LEMMA 4.

Given an alphabet E₂ = {e₂, e₃}, we have

πe2e3(e2∑n=0+∞Sn(−A3)e2nzn)=(e2+e3)+e2e3S1(−A3)z−e22e32S3(−A3)z3(∑n=0+∞Sn(−A3)e2nzn)(∑n=0+∞Sn(−A3)e3nzn)

Proof. We have

πe2e3(e2∑n=0+∞Sn(−A3)e2nzn)=1e2−e3(e22∑n=0+∞Sn(−A3)e2nzn−e32∑n=0+∞Sn(−A3)e3nzn)=(∑n=0+∞Sn(−A3)e22e3n−e32e2ne2−e3n)(∑n=0+∞Sn(−A3)e2nzn)(∑n=0+∞Sn(−A3)e3nzn)=(e2+e3)+e2e3S1(−A3)z−e22e33∑n=3+∞Sn(−A3)Sn−3(e2+e3)zn(∑n=0+∞Sn(−A3)e2nzn)(∑n=0+∞Sn(−A3)e3nzn)=(e2+e3)+e2e3S1(−A3)z−e22e32S3(−A3)z3(∑n=0+∞Sn(−A3)e2nzn)(∑n=0+∞Sn(−A3)e3nzn)

This completes the proof. □

LEMMA 5

Given an alphabet E₂ = {e₂, e₃}, we have

πe2e3(e22∑n=0+∞Sn(−A3)e2nzn)=((e2+e3)2−e2e3)+e2e3(e2+e3)S1(−A3)z+e22e32S2(−A3)z2(∑n=0+∞Sn(−A3)e2nzn)(∑n=0+∞Sn(−A3)e3nzn).

Proof. We have

πe2e3(e22∑n=0+∞Sn(−A3)e2nzn)=1e2−e3(e23∑n=0+∞Sn(−A3)e2nzn−e33∑n=0+∞Sn(−A3)e3nzn)=(∑n=0+∞Sn(−A)e23e3n−e33e2ne2−e3zn)(∑n=0+∞Sn(−A3)e2nzn)(∑n=0+∞Sn(−A3)e3nzn)=((e2+e3)2−e2e3)+e2e3(e2+e3)S1(−A3)z+e22e32S2(−A3)z2−e23e33∑n=4+∞Sn(−A3)Sn−4(e2+e3)zn(∑n=0+∞Sn(−A3)e2nzn)(∑n=0+∞Sn(−A3)e3nzn)=((e2+e3)2−e2e3)+e2e3(e2+e3)S1(−A3)z+e22e32S2(−A3)z2(∑n=0+∞Sn(−A3)e2nzn)(∑n=0+∞Sn(−A3)e3nzn).

This completes the proof. □

In this part, we are now in a position to provide Theorem 1. Also, we derive the new generating functions of the products of some known numbers and polynomials.

THEOREM 1

Let A₃and E₃be two alphabets defined respectively by {a₁, a₂, a₃} and {e₁, e₂, e₃}. Then we have

(3.1)∑n=0+∞Sn(A3)Sn(E3)zn=P1(z)D1(z),

with

P1(z)=1−S2(−A3)S2(−E3)z2+[S3(−A3)(S2(−E3)S1(−E3)−2S3(−E3))+S1(−A3)S3(−E3)]z3−S1(−A3)S1(−E3)S3(−A3)S3(−E3)z4+S32(−A3)S32(−E3)z6,

and

D1(z)=(∑n=0+∞Sn(−A3)e1nzn)(∑n=0+∞Sn(−A3)e2nzn)(∑n=0+∞Sn(−A3)e3nzn).

Proof. We have

∑n=0+∞Sn(A3)e1nzn=1∑n=0+∞Sn(−A3)e1nzn.

Action the operator π_e₂e₃π_e₁e₂ on the identity above gives us

πe2e3πe1e2∑n=0+∞Sn(A3)e1nzn=πe2e3πe1e21∑n=0+∞Sn(A3)e1nzn.

Lemmas 1 and 2 then allow us to write

∑n=0+∞Sn(A3)Sn(E3)zn=[πe2e3(∑n=0+∞Sn(−A3)e2nzn)−1−(e1S2(−A3)z2+e12S3(−A3)z3)πe2ϵ3e2(∑n=0+∞Sn(−A3)e2nzn)−1−e1S3(−A3)z3πe2e3e22(∑n=0+∞Sn(−A3)e2nzn)−1][∑n=0+∞Sn(−A3)e1nzn]−1

after the Lemmas 3, 4, and 5, we obtain

∑n=0+∞Sn(A3)Sn(E3)zn=[1−e2e3(S2(−A3)z2+S3(−A3)S1(e2+e3)z3)−(e1S2(−A3)z2+e12S3(−A3)z3)((e2+e3)+e2e3S1(−A3)z−e22e32S3(−A3)z3)−e1S3(−A3)z3[S2(e2+e3)+e2e3S1(−A3)S1(e2+e3)z+e22e32S2(−A3)z2]]×[(∑n=0+∞Sn(−A3)e1nzn)(∑n=0+∞Sn(−A3)e2nzn)(∑n=0+∞Sn(−A3)e3nzn)]−1

