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\begin{center}
\vskip 1cm{\LARGE\bf Reciprocal Sums of the Tribonacci Numbers}
\vskip 1cm
{\large  Pornpawee Anantakitpaisal and Kantaphon Kuhapatanakul \\
Department of Mathematics \\
Faculty of Science\\
Kasetsart University\\
Bangkok 10900 \\
Thailand\\
\href{mailto:g5714400054@ku.ac.th}{\tt g5714400054@ku.ac.th} \\
\href{mailto:fscikpkk@ku.ac.th}{\tt fscikpkk@ku.ac.th} \\
}
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\vskip .2 in

\begin{abstract}
In this article, we consider infinite sums of the reciprocals of the
tribonacci numbers. Then, by applying the floor function to the
reciprocals of these sums, we obtain a new identity involving the
tribonacci numbers. Further, we give the formulas for an alternating
sum of the reciprocals of the tribonacci numbers and a sum of
reciprocal hypertribonacci numbers.
\end{abstract}

\section{Introduction}

As is well known, the sequences of Fibonacci numbers \seqnum{A000045} $\{F_n\}_{n=0}^\infty$ and tribonacci numbers \seqnum{A000073} $\{T_n\}_{n=0}^\infty$ are defined, respectively,  by
\begin{align*}
F_{n+1}&amp;=F_n+F_{n-1},~F_0=0,~F_1=1\\
T_{n+2}&amp;= T_{n+1}+T_n+T_{n-1},~~T_0=0,~T_1=T_2=1.
\end{align*}

Ohtsuka and Nakamura \cite{on} derived a formula for infinite sums of reciprocal Fibonacci
numbers, as follows:
\begin{equation}    \label{sumf}
{\left\lfloor \left(\sum_{k=n}^\infty \frac{1}{F_k}\right)^{-1}\right\rfloor}=\begin{cases}
                            F_{n-2},  &amp;\text{if}~n~\text{is\ even~and}~n\ge2;\\
                            F_{n-2}-1,\ \ &amp;\text{if}~n~\text{is\ odd~and}~n\ge1,
                           \end{cases}
\end{equation} 
where $\left\lfloor \cdot \right\rfloor$ is the floor function. Holliday and Komatsu \cite{hk} generalized a formula \eqref{sumf} for the generalized Fibonacci numbers. Similar properties were investigated in several different ways; see \cite{ku,wt,wz,zh}. In \cite{ku}, the author gave a similar formula \eqref{sumf} for alternating sums of reciprocal Fibonacci numbers as 
\begin{equation}    \label{sumfi}
{\left\lfloor \left(\sum_{k=n}^\infty \frac{(-1)^k}{F_k} \right)^{-1}\right\rfloor}=(-1)^nF_{n+1}-1\ \ \ \ \ (n&gt;1),
\end{equation}
and the generalized Fibonacci numbers are shown in \cite{ku1}. Liu and Zhao \cite{lz} showed the formula for the infinite sums of reciprocal hyperfibonacci numbers \seqnum{A136431} as:
\begin{equation}    \label{sumhf}
{\left\lfloor \left(\sum_{k=n}^\infty \frac{1}{\sum_{i=1}^k F_i} \right)^{-1}\right\rfloor}=
                            F_n-1\ \ \ \ \ (n\ge3).
\end{equation} 

Komatsu \cite{ko} studied the reciprocals of the tribonacci numbers and proved the following formula: 
$$\left\| \left(\sum_{k=n}^\infty \frac{1}{T_k} \right)^{-1}\right\|=T_n-T_{n-1},$$
where $\|\cdot\|$ is the nearest integer. Komatsu and Laohakosol \cite{kol} extended the above identity on the tribonacci numbers to the higher order linear recurrence.

In this paper, we consider the floor function, to develop similar formulas to \eqref{sumf}, \eqref{sumfi}, and \eqref{sumhf} for the tribonacci numbers. 

