# Some Properties of the(p,q)-Fibonacci and (p,q)-Lucas Polynomials

Authors: GwangYeon Lee; Mustafa Asci
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

Riordan arrays are useful for solving the combinatorial sums by the help of generating functions. Many theorems can be easily proved by Riordan arrays. In this paper we consider the Pascal matrix and define a new generalization of Fibonacci polynomials called(p,q)-Fibonacci polynomials. We obtain combinatorial identities and by using Riordan method we get factorizations of Pascal matrix involving (p,q)-Fibonacci polynomials.


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## Body

## 1. Introduction

Large classes of polynomials can be defined by Fibonacci-like recurrence relation and yield Fibonacci numbers [1]. Such polynomials, called the Fibonacci polynomials, were studied in 1883 by the Belgian Mathematician Eugene Charles Catalan and the German Mathematician E. Jacobsthal. The polynomials fn(x) studied by Catalan are defined by the recurrence relation
(1.1)fn(x)=xfn-1(x)+fn-2(x),
where f1(x)=1,f2(x)=x, and n≥3. Notice that fn(1)=Fn, the nth Fibonacci number. The Fibonacci polynomials studied by Jacobsthal were defined by
(1.2)Jn(x)=Jn-1(x)+xJn-2(x),
where J1(x)=1=J2(x) and n≥3. The Pell polynomials pn(x) are defined by
(1.3)pn(x)=2xpn-1(x)+pn-2(x),
where p0(x)=0,p1(x)=1 and n≥2. The Lucas polynomials Ln(x), originally studied in 1970 by Bicknell, are defined by
(1.4)Ln(x)=xLn-1(x)+Ln-2(x),
where L0(x)=2,L1(x)=x, and n≥2.Horadam [2] introduced the polynomial sequence {wn(x)} defined recursively by
(1.5)wn(x)=p(x)wn-1(x)+q(x)wn-2(x),(n≥2),
where
(1.6)w0(x)=c0,w1(x)=c1xd,p(x)c2xd,q(x)=c3xd
in which c0,c1,c2,c3 are constants and d=0 or 1. Special cases of the w(x) with given initial conditions are given in Table 1:Table 1
Special cases of thew(x) with given initial conditions are given.
p
(
x)
q(x)
w
0(x) =  0, w1(x) = 1
w
0
(
x
)
=
2, w1(x)=x
x
1
Fibonacci Polynomial
Lucas Polynomial
2
x
1
Pell Polynomial
Pell-Lucas Polynomial
1
2x
Jacobsthal Poly
Jacobsthal-Lucas Poly.
3
x
−2
Fermat Polynomial
Fermat-Lucas Polynomial
2
x
−1
Cheby. Poly. 2nd kind
Cheby. Poly. first kindFor a fixedn, Brawer and Pirovino [3] defined the n×n lower triangular Pascal as matrix Pn=[pi,j]i,j=1,2,…,n(1.7)Pi,j={(i-1j-1)ifi≥j,0otherwise.The Pascal matrices have many applications in probability, numerical analysis, surface reconstruction, and combinatorics. In [4] the relationships between the Pascal matrix and the Vandermonde, Frobenius, Stirling matrices are studied. Also in [4] other applications in stability properties of numerical methods for solving ordinary differential equations are shown. In [5–8] the binomial coefficients, fibonomial coefficients, Pascal matrix, and its generalizations are studied. The authors in [9] factorized the Pascal matrix involving the Fibonacci matrix.Lee et al. [10] defined the n×n Fibonacci matrix as follows:
(1.8)ℱn=[fij]={Fi-j+1ifi-j+1≥0,0ifi-j+1<0.
Also in [10] factorizations and eigenvalues of Fibonacci and symmetric Fibonacci matrices are studied. The inverse of this matrix is given as follows;
(1.9)ℱn-1=[1000⋯0-1100⋯0-1-110⋯00-1-11⋱⋮⋮⋱⋱⋱⋱00⋯0-1-11].The Riordan group was introduced by Shapiro et al. in [6] as follows.LetR=[rij]i,j≥0 be an infinite matrix with complex entries. Let ci(t)=∑n≥0∞rn,itn be the generating function of the ith column of R. We call R a Riordan matrix if ci(t)=g(t)[f(t)]i, where
(1.10)g(t)=1+g1t+g2t2+g3t3+⋯,f(t)=t+f2t2+f3t3+⋯.
