On Two New Classes of B-q-bonacci Polynomials

Suchita Arolkar Department of Mathematics and Statistics Dnyanprassarak Mandal’s College and Research Centre Assagao Bardez Goa 403 507, India

              Suchita.golatkar@yahoo.com

Yeshwant Shivrai Valaulikar Department of Mathematics Goa University Taleigao Plateau Goa 403 206, India

                   ysv@unigoa.ac.in

Abstract

In this paper we define two new classes of polynomials associated with generalized Fibonacci polynomials. We call them h(x)-B-q-bonacci polynomials and incomplete h(x)-B-q-bonacci polynomials. We present some identities for the two classes of polynomials, and the convolution property of h(x)-B-q-bonacci polynomials and its applications.

1 Introduction

The Fibonacci sequence, polynomials associated with the Fibonacci sequence, and their extended forms produce interesting and fascinating properties. For details see [8, 14]. Arolkar and Valaulikar introduced the B-tribonacci sequence [2] and B-tribonacci polynomials [1].

The B-tribonacci sequence [2] and B-tribonacci polynomials [1] are further extended to q^{th} order recurrence relations in [5] and [6] respectively. Arolkar and Valaulikar extended and studied the h(x)-Fibonacci polynomials [9] to h(x)-B-tribonacci polynomials [3]. Filipponi [7] introduced the incomplete Fibonacci and Lucas numbers. Ramírez [11] studied various identities related to the incomplete k-Fibonacci and k-Lucas numbers. Ramírez [13] introduced interesting classes of polynomials, namely, the incomplete h(x)-Fibonacci and h(x)-Lucas polynomials. Arolkar and Valaulikar [4] extended the incomplete h(x)-Fibonacci and h(x)-Lucas polynomials. Ramírez and Sirvent [12] defined and studied identities related to the incomplete tribonacci numbers and polynomials. Yilmaz and Taskara [10] obtained identities for the incomplete tribonacci-Lucas numbers and polynomials.

The aim of this paper is to extend two classes of polynomials, namely, the h(x)-B-tribonacci polynomials [3] and incomplete h(x)-B-tribonacci polynomials of [4] to the q^{th} order relations. We call them the h(x)-B-q-bonacci polynomials and incomplete h(x)-B-q-bonacci polynomials. We study some properties of these polynomials.

2 h(x)-B-q-bonacci polynomials

We first define the class of h(x)-B-q-bonacci polynomials.

Definition 1. Let h(x) be a polynomial with real coefficients. The h(x)-B-q-bonacci polynomials, denoted by (^{q}B)_{h,n}(x), n ∈ ℕ ∪ {0}, q ≥ 2, are defined by

(^{q}B)_{h,n+q−1}(x) = ∑_{r=0}^{q−1} \frac{(q − 1)\underline{r}}{r!} h^{q−1−r}(x)(^{q}B)_{h,n+q−2−r}(x), ∀n ≥ 1,  (1)

with (^{q}B)_{h,i}(x) = 0, i = 0, 1, 2, 3, …, q − 2 and (^{q}B)_{h,q−1}(x) = 1, where the coefficients of the terms on the right-hand side are the terms of the binomial expansion of (h(x) + 1)^{q−1} and (^{q}B)_{h,n}(x) is the n^{th} polynomial.

For simplicity, henceforth we denote (^{q}B)_{h,n}(x) by (^{q}B)_{h,n} and h(x) by h. We have the following identities for (^{q}B)_{h,n}.

(1) The n^{th} term (^{q}B)_{h,n} of (1) is given by

(^{q}B)_{h,n} = \begin{aligned}⌊ & \frac{(q−1)(n−(q−1))}{q} & ⌋ \\ & ∑ \\ & r=0 \end{aligned} \frac{((q − 1) (n − (q − 1) − r))\underline{r}}{r!} h^{(q−1)(n−(q−1)−r)−r},  (2)

n ≥ q − 1, where ⌊·⌋ denotes the floor function.

Proof. We prove the identity using induction on n.

For n = q − 1, (2) implies

(^{q}B)_{h,q−1} = ∑_{r=0}^{0} \frac{((q − 1) (−r))\underline{r}}{r!} h^{(q−1)(−r)−r} = 1.