By reorganizing the numerator according to the powers of z, we obtain

∑n=0+∞Sn(A3)Sn(E3)zn=[1−S2(−A3)(e1e2+e1e3+e2e3)z2−(S3(−A3)(e22e3+e2e32+e12e2+e12e3+e1e22+e1e32+e1e2e3)+e1e2e3S1(−A3))z3−S1(−A3)S3(−A3)(e12e2e3+e1e22e3+e1e2e32)z4+e12e22e32S32(−A3)z6]×[(∑n=0+∞Sn(−A3)e1nzn)(∑n=0+∞Sn(−A3)e2nzn)(∑n=0+∞Sn(−A3)e3nzn)]−1

we will have

∑n=0+∞Sn(A3)Sn(E3)zn=P1(z)D1(z).

Thus, this completes the proof. □

Based on the relation (3.1), we have

(3.2)∑n=0+∞Sn−1(A3)Sn−1(E3)zn=P2(z)D1(z),

with

P2(z)=z−S2(−A3)S2(−E3)z3+[S3(−A3)(S2(−E3)S1(−E3)−2S3(−E3))+S1(−A3)S3(−E3)]z4−S1(−A3)S1(−E3)S3(−A3)S3(−E3)z5+S32(−A3)S32(−E3)z7.

(3.3)∑n=0+∞Sn−2(A3)Sn−2(E3)zn=P3(z)D1(z),

with

P3(z)=z2−S2(−A3)S2(−E3)z4+[S3(−A3)(S2(−E3)S1(−E3)−2S3(−E3))+S1(−A3)S3(−E3)]z5−S1(−A3)S1(−E3)S3(−A3)S3(−E3)z6+S32(−A3)S32(−E3)z8.

THEOREM 2

Let A₃ = {a₁, a₂, a₃} and E₃ = {e₁, e₂, e₃} be two alphabets. Then we have

(3.4)∑n=0+∞Sn(A3)Sn−1(E3)zn=P4(z)D1(z),

with

P4(z)=−S1(−A3)z+S2(−A3)S1(−E3)z2−S3(−A3)[S12(−E3)−S2(−E3)]z3+[S1(−A3)S3(−A3)S3(−E3)−S22(−A3)S3(−E3)]z4+S2(−A3)S3(−A3)S1(−E3)S3(−E3)z5−S32(−A3)S2(−E3)S3(−E3)z6.

Proof. We have (see[16])

∑n=0+∞Sn(A3)Sn−1(e1+e2)zn=−S1(−A3)z−(e1+e2)S2(−A3)z2−((e1+e2)2−e1e2)S3(−A3)z3(∑n=0+∞Sn(−A3)e1nzn)(∑n=0+∞Sn(−A3)e2nzn).

Suffices to apply the operator π_e₂e₃ to the identity above, we obtain

πe2e3∑n=0+∞Sn(A3)Sn−1(e1+e2)zn=πe2e3−S1(−A3)z−(e1+e2)S2(−A3)z2−((e1+e2)2−e1e2)S3(−A3)z3(∑n=0+∞Sn(−A3)e1nzn)(∑n=0+∞Sn(−A3)e2nzn),∑n=0+∞Sn(A3)πe2e3Sn−1(e1+e2)zn=e2−S1(−A3)z−(e1+e2)S2(−A3)z2−((e1+e2)2−e1e2)S3(−A3)z3(e2−e3)(∑n=0+∞Sn(−A3)e1nzn)(∑n=0+∞Sn(−A3)e2nzn)−e3−S1(−A3)z−(e1+e3)S2(−A3)z2−((e1+e3)2−e1e3)S3(−A3)z3(e2−e3)(∑n=0+∞Sn(−A3)e1nzn)(∑n=0+∞Sn(−A3)e3nzn)∑n=0+∞Sn(A3)Sn−1(e1+e2+e3)zn=[−(e2−e3)S1(−A3)z−(e2−e3)(e1+e2+e3)S2(−A3)z2−(e2−e3)S3(−A3)((e1+e2+e3)2−(e1e2+e1e3+e2e3))z3−(e2−e3)e1e2e3(S1(−A3)S3(−A3)−S22(−A3))z4+(e2−e3)e1e2e3(e1+e2+e3)S2(−A3)S3(−A3)z5+(e2−e3)e1e2e3(e1e3+e1e2+e2e3)S32(−A3)z6]×[(e2−e3)(∑n=0+∞Sn(−A3)e1nzn)(∑n=0+∞Sn(−A3)e2nzn)(∑n=0+∞Sn(−A3)e3nzn)]−1∑n=0+∞Sn(A3)Sn−1(E3)zn=[−S1(−A3)z+S2(−A3)S1(−E3)z2−S3(−A3)(S12(−E3)−S2(−E3))z3+S3(−E3)(S1(−A3)S3(−A3)−S22(−A3))z4+S2(−A3)S3(−A3)S1(−E3)S3(−E3)z5−S32(−A3)S2(−E3)S3(−E3)z6]×[(∑n=0+∞Sn(−A3)e1nzn)(∑n=0+∞Sn(−A3)e2nzn)(∑n=0+∞Sn(−A3)e3nzn)]−1

Therefore

∑n=0+∞Sn(A3)Sn−1(E3)zn=P4(z)D1(z).

