\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} \usepackage{array} \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}}} \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 Incomplete Tribonacci Numbers and \vskip .1in Polynomials } \vskip 1cm \large Jos\'e L. Ram\'irez\footnote{The first author was partially supported by Universidad Sergio Arboleda.} \\ Instituto de Matem\'aticas y sus Aplicaciones\\ Calle 74 No.~14 - 14 \\ Bogot\'a \\ Colombia \\ \href{mailto:josel.ramirez@ima.usergioarboleda.edu.co}{\tt josel.ramirez@ima.usergioarboleda.edu.co} \\ \ \\ V\'{\i}ctor F. Sirvent\footnote{The second author was partially supported by FWF project Nr.\ P23990.}\\ Universidad Sim\'{o}n Bol\'{i}var\\ Departamento de Matem\'aticas\\ Apartado 89000 \\ Caracas 1086-A \\ Venezuela \\ \href{mailto:vsirvent@usb.ve}{\tt vsirvent@usb.ve} \end{center} \vskip .2 in \begin{abstract} We define the incomplete tribonacci sequence of numbers and polynomials. We study recurrence relations, some properties of these numbers and polynomials, and the generating function of the incomplete tribonacci numbers. \end{abstract} \section{Introduction} Fibonacci numbers and their generalizations have many interesting properties and applications in many fields of science and art~(cf.~\cite{koshy}). The Fibonacci numbers $F_n$ are defined by the recurrence relation \begin{align*} F_{0}=0, \ \ F_{1}=1, \ \ F_{n+1}=F_{n}+F_{n-1}, \ n\geqslant 1. \end{align*} The first few terms are 0,1,1,2,3,5,8,13,$\dots$ (sequence \seqnum{A000045})\footnote{Many integer sequences and their properties are given in the {\it On-Line Encyclopedia of Integer Sequences} \cite{OEIS}.}. Another important sequence is the Lucas sequence. These numbers are defined by the recurrence relation \begin{align*} L_{0}=2, \ \ L_{1}=1, \ \ L_{n+1}=L_{n}+L_{n-1}, \ n\geqslant 1. \end{align*} The first few terms are 2,1,3,7,11,18,29,37,$\dots$ (sequence \seqnum{A000032}). The Fibonacci and Lucas numbers have been studied extensively. In particular, there is a beautiful combinatorial identity for Fibonacci numbers and Lucas numbers~(cf.~\cite{koshy}): \begin{align} F_{n}=\sum_{i=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor}\binom{n-i-1}{i}, \hspace{1cm} L_{n}=\sum_{i=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\frac{n}{n-j}\binom{n-j}{j} . \label{eccomb} \end{align} \smallskip In analogy with (\ref{eccomb}), Filipponi \cite{FILI} introduced the incomplete Fibonacci numbers $F_n(s)$ and the incomplete Lucas numbers $L_n(s)$. They are defined by \begin{align*} F_n(s)=\sum_{j=0}^{s}\binom{n-1-j}{j} , \ \ \left(n=1, 2, 3, \ldots; 0 \leq s \leq \left\lfloor\frac{n-1}{2}\right\rfloor\right); \end{align*} and \begin{align*} L_n(s)=\sum_{j=0}^{s}\frac{n}{n-j}\binom{n-j}{j}, \ \ \left(n=1, 2, 3, \ldots; 0 \leq s \leq \left\lfloor\frac{n}{2}\right\rfloor\right). \end{align*} \smallskip Generating functions of the incomplete Fibonacci and Lucas numbers were determined by Pint\'er and Srivastava~\cite{PINTER}. Djordjevi\'{c} \cite{djor} defined and studied incomplete generalized Fibonacci and Lucas numbers. Djordjevi\'{c} and Srivastava~\cite{djor2} defined incomplete generalized Jacobsthal and Jacobsthal-Lucas numbers. Tasci and Cetin Firengiz~\cite{durs} defined the incomplete Fibonacci and Lucas $p$-numbers. Tasci et al.~\cite{durs2} defined the incomplete bivariate Fibonacci and Lucas $p$-polynomials. Ram\'irez~\cite{RAM} introduced the incomplete $k$-Fibonacci and $k$-Lucas numbers, the incomplete $h(x)$-Fibonacci and $h(x)$-Lucas polynomials~\cite{RAM2}, and the bi-periodic incomplete Fibonacci sequences~\cite{RAM3}. \smallskip The tribonacci numbers are defined by the recurrence relation: \begin{align} t_{0}=0, \ \ \ \ t_{1}=t_2=1, \ \ \ t_{n+2}=t_{n+1}+t_{n}+t_{n-1} \ \ \text{ for } \ n\geqslant 1. \label{eq1} \end{align} The first few terms of the tribonacci numbers are 0, 1, 1, 2, 4, 7, 13, 24, 44, 81,$\dots$, (sequence \seqnum{A000073}). The tribonacci numbers have been studied in different contexts; see \cite{tribo1, tribo7, tribo6, tribo2, tribo8, tribo3, tribo4, tribo5, rauzy}. Alladi and Hoggatt~\cite{tribo1} defined the tribonacci triangle; see Table~\ref{table2}. It was used to derive the expansion of the tribonacci numbers and each element is defined in similar way as in the Pascal Triangle~(cf.~\cite{koshy}). \begin{table}[ht] \centering \begin{tabular}{c|ccccccccc} &0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & $\cdots$ \\ \hline 0 &1 & & & & & &&&\\ 1 & 1 &1&&&&&&&\\ 2& 1& 3& 1&&&&&&\\ 3& 1& 5& 5& 1&&&&&\\ 4 &1& 7& 13& 7& 1&&&&\\ 5& 1& 9& 25& 25& 9& 1&&&\\ 6& 1& 11& 41& 63& 41& 11 &1&&\\ 7 &1& 13& 61& 129& 129 &61 &13 &1&\\ $\vdots$ &&&&$\vdots$&&&& \end{tabular} \smallskip \caption{Tribonacci triangle} \label{table2} \end{table} Let $B(n,i)$ be the element in the $n$-th row and $i$-th column of the tribonacci triangle. By the definition of the triangle, we have \begin{equation}\label{eqn:triangule} B(n + 1, i) = B(n, i) + B(n, i -1) + B(n - 1, i -1), \end{equation} where $B(n, 0) = B(n, n) = 1$. The sum of elements on the rising diagonal lines in the tribonacci triangle is the tribonacci number $t_n$~(cf.~\cite{tribo1}), i.e., $$t_{n}=\sum_{i=0}^{\lfloor\frac{n-1}{2}\rfloor}B(n-1-i,i).