Hence (2) is true for n = q − 1. Assume that (2) is true for n ≤ m. We divide the result into q cases, namely, m = qk, qk + 1, qk + 2, …, qk + (q − 1), for some k ≥ 1. Case (i): Let m = qk and t = ⌊\frac{(q−1)(qk−s−(q−1))}{q}⌋. Then

∑_{s=0}^{q−1} \frac{(q − 1)\underline{s}}{s!} h^{q−1−s}(^{q}B)_{h,qk−s}

= ∑_{s=0}^{q−1} \frac{(q − 1)\underline{s}}{s!} ∑_{r=0}^{t} \frac{((q − 1) (qk − (q − 1) − (r + s)))\underline{r}}{r!} h^{(q−1)(qk+1−(q−1)−(r+s))−(r+s)}

= ∑_{s=0}^{q−1} \frac{(q − 1)\underline{s}}{s!} ∑_{p=s}^{(q−1)k−(q−2)} \frac{((q − 1) (qk − (q − 1) − p))\underline{p−s}}{(p − s)!} h^{(q−1)(qk+1−(q−1)−p)−p}

= (\frac{(q − 1)\underline{0}}{0!} ∑_{p=0}^{(q−1)k−(q−2)} \frac{((q − 1) (qk − (q − 1) − p))\underline{p}}{p!}

+ \frac{(q − 1)\underline{1}}{1!} ∑_{p=1}^{(q−1)k−(q−2)} \frac{((q − 1) (qk − (q − 1) − p))\underline{p−1}}{(p − 1)!} + ⋯

+ \frac{(q − 1)\underline{(q−1)}}{(q − 1)!} ∑_{p=q−1}^{(q−1)k−(q−2)} \frac{((q − 1) (qk − (q − 1) − p))\underline{p−(q−1)}}{(p − (q − 1))!} ) h^{(q−1)(qk+1−(q−1)−p)−p}

= (\frac{((q − 1) (qk − (q − 1)))\underline{0}}{0!}

+ (\frac{((q − 1) (qk − (q − 1) − 1))\underline{1}}{1!} + \frac{(q − 1)\underline{1}}{1!} \frac{((q − 1) (qk − (q − 1) − 1))\underline{0}}{0!} ) + ⋯

+ ∑_{p=q−1}^{(q−1)k−(q−2)} ∑_{s=0}^{q−1} \frac{(q − 1)\underline{s}}{s!} \frac{((q − 1) (qk − (q − 1) − p))\underline{p−s}}{(p − s)!} ) h^{(q−1)(qk+1−(q−1)−p)−p}.

Therefore, using ∑_{s=0}^{q−1} \frac{(q−1)\underline{s}}{s!} \frac{n\underline{p−s}}{(p−s)!} = \frac{(n+(q−1))\underline{p}}{p!} , we have

∑_{s=0}^{q−1} \frac{(q − 1)\underline{s}}{s!} h^{q−1−s}(^{q}B)_{h,qk−s}

= ∑_{p=0}^{(q−1)k−(q−2)} \frac{((q − 1) (qk + 1 − (q − 1) − p))\underline{p}}{p!} h^{(q−1)(qk+1−(q−1)−p)−p}

= (^{q}B)_{h,qk+1}.

Thus, assuming the result for m = qk, we have proved it for m = qk + 1. Similarly, we can prove the other cases. We conclude that ∑_{s=0}^{q−1} \frac{(q−1)\underline{s}}{s!} h^{q−1−s}(^{q}B)_{h,m−s} = (^{q}B)_{h,m+1}. Hence, by induction, the result follows.

(2) The sum of the first n + 1 terms of (1) is given by

∑_{r=0}^{n}(^{q}B)_{h,r} = \frac{(^{q}B)_{h,n+1} + ∑_{i=0}^{q−2} ∑_{r=1+i}^{q−1} \frac{(q−1)\underline{r}}{r!} h^{q−1−r} (^{q}B)_{h,n−i} − 1}{(h + 1)^{q−1} − 1},  (3)

provided \begin{cases}h ≠ 0, & if q is even; \\ h ≠ 0, −2, & if q is odd. \end{cases}

Proof. We obtain the result by induction on n. For n = q − 1, ∑_{r=0}^{q−1}(^{q}B)_{h,r} = (^{q}B)_{h,q−1} = 1. Also,

\frac{(^{q}B)_{h,q} + ∑_{i=0}^{q−2} ∑_{r=1+i}^{q−1} \frac{(q−1)\underline{r}}{r!} h^{q−1−r} (^{q}B)_{h,q−1−i} − 1}{(h + 1)^{q−1} − 1}

= \frac{h^{q−1} + ∑_{r=1}^{q−1} \frac{(q−1)\underline{r}}{r!} h^{q−1−r} (^{q}B)_{h,q−1} − 1}{(h + 1)^{q−1} − 1} = 1.

