\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics,amsmath,amssymb} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.1in} \setlength{\textheight}{8.4in} \newcommand{\seqnum}[1]{\href{http://oeis.org/#1}{\underline{#1}}} \newcommand{\N}{{\mathbb N}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \begin{center} \vskip 1cm{\LARGE\bf On Two New Classes of $B$-$q$-bonacci \vskip .1in Polynomials} \vskip 1cm \large Suchita Arolkar\\ Department of Mathematics and Statistics\\ Dnyanprassarak Mandal's College and Research Centre\\ Assagao Bardez Goa\\ 403 507, India\\ \href{mailto:Suchita.golatkar@yahoo.com}{\tt Suchita.golatkar@yahoo.com}\\ \ \\ Yeshwant Shivrai Valaulikar\\ Department of Mathematics \\ Goa University \\ Taleigao Plateau Goa\\ 403 206, India\\ \href{mailto:ysv@unigoa.ac.in}{\tt ysv@unigoa.ac.in } \end{center} \vskip .2 in \begin{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. \end{abstract} \section{Introduction} The Fibonacci sequence, polynomials associated with the Fibonacci sequence, and their extended forms produce interesting and fascinating properties. For details see \cite{Thomas, Vajda}. Arolkar and Valaulikar introduced the $B$-tribonacci sequence \cite{Sa1} and $B$-tribonacci polynomials \cite{Sa2}. The $B$-tribonacci sequence \cite{Sa1} and $B$-tribonacci polynomials \cite{Sa2} are further extended to $q^{th}$ order recurrence relations in \cite{Sa5} and \cite{Sa6} respectively. Arolkar and Valaulikar extended and studied the $h(x)$-Fibonacci polynomials \cite{Nalli} to $h(x)$-$B$-tribonacci polynomials \cite{Sa4}. Filipponi \cite{Filipponi} introduced the incomplete Fibonacci and Lucas numbers. Ram\'{i}rez \cite{Jose} studied various identities related to the incomplete $k$-Fibonacci and $k$-Lucas numbers. Ram\'{i}rez \cite{Jose1} introduced interesting classes of polynomials, namely, the incomplete $h(x)$-Fibonacci and $h(x)$-Lucas polynomials. Arolkar and Valaulikar \cite{Sa3} extended the incomplete $h(x)$-Fibonacci and $h(x)$-Lucas polynomials. Ram\'{i}rez and Sirvent \cite{victor} defined and studied identities related to the incomplete tribonacci numbers and polynomials. Yilmaz and Taskara \cite{Nazmiye} 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 \cite{Sa4} and incomplete $h(x)$-$B$-tribonacci polynomials of \cite{Sa3} 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. \section{$h(x)$-$B$-$q$-bonacci polynomials}\label{sec:2} We first define the class of $h(x)$-$B$-$q$-bonacci polynomials. \begin{definition} Let $h(x)$ be a polynomial with real coefficients. The $h(x)$-$B$-$q$-bonacci polynomials, denoted by ${({}^qB)_{h,n}}(x), n \in {\N} \cup \{0\}, q \geq 2 $, are defined by \begin{equation}\label{eq:1.1} ({}^qB)_{h,n+q-1}(x)= \sum_{r=0}^{q-1}\frac{(q-1)^{\underline{r}}}{r!