\section{Results}

We begin by noting that the sequence of tribonacci numbers can be defined for negative values of $n$ by using the definition and the given initial conditions. The first forty negative terms are shown in the following table.
\[
\begin{tabular}{|c|c|}
\hline 
  $n$   &amp;$T_{-n}$\\
\hline   
  $1$  &amp;$0$\\
\hline   
  $2$  &amp;$1$\\
\hline   
  $3$  &amp;$-1$\\
\hline   
  $4$  &amp;$0$\\
\hline   
  $5$  &amp;$2$\\
\hline   
  $6$  &amp;$-3$\\  
\hline
  $7$  &amp;$1$\\  
\hline
  $8$  &amp;$4$\\  
\hline
  $9$   &amp;$-8$\\
\hline   
  $10$   &amp;$5$\\
\hline 
\end{tabular} \ \ \ \ \ \ \ \ \ \ \ \ 
\begin{tabular}{|c|c|}
\hline 
   $n$   &amp;$T_{-n}$\\
\hline   
  $11$  &amp;$7$\\
\hline    
  $12$  &amp;$-20$\\
\hline   
  $13$  &amp;$18$\\  
\hline
  $14$  &amp;$9$\\  
\hline
  $15$  &amp;$-47$\\   
\hline
  $16$  &amp;$56$\\
\hline   
  $17$  &amp;$0$\\
\hline   
  $18$  &amp;$-103$\\
\hline   
  $19$  &amp;$159$\\
\hline   
  $20$  &amp;$-56$\\  
\hline
\end{tabular}  \ \ \ \ \ \ \ \ \ \ \ \ 
\begin{tabular}{|c|c|}
\hline 
  $n$   &amp;$T_{-n}$\\
\hline
  $21$  &amp;$-206$\\ 
\hline   
  $22$  &amp;$421$\\
\hline   
  $23$  &amp;$-271$\\
\hline   
  $24$  &amp;$-356$\\
\hline   
  $25$  &amp;$1048$\\
\hline   
  $26$  &amp;$-963$\\  
\hline
  $27$  &amp;$-441$\\  
\hline
  $28$  &amp;$2452$\\  
\hline
  $29$  &amp;$-2974$\\
\hline   
  $30$  &amp;$81$\\  
\hline
\end{tabular}\ \ \ \ \ \ \ \ \ \ \ \ 
\begin{tabular}{|c|c|}
\hline 
   $n$   &amp;$T_{-n}$\\
\hline
  $31$  &amp;$5345$\\ 
\hline    
  $32$  &amp;$-8400$\\
\hline   
  $33$  &amp;$3136$\\  
\hline
  $34$  &amp;$10609$\\  
\hline
  $35$  &amp;$-22145$\\   
\hline
  $36$  &amp;$14672$\\
\hline   
  $37$  &amp;$18082$\\
\hline   
  $38$  &amp;$-54899$\\
\hline   
  $39$  &amp;$51489$\\
\hline   
  $40$  &amp;$21492$\\   
\hline
\end{tabular}
\]

We first provide three lemmas which will be used in the proofs of the main theorems.
\begin{lemma}   \label{ident}
Let $n$ be a positive integer. Then 
\begin{itemize}
    \item [(i)] $T_n^2-T_{n-1}T_{n+1}=T_{-(n+1)}$ 
    \item[(ii)]$\sum_{i=1}^nT_i= (T_{n+2}+T_n-1)/2$
    \item[(iii)]$T_n&gt; T_{-(n+3)}\ \ \ \ (n\ge3).$
\end{itemize}