This completes the proof. □

From (3.4) we conclude the following relationships

(3.5)∑n=0+∞Sn−1(A3)Sn−2(E3)zn=P5(z)D1(z),

with

P5(z)=−S1(−A3)z2+S2(−A3)S1(−E3)z3−S3(−A3)[S12(−E3)−S2(−E3)]z4+[S1(−A3)S3(−A3)S3(−E3)−S22(−A3)S3(−E3)]z5+S2(−A3)S3(−A3)S1(−E3)S3(−E3)z6−S32(−A3)S2(−E3)S3(−E3)z7.

THEOREM 3

Let A₃and E₃be two alphabets, defined by {a₁, a₂, a₃}, {e₁, e₂, e₃}, respectively. Then we have

(3.6)∑n=0+∞Sn−1(A3)Sn(E3)zn=P6(z)D1(z),

with

P6(z)=−S1(−E3)z+S1(−A3)S2(−E3)z2−S3(−E3)(S12(−A3)−S2(−A3))z3−S3(−A3)(S22(−E3)−S1(−E3)S3(−E3))z4+S1(−A3)S3(−A3)S2(−E3)S3(−E3)z5−S32(−E3)S2(−A3)S3(−A3)z6.

Proof. By the same method given in Theorem 2, the proof can be easily made. □

Based to the relation (3.6), we obtain

(3.7)∑n=0+∞Sn−2(A3)Sn−1(E3)zn=P7(z)D1(z),

with

P7(z)=−S1(−E3)z2+S1(−A3)S2(−E3)z3−S3(−E3)(S12(−A3)−S2(−A3))z4−S3(−A3)(S22(−E3)−S1(−E3)S3(−E3))z5+S1(−A3)S3(−A3)S2(−E3)S3(−E3)z6−S32(−E3)S2(−A3)S3(−A3)z7.

THEOREM 4

Let A₃ = {a₁, a₂, a₃} and E₃ = {e₁, e₂, e₃} be two alphabets. Then we have

(3.8)∑n=0+∞Sn−2(A3)Sn(E3)zn=P8(z)D1(z),

with

P8(z)=(S12(−E3)−S2(−E3))z2−S1(−A3)(S1(−E3)S2(−E3)−S3(−E3))z3+[S12(−A3)S1(−E3)S3(−E3)+S2(−A3)(S22(−E3)−2S1(−E3)S3(−E3))]z4−S2(−E3)S3(−E3)[S1(−A3)S2(−A3)−S3(−A3)]z5+[S22(−A)−S1(−A3)S3(−A3)]S32(−E3)z6.

Proof. We have (see [16])

∑n=0+∞Sn−2(A3)Sn(e1+e2)zn=((e1+e2)2−e1e2)z2+e1e2(e1+e2)S1(−A3)z3+e12e22S2(−A3)z4(∑n=0+∞Sn(−A3)e1nzn)(∑n=0+∞Sn(−A3)e2nzn).

By following the same demonstration steps as those of identity (3.4), we get (3.8). □

From (3.8), we have

(3.9)∑n=0+∞Sn(A3)Sn−2(E3)zn=P9(z)D1(z),

with

P9(z)=(S12(−A3)−S2(−A3))z2−S1(−E3)(S1(−A3)S2(−A3)−S3(−A3))z3+[S12(−E3)S1(−A3)S3(−A3)+S2(−E3)(S22(−A3)−2S1(−A3)S3(−A3))]z4−S2(−A3)S3(−A3)[S1(−E3)S2(−E3)−S3(−E3)]z5+[S22(−E)−S1(−E3)S3(−E3)]S32(−A3)z6,

and

D1(z)=1−[S1(−E3)S1(−A3)+S2(−E3)S12(−A3)]z+[S2(−A3)S12(−E3)−2S2(−E3)S2(−A3)]z2−[S13(−E3)S3(−A3)+S3(−E3)(S13(−A3)−6S3(−A3))+(S1(−A3)S2(−A3)−3S3(−A3))(S1(−E3)S2(−E3)−3S3(−E3))]z3+[S3(−E3)S1(−E3)S2(−A3)S12(−A3)+(S22(−A3)−2S3(−A3)S1(−A3))(S22(−E3)−2S3(−E3)S1(−E3))+S3(−A3)S1(−A3)(S12(−E3)S2(−E3)−5S3(−E3)S1(−E3))]z4−[S3(−E3)S2(−E3)(S1(−A3)S22(−A3)+2S3(−A3)S1(−A3)−S3(−A3)S2(−A3))+S3(−E3)S12(−E3)(S3(−A3)S12(−A3)−2S3(−A3)S2(−A3))+S1(−E3)S2(−E3)S2(−A3)S3(−A3)]z5+[S32(−E3)S23(−A3)+S3(−E3)S3(−A3)(S1(−A3)S2(−A3)−3S3(−A3))×(S1(−E3)S2(−E3)−3S3(−E3))+S32(−A3)(S23(−E3)−6S32(−E3))]z6−[S32(−E3)S1(−E3)S3(−A3)(S22(−A3)−2S3(−A3)S1(−A3))+S32(−A3)S1(−A3)S3(−E3)S2(−E3)]z7+S32(−E3)S2(−E3)S32(−A3)S2(−A3)z8−S33(−E3)S33(−A3)z9.