}\ \ h^{q-1-r}(x) ({}^qB)_{h,n+q-2-r}(x),\forall n \geq 1, \end{equation} with $({}^qB)_{h, i}(x) = 0,\ i = 0,1,2,3,\ldots,q-2$ and $({}^qB)_{h,q-1}(x) = 1$, 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 $({}^qB)_{h,n}(x)$ is the $n^{th}$ polynomial. \end{definition} For simplicity, henceforth we denote $({}^qB)_{h,n}(x)$ by $({}^qB)_{h,n}$ and $h(x)$ by $h$. We have the following identities for $({}^qB)_{h,n}$. \begin{enumerate} \item [(1)] The $n^{th}$ term $({}^qB)_{h,n}$ of (\ref{eq:1.1}) is given by \begin{equation}\label{eq:2.1} ({}^qB)_{h,n} = \sum_{r=0}^{\left \lfloor \frac{(q-1)\left(n-(q-1)\right)}{q}\right \rfloor} \frac{\left((q-1)\left(n-(q-1)-r\right)\right)^{\underline{r}}}{r!}\ \ h^{(q-1)(n-(q-1)-r)-r}, \end{equation} $n \geq q-1,$ where $\lfloor\cdot\rfloor$ denotes the floor function. \begin{proof} We prove the identity using induction on $n$. For $n = q-1$, $(\ref{eq:2.1})$ implies \begin{align*} ({}^qB)_{h,q-1} = \sum_{r=0}^{0} \frac{\left((q-1)\left(-r\right)\right)^{\underline{r}}}{r!}\ \ h^{(q-1)(-r)-r} =1. \end{align*} Hence $(\ref{eq:2.1})$ is true for $n=q-1$. Assume that $(\ref{eq:2.1})$ is true for $n \leq m$. We divide the result into $q$ cases, namely, $m = qk, qk+1, qk+2, \ldots , qk+(q-1)$, for some $k \geq 1$. \noindent Case (i): Let $m=qk$ and $t=\left\lfloor\frac{(q-1)\left(qk-s-(q-1)\right)}{q}\right \rfloor$. Then \begin{align*} &\sum_{s=0}^{q-1}\frac{(q-1)^{\underline{s}}}{s!}\ h^{q-1-s} ({}^qB)_{h,qk-s}\\ &=\sum_{s=0}^{q-1}\frac{(q-1)^{\underline{s}}}{s!}\sum_{r=0}^{t} \frac{\left((q-1)\left(qk-(q-1)-(r+s)\right)\right)^{\underline{r}}}{r!} \ h^{(q-1)(qk+1-(q-1)-(r+s))-(r+s)}\\ &= \sum_{s=0}^{q-1}\frac{(q-1)^{\underline{s}}}{s!}\;\sum_{p=s}^{(q-1)k-(q-2) } \frac{\left((q-1)\left(qk-(q-1)-p\right)\right)^{\underline{p-s}}}{(p-s)!} \ h^{(q-1)(qk+1-(q-1)-p)-p}\\ &=\Big( \frac{(q-1)^{\underline{0}}}{0!} \sum_{p=0}^{(q-1)k-(q-2)} \frac{\left((q-1)\left(qk-(q-1)-p\right)\right)^{\underline{p}}}{p!}\\ &+ \frac{(q-1)^{\underline{1}}}{1!} \sum_{p=1}^{(q-1)k-(q-2)} \frac{\left((q-1)\left(qk-(q-1)-p\right)\right)^{\underline{p-1}}}{(p-1)!} +\cdots\\ &+\frac{(q-1)^{\underline{(q-1)}}}{(q-1)!} \sum_{p=q-1}^{(q-1)k-(q-2)} \frac{\left((q-1)\left(qk-(q-1)-p\right)\right)^{\underline{p-(q-1)}}}{(p-(q-1))!