Thus, (3) is true for n = q − 1. Assume that (3) is true for n ≤ m. Then

∑_{r=0}^{m+1}(^{q}B)_{h,r} = ∑_{r=0}^{m}(^{q}B)_{h,r} + (^{q}B)_{h,m+1}

= \frac{(^{q}B)_{h,m+1} + ∑_{i=0}^{q−2} ∑_{r=1+i}^{q−1} \frac{(q−1)\underline{r}}{r!} h^{q−1−r} (^{q}B)_{h,m−i} − 1}{(h + 1)^{q−1} − 1} + (^{q}B)_{h,m+1}

= \frac{(^{q}B)_{h,m+2} + ∑_{i=0}^{q−2} ∑_{r=1+i}^{q−1} \frac{(q−1)\underline{r}}{r!} h^{q−1−r} (^{q}B)_{h,m+1−i} − 1}{(h + 1)^{q−1} − 1}.

Thus, the result is true for n = m + 1. Hence, by induction, the result follows.

(3) The generating function for (1) is given by

(^{q}G_{(B)})_{h}(z) = \frac{1}{1 − z(h + z)^{q−1}} ,  (4)

provided |(z(h + z)^{q−1})| < 1.

Proof. Let t = ⌊\frac{(q−1)(n−(q−1))}{q}⌋. Then

(^{q}G_{(B)})_{h}(z) = ∑_{n=0}^{∞}(^{q}B)_{h,n} z^{n−(q−1)}

= ∑_{n=0}^{q−2}(^{q}B)_{h,n} z^{n−(q−1)} + ∑_{n=q−1}^{∞}(^{q}B)_{h,n} z^{n−(q−1)}

= ∑_{n=q−1}^{∞}(^{q}B)_{h,n} z^{n−(q−1)}, since (^{q}B)_{h,i} = 0, i = 0, 1, 2, 3, …, q − 2

= ∑_{n=q−1}^{∞} ∑_{r=0}^{t} \frac{((q − 1) (n − (q − 1) − r))\underline{r}}{r!} h^{(q−1)(n−(q−1)−r)−r} z^{n−(q−1)}

= 1 + h^{q−1}z + (h^{2(q−1)} + \frac{(q − 1)\underline{1}}{1!} h^{q−2}) z^{2} + …

= 1 + z(h + z)^{q−1} + z^{2}(h + z)^{2(q−1)} + …

= \frac{1}{1 − z(h + z)^{q−1}} , provided |(z(h + z)^{q−1})| < 1.

We now obtain the following property. Theorem 2. (Convolution property for (^{q}B)_{h,n})

For all n ≥ q − 1, we have

\frac{d}{dx} ((^{q}B)_{h,n}) = (q − 1)\frac{dh}{dx} (∑_{r=0}^{q−2} \frac{(q − 2)\underline{r}}{r!} h^{(q−2)−r} ∑_{i=0}^{n+q−2−r}(^{q}B)_{h,i} (^{q}B)_{h,n+q−2−r−i}) . (5)

Proof. Equation (4) implies

∑_{n=0}^{∞}(^{q}B)_{h,n} z^{n−(q−1)} = \frac{1}{1 − z(h + z)^{q−1}} .

Differentiating both sides with respect to x, we get

∑_{n=0}^{∞} \frac{d}{dx} ((^{q}B)_{h,n}) z^{n−(q−1)} = z(q − 1)(h + z)^{q−2} \frac{1}{(1 − z(h + z)^{q−1})^{2}} \frac{dh}{dx}

= ((q − 1) ∑_{r=0}^{q−2} \frac{(q − 2)\underline{r}}{r!} h^{(q−2)−r}z^{r+1} (∑_{n=0}^{∞}(^{q}B)_{h,n} z^{n−(q−1)})^{2}) \frac{dh}{dx}

= ((q − 1) ∑_{r=0}^{q−2} \frac{(q − 2)\underline{r}}{r!} h^{(q−2)−r} z^{−2(q−1)+r+1} (∑_{n=0}^{∞}(^{q}B)_{h,n} z^{n})^{2}) \frac{dh}{dx}

= (q − 1)\frac{dh}{dx} ∑_{r=0}^{q−2} \frac{(q − 2)\underline{r}}{r!} h^{(q−2)−r} ∑_{n=0}^{∞} (∑_{i=0}^{n}(^{q}B)_{h,i}(^{q}B)_{h,n−i} z^{n−2(q−1)+r+1}) .