4. Applications

All the identities appearing in the following section are corollaries of (3.1), (3.2), (3.3), (3.4), (3.5), (3.6), (3.7), (3.8) and (3.9).

4.1. Generating functions of square of some known numbers

In this part, we now derive the new generating functions of the square of some known numbers. This case consists of four related parts. Firstly, the substitutions

{S1(−A3)=−1S2(−A3)=−1S3(−A3)=−1and{S1(−E3)=−1S2(−E3)=−1S3(−E3)=−1

in (3.1)–(3.5) and (3.9), we have

(4.1)∑n=0+∞Sn(A3)Sn(E3)zn=1−z2−2z3−z4+z61−z−4z2−12z3−6z4−4z5+8z6+2z7+z8−z9,

(4.2)∑n=0+∞Sn−1(A3)Sn−1(E3)zn=z−z3−2z4−z5+z71−z−4z2−12z3−6z4−4z5+8z6+2z7+z8−z9,

(4.3)∑n=0+∞Sn−2(A3)Sn−2(E3)zn=z2−z4−2z5−z6+z81−z−4z2−12z3−6z4−4z5+8z6+2z7+z8−z9,

(4.4)∑n=0+∞Sn(A3)Sn−1(E3)zn=z+z2+2z3+z5−z61−z−4z2−12z3−6z4−4z5+8z6+2z7+z8−z9,

(4.5)∑n=0+∞Sn−1(A3)Sn−2(E3)zn=z2+z4+2z5+z6−z71−z−4z2−12z3−6z4−4z5+8z6+2z7+z8−z9,

(4.6)∑n=0+∞Sn(A3)Sn−2(E3)zn=2z2+2z3+2z4−2z51−z−4z2−12z3−6z4−4z5+8z6+2z7+z8−z9.

Then, we have the following proposition and theorem.

PROPOSITION 3

For n ∈ ℕ, the new generating function of the square of Tribonacci numbers is given by

∑n=0+∞Tn2zn=1−z2−2z3−z4+z61−z−4z2−12z3−6z4−4z5+8z6+2z7+z8−z9.

We can state the following corollary.

COROLLARY 2

The following identity holds true:

Tn2=Sn(A3)Sn(E3).

THEOREM 5

For n ∈ ℕ, the new generating function of the square of Tribonacci Lucas numbers is given by

∑n=0+∞Kn2zn=9−8z−28z2−54z3−22z4−6z5+25z6+z81−z−4z2−12z3−6z4−4z5+8z6+2z7+z8−z9.

Proof. In [10], we have K_n = 3S_n(A₃) – 2S_n–1(A₃) – S_n–2(A₃), then we see that

∑n=0+∞Kn2zn=∑n=0+∞[3Sn(A3)−2Sn−1(A3)−Sn−2(A3)][3Sn(E3)−2Sn−1(E3)−Sn−2(E3)]zn=9∑n=0+∞Sn(A3)Sn(E3)zn+4∑n=0+∞Sn−1(A3)Sn−1(E3)zn+∑n=0+∞Sn−2(A3)Sn−2(E3)zn−12∑n=0+∞Sn(A3)Sn−1(E3)zn−6∑n=0+∞Sn(A3)Sn−2(E3)zn+4∑n=0+∞Sn−1(A3)Sn−2(E3)zn=9∑n=0+∞Tn2zn+4∑n=0+∞Sn−1(A3)Sn−1(E3)zn+∑n=0+∞Sn−2(A3)Sn−2(E3)zn−12∑n=0+∞Sn(A3)Sn−1(E3)zn−6∑n=0+∞Sn(A3)Sn−2(E3)zn+4∑n=0+∞Sn−1(A3)Sn−2(E3)zn.

By using (4.1)–(4.6), we obtain

∑n=0+∞Kn2zn=9−8z−28z2−54z3−22z4−6z5+25z6+z81−z−4z2−12z3−6z4−4z5+8z6+2z7+z8−z9,

which completes the proof of this theorem. □

Secondly, let us now consider the following conditions for Eqs. (3.1)–(3.5) and (3.9):

{S1(−A3)=−1S2(−A3)=−1S3(−A3)=−2and{S1(−E3)=−1S2(−E3)=−1S3(−E3)=−2.

Then it yields

(4.7)∑n=0+∞Sn(A3)Sn(E3)zn=1−z2−8z3−4z4+16z61−z−4z2−29z3−15z4−16z5+92z6+32z7+16z8−64z9,

(4.8)∑n=0+∞Sn−1(A3)Sn−1(E3)zn=z−z3−8z4−4z5+16z71−z−4z2−29z3−15z4−16z5+92z6+32z7+16z8−64z9,

(4.9)∑n=0+∞Sn−2(A3)Sn−2(E3)zn=z2−z4−8z5−4z6+16z81−z−4z2−29z3−15z4−16z5+92z6+32z7+16z8−64z9,

(4.10)∑n=0+∞Sn(A3)Sn−1(E3)zn=z+z2+4z3−2z4+4z5−8z61−z−4z2−29z3−15z4−16z5+92z6+32z7+16z8−64z9,

(4.11)∑n=0+∞Sn−1(A3)Sn−2(E3)zn=z2+z3+4z4−2z5+4z6−8z71−z−4z2−29z3−15z4−16z5+92z6+32z7+16z8−64z9,

(4.12)∑n=0+∞Sn(A3)Sn−2(E3)zn=2z2+3z3+5z4−6z5−4z61−z−4z2−29z3−15z4−16z5+92z6+32z7+16z8−64z9.