In this case we write R=(g(t),f(t)) and we denote by ℛ the set of Riordan matrices. Then the set ℛ is a group under matrix multiplication *, with the following properties:(R1)
(g(t),f(t))*(h(t),l(t))=(g(t)h(f(t)),l(f(t))),(R2)
I=(1,t) is the identity element,(R3)
the inverse ofR is given by R-1=(1/g(f¯(t)),f¯(t)), where f¯(t) is the compositional inverse of f(t), that is, f(f¯(t))=f¯(f(t))=t,(R4)
if(a0,a1,a2,…)T is a column vector with the generating function A(t), then multiplying R=(g(t),f(t)) on the right by this column vector yields a column vector with the generating function B(t)=g(t)A(f(t)).This group has many applications. Three of them are given in [6] such as Euler's problem of the King walk, binomial and inverse identities, and Bessel-Neumann expansion.Riordan arrays are also useful for solving the combinatorial sums by the help of generating functions. For example, in [11], Cheon, Kim, and Shapiro have many results including a generalized Lucas polynomial sequences from Riordan array and combinatorial interpretations for a pair of generalized Lucas polynomial sequences.


## 2. The(p,q)-Fibonacci and (p,q)-Lucas Polynomials with Some Properties

In [12], the authors introduced the h(x)-Fibonacci polynomials, where h(x) is a polynomial with real coefficients. The h(x)-Fibonacci polynomials {Fh,n(x)}n=0∞ are defined by the recurrence relation
(2.1)Fh,n+1(x)=h(x)Fh,n(x)+Fh,n-1(x),n≥1,
with initial conditions Fh,0(x)=0, Fh,1(x)=1.In this paper, we introduce a generalization of theh(x)-Fibonacci polynomials.Definition 2.1.
Letp(x) and q(x) be polynomials with real coefficients. The (p,q)-Fibonacci polynomials {Fp,q,n(x)}n=0∞ are defined by the recurrence relation
(2.2)Fp,q,n+1(x)=p(x)Fp,q,n(x)+q(x)Fp,q,n-1(x),n≥1
with the initial conditions Fp,q,0(x)=0 and Fp,q,1(x)=1.For later useFp,q,2(x)=p(x), Fp,q,3(x)=p2(x)+q(x), Fp,q,4(x)=p3(x)+2p(x)q(x), Fp,q,5(x)=p4(x)+3p2(x)q(x)+q2(x)⋯.Now, we introduce(p,q)-Lucas polynomials {Lp,q,n(x)}n=0∞ as the following definition.Definition 2.2.
The(p,q)-Lucas polynomials {Lp,q,n(x)}n=0∞ are defined by the recurrence relation
(2.3)Lp,q,n(x)=Fp,q,n+1(x)+q(x)Fp,q,n-1(x).Also for later useLp,q,0(x)=2, Lp,q,1(x)=p(x), Lp,q,2(x)=p2(x)+2q(x), Lp,q,3(x)=p3(x)+3p(x)q(x), Lp,q,4(x)=p4(x)+4p2(x)q(x)+2q2(x)⋯.In [12], the authors defined h(x)-Lucas polynomials as follows:
(2.4)Lh,n+1(x)=h(x)Lh,n(x)+Lh,n-1(x),n≥1,
with initial conditions Lh,0(x)=2, Lh,1(x)=h(x). However, we defined (p,q)-Lucas polynomials in the Definition 2.2 which is different from h(x)-Lucas polynomials. From the Definition 2.2, for p(x)=1 and q(x)=1, we obtain the usual Lucas numbers. And, for p(x)=h(x) and q(x)=1, we obtain the h(x)-Lucas polynomials.For the special cases ofp(x) and q(x), we can get the polynomials given in Table 1.The generating functiongF(t) of the sequence {Fp,q,n(x)} is defined by
(2.5)gF(t)=∑n=0∞‍Fp,q,n(x)tn.
We know that the generating function gF(t) is a convergence formal series.Theorem 2.3.
LetgF(t) be the generating function of the (p,q)-Fibonacci polynomial sequence Fp,q,n(x). Then
(2.6)gF(t)=t1-p(x)t-q(x)t2.Proof.
