\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 $i<h$. We prove it for $h$:
\begin{align*}
\sum_{i=0}^{h}t_{n+i}(s)&=\sum_{i=0}^{h-1}t_{n+i}(s)+ t_{n+h}(s)\\
&=\frac{1}{2}\left(t_{n+h+2}(s+1)-t_{n+2}(s+1)+t_{n}(s)-t_{n+h}(s) \right) + t_{n+h}(s)\\
&=\frac{1}{2}\left(t_{n+h+2}(s+1)-t_{n+2}(s+1)+t_{n}(s)+ t_{n+h}(s)\right)\\
&=\frac{1}{2}\left(t_{n+h+3}(s+1)-t_{n+2}(s+1)+t_{n}(s)- t_{n+h+1}(s)\right).
\end{align*}
\end{proof}

\smallskip

The following proposition shows the sum of the $n$-th row of the  Table \ref{table1p}.
\begin{proposition}\label{propofibo}
The following equality holds:
\begin{align}
\sum_{s=0}^{l}T_{n}^{(s)}(x)=(l+1)T_n(x)-\sum_{i=0}^{l}\sum_{j=0}^{i} i\binom{i}{j}\binom{n-i-j-1}{i}x^{2n-3(i+j)-2}, \label{ec8}
 \end{align}
 where $l=\left\lfloor\frac{n-1}{2}\right\rfloor$.
\end{proposition}

\begin{proof}
We know
$$
T_{n}^{(s)}(x)=\sum_{i=0}^s B(n-1-i,i)(x).
$$
Then, we have
\begin{align*}
\sum_{s=0}^{l}T_{n}^{(s)}(x)&=
B(n-1-0,0)(x)+\left(B(n-1-0,0)(x)+B(n-1-1,1)(x)\right) + \cdots \\
&+\left(B(n-1-0,0)(x)+B(n-1-1,1)(x) + \cdots + B(n-1-l,l)(x)\right)\\
&=(l+1)B(n-1-0,0)(x) +  lB(n-1-1,1)(x) + \cdots +B(n-1-l,l)(x).
\end{align*}
Hence
\begin{align*}
\sum_{s=0}^{l}T_{n}^{(s)}(x)&=\sum_{i=0}^{l}(l+1-i)B(n-i-1,i)(x) \\
&=\sum_{i=0}^{l}(l+1)B(n-i-1,i)(x) - \sum_{i=0}^{l}iB(n-i-1,i)(x)\\
&=(l+1)T_n(x)-\sum_{i=0}^{l}iB(n-i-1,i)(x).
\end{align*}
\end{proof}

%\begin{align*}
%\sum_{s=0}^{l}T_{n}^{(s)}(x)
%&=T_{n}^{(0)}(x)+T_{n}^{(1)}(x)+\cdots+T_{n}^{(l)}(x)\\
%&=B(n-1-0,0)(x)+\left(B(n-1-0,0)(x)+B(n-1-1,1)(x)\right) + \cdots \\
%&+\left(B(n-1-0,0)(x)+B(n-1-1,1)(x) + \cdots + B(n-1-l,l)(x)\right)\\
%&=(l+1)B(n-1-0,0)(x) +  lB(n-1-1,1)(x) + \cdots +B(n-1-l,l)(x)\\
%&=\sum_{i=0}^{l}(l+1-i)B(n-i-1,i)(x) \\
%&=\sum_{i=0}^{l}(l+1)B(n-i-1,i)(x) - \sum_{i=0}^{l}iB(n-i-1,i)(x)\\
%&=(l+1)T_n(x)-\sum_{i=0}^{l}iB(n-i-1,i)(x).
%\end{align*}
%\end{proof}

\smallskip

The following corollary shows the sum of the $n$-th row of the  Table \ref{table1}. It is obtained   from (\ref{ec8}) with $x=1$.
\begin{corollary}\label{propofibon}
The following equality holds:
\begin{align*}
\sum_{s=0}^{l}t_{n}(s)=(l+1)t_n-\sum_{i=0}^{l}\sum_{j=0}^{i} i\binom{i}{j}\binom{n-i-j-1}{i},
 \end{align*}
 where $l=\left\lfloor\frac{n-1}{2}\right\rfloor$.
\end{corollary}

Let $$a_n=\sum_{i=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor}\sum_{j=0}^{i} i\binom{i}{j}\binom{n-i-j-1}{i}.$$
We conjecture that the generating function of $\left(a_n\right)_{n=0}^{\infty}$  is
$$\frac{z^3+z^4}{(1-z-z^2-z^3)^2}=(z^3+z^4)T^2(z),$$
where $T(z)$ is the generating function of the tribonacci numbers and $T^2(z)$ is the generating function of the convolution of tribonacci sequence; see sequence \seqnum{A073778}.
If this conjecture is true, then
\begin{align*}
\sum_{s=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor}t_{n}(s)=(l+1)t_n-\sum_{j=0}^{n}(t_{j-2}+t_{j-3})t_{n-j+1}.
 \end{align*}