Comparing the coefficients of z^{n−(q−1)}, we get

\frac{d}{dx} ((^{q}B)_{h,n}) = (q − 1)\frac{dh}{dx} ( ∑_{r=0}^{q−2} \frac{(q − 2)\underline{r}}{r!} h^{(q−2)−r} ∑_{i=0}^{n+q−2−r}(^{q}B)_{h,i}(^{q}B)_{h,n+q−2−r−i}) .

We now give an application of the convolution property. It is required to prove an identity in the next section. Theorem 3. For n ≥ q − 1,

\begin{aligned}⌊ & \frac{(q−1)(n−(q−1))}{q} & ⌋ \\ & ∑ \\ & r=0 \end{aligned} r \frac{((q − 1)(n − (q − 1) − r))\underline{r}}{r!} h^{(q−1)(n−(q−1))−qr}

= \frac{(q − 1) (n − (q − 1))}{q} (^{q}B)_{h,n}

−\frac{h}{q} (q − 1) (∑_{r=0}^{q−2} \frac{(q − 2)\underline{r}}{r!} h^{(q−2)−r} ∑_{i=0}^{n+q−2−r}(^{q}B)_{h,i} (^{q}B)_{h,n+q−2−r−i}) .  (6)

Proof. Equation (2) implies

(^{q}B)_{h,n} = \begin{aligned}⌊ & \frac{(q−1)(n−(q−1))}{q} & ⌋ \\ & ∑ \\ & r=0 \end{aligned} \frac{((q − 1)(n − (q − 1) − r))\underline{r}}{r!} h^{(q−1)(n−(q−1))−qr}.

Differentiating both sides with respect to x and simplifying, we get

\frac{d}{dx}((^{q}B)_{h,n})h = ((q − 1)(n − (q − 1))) (^{q}B)_{h,n} \frac{dh}{dx}

− q \frac{dh}{dx} \begin{aligned}⌊ & \frac{(q−1)(n−(q−1))}{q} & ⌋ \\ & ∑ \\ & r=0 \end{aligned} r \frac{((q − 1)(n − (q − 1) − r))\underline{r}}{r!} h^{(q−1)(n−(q−1))−qr}.

Thus,

\frac{dh}{dx} \begin{aligned}⌊ & \frac{(q−1)(n−(q−1))}{q} & ⌋ \\ & ∑ \\ & r=0 \end{aligned} r \frac{((q − 1)(n − (q − 1) − r))\underline{r}}{r!} h^{(q−1)(n−(q−1))−qr}

= \frac{(q − 1)(n − (q − 1))}{q} (^{q}B)_{h,n} \frac{dh}{dx} − \frac{h}{q} \frac{d}{dx} ((^{q}B)_{h,n}) .

Hence (5) implies

\begin{aligned}⌊ & \frac{(q−1)(n−(q−1))}{q} & ⌋ \\ & ∑ \\ & r=0 \end{aligned} r \frac{((q − 1)(n − (q − 1) − r))\underline{r}}{r!} h^{(q−1)(n−(q−1))−qr}

= \frac{(q − 1)(n − (q − 1))}{q} (^{q}B)_{h,n}

− \frac{h}{q} (q − 1) ( ∑_{r=0}^{q−2} \frac{(q − 2)\underline{r}}{r!} h^{(q−2)−r} ∑_{i=0}^{n+q−2−r}(^{q}B)_{h,i} (^{q}B)_{h,n+q−2−r−i}) .

3 Incomplete h(x)-B-q-bonacci Polynomials

In this section we define the class of incomplete h(x)-B-q-bonacci polynomials and discuss some of its properties. Definition 4. The incomplete h(x)-B-q-bonacci polynomials are defined by

(^{q}B)_{h,n}^{l}(x) = ∑_{r=0}^{l} \frac{((q − 1)(n − (q − 1) − r))\underline{r}}{r!} h^{(q−1)(n−(q−1)−r)−r}(x),  (7)

∀ 0 ≤ l ≤ ⌊\frac{(q − 1)(n − (q − 1))}{q}⌋ and n ≥ q − 1.