}\Big)\ h^{(q-1)(qk+1-(q-1)-p)-p}\\ &=\Bigg(\frac{\left((q-1)\left(qk-(q-1)\right)\right)^{\underline{0}}}{0!}\\ &+ \left(\frac{\left((q-1)\left(qk-(q-1)-1\right)\right)^{\underline{1}}}{1!}+ \frac{(q-1)^{\underline{1}}}{1!} \frac{\left((q-1)\left(qk-(q-1)-1\right)\right)^{\underline{0}}}{0!}\right)+\cdots\\ &+ \sum_{p= q-1}^{(q-1)k-(q-2) } \sum_{s=0}^{q-1}\frac{(q-1)^{\underline{s}}}{s!} \frac{\left((q-1)\left(qk-(q-1)-p\right)\right)^{\underline{p-s}}}{(p-s)!}\Bigg)\ h^{(q-1)(qk+1-(q-1)-p)-p}. \end{align*} Therefore, using $\sum_{s=0}^{q-1}\frac{(q-1)^{\underline{s}}}{s!} \frac{n^{\underline{p-s}}}{(p-s)!} = \frac{(n+(q-1))^{\underline{p}}}{p!},$ we have \begin{align*} &\sum_{s=0}^{q-1}\frac{(q-1)^{\underline{s}}}{s!}\;h^{q-1-s} ({}^qB)_{h,qk-s}\\ &=\sum_{p= 0}^{(q-1)k-(q-2) } \frac{\left((q-1)\left(qk+1-(q-1)-p\right)\right)^{\underline{p}}}{p!}\ h^{(q-1)(qk+1-(q-1)-p)-p}\\ &=({}^qB)_{h,qk+1}. \end{align*} 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 $\sum_{s=0}^{q-1}\frac{(q-1)^{\underline{s}}}{s!}\;h^{q-1-s} ({}^qB)_{h,m-s} =({}^qB)_{h,m+1}$. Hence, by induction, the result follows. \end{proof} \item [(2)] The sum of the first $n+1$ terms of (\ref{eq:1.1}) is given by \begin{equation}\label{eq:2.2} \sum_{r=0}^{n} ({}^qB)_{h,r} = \frac {({}^qB)_{h,n+1} +\sum_{i=0}^{q-2}\sum_{r=1+i}^{q-1} \frac{(q-1)^{\underline{r}}}{r!}\; h^{q-1-r}\ \ ({}^qB)_{h,n-i} -1}{(h+1)^{q-1}-1},\ \ \end{equation} provided $ \begin{cases} h \neq 0, & \text{if $q$ is even;}\\ h \neq 0,-2, & \text{if $q$ is odd.} \end{cases} $ \begin{proof} We obtain the result by induction on $n$. For $n=q-1,$ $\sum_{r=0}^{q-1} ({}^qB)_{h,r} = ({}^qB)_{h,q-1} =1$. Also, \begin{align*} &\frac {({}^qB)_{h,q} +\sum_{i=0}^{q-2}\sum_{r=1+i}^{q-1} \frac{(q-1)^{\underline{r}}}{r!}\ h^{q-1-r}\ \ ({}^qB)_{h,q-1-i} -1}{(h+1)^{q-1}-1}\\ & = \frac {h^{q-1} + \sum_{r=1}^{q-1} \frac{(q-1)^{\underline{r}}}{r!}\ h^{q-1-r}\;({}^qB)_{h,q-1} -1}{(h+1)^{q-1}-1}=1. \end{align*} Thus, (\ref{eq:2.2}) is true for $n=q-1$. Assume that (\ref{eq:2.2}) is true for $n\leq m$. Then \begin{align*} &\sum_{r=0}^{m+1} ({}^qB)_{h,r} = \sum_{r=0}^{m} ({}^qB)_{h,r} + ({}^qB)_{h,m+1} \\ &=\frac {({}^qB)_{h,m+1} +\sum_{i=0}^{q-2}\sum_{r=1+i}^{q-1} \frac{(q-1)^{\underline{r}}}{r!}\ h^{q-1-r}\ ({}^qB)_{h,m-i} -1}{(h+1)^{q-1}-1} + ({}^qB)_{h,m+1}\\ &=\frac {({}^qB)_{h,m+2} +\sum_{i=0}^{q-2}\sum_{r=1+i}^{q-1} \frac{(q-1)^{\underline{r}}}{r!