$$ Moreover, Barry~\cite{Barry} showed that these coefficients satisfy the relation $$ B(n,i)=\sum_{j=0}^{i}\binom{i}{j}\binom{n-j}{i}. $$ Therefore, \begin{align} t_{n}=\sum_{i=0}^{\lfloor \frac{n-1}{2}\rfloor}B(n-1-i,i)=\sum_{i=0}^{\lfloor \frac{n-1}{2}\rfloor}\sum_{j=0}^{i}\binom{i}{j}\binom{n-1-i-j}{i}. \label{ec20} \end{align} \smallskip A large class of polynomials can also be defined by Fibonacci-like recurrences and tribonacci-like recurrences, such that yield Fibonacci numbers and tribonacci numbers. Such polynomials are called Fibonacci polynomials~\cite{koshy} and tribonacci polynomials~\cite{HOG}, respectively. \smallskip In 1883, Catalan and Jacobsthal introduced Fibonacci polynomials~(cf.~\cite{koshy}). The polynomials $F_n(x)$, studied by Catalan, are defined by the recurrence relation \begin{align*} F_{0}(x)=0, \ \ F_{1}(x)=1, \ \ F_{n+1}(x)=xF_{n}(x)+F_{n-1}(x), \ n\geqslant 1. \end{align*} The Fibonacci polynomials, studied by Jacobsthal, are defined by \begin{align*} J_{0}(x)=1, \ \ J_{1}(x)=1, \ \ J_{n+1}(x)=J_{n}(x)+xJ_{n-1}(x), \ n\geqslant 1. \end{align*} The Lucas polynomials $L_n(x)$, originally studied in 1970 by Bicknell, are defined by \begin{align*} L_{0}(x)=2, \ \ L_{1}(x)=x, \ \ L_{n+1}(x)=xL_{n}(x)+L_{n-1}(x), \ n\geqslant 1. \end{align*} \smallskip Hoggatt and Bicknell \cite{HOG} introduced tribonacci polynomials. The tribonacci polynomials $T_n(x)$ are defined by the recurrence relation \begin{align*} T_{0}(x)=0, \ \ T_{1}(x)=1, \ \ T_2(x)=x^2, \ \ T_{n+2}(x)=x^2T_{n+1}(x)+xT_{n}(x)+T_{n-1}(x), \ n\geqslant 2. \end{align*} Note that $T_n(1)=t_n$ for all integer positive $n$. The first few tribonacci polynomials are $$ \begin{array}{ll} T_1(x)=1, & T_5(x)=x^8+3x^5+3x^2,\\ T_2(x)=x^2, & T_6(x)=x^{10}+4x^7+6x^4+2x,\\ T_3(x)=x^4+x, & T_7(x)=x^{12}+5x^9+10x^6+7x^3+1,\\ T_4(x)=x^6+2x^3+1, & T_8(x)=x^{14}+6x^{11}+15x^8+16x^5+6x^2. \end{array} $$ \smallskip In analogy with the tribonacci triangle, we define the tribonacci polynomial triangle; see Table \ref{tablep}. \begin{table}[ht] \centering \begin{tabular}{c|cccccc} &0 & 1 & 2 & 3 & 4 & 5 $\cdots$ \\ \hline 0 &1 & & & & & \\ 1 & $x^2$ &$x$&&&&\\ 2& $x^4$ & $2x^3+1$& $x^2$&&&\\ 3& $x^6$& $3x^5+2x^2$& $3x^4+2x$& $x^3$&&\\ 4 &$x^8$& $4x^7+3x^4$& $6x^6+6x^3+1$& $4x^5+3x^2$& $x^4$&\\ 5& $x^{10}$& $5x^9+4x^6$& $10x^8+12x^5+3x^2$& $10x^7+12x^4+3x$& $5x^6+4x^3$& $x^5$\\ $\vdots$ &&&&$\vdots$& \end{tabular} \smallskip \caption{Tribonacci polynomial triangle} \label{tablep} \end{table} Let $B(n,i)(x)$ be the element in the $n$-th row and $i$-th column of the tribonacci polynomial triangle. Then $$B(n + 1, i)(x) = x^2B(n, i)(x) + xB(n, i -1)(x) + B(n - 1, i -1)(x),$$ where $B(n, 0)(x) =x^{2n}$ and $B(n, n) = x^n$. \smallskip It can be proved by induction on $n$, that the sum of elements on the rising diagonal lines in