\end{lemma}
\begin{proof}
For any integer $k$, Shannon and Horadam \cite{sh} gave the following matrix equation: 
$$\begin{bmatrix}
          T_{n+k}\\
          T_{n+k-1}\\
          T_{n+k-2}
    \end{bmatrix} = \begin{bmatrix}
          1   &amp;1   &amp;1\\
          1   &amp;0   &amp;0\\
          0   &amp;1   &amp;0
    \end{bmatrix}^n\begin{bmatrix}
          T_{k}\\
          T_{k-1}\\
          T_{k-2}
    \end{bmatrix}=A^n\begin{bmatrix}
          T_{k}\\
          T_{k-1}\\
          T_{k-2}
    \end{bmatrix}.$$ 
We have
\begin{align*}
T_n^2-T_{n-1}T_{n+1}&amp;=\begin{vmatrix}
          T_{n+1}   &amp;T_{n+2}   &amp;1\\
          T_n       &amp;T_{n+1}   &amp;0\\
          T_{n-1}   &amp;T_n       &amp;0
                       \end{vmatrix}\\
          &amp;=\begin{vmatrix}
             A^n\begin{bmatrix}
             T_1\\
             T_0\\
             T_{-1} \end{bmatrix}
            &amp;A^n\begin{bmatrix}
             T_2\\
             T_1\\
             T_0 \end{bmatrix} 
           &amp;A^nA^{-n}\begin{bmatrix}
            T_1\\
            T_0\\
            T_{-1} \end{bmatrix}\end{vmatrix}\\
          &amp;=|A^n|\begin{vmatrix}
                  1     &amp;1   &amp;T_{-n+1}\\
                  0     &amp;1   &amp;T_{-n}\\
                  0     &amp;0   &amp;T_{-n-1}
                 \end{vmatrix}\\
          &amp;=T_{-(n+1)},
\end{align*}
which completes the proof of part (i). Since $T_{n+2}-T_{n+1}=T_n+T_{n-1}$, evaluating a partial sum by using a telescopic sum, we get part (ii).
For part (iii), a straightforward proof may be carried out by induction.
\end{proof}

\begin{lemma}   \label{recip1}
Let $n &gt;1$ be a positive integer. Then 
\begin{itemize}
   \item[(i)]$\dfrac{1}{T_n-T_{n-1}+1}&lt;\displaystyle\sum_{k=n}^\infty\dfrac{1}{T_k}&lt;\frac{1}{T_n-T_{n-1}-1}$
   \item[(ii)]$\dfrac{1}{(-1)^n(T_n+T_{n-1})+1}&lt;\displaystyle\sum_{k=n}^\infty\dfrac{(-1)^k}{T_k}&lt;\frac{1}{(-1)^n(T_n+T_{n-1})-1}.$
\end{itemize}
\end{lemma}
\begin{proof}
Since the proofs of both parts (i) and (ii) are quite similar,
we only give a proof for part (i).
Using Lemma \ref{ident} (i), we have 
\begin{align*}
\frac{1}{T_n-T_{n-1}-1}-\frac{1}{T_{n+1}-T_n-1}-\frac{1}{T_n}
&amp;=\frac{T_{n-1}+1}{T_n(T_n-T_{n-1}-1)}-\frac{1}{T_{n+1}-T_n-1}\\
&amp;=\frac{T_{n-1}T_{n+1}-T^2_n+T_{n+1}-T_{n-1}-1}{T_n(T_n-T_{n-1}-1)(T_{n+1}-T_n-1)}\\
&amp;=\frac{T_{n+1}-T_{n-1}-T_{-(n+1)}-1}{T_n(T_n-T_{n-1}-1)(T_{n+1}-T_n-1)}\\
&amp;=\frac{T_n+T_{n-2}-T_{-(n+1)}-1}{T_n(T_n-T_{n-1}-1)(T_{n+1}-T_n-1)}.
\end{align*}
By Lemma \ref{ident} (iii), the numerator of the right-hand side of the above identity is positive, 
$$\frac{1}{T_n-T_{n-1}-1}&gt; \frac{1}{T_n}+\frac{1}{T_{n+1}-T_n-1}.$$
By applying the above inequality repeatedly, we obtain 
\begin{align*}
\frac{1}{T_n-T_{n-1}-1}&amp;&gt; \frac{1}{T_n}+\frac{1}{T_{n+1}-T_n-1}\\
                                       &amp;&gt; \frac{1}{T_n}+\frac{1}{T_{n+1}}+\frac{1}{T_{n+2}-T_{n+1}-1}\\
                                       &amp;&gt; \frac{1}{T_n}+\frac{1}{T_{n+1}}+\frac{1}{T_{n+2}}+\cdots,
\end{align*}
so
$$\sum_{k=n}^\infty\frac{1}{T_k}&lt;\frac{1}{T_n-T_{n-1}-1}.$$
Similarly, we can show that 
$$\frac{1}{T_n-T_{n-1}+1}
   &lt;\frac{1}{T_n}+\frac{1}{T_{n+1}-T_n+1}.$$
Repeating the above inequality, we get the left inequality of part (i).
\end{proof}