LetgF(t) be the generating function of the (p,q)-Fibonacci polynomial sequence Fp,q,n(x), then
(2.7)gF(t)=∑n=0∞‍Fp,q,n(x)tn=Fp,q,1(x)t+Fp,q,2(x)t2+∑n=3∞‍Fp,q,n(x)tn=t+t2p(x)+∑n=3∞‍[p(x)Fp,q,n-1(x)+q(x)Fp,q,n-2(x)]tn=t+t2p(x)+∑n=3∞‍p(x)Fp,q,n-1(x)tn+∑n=3∞‍q(x)Fp,q,n-2(x)tn=t+t2p(x)+t∑n=3∞‍p(x)Fp,q,n-1(x)tn-1+t2∑n=3∞‍q(x)Fp,q,n-2(x)tn-2=t+t2p(x)+tp(x)∑n=2∞‍Fp,q,n(x)tn+t2q(x)∑n=1∞‍Fp,q,n(x)tn=t+t2p(x)+tp(x)[∑n=1∞‍Fp,q,n(x)tn-t]+t2q(x)∑n=1∞‍Fp,q,n(x)tn=t+t2p(x)+tp(x)[gF(t)-t]+t2q(x)gF(t).
By taking gF(t) parenthesis we get
(2.8)gF(t)=t1-p(x)t-q(x)t2.
The proof is completed.Corollary 2.4.
LetgL(t) be the generating function of the (p,q)-Lucas polynomial sequence Lp,q,n(x). Then
(2.9)gL(t)=2-p(x)t1-p(x)t-q(x)t2.The Binet formula is also very important in Fibonacci numbers theory. Now we can get the Binet formula of(p,q)-Fibonacci polynomials. Let α(x) and β(x) be the roots of the characteristic equation
(2.10)t2-p(x)t-q(x)=0
of the recurrence relation (2.2). Then
(2.11)α(x)=p(x)+p2(x)+4q(x)2,β(x)=p(x)-p2(x)+4q(x)2.Note thatα(x)+β(x)=p(x) and α(x)β(x)=-q(x). Now we can give the Binet formula for the (p,q)-Fibonacci and (p,q)-Lucas polynomials.Theorem 2.5.
Forn≥0(2.12)Fp,q,n(x)=αn(x)-βn(x)α(x)-β(x),Lp,q,n(x)=αn(x)+βn(x).Proof.
The theorem can be proved by mathematical induction onn.Lemma 2.6.
Forn≥1,
(2.13)tn=Fp,q,n(x)t+q(x)Fp,q,n-1(x).Proof.
From the characteristic equation of the(p,q)-Fibonacci polynomials we have
(2.14)t2=p(x)t+q(x)=Fp,q,2(x)t+q(x)Fp,q,1(x).
By induction on n we get
(2.15)tn+1=tnt=(Fp,q,n(x)t+q(x)Fp,q,n-1(x))t=Fp,q,n(x)t2+q(x)tFp,q,n-1(x)=Fp,q,n(x)(p(x)t+q(x))+q(x)tFp,q,n-1(x)=(Fp,q,n(x)p(x)+q(x)Fp,q,n-1(x))t+q(x)Fp,q,n(x)=Fp,q,n+1(x)t+q(x)Fp,q,n(x).
Thus we have
(2.16)tn=Fp,q,n(x)t+q(x)Fp,q,n-1(x).Theorem 2.7.
LetLp,q,n(x)=Fp,q,n+1(x)+q(x)Fp,q,n-1(x). Then for n≥3,
(2.17)Lp,q,n(x)=p(x)Lp,q,n-1(x)+q(x)Lp,q,n-2(x).Proof.
Ifn=3 then
(2.18)Lp,q,3=p3(x)+3p(x)q(x)=p(x)Lp,q,2(x)+q(x)Lp,q,1(x).
By induction on n we have
(2.19)p(x)Lp,q,n(x)+q(x)Lp,q,n-1(x)=p(x)(Fp,q,n+1(x)+q(x)Fp,q,n-1(x))+q(x)(Fp,q,n(x)+q(x)Fp,q,n-2(x))=p(x)Fp,q,n+1(x)+p(x)q(x)Fp,q,n-1(x)+q(x)Fp,q,n(x)+q2(x)Fp,q,n-2(x)=p(x)Fp,q,n+1(x)+q(x)Fp,q,n(x)+q(x)(p(x)Fp,q,n-1(x)+q(x)Fp,q,n-2(x))=Fp,q,n+2(x)+q(x)Fp,q,n(x)=Lp,q,n+1(x).In [12], the author introduced the matrix Qh(x) that plays the role of the Q-matrix. The Q-matrix is associated with the Fibonacci numbers and is defined as
(2.20)Q=[1110].