the tribonacci polynomial triangle is the tribonacci polynomial $T_n(x)$, i.e., $$ T_{n}(x)=\sum_{i=0}^{\lfloor\frac{n-1}{2}\rfloor}B(n-1-i,i)(x). $$ Moreover, $$ B(n,i)(x)=\sum_{j=0}^{i}\binom{i}{j}\binom{n-j}{i}x^{2n-i-3j}. $$ Therefore, \begin{align} T_{n}(x)=\sum_{i=0}^{\lfloor \frac{n-1}{2}\rfloor}B(n-1-i,i)(x)=\sum_{i=0}^{\lfloor \frac{n-1}{2}\rfloor}\sum_{j=0}^{i}\binom{i}{j}\binom{n-1-i-j}{i}x^{2n-3(i+j)-2}. \label{ec201} \end{align} From Equations~(\ref{ec20}) and~(\ref{ec201}), we introduce the incomplete tribonacci polynomials and incomplete tribonacci numbers. In this way we obtain new recurrence relations, new identities and the generating function of the incomplete tribonacci numbers. \section{Incomplete tribonacci polynomials and incomplete tribonacci numbers} \begin{definition}\label{def1} For $n\geqslant 1$, {\em incomplete tribonacci polynomials} are defined as \begin{align} T_{n}^{(s)}(x)&=\sum_{i=0}^{s}B(n-1-i,i)(x)\\ &=\sum_{i=0}^{s}\sum_{j=0}^{i}\binom{i}{j}\binom{n-i-j-1}{i}x^{2n-3(i+j)-2}, \ 0\leqslant s \leqslant \left\lfloor \frac{n-1}{2}\right\rfloor. \label{ec1} \end{align} We define the {\em incomplete tribonacci numbers}, $t_n(s)$ as the value of $T_{n}^{(s)}(x)$ at $x=1$, i.e., $t_n(s)=T_{n}^{(s)}(1)$. \end{definition} In Tables \ref{table1p} and \ref{table1}, some values of incomplete tribonacci polynomials and incomplete tribonacci numbers are provided. \begin{table}[ht] \centering \footnotesize \begin{tabular}{|>{$}c<{$}|>{$}c<{$}| >{$}c<{$}|>{$}c<{$}|>{$}c<{$}|} \hline n/s & 0 & 1 & 2 & 3 \\ \hline 1 & 1 & \text{} & \text{} & \text{} \\ 2 & x^2 & \text{} & \text{} & \text{} \\ 3 & x^4 & x^4+x & \text{} & \text{} \\ 4 & x^6 & x^6+2 x^3+1 & \text{} & \text{} \\ 5 & x^8 & x^8+3 x^5+2 x^2 & x^8+3 x^5+3 x^2 & \text{} \\ 6 & x^{10} & x^{10}+4 x^7+3 x^4 & x^{10}+4 x^7+6 x^4+2 x & \text{} \\ 7 & x^{12} & x^{12}+5 x^9+4 x^6 & x^{12}+5 x^9+10 x^6+6 x^3+1 & x^{12}+5 x^9+10 x^6+7 x^3+1 \\ 8 & x^{14} & x^{14}+6 x^{11}+5 x^8 & x^{14}+6 x^{11}+15 x^8+12 x^5+3 x^2 & x^{14}+6 x^{11}+15 x^8+16 x^5+6 x^2 \\ \hline \end{tabular} \medskip \caption{Polynomials $T_{n}^{(s)}(x)$, for $1\leqslant n \leqslant 8$ and $0\leqslant s \leqslant 3$.} \label{table1p} \end{table} \smallskip From the definitions follow \begin{align*} T_n^{\left( \left\lfloor \frac{n-1}{2}\right\rfloor\right)}(x)=T_n(x) \ \ \text{and} \ \ t_n\left( \left\lfloor \frac{n-1}{2}\right\rfloor\right)=t_n. \end{align*} \begin{table}[ht] \centering \begin{tabular}{|c|c|c|c|c|c|c|c|c|}\hline $n/s$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline 1&1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 2&1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 3&1 & 2 