Therefore, we state the following proposition and theorem.

PROPOSITION 4

For n ∈ ℕ, the new generating function of square of third order of Jacobsthal numbers is given by

∑n=0+∞(Jn(3))2zn=z−z3−8z4−4z5+16z71−z−4z2−29z3−15z4−16z5+92z6+32z7+16z8−64z9.

We can state the following corollary.

COROLLARY 3

The following identity holds true:

(Jn(3))2=Sn−1(A3)Sn−1(E3).

THEOREM 6

For n ∈ ℕ, the new generating function of square of third order of Jacobsthal Lucas numbers is given by

∑n=0+∞(Jn(3))2zn=4−3z−8z3−29z3−4z4−92z5+32z6+48z7+64z81−z−4z2−29z3−15z4−16z5+92z6+32z7+16z8−64z9.

Proof. By referred to [10], we have Jn(3)=2Sn(A3)−Sn−1(A3)+2Sn−2(A3), then

∑n=0+∞(jn(3))2zn=∑n=0+∞[2Sn(A3)−Sn−1(A3)+2Sn−2(A3)][2Sn(E3)−Sn−1(E3)+2Sn−2(E3)]zn=4∑n=0+∞Sn(A3)Sn(E3)zn−4∑n=0+∞Sn(A3)Sn−1(E3)zn+8∑n=0+∞Sn(A3)Sn−2(E3)zn−4∑n=0+∞Sn−1(A3)Sn−2(E3)zn+∑n=0+∞Sn−1(A3)Sn−1(E3)zn+4∑n=0+∞Sn−2(A3)Sn−2(E3)zn=4∑n=0+∞Sn(A3)Sn(E3)zn−4∑n=0+∞Sn(A3)Sn−1(E3)zn+8∑n=0+∞Sn(A3)Sn−2(E3)zn−4∑n=0+∞Sn−1(A3)Sn−2(E3)zn+∑n=0+∞(Jn(3))2zn+4∑n=0+∞Sn−2(A3)Sn−2(E3)zn.

By using (4.7), (4.8), (4.9), (4.10), (4.11), and (4.12), we obtain

∑n=0+∞(Jn(3))2zn=4−3z+8z2+29z3+4z4−100z5+32z6+48z7+64z81−z−4z2−29z3−15z4−16z5+92z6+32z7+16z8−64z9,

which completes the proof of this theorem. □

Thirdly, the substitutions

{S1(−A3)=−1S2(−A3)=0S3(−A3)=−1and{S1(−E3)=−1S2(−E3)=0S3(−E3)=−1

in (3.2), we obtain the following proposition.

PROPOSITION 5

For n ∈ ℕ, the new generating function of square of Narayana numbers is given by

∑n=0+∞Nn2zn=z−z4−z5+z71−z−5z3−z4−z5+3z6+2z7−z9.

We can state the following corollary.

COROLLARY 4

The following identity holds true:

Nn2=Sn−1(A3)Sn−1(E3).

Fourthly, the substitutions

{S1(−A3)=0S2(−A3)=−1S3(−A3)=−1and{S1(−E3)=0S2(−E3)=−1S3(−E3)=−1

in (3.3) we deduce the following proposition.

PROPOSITION 6

For n ∈ ℕ, the new generating function of square of Padovan Perin numbers is given by

∑n=0+∞Sn2zn=z2−z4−2z5+z81−2z2−3z3+z4+z5+z6+z8−z9.

COROLLARY 5

The following identity holds true:

Sn2=Sn−2(A3)Sn−2(E3).

4.2. Generating functions of product of some known polynomials

In this part, we now derive the new generating functions of the product of some known polynomials. This case consists of three related parts. Firstly, let us now consider the following conditions for equations (3.1)–(3.9):

{S1(−A3)=−x2S2(−A3)=−xS3(−A3)=−1and{S1(−E3)=−y2S2(−E3)=−yS3(−E3)=−1.