Actually, in [12], Nalli and Pentti defined the matrix Qh(x) as follows
(2.21)Qh(x)=[h(x)110].We now introduce the matrixQp,q(x) which is a generalization of the Qh(x).Definition 2.8.
LetQp,q(x) denote the 2×2 matrix defined as
(2.22)Qp,q(x)=[p(x)q(x)10].Theorem 2.9.
Letn≥1. Then
(2.23)Qp,qn(x)=[Fp,q,n+1(x)q(x)Fp,q,n(x)Fp,q,n(x)q(x)Fp,q,n-1(x)].Proof.
We can prove the theorem by induction onn. The result holds for n=1. Suppose that it holds for n=m  (m≥1). Then
(2.24)Qp,qm+1(x)=Qp,qm(x)·Qp,q(x)=[Fp,q,m+1(x)q(x)Fp,q,m(x)Fp,q,m(x)q(x)Fp,q,m-1(x)][p(x)q(x)10]=[Fp,q,m+2(x)q(x)Fp,q,m+1(x)Fp,q,m+1(x)q(x)Fp,q,m(x)]
which completes the proof.Corollary 2.10.
Letm,n≥0. Then
(2.25)Fp,q,m+n+1(x)=Fp,q,m+1(x)Fp,q,n+1(x)+q(x)Fp,q,m(x)Fp,q,n(x).If an integera≠0 divides an integer b, we denote a∣b.Corollary 2.11.
Fork≥1,
(2.26)Fp,q,n(x)∣Fp,q,kn(x).Corollary 2.12.
The roots of characteristic equation ofQp,qn(x) are αn(x) and βn(x).Corollary 2.13.
Forn≥1(2.27)Fp,q,n(x)=∑n=0⌈(n-1)/2⌉‍(n-ii)pn-2i(x)qi(x).The following identities of which originated from Koshy (1998) [1] are a generalization of Koshy's results.Theorem 2.14.
Fork≥1, one has
(2.28)∑i=0n‍Fp,q,ki+j(x)={Fp,q,nk+j(x)pk(x)+Fp,q,j(x)-Fp,q,nk+k+j(x)+Fp,q,k-j(x)(-q(x))jpk(x)-Lp,q,k(x)+1,j<k,Fp,q,nk+j(x)pk(x)+Fp,q,j(x)-Fp,q,nk+k+j(x)+Fp,q,j-k(x)(-q(x))kpk(x)-Lp,q,k(x)+1,otherwise.Proof.
We know that(2.29)α(x)-β(x)=p2(x)+4q(x),Lp,q,n(x)=αn(x)+βn(x).
Since Fp,q,n(x)=(αn(x)-βn(x))/p2(x)+4q(x), from Theorem 2.5, we have
(2.30)∑i=0n‍Fp,q,ki+j(x)=∑i=0n‍αki+j(x)-βki+j(x)p2(x)+4q(x)=1p2(x)+4q(x)(αj(x)∑i=0n‍αki(x)-βj(x)∑i=0n‍βki(x))=1p2(x)+4q(x)(αj(x)αnk+k(x)-1αk(x)-1-βj(x)βnk+k(x)-1βk(x)-1)=1p2(x)+4q(x)(αnk+k+j(x)-αj(x)αk(x)-1-βnk+k+j(x)-βj(x)βk(x)-1).
Set 1/p2(x)+4q(x)=A(x) and
(2.31)C(x)={Fp,q,k-j(x)(-q(x))jifj<k,Fp,q,j-k(x)(-q(x))kotherwise.
Then we have
(2.32)∑i=0n‍Fp,q,ki+j(x)=A(x)(α(x)β(x))k-(αk(x)+βk(x))+1×(αnk+k+j(x)βk(x)-αj(x)βk(x)-αnk+k+j(x)+αj(x)-βnk+k+j(x)αk(x)+αk(x)βj(x)+βn(x))=A(x)pk(x)-Lp,q,k(x)+1×(Fp,q,nk+jpk(x)1A(x)+Fp,q,j(x)1A(x)-Fp,q,nk+k+j(x)1A(x)+αk(x)βj(x)-αj(x)βk(x)1A(x))=Fp,q,nk+j(x)pk(x)+Fp,q,j(x)-Fp,q,nk+k+j(x)+C(x)pk(x)-Lp,q,k(x)+1.