}\ h^{q-1-r}\ \ ({}^qB)_{h,m+1-i} -1}{(h+1)^{q-1}-1}. \end{align*} Thus, the result is true for $ n= m+1$. Hence by induction the result follows. \end{proof} \item [(3)] The generating function for (\ref{eq:1.1}) is given by \begin{equation}\label{eq:2.3} ({}^qG_{(B)})_{h}(z) = \frac{1}{1-z(h+z)^{q-1}}, \end{equation} provided $|(z(h+z)^{q-1})|<1$. \begin{proof} Let $t = \left \lfloor\frac{(q-1)\left(n-(q-1)\right)}{q}\right \rfloor$ \begin{align*} ({}^qG_{(B)})_{h}(z)&= \sum_{n=0}^{\infty}({}^qB)_{h,n}\ z^{n-(q-1)}\\ &=\sum_{n=0}^{q-2}({}^qB)_{h,n}\ z^{n-(q-1)}+\sum_{n=q-1}^{\infty}({}^qB)_{h,n}\ z^{n-(q-1)}\\ &=\sum_{n=q-1}^{\infty}({}^qB)_{h,n}\;z^{n-(q-1)},\ \text{since}\ ({}^qB)_{h, i} = 0,\ i = 0,1,2,3,\ldots,q-2\\ &=\sum_{n=q-1}^{\infty} \sum_{r=0}^{t} \frac{\left((q-1)\left(n-(q-1)-r\right)\right)^{\underline{r}}}{r!}\ \ h^{(q-1)(n-(q-1)-r)-r}\ z^{n-(q-1)}\\ &=1+ h^{q-1} z+ \left(h^{2(q-1)}+ \frac{(q-1)^{\underline{1}}}{1!} h^{q-2}\right) z^{2}+\dots\\ &=1+z(h+z)^{q-1}+z^{2}(h+z)^{2(q-1)}+\dots\\ &= \frac{1}{1-z(h+z)^{q-1}},\ \text{provided}\ |(z(h+z)^{q-1})|<1. \end{align*} \end{proof} \end{enumerate} We now obtain the following property. \begin{theorem}\textbf{$($Convolution property for $({}^qB)_{h,n}$$)$} For all $n \geq q-1,$ we have \begin{equation}\label{eq:2.5} \frac{d}{dx}\left(({}^qB)_{h,n}\right)= (q-1) \frac{dh}{dx}\left( \sum_{r=0}^{q-2} \frac{(q-2)^{\underline{r}}}{r!}\ h^{(q-2)-r}\ \sum_{i=0}^{n+q-2-r}({}^qB)_{h,i}\ ({}^qB)_{h,n+q-2-r-i}\right). \end{equation} \end{theorem} \begin{proof} Equation (\ref{eq:2.3}) implies \begin{align*} \sum_{n=0}^{\infty}({}^qB)_{h,n}\ z^{n-(q-1)} = \frac{1}{1-z(h+z)^{q-1}}. \end{align*} Differentiating both sides with respect to $x,$ we get \begin{align*} &\sum_{n=0}^{\infty}\frac{d}{dx}\left(({}^qB)_{h,n}\right)z^{n-(q-1)} =z(q-1)(h+z)^{q-2}\ \frac{1}{\left(1-z(h+z)^{q-1}\right)^{2}}\ \frac{dh}{dx}\\ &=\left((q-1)\ \sum_{r=0}^{q-2} \frac{(q-2)^{\underline{r}}}{r!}\ h^{(q-2)-r}z^{r+1}\ \left(\sum_{n=0}^{\infty}({}^qB)_{h,n}\ z^{n-(q-1)}\right)^{2}\right)\ \frac{dh}{dx}\\ &= \left((q-1)\ \sum_{r=0}^{q-2} \frac{(q-2)^{\underline{r}}}{r!}\ h^{(q-2)-r}\ z^{-2(q-1)+r+1}\ \left( \sum_{n=0}^{\infty}({}^qB)_{h,n}\ z^{n}\right)^{2}\right)\ \frac{dh}{dx}\\ &= (q-1)\frac{dh}{dx}\ \sum_{r=0}^{q-2} \frac{(q-2)^{\underline{r}}}{r!}\ h^{(q-2)-r}\ \sum_{n=0}^{\infty}\left( \sum_{i=0}^{n}({}^qB)_{h,i}({}^qB)_{h,n-i}\ z^{n-2(q-1)+r+1}\right). \end{align*} Comparing the coefficients of $z^{n-(q-1)},$ we get \begin{eqnarray*} \frac{d}{dx}\left(({}^qB)_{h,n}\right) = (q-1)\frac{dh}{dx} \left(\;\sum_{r=0}^{q-2} \frac{(q-2)^{\underline{r}}}{r!