Then it yields

(4.13)∑n=0+∞Sn(A3)Sn(E3)zn=1−xyz2+(x2−y3−2)z3−x2y2z4+z6D2(z),

(4.14)∑n=0+∞Sn−1(A3)Sn−1(E3)zn=z−xyz3+(x2−y3−2)z4−x2y2z5+z7D2(z),

(4.15)∑n=0+∞Sn−2(A3)Sn−2(E3)zn=z2−xyz4+(x2−y3−2)z5−x2y2z6+z8D2(z),

(4.16)∑n=0+∞Sn(A3)Sn−1(E3)zn=x2z+xy2z2+y(1+y3)z3+xy2z5−yz6D2(z),

(4.17)∑n=0+∞Sn−1(A3)Sn−2(E3)zn=x2z2+xy2z3+(1+y3)yz4+xy2z6−yz7D2(z),

(4.18)∑n=0+∞Sn−1(A3)Sn(E3)zn=y2z+x2yz2+(1+x3)xz3+x2yz5−xz6D2(z),

(4.19)∑n=0+∞Sn−2(A3)Sn−1(E3)zn=y2z2+x2yz3+(1+x3)xz4+x2yz6−xz7D2(z),

(4.20)∑n=0+∞Sn−2(A3)Sn(E3)zn=(1+y3)yz2+(1+y3)x2z3+xy2(1+x3)z4−(1+x3)yz5D2(z),

(4.21)∑n=0+∞Sn(A3)Sn−2(E3)zn=(1+x3)xz2+(1+x3)y2z3+(1+y3)x2yz4−(1+y3)xz5D2(z),

with

D2(z)=1−x2y2z−xy(x3+y3+2)z2−(x6+y6+x3y3+3x3+3y3+3)z3−x2y2(x3+y3+4)z4+xy(x3−2x+1−y2((x3+2)y+1))z5+(2y3+2x3+x3y3+3)z6+x2y(y+1)z7+xyz8−z9.

Therefore, we state the following results.

PROPOSITION 7

Let n be a natural number. Then we have the new generating function for the product of Tribonacci polynomials:

∑n=0+∞Tn(x)Tn(y)zn=[1−xyz2+(x2−y3−2)z3−x2y2z4+z6]×[1−x2y2z−xy(x3+y3+2)z2−(x6+y6+x3y3+3x3+3y3+3)z3−x2y2(x3+y3+4)z4+xy(x3−2x+1−y2((x3+2)y+1))z5+(2y3+2x3+x3y3+3)z6+x2y(y+1)z7+xyz8−z9]−1.

We can state the following corollary.

COROLLARY 6

The following identity holds true:

Tn(x)Tn(y)=Sn(A3)Sn(E3).

THEOREM 7

Let n be a natural number. Then we have the new generating function for the product of Tribonacci Lucas polynomials:

∑n=0+∞Kn(x)Kn(y)zn=P10(z)D2(z),

and

P9(z)=9−8x2y2z−7xy(2+x3+y3)z2+3(3x2−3x3−2x6−6y3−2y6−2x3y3−6)z3−x2y2(20+5y3+x3−4x2)z4−xy(4x3y3−x2+4y3+3x3−4)z5+3(3+x3y3+2x3+2y3)z6+xyz8.

Proof. Recall that, we have (see [16]) K_n(x) = 3S_n(A₃) – 2x²S_n–1(A₃) – xS_n–2(A₃).

∑n=0+∞Kn(x)Kn(y)zn=∑n=0+∞[3Sn(A3)−2x2Sn−1(A3)−xSn−2(A3)]×[3Sn(E3)−2y2Sn−1(E3)−ySn−2(E3)]zn=9∑n=0+∞Sn(A3)Sn(E3)zn+4x2y2∑n=0+∞Sn−1(A3)Sn−1(E3)zn+xy∑n=0+∞Sn−2(A3)Sn−2(E3)zn−6y2∑n=0+∞Sn(A3)Sn−1(E3)zn−3y∑n=0+∞Sn(A3)Sn−2(E3)zn−6x2∑n=0+∞Sn−1(A3)Sn(E3)zn+2x2y∑n=0+∞Sn−1(A3)Sn−2(E3)zn−3x∑n=0+∞Sn−2(A3)Sn(E3)zn+2xy2∑n=0+∞Sn−2(A3)Sn−1(E3)zn=9∑n=0+∞Tn(x)Tn(y)zn+4x2y2∑n=0+∞Sn−1(A3)Sn−1(E3)zn+xy∑n=0+∞Sn−2(A3)Sn−2(E3)zn−6y2∑n=0+∞Sn(A3)Sn−1(E3)zn−3y∑n=0+∞Sn(A3)Sn−2(E3)zn−6x2∑n=0+∞Sn−1(A3)Sn(E3)zn+2x2y∑n=0+∞Sn−1(A3)Sn−2(E3)zn−3x∑n=0+∞Sn−2(A3)Sn(E3)zn+2xy2∑n=0+∞Sn−2(A3)Sn−1(E3)zn.

By using (4.13)–(4.21), we obtain

∑n=0+∞Kn(x)Kn(y)zn=P10(z)D2(z),

which completes the proof of this theorem. □

Secondly, the substitutions

{S1(−A3)=−1S2(−A3)=−xS3(−A3)=−x2and{S1(−E3)=−1S2(−E3)=−yS3(−E3)=−y2

in (3.1), we have the following proposition.

PROPOSITION 8

We have the following new generating function of the product of Tricobsthal polynomials:

∑n=0+∞Jn(3)(x)Jn(3)(y)zn=1−xyz2+y(y−x2(1+2y))z3−x2y2z4+x4y4z6D3(z),

with

D3(z)=1−z−(x+y+2xy)z2−(y2+x2+3x2y2+xy+3x2y+3xy2)z3−xy(x+4xy+y)z4−x2y(x+y+2xy−xy2+y2)z5+x3y3(2x+2y+3xy+1)z6+x4y3(1+y)z7+x5y5z8−x6y6z9.