Thus the proof is completed.Theorem 2.15.
Forn≥0 one has
(2.33)∑k=0n‍(nk)pk(x)qn-k(x)Fp,q,k(x)=Fp,q,2n(x).Proof.
By Theorem2.5 we have
(2.34)∑k=0n‍(nk)pk(x)qn-k(x)Fp,q,k(x)=∑k=0n‍(nk)pk(x)qn-k(x)αk(x)-βk(x)α(x)-β(x)=1α(x)-β(x)(∑k=0n‍(nk)pk(x)qn-k(x)(αk(x)-βk(x)))=1α(x)-β(x)(∑k=0n‍(nk)(p(x)α(x))kqn-k(x)-∑k=0n‍(nk)(p(x)β(x))kqn-k(x))=1α(x)-β(x)((p(x)α(x)+q(x))n+(p(x)β(x)+q(x))n).
Sinceα(x) and β(x) are the solutions of the equation t2-p(x)t-q(x)=0,
(2.35)α2(x)=p(x)α(x)+q(x),β2(x)=p(x)β(x)+q(x).
Thus we have
(2.36)∑k=0n‍(nk)pk(x)qn-k(x)Fp,q,k(x)=(α2(x))n-(β2(x))nα(x)-β(x)=Fp,q,2n(x).
The proof is completed.Theorem 2.16.
Forn≥0, one has
(2.37)∑k=0n‍(nk)pk(x)(-q(x))n-kFp,q,k(x)=∑k=0n‍(nk)(-2q(x))kFp,q,2(n-k)(x).Proof.
By Theorem2.5 we have
(2.38)∑k=0n‍(nk)pk(x)(-q(x))n-kFp,q,k(x)=∑k=0n‍(nk)pk(x)(-q(x))n-kαk(x)-βk(x)α(x)-β(x)=1α(x)-β(x)(∑k=0n‍(nk)pk(x)(-q(x))n-k(αk(x)-βk(x)))=1α(x)-β(x)(∑k=0n‍(nk)(p(x)α(x))k(-q(x))n-k-∑k=0n‍(nk)(p(x)β(x))k(-q(x))n-k)=1α(x)-β(x)((p(x)α(x)-q(x))n+(p(x)β(x)-q(x))n).
Since α(x) and β(x) are the solutions of the equation t2-p(x)t-q(x)=0,
(2.39)p(x)α(x)-q(x)=α2(x)-2q(x),p(x)β(x)-q(x)=β2(x)-2q(x).
Thus we have
(2.40)∑k=0n‍(nk)pk(x)(-q(x))n-kFp,q,k(x)=(α2(x)-2q(x))n-(β2(x)-2q(x))nα(x)-β(x)=∑k=0n‍(nk)(-2q(x))kFp,q,2(n-k)(x).
The proof is completed.Corollary 2.17.
Forn≥0 one has
(2.41)∑k=0n‍(nk)pn-k(x)(-1)kLp,q,k(x)=Lp,q,n(x).Proof.
Sincep(x)-α(x)=-q(x)/α(x) and p(x)-β(x)=-q(x)/β(x), we have
(2.42)∑k=0n‍(nk)pn-k(x)(-1)kLp,q,k(x)=∑k=0n‍(nk)pn-k(x)(-1)k(αk(x)+βk(x))=∑k=0n‍(nk)pn-k(x)(-1)k(-α(x))k+∑k=0n‍(nk)pn-k(x)(-1)k(-β(x))k=(p(x)-α(x))n+(p(x)-β(x))n=(-q(x)α(x))n+(-q(x)β(x))n=(-q(x))nαn(x)+βn(x)(α(x)β(x))n=Lp,q,n(x).Corollary 2.18.
Forn≥0,
(2.43)Fp,q,2n(x)=Fp,q,n(x)Lp,q,n(x).Corollary 2.19.
Forn≥m,
(2.44)Fp,q,n+m(x)-(-q(x))mFp,q,n-m(x)=Fp,q,m(x)Lp,q,n(x)Proof.
From the Binet formula of the(p,q)-Fibonacci and Lucas polynomials we have
(2.45)Fp,q,n+m(x)-(-q(x))mFp,q,n-m(x)=αn+m(x)-βn+m(x)α(x)-β(x)-(-q(x))mαn-m(x)-βn-m(x)α(x)-β(x)=αn+m(x)-βn+m(x)-(-q(x))m(αn-m(x)-βn-m(x))α(x)-β(x).