\section{Generating function of the incomplete tribonacci numbers}

In this section, we give the generating functions of  the incomplete tribonacci numbers.

\begin{lemma}\label{lem4}
Let $\left(s_n\right)_{n=0}^{\infty}$ be a complex sequence satisfying the following non-homogeneous and non-linear recurrence relation
\begin{align}
s_n=s_{n-1}+s_{n-2}+s_{n-3}+r_n,   \ (n>2),
\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. Hoggatt Jr., On tribonacci numbers and related functions, \emph{Fibonacci Quart.} \textbf{15} (1977), 42--45.

\bibitem{Barry} P. Barry, On integer-sequence-based constructions of
generalized Pascal triangles, \emph{J. Integer Seq.} \textbf{9} (2006),
\href{https://cs.uwaterloo.ca/journals/JIS/VOL9/Barry/barry91.html}{Article 06.2.4}.

 \bibitem{djor} G. B. Djordjevi\'{c}, Generating functions of the incomplete generalized Fibonacci and generalized Lucas numbers, \emph{Fibonacci Quart.} \textbf{42} (2004), 106--113.

\bibitem{djor2} G. B. Djordjevi\'{c} and H. M. Srivastava, Incomplete generalized Jacobsthal and Jacobsthal-Lucas numbers, \emph{Math. Comput. Modelling} \textbf{42} (2005),  1049--1056.

\bibitem{tribo7} J. Feng, More identities on the tribonacci numbers, \emph{Ars Comb.} \textbf{100} (2011), 73--78.

\bibitem{FILI} P. Filipponi, Incomplete Fibonacci and Lucas numbers, \emph{Rend. Circ. Mat. Palermo} \textbf{45} (1996), 37--56.

\bibitem{HOG} V. E. Hoggatt Jr. and M. Bicknell, Generalized Fibonacci polynomials, \emph{Fibonacci Quart.} \textbf{11} (1973), 457--465.

\bibitem{tribo6} N. Irmak and M. Alp, Tribonacci numbers with indices in arithmetic progression and their sums, \emph{Miskolc Math. Notes} \textbf{14} (2013), 125--133.

\bibitem{tribo2} E. Kili\c{c}, Tribonacci sequences with certain indices and their sums, \emph{Ars Comb.} \textbf{86} (2008), 13--22.

\bibitem{koshy} T. Koshy, \emph{Fibonacci and Lucas Numbers with Applications}, Wiley-Interscience, 2001.

\bibitem{tribo8} K. Kuhapatanakul, Some connections between  a generalized tribonacci triangle and a generalized Fibonacci sequence, \emph{Fibonacci Quart.} \textbf{50} (2012), 44--50.

\bibitem{tribo3} P.-Y. Lin, De Moivre-type identities for the tribonacci numbers,  \emph{Fibonacci Quart.} \textbf{26} (1988), 131--134.

\bibitem{tribo4} C. P. McCarty, A formula for tribonacci numbers, \emph{Fibonacci Quart.} \textbf{19} (1981), 391--393.

\bibitem{tribo5} S. Pethe, Some identities for tribonacci sequences, \emph{Fibonacci Quart.} \textbf{26} (1988), 144--151.

\bibitem{PINTER} \'A. Pint\'er and  H. M. Srivastava, Generating functions of the incomplete Fibonacci and Lucas numbers, \emph{Rend. Circ. Mat. Palermo} \textbf{48} (1999), 591--596.

\bibitem{RAM3} J. Ram\'irez, Bi-periodic incomplete Fibonacci sequence, \emph{Ann. Math. Inform.} \textbf{42} (2013), 83--92.

\bibitem{RAM2} J. Ram\'irez, Incomplete generalized Fibonacci and Lucas
polynomials, preprint, \url{http://arxiv.org/abs/1308.4192}.

\bibitem{RAM} J. Ram\'irez, Incomplete $k$-Fibonacci and $k$-Lucas Numbers, \emph{Chinese J. of Math.} (2013), Article ID 107145.

\bibitem{rauzy} G. Rauzy, Nombres alg\'{e}briques et substitutions,  \emph{Bull. Soc. Math. France} \textbf{110} (1982), 147--178.

\bibitem{OEIS}  N. J. A. Sloane,  The On-Line Encyclopedia of Integer Sequences, \url{https://oeis.org/}.

  \bibitem{durs} D. Tasci and M. 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} 

.