}\; h^{(q-2)-r}\; \sum_{i=0}^{n+q-2-r}({}^qB)_{h,i}({}^qB)_{h,n+q-2-r-i}\right). \end{eqnarray*} \end{proof} We now give an application of the convolution property. It is required to prove an identity in the next section. \begin{theorem}\label{thm:3} For $n\geq q-1$, \begin{align*} \sum_{r=0}^{\left \lfloor\frac{(q-1)(n-(q-1))}{q}\right \rfloor} r\;\frac{\left((q-1)(n-(q-1)-r)\right)^{\underline{r}}}{r!}\;\;h^{(q-1)(n-(q-1))-qr} \end{align*} \begin{align*} =\frac{(q-1)\left(n-(q-1)\right)}{q}\ ({}^qB)_{h,n} \end{align*} \begin{equation}\label{eq:2.6} -\frac{h}{q}\ (q-1) \left(\sum_{r=0}^{q-2} \frac{(q-2)^{\underline{r}}}{r!}\ h^{(q-2)-r}\ \sum_{i=0}^{n+q-2-r}({}^qB)_{h,i}\ ({}^qB)_{h,n+q-2-r-i}\right). \end{equation} \end{theorem} \begin{proof} Equation (\ref{eq:2.1}) implies \begin{align*} ({}^qB)_{h,n} = \sum_{r=0}^{\left \lfloor\frac{(q-1)(n-(q-1))}{q} \right \rfloor} \frac{\left((q-1)(n-(q-1)-r)\right)^{\underline{r}}}{r!}\ h^{(q-1)(n-(q-1))-qr} . \end{align*} Differentiating both sides with respect to $x$ and simplifying, we get \begin{align*} &\frac{d}{dx}\big(({}^qB)_{h,n}\big)h = ((q-1)(n-(q-1)))\ ({}^qB)_{h,n}\ \frac{dh}{dx}\\ &-q \ \frac{dh}{dx}\sum_{r=0}^{\left \lfloor\frac{(q-1)(n-(q-1))}{q}\right \rfloor}r\ \frac{((q-1)(n-(q-1)-r))^{\underline{r}}}{r!}\ h^{(q-1)(n-(q-1))-qr}. \end{align*} Thus, \begin{align*} &\frac{dh}{dx}\ \sum_{r=0}^{\left \lfloor\frac{(q-1)(n-(q-1))}{q}\right \rfloor} 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}\ ({}^qB)_{h,n}\ \frac{dh}{dx}-\frac{h}{q}\ \frac{d}{dx}\left(({}^qB)_{h,n}\right). \end{align*} Hence (\ref{eq:2.5}) implies \begin{align*} &\sum_{r=0}^{\left \lfloor\frac{(q-1)(n-(q-1))}{q}\right \rfloor} 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}\ ({}^qB)_{h,n}\\ &-\frac{h}{q}\ (q-1) \left(\ \sum_{r=0}^{q-2} \frac{(q-2)^{\underline{r}}}{r!}\ h^{(q-2)-r}\ \sum_{i=0}^{n+q-2-r}({}^qB)_{h,i}\ ({}^qB)_{h,n+q-2-r-i}\right). \end{align*} \end{proof} \section{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. \begin{definition} The incomplete $h(x)$-$B$-$q$-bonacci polynomials are defined by \begin{equation}\label{eq:3.1} ({}^qB)^{l}_{h,n}(x) = \sum_{r=0}^{l} \frac{\left((q-1)(n-(q-1)-r)\right)^{\underline{r}}}{r!}\ \ h^{(q-1)(n-(q-1)-r)-r}(x),\\ \end{equation} \begin{align*} \forall \ 0 \leq l \leq \left \lfloor\frac{(q-1)(n-(q-1))}{q}\right \rfloor \ \text{and}\ n \geq q-1.\\ \end{align*} \end{definition} Note that $({}^qB)^{\left \lfloor\frac{(q-1)(n-(q-1))}{q}\right \rfloor}_{h,n} (x)= ({}^qB)_{h,n}(x)$. For simplicity, we use $({}^qB)^{l}_{h,n}(x) = ({}^qB)^{l}_{h,n}, ({}^qB)_{h,n}(x) = ({}^qB)_{h,n}$ and $h(x)=h$. We prove identities related to the recurrence relation for $({}^qB)^{l}_{h,n}$. \begin{theorem} The recurrence relation for $({}^qB)^{l}_{h,n}$ is given by \begin{equation}\label{eq:3.2} ({}^qB)^{l+q-1}_{h,n+q} = \sum_{r=0}^{q-1}\frac{(q-1)^{\underline{r}}}{r!