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 4&1 & 4 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 5&1 & 6 & 7 & \text{} & \text{} & \text{} & \text{} & \text{} \\ 6&1 & 8 & 13 & \text{} & \text{} & \text{} & \text{} & \text{} \\ 7&1 & 10 & 23 & 24 & \text{} & \text{} & \text{} & \text{} \\ 8&1 & 12 & 37 & 44 & \text{} & \text{} & \text{} & \text{} \\ 9&1 & 14 & 55 & 80 & 81 & \text{} & \text{} & \text{} \\ 10&1 & 16 & 77 & 140 & 149 & \text{} & \text{} & \text{} \\ 11&1 & 18 & 103 & 232 & 273 & 274 & \text{} & \text{} \\ 12&1 & 20 & 133 & 364 & 493 & 504 & \text{} & \text{} \\ 13&1 & 22 & 167 & 544 & 865 & 926 & 927 & \text{} \\ 14&1 & 24 & 205 & 780 & 1461 & 1692 & 1705 & \text{} \\ 15&1 & 26 & 247 & 1080 & 2369 & 3050 & 3135 & 3136 \\ 16&1 & 28 & 293 & 1452 & 3693 & 5376 & 5753 & 5768 \\ \hline \end{tabular} \medskip \caption{Numbers $t_{n}(s)$, for $1\leqslant n \leqslant 16$ and $0\leqslant s \leqslant 7$.} \label{table1} \end{table} Some special cases of (\ref{ec1}) are \begin{align} &T_{n}^{(0)}(x)=x^{2n-2}, \ (n\geq1);\\ &T_{n}^{(1)}(x)=x^{2n-2}+(n-2)x^{2n-5}+(n-3)x^{2n-8}, \ (n\geq3);\\ &T_{n} ^{\left(\left\lfloor \frac{n-1}{2} \right\rfloor\right)}(x)=T_{n}(x), \ (n\geq1);\\ &T_{n}^{\left(\left\lfloor \frac{n-3}{2} \right\rfloor\right)}(x)=\begin{cases} T_{n}(x) - \left(\frac{n}{2}x^{\frac{n+2}{2}}+\frac{n-2}{2}x^{\frac{n-4}{2}} \right), & \ \text{if $n\geq 3$ and even}; \vspace{0.15cm}\\ T_n(x) - x^{\frac{n-1}{2}}, & \ \text{if $n\geq 3$ and odd.} \end{cases} \end{align} Throughout the rest of this section, we describe some recurrence properties of the polynomials $T_{n}^{(s)}(x)$ and numbers $t_n(s)$. \begin{proposition} The non-linear recurrence relation of the incomplete tribonacci polynomials $T_{n}^{(s)}(x)$ is \begin{align} T_{n+3}^{(s+1)}(x)&=x^2T_{n+2}^{(s+1)}(x)+xT_{n+1}^{(s)}(x)+T_n^{(s)}(x), \ 0\leqslant s \leqslant \left\lfloor \frac{n-1}{2}\right\rfloor. \label{ec3} \end{align} The relation (\ref{ec3}) can be transformed into the non-homogeneous recurrence relation \begin{align} T_{n+3}^{(s)}(x)&=x^2T_{n+2}^{(s)}(x)+xT_{n+1}^{(s)}(x)+T_n^{(s)}(x)-\left(xB(n-s,s)(x)+B(n-1-s,s)(x) \right). \label{ec15} \end{align} \end{proposition} \begin{proof} We use the Definition \ref{def1} to rewrite the right-hand side of (\ref{ec3}) as $$ x^2\sum_{i=0}^{s+1}B(n+1-i,i)(x) + x\sum_{i=0}^{s}B(n-i,i)(x) + \sum_{i=0}^{s}B(n-1-i,i)(x). $$ Therefore, \begin{align*} x^2T_{n+2}^{(s+1)}&(x)+xT_{n+1}^{(s)}(x)+T_n^{(s)}(x)=&\\ %&=x^2\sum_{i=0}^{s+1}B(n+1-i,i)(x) + x\sum_{i=0}^{s}B(n-i,i)(x) + \sum_{i=0}^{s}B(n-1-i,i)(x)\\ &=x^2\sum_{i=0}^{s+1}B(n+1-i,i)(x)+ x\sum_{i=1}^{s+1}B(n-i+1,i-1)(x) + \sum_{i=1}^{s+1}B(n-i,i-1)(x)\\ &= \sum_{i=0}^{s+1}\left(x^2B(n+1-i,i)(x) + xB(n-i+1,i-1)(x) + B(n-i,i-1)(x)\right) \\ &- xB(n+1,-1)(x)-B(n,-1)(x)\\ &= \sum_{i=0}^{s+1}B(n+2-i,i)(x)=T_{n+3}^{(s+1)}(x). \end{align*} \end{proof} \begin{corollary} The non-linear recurrence relation of the incomplete tribonacci numbers $t_{n}(s)$ is \begin{align} t_{n+3}(s+1)&=t_{n+2}(s+1)+t_{n+1}(s)+t_n(s), \ 0\leqslant s \leqslant \left \lfloor \frac{n-1}{2}\right\rfloor. \label{ec3n} \end{align} The relation (\ref{ec3n}) can be transformed into the non-homogeneous recurrence relation \begin{align} t_{n+3}(s)&=t_{n+2}(s)+t_{n+1}(s)+t_n(s)-\left( B(n-s,s)+B(n-1-s,s) \right). \label{ec15n} \end{align} \end{corollary} \begin{corollary} For $n\geq 2s+2$, \begin{align} \sum_{i=0}^{h-1}t_{n+i}(s)=\frac{1}{2}\left(t_{n+h+2}(s+1)-t_{n+2}(s+1)+t_{n}(s)-t_{n+h}(s)\right). \label{ec5n} \end{align} \end{corollary} \begin{proof} We proceed by induction on $h$. The sum (\ref{ec5n}) clearly holds for $h=1$; see (\ref{ec3n}). Now suppose that the result is true for all $i2), \end{align} where $\left(r_n\right)_{n=0}^{\infty}$ is a given complex sequence. Then the generating function $U(t)$ of the sequence $\left(s_n\right)_{n=0}^{\infty}$ is \begin{align} U(t)=\dfrac{G(t)+s_0-r_0+(s_1-s_0-r_1)t+(s_2-s_1-s_0-r_2)t^2}{1-t-t^2-t^3} \label{ecgenf1} \end{align} where $G(t)$ denotes the generating function of $\left(r_n\right)_{n=0}^{\infty}$. \end{lemma} \begin{proof} Since $U(t)$ and $G(t)$ are the generating functions of $\left(s_n\right)_{n=0}^{\infty}$ and $\left( r_n\right)_{n=0}^{\infty}$, respectively. Their power series representations are \begin{align*} U(t)=s_0+s_1t+s_2t^2+\cdots +s_kt^k + \cdots ,\\ G(t)=r_0+r_1t+r_2t^2+\cdots +r_kt^k + \cdots . \end{align*} Note that, \begin{align*} tU(t)=s_0t+s_1t^2+s_2t^3+\cdots +s_kt^{k+1} + \cdots ,\\ t^2U(t)=s_0t^2+s_1t^3+s_2t^4+\cdots +s_kt^{k+2} + \cdots,\\ t^3U(t)=s_0t^3+s_1t^4+s_2t^5+\cdots +s_kt^{k+3} + \cdots. \end{align*} Therefore \begin{align*} (1-t-t^2-t^3)U(t)-G(t)=(s_0-r_0)+(s_1-s_0-r_1)t+(s_2-s_1-s_0-r_2)t^2. \end{align*} Then the Equation~(\ref{ecgenf1}) follows. \end{proof} \begin{theorem}\label{teo1} Let $Q_s(z)$ be the generating function of the incomplete tribonacci numbers $t_n(s)$. Then $$ Q_{s}(z)=\dfrac{t_{2s+1} + (t_{2s+2}-t_{2s+1})z+(t_{2s+3}-t_{2s+2}-t_{2s+1}-2)z^2 -(z^2+z^3)\dfrac{(1+z)^s}{(1-z)^{s+1}}}{1-z-z^2-z^3}. $$ \end{theorem} \begin{proof} Let $s$ be a fixed positive integer and $t_n(s)$ the $n$-th incomplete tribonacci number. Since $Q_s(z)$ is the generating function of the $t_n(s)$, we have $Q_s(z)=\sum_{i=0}^{\infty}t_{i}(s)z^i$. From (\ref{ec1}) with $x=1$ and (\ref{ec15n}), we get $t_{n}(s)=0$ for $0\leq n < 2s+1$, $t_{2s+1}(s)=t_{2s+1}, t_{2s+2}(s)=t_{2s+2}$ and $t_{2s+3}(s)=t_{2s+3}-1$, and that \begin{align} t_{n}(s)&=t_{n-1}(s)+t_{n-2}(s)+t_{n-3}(s)-\left( B(n-3-s,s)+B(n-4-s,s) \right). \end{align} Now let \begin{align*} s_0=t_{2s+1}(s), \ \ \ s_1=t_{2s+2}(s), \ \ \ s_2=t_{2s+3}(s) \ \ \ \text{and} \end{align*} \begin{align*} s_n=t_{n+2s+1}(s). \end{align*} Also let $r_0=r_1=0, r_1=1$ and \begin{align*} r_n&=B(n+s-2,s)+B(n+s-3,s)\\ &=\sum_{j=0}^{s}\binom{s}{j}\binom{n+s-2-j}{s}+\sum_{j=0}^{s}\binom{s}{j}\binom{n+s-3-j}{s}. \end{align*} The generating function of the sequence $\left(r_n\right)_{n\geq 0}$ is computed using the methods expounded in \cite[page 127]{wilf}. Hence the generating function is equal to $$ (z^2+z^3)\frac{(1+z)^s}{(1-z)^{s+1}}. $$ Thus, from Lemma \ref{lem4}, we get the generating function $Q_{s}(z)$ of sequence $\left(t_n(s)\right)_{n=0}^{\infty}$. \end{proof} \begin{example} The generating functions of the incomplete tribonacci numbers for $s=1,2,3,4$ are \begin{eqnarray*} Q_1(z) &=&\dfrac{2}{(z-1)^2}=2 + 4z + 6z^2 + 8z^3 + 10z^4 + 12z^5 + 14z^6 + 16z^7 + 18z^8+\cdots\\ & &\\ Q_2(z)&=&\dfrac{-5z^2+8z-7}{(z-1)^3}= 7 + 13z + 23z^2 + 37z^3 + 55z^4 + 77z^5 + 103z^6 + 133z^7 + \cdots\\ & &\\ Q_3(z)&=&\dfrac{-12z^{3}+48z^2-52z+24}{(z-1)^4}=24 + 44z + 80z^2 + 140z^3 + 232z^4 + 364z^5 +\cdots \\ & & \\ Q_4(z)&=&\dfrac{-45z^4+192z^3-338z^2+256z-81}{(z-1)^5}=81 + 149z + 273z^2 + 493z^3 + 865z^4 + \cdots \end{eqnarray*} \end{example} \smallskip An open problem is to find the generating function of the incomplete tribonacci polynomials. \begin{thebibliography}{99} \bibitem{tribo1} K. Alladi and V. E. 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Cetin Firengiz, Incomplete Fibonacci and Lucas $p$-numbers, \emph{Math. Comput. Modelling} \textbf{52} (2010), 1763--1770. \bibitem{durs2} D. Tasci, M. Cetin Firengiz, and N. Tuglu, Incomplete bivariate Fibonacci and Lucas $p$-polynomials, \emph{Discrete Dynam. Nat. Soc.} (2012), Article ID 840345. \bibitem{wilf} H. Wilf, \emph{generatingfunctionology}, CRC Press, third edition, 2005. \end{thebibliography} \bigskip \hrule \bigskip \noindent 2010 {\it Mathematics Subject Classification}: Primary 11B39; Secondary 11B83, 05A15. \noindent \emph{Keywords: } incomplete tribonacci number, incomplete tribonacci polynomial, tribonacci number. \bigskip \hrule \bigskip \noindent (Concerned with sequences \seqnum{A000045}, \seqnum{A000032}, \seqnum{A000073}, and \seqnum{A073778}.) \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received November 7 2013; revised versions received December 10 2013; January 22 2014. Published in {\it Journal of Integer Sequences}, February 16 2014. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document}