Since α(x)β(x)=-q(x) then
(2.46)Fp,q,n+m(x)-(-q(x))mFp,q,n-m(x)=αn+m(x)-βn+m(x)-(α(x)β(x))m(αn-m(x)-βn-m(x))α(x)-β(x)=αn+m(x)-βn+m(x)-(αn(x)βm(x)-αm(x)βn(x))α(x)-β(x)=(αm(x)-βm(x))-(αn(x)+βn(x))α(x)-β(x)=αm(x)-βm(x)α(x)-β(x)((αn(x)+βn(x)))=Fp,q,m(x)Lp,q,n(x).Corollary 2.20.
Forn≥0 one has
(2.47)∑k=0n‍(nk)pk(x)qn-k(x)Lp,q,k(x)=Lp,q,2n(x).Theorem 2.21.
Forn≥1, one has
(2.48)Fp,q,n-1(x)Fp,q,n+1(x)-Fp,q,n2(x)=(-1)nqn-1(x).Proof.
We will prove the theorem by mathematical induction onn. Since
(2.49)Fp,q,0(x)Fp,q,1(x)-Fp,q,12(x)=0p(x)-12=0-1=(-1)1(q0(x)),
the given statement is true when n=1.
Now, we assume that it is true for an arbitrary positive integerk, that is,
(2.50)Fp,q,k-1(x)Fp,q,k+1(x)-Fp,q,k2(x)=(-1)kqk-1(x).
Then
(2.51)Fp,q,k(x)Fp,q,k+2(x)-Fp,q,k+12(x)=1p(x)(Fp,q,k+1(x)-q(x)Fp,q,k-1(x))×(p(x)Fp,q,k+1(x)+q(x)Fp,q,k(x))-Fp,q,k+12(x)=1p(x)(q(x)Fp,q,k+1(x)Fp,q,k(x)+p(x)Fp,q,k+12(x)-q2(x)Fp,q,k(x)Fp,q,k-1(x)-p(x)q(x)Fp,q,k-1(x)Fp,q,k+1(x))-Fp,q,k+12(x)=q(x)p(x)Fp,q,k+1(x)Fp,q,k(x)+Fp,q,k+12(x)-q2(x)p(x)Fp,q,k(x)Fp,q,k-1(x)-q(x)Fp,q,k-1(x)Fp,q,k+1(x)-Fp,q,k+12(x)=q(x)p(x)Fp,q,k+1(x)Fp,q,k(x)-q2(x)p(x)Fp,q,k(x)Fp,q,k-1(x)-q(x)Fp,q,k-1(x)Fp,q,k+1(x).
By assumption we have
(2.52)Fp,q,k(x)Fp,q,k+2(x)-Fp,q,k+12(x)=q(x)p(x)Fp,q,k+1(x)Fp,q,k(x)-q2(x)p(x)Fp,q,k(x)Fp,q,k-1(x)-q(x)((-1)kqk(x)+Fp,q,k2(x))=q(x)p(x)Fp,q,k+1(x)Fp,q,k(x)-q2(x)p(x)Fp,q,k(x)Fp,q,k-1(x)-q(x)Fp,q,k2(x)-(-1)kqk(x)=q(x)p(x)Fp,q,k+1(x)Fp,q,k(x)-q(x)Fp,q,k(x)×(q(x)p(x)Fp,q,k-1(x)+Fp,q,k(x))+(-1)k+1qk(x)=q(x)p(x)Fp,q,k+1(x)Fp,q,k(x)-q(x)Fp,q,k(x)1p(x)Fp,q,k+1(x)+(-1)k+1qk(x)=(-1)k+1qk(x).
Thus the formula works for n=k+1. So by mathematical induction, the statement is true for every integer n≥1.


## 3. The Infinite(p,q)-Fibonacci and (p,q)-Lucas Polynomial Matrix

In this section we define a new matrix which we call(p,q)-Fibonacci polynomials matrix. The infinite (p,q)-Fibonacci polynomials matrix
(3.1)ℱ(x)=[Fp,q,i,j(x)]
is defined as follows:
(3.2)ℱ(x)=[1000p(x)100⋮p2(x)+q(x)p(x)10⋮p3(x)+2p(x)q(x)p2(x)+q(x)p(x)1⋯⋯]=(gℱ(x)(t),fℱ(x)(t)).