}\;\;h^{q-1-r} ({}^qB)^{l+q-1-r}_{h,n+q-1-r},\; 0 \leq l\leq \left \lfloor \frac{(q-1)(n-q)}{q} \right \rfloor, \forall n \geq q. \end{equation} \end{theorem} \begin{proof} Consider, \begin{align*} &\sum_{r=0}^{q-1}\frac{(q-1)^{\underline{r}}}{r!}\;h^{q-1-r}\; ({}^qB)^{l+q-1-r}_{h,n+q-1-r}\\ &=\sum_{r=0}^{q-1}\frac{(q-1)^{\underline{r}}}{r!}\;h^{q-1-r}\\ &\sum_{i=0}^{l+q-1-r} \frac{\left((q-1)(n+q-1-r-(q-1)-i)\right)^{\underline{i}}}{i!}\;\;h^{(q-1)(n+q-1-r-(q-1)-i)-i}\\ &=\sum_{r=0}^{q-1}\frac{(q-1)^{\underline{r}}}{r!}\;h^{q-1-r}\;\sum_{i=0}^{l+q-1-r} \frac{\left((q-1)(n-r-i)\right)^{\underline{i}}}{i!}\ h^{(q-1)(n-r)-qi}\\ &=\sum_{r=0}^{q-1}\frac{(q-1)^{\underline{r}}}{r!}\ \sum_{i=0}^{l+q-1-r} \frac{\left((q-1)(n-r-i)\right)^{\underline{i}}}{i!}\ h^{(q-1)(n+1)-qr-qi}\\ &=\sum_{r=0}^{q-1}\frac{(q-1)^{\underline{r}}}{r!}\ \sum_{i=0}^{l+q-1-r} \frac{\left((q-1)(n-(r+i))\right)^{\underline{i}}}{i!}\ h^{(q-1)(n+1)-q(r+i)}. \end{align*} Taking $j = i+r,$ we get \begin{align*} &\sum_{r=0}^{q-1}\frac{(q-1)^{\underline{r}}}{r!}\ ({}^qB)^{l+q-1-r}_{h,n+q-1-r}\ h^{q-1-r}\\ &=\sum_{r=0}^{q-1}\frac{(q-1)^{\underline{r}}}{r!}\ \sum_{j=r}^{l+q-1} \frac{\left((q-1)(n-j)\right)^{\underline{j-r}}}{(j-r)!}\ h^{(q-1)(n+1)-qj}\\ &=\sum_{j=0}^{l+q-1} \frac{\left((q-1)(n+1-j)\right)^{\underline{j}}}{j!}\ h^{(q-1)(n+1)-qj}\\ &=({}^qB)^{l+q-1}_{h,n+q}. \end{align*} \end{proof} \begin{theorem} For $s \geq 1,$ \begin{equation}\label{eq:3.3} ({}^qB)_{h,n+qs}^{l+(q-1)s}=\sum_{i=0}^{(q-1)s}\frac{((q-1)s)^{\underline{i}}}{i!}\; ({}^qB)_{h,n+i}^{l+i}\;h^{i}, \end{equation} $0 \leq l\leq \left \lfloor \frac{(q-1)(n-s-(q-1))}{q} \right \rfloor.$ \end{theorem} \begin{proof} Follows using induction. \end{proof} \begin{theorem} For $n\geq \left \lfloor \frac{ql+2(q-1)}{q-1} \right \rfloor,$ $({}^qB)_{h,n+(q-1)+s}^{l+(q-1)}-h^{(q-1)s}({}^qB)_{h,n+q-1}^{l+q-1}$ \begin{equation}\label{eq:3.4} =\sum_{i=0}^{s-1}\sum_{r=1}^{q-1} \frac{(q-1)^{\underline{r}}}{r!} h^{(q-1)s-(q-1)i-r}\ ({}^qB)_{h,n+(q-1)+i-r}^{l+(q-1)-r}. \end{equation} \end{theorem} \begin{proof} Follows using induction. \end{proof} The next theorem is related to the sum of incomplete $h(x)$-$B$-$q$-bonacci polynomials $({}^qB)^{l}_{h,n}.