COROLLARY 7

The following identity holds true:

Jn(3)(x)Jn(3)(y)=Sn(A3)Sn(E3).

Thirdly, the substitutions

{S1(−A3)=−xS2(−A3)=αS3(−A3)=−γand{S1(−E3)=−yS2(−E3)=αS3(−E3)=−γ

in (3.1), we get the following proposition.

PROPOSITION 9

Let n be a natural number. Then we have the new generating function for the product of 2-Chebyshev polynomials of first kind:

∑n=0+∞T^n(x)T^n(y)zn=1−α2z2+γ(x−2γ+αy)z3−γ2xyz4+γ4z6D4(z),

with

D4(z)=1−xyz+α(x2+y2−2α)z2−(γy3+γx3+3γ2+α2xy−3αγy−3αγx)z3+(αγx2y+αγxy2−γ2xy−2α2γx−2α2γy+α4)z4−(αγx(α2−2γ)+γ2y2x2−α2γ2+αγ2y(α−2γ))z5+(2α3γ2+γ2(αy−3γ)(αx−3γ)−6γ4)z6−γ3(y(α2−2γx)+αx)z7+α2γ4z8−γ6z9.

We can state the following corollary.

COROLLARY 8

The following identity holds true:

T^n(x)T^n(y)=Sn(A3)Sn(E3).

4.3. The case A₃ = {1, 2, 3} and E₃ = {1, 2, 3}

Let us now consider the following conditions for equations (3.1)–(3.5) and (3.9):

{S1(−A3)=S1(−E3)=−6S2(−A3)=S2(−E3)=11S3(−A3)=S3(−E3)=−6.

We have the following results.

PROPOSITION 10

For n ∈ ℕ, the new generating functions of Stirling numbers of the first kind is given by

∑n=0+∞(3n+2+12−2n+2)2zn=1−121z2+360z3−1296z4+1296z6(1−z)(1−2z)2(1−3z)2(1−4z)(1−6z)2(1−9z),∑n=0+∞(3n+1+12−2n+1)2zn=z−121z3+360z4−1296z5+1296z7(1−z)(1−2z)2(1−3z)2(1−4z)(1−6z)2(1−9z).

PROPOSITION 11

Let n be a natural number. Then we have the new generating functions of Stirling numbers of the first type:

∑n=0+∞(3n+2+12−2n)2zn=z−121z4+360z5−1296z6+1296z8(1−z)(1−4z)(1−9z)(1−2z)2(1−3z)2(1−6z)2,∑n=0+∞(3n+2+12−2n+2)2(3n+1+12−2n+1)zn=6z−66z2+150z3−510z4−2376z5+2376z6(1−z)(1−4z)(1−9z)(1−2z)2(1−3z)2(1−6z)2.

PROPOSITION 12

For n ∈ ℕ, the new generating functions of Stirling numbers of the first type is given by

∑n=0+∞(3n+1+12−2n+1)(3n+12−2n)zn=6z2−66z3+150z4+510z5−2376z6+2376z7(1−z)(1−4z)(1−9z)(1−2z)2(1−3z)2(1−6z)2,∑n=0+∞(3n+2+12−2n+2)(3n+12−2n)zn=25z2−360z3+1835z4−3960z5−3060z6(1−z)(1−4z)(1−9z)(1−2z)2(1−3z)2(1−6z)2.

4.4. The case A₃ = {1, 1, 1} and E₃ = {1, 1, 1}

Let us now consider the following conditions for equations (3.1)–(3.5) and (3.9):

{S1(−A3)=S1(−E3)=−3S2(−A3)=S2(−E3)=3S3(−A3)=S3(−E3)=−1.

Using the same procedure, we obtain the new generating functions:

∑n=0+∞(2n+1n)2zn=1−9z2+10z3−9z4+z61−9z+36z2−84z3+126z4−126z5+84z6−36z7+9z8−z9,∑n=0+∞(2nn−1)2zn=z−9z3+10z4−9z5+z71−9z+36z2−84z3+126z4−126z5+84z6−36z7+9z8−z9,∑n=0+∞(2n−1n−2)2zn=z2−9z4+10z5−9z6+z81−9z+36z2−84z3+126z4−126z5+84z6−36z7+9z8−z9,∑n=0+∞(2n+1n)(2nn−1)zn=3z−9z2+6z3+6z4−9z5+3z61−9z+36z2−84z3+126z4−126z5+84z6−36z7+9z8−z9,∑n=0+∞(2nn−1)(2n−1n−2)zn=3z2−9z3+6z4+6z5−9z6+3z71−9z+36z2−84z3+126z4−126z5+84z6−36z7+9z8−z9,∑n=0+∞(2n+1n−1)(2n−1n−2)zn=6z2−24z3+36z4−24z5+6z61−9z+36z2−84z3+126z4−126z5+84z6−36z7+9z8−z9.