The matrix ℱ(x) is an element of the set of Riordan matrices. Since the first column of ℱ(x) is
(3.3)(1,p(x),p2(x)+q(x),p3(x)+2p(x)q(x),p4(x)+3p2(x)q(x)+q2(x),…)T,
then it is obvious that gℱ(x)(t)=∑n=0∞‍Fp,q,n(x)tn=1/(1-p(x)t-q(x)t2). In the matrix ℱ(x) each entry has a rule with the upper two rows, that is,
(3.4)Fp,q,n+1,j(x)=p(x)Fp,q,n,j(x)+q(x)Fp,q,n-1,j(x).
Then fF(t)=t, that is,
(3.5)ℱ(x)=(gℱ(x)(t),fℱ(x)(t))=(11-p(x)t-q(x)t2,t);
hence ℱ(x) is in ℛ.Similarly we can define the(p,q)-Lucas polynomials matrix. The infinite (p,q)-Lucas polynomials matrix
(3.6)ℒ(x)=[Lp,q,i,j(x)]
can be written as
(3.7)ℒ(x)=(gℒ(x)(t),fℒ(x)(t))=(2-p(x)t1-p(x)t-q(x)t2,t).In this section  we give two factorizations of Pascal Matrix involving the(p,q)-Fibonacci polynomial matrix. For these factorizations we need to define two matrices. Firstly we define an infinite matrix M(x)=(mij(x)) as follows:
(3.8)mij(x)=(i-1j-1)-p(x)(i-2j-1)-q(x)(i-3j-1).We have the infinite matrixM(x) as follows:
(3.9)M(x)=[10001-p(x)1001-p(x)-q(x)2-p(x)10⋮1-p(x)-q(x)3-2p(x)-q(x)3-p(x)1⋯].Now we can give the first factorization of the infinite Pascal matrix via the infinite(p,q)-Fibonacci polynomial matrix and the infinite matrix M(x) defined in (3.8) by the following theorem.Theorem 3.1.
LetM(x) be the infinite matrix as in (3.8) and ℱ(x) be the infinite (p,q)-Fibonacci polynomial matrix; then,
(3.10)P(x)=ℱ(x)*M(x),
where P is the usual Pascal matrix.Proof.
From the definitions of the infinite Pascal matrix and the infinite(p,q)-Fibonacci polynomial matrix we have the following Riordan representations:
(3.11)P=(11-t,t1-t),ℱ(x)=(11-p(x)t-q(x)t2,t).
Now we can find the Riordan representation of the infinite matrix
(3.12)M(x)=(gM(x)(t),fM(x)(t))
as follows:
(3.13)M(x)=[10001-p(x)1001-p(x)-q(x)2-p(x)10⋮1-p(x)-q(x)3-2p(x)-q(x)3-p(x)1⋯].
From the first column of the matrix M(x) we obtain
(3.14)gM(x)(t)=1-p(x)t-q(x)t21-t.
From the rule of the matrix M(x),fM(x)(t)=t/(1-t). Thus
(3.15)M(x)=(1-p(x)t-q(x)t21-t,t1-t).
Finally by the Riordan representations of the matrices ℱ(x) and M(x) we complete the proof.Now we define then×n matrix R(x)=(rij(x)) as follows:
(3.16)(rij(x))=(i-1j-1)-p(x)(i-1j)-q(x)(i-1j+1).
We have the infinite matrix R(x) as follows.
(3.17)R(x)=[10001-p(x)1001-2p(x)-q(x)2-p(x)10⋮1-3p(x)-3q(x)3-3p(x)-q(x)3-p(x)1⋯].Now we can give second factorization of Pascal matrix via the infinite(p,q)-Fibonacci polynomial matrix by the following corollary:Corollary 3.2.
LetR(x) be the matrix as in (3.16). Then
(3.18)P=R(x)*ℱ(x).We can find the inverses of the matrices by using the Riordan representations of the matrices easily.Corollary 3.3.
One has(3.19)ℱ-1(x)=(1-p(x)t-q(x)t2,t)M-1(x)=(1+t1+(2-p(x))t-(1-p(x)-q(x))t2,t1+t)ℒ-1=(1-p(x)t-q(x)t22-p(x)t,t).



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*Source: 264842-2012-09-06.xml*