Note that (^{q}B)⌊\frac{(q−1)(n−(q−1))}{q}⌋_{h,n}(x) = (^{q}B)_{h,n}(x).

For simplicity, we use (^{q}B)_{h,n}^{l}(x) = (^{q}B)_{h,n}^{l}, (^{q}B)_{h,n}(x) = (^{q}B)_{h,n} and h(x) = h. We prove identities related to the recurrence relation for (^{q}B)_{h,n}^{l}.

Theorem 5. The recurrence relation for (^{q}B)_{h,n}^{l} is given by

(^{q}B)_{h,n+q}^{l+q−1} = ∑_{r=0}^{q−1} \frac{(q − 1)\underline{r}}{r!} h^{q−1−r}(^{q}B)_{h,n+q−1−r}^{l+q−1−r}, 0 ≤ l ≤ ⌊\frac{(q − 1)(n − q)}{q}⌋ , ∀n ≥ q. (8)

Proof. Consider,

∑_{r=0}^{q−1} \frac{(q − 1)\underline{r}}{r!} h^{q−1−r} (^{q}B)_{h,n+q−1−r}^{l+q−1−r}

= ∑_{r=0}^{q−1} \frac{(q − 1)\underline{r}}{r!} h^{q−1−r}

∑_{i=0}^{l+q−1−r} \frac{((q − 1)(n + q − 1 − r − (q − 1) − i))\underline{i}}{i!} h^{(q−1)(n+q−1−r−(q−1)−i)−i}

= ∑_{r=0}^{q−1} \frac{(q − 1)\underline{r}}{r!} h^{q−1−r} ∑_{i=0}^{l+q−1−r} \frac{((q − 1)(n − r − i))\underline{i}}{i!} h^{(q−1)(n−r)−qi}

= ∑_{r=0}^{q−1} \frac{(q − 1)\underline{r}}{r!} ∑_{i=0}^{l+q−1−r} \frac{((q − 1)(n − r − i))\underline{i}}{i!} h^{(q−1)(n+1)−qr−qi}

= ∑_{r=0}^{q−1} \frac{(q − 1)\underline{r}}{r!} ∑_{i=0}^{l+q−1−r} \frac{((q − 1)(n − (r + i)))\underline{i}}{i!} h^{(q−1)(n+1)−q(r+i)}.

Taking j = i + r, we get

∑_{r=0}^{q−1} \frac{(q − 1)\underline{r}}{r!} (^{q}B)_{h,n+q−1−r}^{l+q−1−r} h^{q−1−r}

= ∑_{r=0}^{q−1} \frac{(q − 1)\underline{r}}{r!} ∑_{j=r}^{l+q−1} \frac{((q − 1)(n − j))\underline{j−r}}{(j − r)!} h^{(q−1)(n+1)−qj}

= ∑_{j=0}^{l+q−1} \frac{((q − 1)(n + 1 − j))\underline{j}}{j!} h^{(q−1)(n+1)−qj}

= (^{q}B)_{h,n+q}^{l+q−1}.

Theorem 6. For s ≥ 1,

(^{q}B)_{h,n+qs}^{l+(q−1)s} = ∑_{i=0}^{(q−1)s} \frac{((q − 1)s)\underline{i}}{i!} (^{q}B)_{h,n+i}^{l+i} h^{i},  (9)

0 ≤ l ≤ ⌊\frac{(q−1)(n−s−(q−1))}{q}⌋ . Proof. Follows using induction. Theorem 7. For n ≥ ⌊\frac{ql+2(q−1)}{q−1}⌋ , (^{q}B)_{h,n+(q−1)+s}^{l+(q−1)} − h^{(q−1)s}(^{q}B)_{h,n+q−1}^{l+q−1}

= ∑_{i=0}^{s−1} ∑_{r=1}^{q−1} \frac{(q − 1)\underline{r}}{r!} h^{(q−1)s−(q−1)i−r} (^{q}B)_{h,n+(q−1)+i−r}^{l+(q−1)−r}.  (10)

Proof. Follows using induction.