4.5. The case A₃ = {1, 1, 1} and E₃ = {1, 2, 3}

Let us now consider the following conditions for equations (3.1)–(3.9) :

{S1(−A3)=−3S2(−A3)=3S3(−A3)=−1and{S1(−E3)=−6S2(−E3)=11S3(−E3)=−6.

We have the following propositions.

PROPOSITION 13

For n ∈ ℕ, the new generating functions is given by

∑n=0+∞(2n+1n)(3n+2+12−2n+2)zn=1−33z2+72z3−108z4+36z61−18z+141z2−630z3+1767z4−2034z5+3815z6−846z7+1188z8−216z9,∑n=0+∞(2nn−1)(3n+1+12−2n+1)zn=z−33z3+72z4−108z5+36z71−18z+141z2−630z3+1767z4−2034z5+3815z6−846z7+1188z8−216z9.

PROPOSITION 14

Let n be a natural number. Then we have the new generating functions

∑n=0+∞(2n−1n−2)(3n+12−2n)zn=z2−33z4+72z5−108z6+36z81−18z+141z2−630z3+1767z4−2034z5+3815z6−846z7+1188z8−216z9,∑n=0+∞(2n+1n)(3n+1+12−2n+1)zn=3z−18z2+25z3+36z4−108z5+66z61−18z+141z2−630z3+1767z4−2034z5+3815z6−846z7+1188z8−216z9.

PROPOSITION 15

For n ∈ ℕ, it is given the new generating functions by

∑n=0+∞(2nn−1)(3n+2+12−2n+2)zn=6z−33z2+36z3+85z4−198z5+108z61−18z+141z2−630z3+1767z4−2034z5+3815z6−846z7+1188z8−216z9,∑n=0+∞(2n−1n−2)(3n+1+12−2n+1)zn=6z2−33z3+36z4+85z5−198z6+108z71−18z+141z2−630z3+1767z4−2034z5+3815z6−846z7+1188z8−216z9.

PROPOSITION 16

Let n be a natural number. Then we have the new generating functions

∑n=0+∞(2n−1n−2)(3n+2+12−2n+2)zn=25z2−108z3+471z4−528z5+216z61−18z+141z2−630z3+1767z4−2034z5+3815z6−846z7+1188z8−216z9,∑n=0+∞(2n+1n)(3n+1+12−2n)zn=6z2−48z3+141z4−180z5+85z61−18z+141z2−630z3+1767z4−2034z5+3815z6−846z7+1188z8−216z9.

5. Conclusion

In this paper, we have derived new theorems in order to determine new generating functions of some numbers and 2-orthogonal polynomials. The derived theorems and corollaries are based on symmetric functions.

This work was supported by Directorate General for Scientific Research and Technological Development (DGRSDT), Algeria.

(Communicated by Istvàn Gaàl)

Acknowledgement

Authors are grateful to the Editor-In-Chief of the Journal and the anonymous reviewers for their constructive comments which improved the quality and the presentation of the paper.

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Received: 2020-12-16

Accepted: 2021-02-11

Published Online: 2022-02-16

Published in Print: 2022-02-16

This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/ms-2022-0002

Keywords for this article

Symmetric functions; generating functions; Tribonacci polynomials; Narayana numbers

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Ordinary Generating Functions of Binary Products of Third-Order Recurrence Relations and 2-Orthogonal Polynomials

Article

Abstract

1. Introduction

DEFINITION 1.

2. Definitions and some properties

DEFINITION 2 ([1]).

Remark 1.

DEFINITION 3 ([2])

DEFINITION 4 ([3,6])

DEFINITION 5 ([4,5])

3. Lemmas and main results

DEFINITION 6

PROPOSITION 1

PROPOSITION 2

COROLLARY 1

LEMMA 1.

LEMMA 2

LEMMA 3

LEMMA 4.

LEMMA 5

THEOREM 1

THEOREM 2

THEOREM 3

THEOREM 4

4. Applications

4.1. Generating functions of square of some known numbers

PROPOSITION 3

COROLLARY 2

THEOREM 5

PROPOSITION 4

COROLLARY 3

THEOREM 6

PROPOSITION 5

COROLLARY 4

PROPOSITION 6

COROLLARY 5

4.2. Generating functions of product of some known polynomials

PROPOSITION 7

COROLLARY 6

THEOREM 7

PROPOSITION 8

COROLLARY 7

PROPOSITION 9

COROLLARY 8

4.3. The case A3 = {1, 2, 3} and E3 = {1, 2, 3}

PROPOSITION 10

PROPOSITION 11

PROPOSITION 12

4.4. The case A3 = {1, 1, 1} and E3 = {1, 1, 1}

4.5. The case A3 = {1, 1, 1} and E3 = {1, 2, 3}

PROPOSITION 13

PROPOSITION 14

PROPOSITION 15

PROPOSITION 16

5. Conclusion

Acknowledgement

References

Articles in the same Issue

Articles in the same Issue

Articles in the same Issue

4.3. The case A₃ = {1, 2, 3} and E₃ = {1, 2, 3}

4.4. The case A₃ = {1, 1, 1} and E₃ = {1, 1, 1}

4.5. The case A₃ = {1, 1, 1} and E₃ = {1, 2, 3}