$ \begin{theorem} For all $n \geq q-1,$ \begin{align*} \sum_{l=0}^{\left \lfloor \frac{(q-1)(n-(q-1))}{q} \right \rfloor} ({}^qB)^{l}_{h,n}=\left(\left \lfloor \frac{(q-1)\left(n-(q-1)\right)}{q}\right \rfloor+\frac{q-(q-1)\left(n-(q-1)\right)}{q}\right)({}^qB)_{h,n} \end{align*} \begin{equation}\label{eq:5.65} +\frac{h}{q}\ (q-1) \left(\sum_{r=0}^{q-2} \frac{(q-2)^{\underline{r}}}{r!}\ h^{(q-2)-r}\ \sum_{i=0}^{n+q-2-r}({}^qB)_{h,i}\ ({}^qB)_{h,n+q-2-r-i}\right). \end{equation} \end{theorem} \begin{proof} $\sum_{l=0}^{\left \lfloor \frac{(q-1)(n-(q-1))}{q} \right \rfloor} ({}^qB)^{l}_{h,n}$ \begin{align*} &= ({}^qB)^{0}_{h,n}+({}^qB)^{1}_{h,n}+\cdots+({}^qB)^{r}_{h,n}+\cdots+({}^qB)^{\left \lfloor \frac{(q-1)(n-(q-1))}{q}\right \rfloor}_{h,n}\\ &=\frac{((q-1)(n-(q-1)))^{\underline{0}}}{0!}\ h^{(q-1)(n-(q-1))}\\ &+\left(\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}\right)\\ &+\cdots+\\ &\left(\frac{((q-1)(n-(q-1)))^{\underline{0}}}{0!}\ h^{(q-1)(n-(q-1))}+\cdots +\frac{((q-1)(n-(q-1)-r))^{\underline{r}}}{r!}\ h^{(q-1)(n-(q-1))-qr}\right)\\ &+\cdots\\ \end{align*} \begin{align*} &+\bigg(\frac{((q-1)(n-(q-1)))^{\underline{0}}}{0!}\ h^{(q-1)(n-(q-1))}+\cdots\\ &+\frac{((q-1)(n-(q-1)-r))^{\underline{r}}}{r!}\ h^{(q-1)(n-(q-1))-qr}+\cdots\\ &+\frac{\left((q-1)\left(n-(q-1)-\lfloor \frac{(q-1)(n-(q-1))}{q}\rfloor\right)\right)^{\underline{\lfloor \frac{(q-1)(n-(q-1))}{q}\rfloor}}}{(\lfloor \frac{(q-1)(n-(q-1))}{q}\rfloor)!}\ h^{(q-1)(n-(q-1))- q\lfloor \frac{(q-1)(n-(q-1))}{q}\rfloor}\bigg)\\ &=\sum_{r=0}^{\left \lfloor \frac{(q-1)(n-(q-1))}{q}\right \rfloor}\left(\lfloor \frac{(q-1)(n-(q-1))}{q}\rfloor+1-r\right)\frac{((q-1)(n-(q-1)-r))^{\underline{r}}}{r!}\ h^{(q-1)(n-(q-1))-qr}\\ &=\sum_{r=0}^{\left \lfloor \frac{(q-1)(n-(q-1))}{q}\right \rfloor}\left(\left \lfloor \frac{(q-1)(n-(q-1))}{q}\right \rfloor+1\right)\frac{((q-1)(n-(q-1)-r))^{\underline{r}}}{r!}\ h^{(q-1)(n-(q-1))-qr}\\ &- \sum_{r=0}^{\left \lfloor \frac{(q-1)(n-(q-1))}{q}\right \rfloor} r\frac{((q-1)(n-(q-1)-r))^{\underline{r}}}{r!}\ h^{(q-1)(n-(q-1))-qr}\\ &=\left(\left \lfloor \frac{(q-1)(n-(q-1))}{q}\right \rfloor+1-\frac{(q-1)(n-(q-1))}{q}\right)({}^qB)_{h,n}\\ &+\frac{h}{q}(q-1) \left(\sum_{r=0}^{q-2} \frac{(q-2)^{\underline{r}}}{r!}\ h^{(q-2)-r}\ \sum_{i=0}^{n+q-2-r}({}^qB)_{h,i}\ ({}^qB)_{h,n+q-2-r-i}\right). \end{align*} Using (\ref{eq:2.6}) of Theorem \ref{thm:3} in Section \ref{sec:2}, the result follows. \end{proof} \section{Acknowledgments} The authors thank the reviewers/referees for their comments and suggestions that helped to improve the article. \begin{thebibliography}{99} \bibitem{Sa2} S. 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