The next theorem is related to the sum of incomplete h(x)-B-q-bonacci polynomials (^{q}B)_{h,n}^{l}. Theorem 8. For all n ≥ q − 1,

\begin{matrix}⌊ & \frac{(q−1)(n−(q−1))}{q} & ⌋ \\ & ∑ \\ & l=0 \end{matrix}(^{q}B)_{h,n}^{l} = (⌊\frac{(q − 1) (n − (q − 1))}{q}⌋ + \frac{q − (q − 1) (n − (q − 1))}{q}) (^{q}B)_{h,n}

+\frac{h}{q} (q − 1) (∑_{r=0}^{q−2} \frac{(q − 2)\underline{r}}{r!} h^{(q−2)−r} ∑_{i=0}^{n+q−2−r}(^{q}B)_{h,i} (^{q}B)_{h,n+q−2−r−i}) .  (11)

Proof. ∑⌊\frac{(q−1)(n−(q−1))}{q}⌋_{l=0}(^{q}B)_{h,n}^{l}

= (^{q}B)_{h,n}^{0} + (^{q}B)_{h,n}^{1} + ⋯ + (^{q}B)_{h,n}^{r} + ⋯ + (^{q}B)⌊\frac{(q−1)(n−(q−1))}{q}⌋_{h,n}

= \frac{((q − 1)(n − (q − 1)))\underline{0}}{0!} h^{(q−1)(n−(q−1))}

+ (\frac{((q − 1)(n − (q − 1)))\underline{0}}{0!} h^{(q−1)(n−(q−1))} + \frac{(q − 1)(n − (q − 1) − 1))\underline{1}}{1!} h^{(q−1)(n−(q−1))−q})

+ ⋯ +

(\frac{((q − 1)(n − (q − 1)))\underline{0}}{0!} h^{(q−1)(n−(q−1))} + ⋯ + \frac{((q − 1)(n − (q − 1) − r))\underline{r}}{r!} h^{(q−1)(n−(q−1))−qr})

+ ⋯

+ (\frac{((q − 1)(n − (q − 1)))\underline{0}}{0!} h^{(q−1)(n−(q−1))} + ⋯

+ \frac{((q − 1)(n − (q − 1) − r))\underline{r}}{r!} h^{(q−1)(n−(q−1))−qr} + ⋯

+ \frac{((q − 1) (n − (q − 1) − ⌊\frac{(q−1)(n−(q−1))}{q}⌋))\underline{⌊ \frac{(q−1)(n−(q−1))}{q} ⌋}}{(⌊\frac{(q−1)(n−(q−1))}{q}⌋)!} h^{(q−1)(n−(q−1))−q⌊ \frac{(q−1)(n−(q−1))}{q} ⌋})

= \begin{aligned}⌊ & \frac{(q−1)(n−(q−1))}{q} & ⌋ \\ & ∑ \\ & r=0 \end{aligned} (⌊\frac{(q − 1)(n − (q − 1))}{q}⌋ + 1 − r) \frac{((q − 1)(n − (q − 1) − r))\underline{r}}{r!} h^{(q−1)(n−(q−1))−qr}

= \begin{aligned}⌊ & \frac{(q−1)(n−(q−1))}{q} & ⌋ \\ & ∑ \\ & r=0 \end{aligned} (⌊\frac{(q − 1)(n − (q − 1))}{q}⌋ + 1) \frac{((q − 1)(n − (q − 1) − r))\underline{r}}{r!} h^{(q−1)(n−(q−1))−qr}

− \begin{aligned}⌊ & \frac{(q−1)(n−(q−1))}{q} & ⌋ \\ & ∑ \\ & r=0 \end{aligned} r\frac{((q − 1)(n − (q − 1) − r))\underline{r}}{r!} h^{(q−1)(n−(q−1))−qr}

= (⌊\frac{(q − 1)(n − (q − 1))}{q}⌋ + 1 − \frac{(q − 1)(n − (q − 1))}{q}) (^{q}B)_{h,n}

+ \frac{h}{q}(q − 1) (∑_{r=0}^{q−2} \frac{(q − 2)\underline{r}}{r!} h^{(q−2)−r} ∑_{i=0}^{n+q−2−r}(^{q}B)_{h,i} (^{q}B)_{h,n+q−2−r−i}) .

Using (6) of Theorem 3 in Section 2, the result follows.

4 Acknowledgments

The authors thank the reviewers/referees for their comments and suggestions that helped to improve the article.

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2010 Mathematics Subject Classification: Primary 11B39; Secondary 11B83. Keywords: Fibonacci polynomial, h(x)-B-q-bonacci polynomial, incomplete h(x)-B-q-bonacci polynomial.

Received October 3 2017; revised versions received October 5 2017; March 16 2018; April 4 2018; April 6 2018. Published in Journal of Integer Sequences, May 7 2018.

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