\begin{lemma}    \label{recip2}
Let $n&gt;2$ be a positive integer. Then 
\begin{itemize}
 \item [(i)] If $T_{-(n+1)}&lt;0$, then $\sum_{k=n}^\infty\frac{1}{T_k}&lt;\frac{1}{T_n-T_{n-1}}$ 
 \item [(ii)] If $T_{-(n+1)}&gt;0$, then $\sum_{k=n}^\infty\frac{1}{T_k}&gt;\frac{1}{T_n-T_{n-1}}.$ 
\end{itemize}
\end{lemma}
\begin{proof}
By Lemma \ref{ident} (i), we have
\begin{align*}
\frac{1}{T_n}-\frac{1}{T_n-T_{n-1}}+\frac{1}{T_{n+1}-T_n}&amp;=\frac{-T_{n-1}}{T_n(T_n-T_{n-1})}+\frac{1}{T_{n+1}-T_n}\\
&amp;=\frac{T^2_n-T_{n+1}T_{n-1}}{T_n(T_n-T_{n-1})(T_{n+1}-T_n)}\\
&amp;=\frac{T_{-(n+1)}}{T_n(T_n-T_{n-1})(T_{n+1}-T_n)}.
\end{align*}
If $T_{-(n+1)}&lt;0$, then  
$$\frac{1}{T_n-T_{n-1}}&gt;\frac{1}{T_n}+\frac{1}{T_{n+1}-T_n},$$
repeating the above inequality, we get
$$\sum_{k=n}^\infty\frac{1}{T_k}&lt;\frac{1}{T_n-T_{n-1}}.$$
We can prove part (ii) in a similar way. 
\end{proof}

à¹ŒNow we are ready to state and prove the main results.
\begin{theorem}    \label{main}
Let $n$ be a positive integer. Then 
$$\left\lfloor \left(\displaystyle\sum_{k=n}^\infty \frac{1}{T_k} \right)^{-1}\right\rfloor =\begin{cases}
                            T_n-T_{n-1},
              &amp;\text{if}~T_{-(n+1)}&lt;0;\\
                            T_n-T_{n-1}-1,\ \ &amp;\text{if}~T_{-(n+1)}&gt; 0.
                           \end{cases}$$                        
\end{theorem}
\begin{proof} 
If $n=1$,  then $T_{-2}=1&gt;0$ and $$\sum_{k=1}^\infty\frac{1}{T_n}&gt;\frac{1}{T_1}=1,$$
so 
$$0&lt;\left(\sum_{k=1}^\infty\frac{1}{T_k}\right)^{-1}&lt;1.$$
Therefore, 
$$\left\lfloor \left(\displaystyle\sum_{k=1}^\infty \frac{1}{T_k} \right)^{-1}\right\rfloor = 0=T_1-T_0-1.$$
The case $n=2$ can be similarly verified.\\
If $n&gt;2$, by combining the inequalities of Lemma \ref{recip1} (i) and Lemma \ref{recip2}, we get
 $$\frac{1}{T_n-T_{n-1}+1}&lt;\sum_{k=n}^\infty\frac{1}{T_n}&lt;\frac{1}{T_n-T_{n-1}}\ \ \ \ \ (T_{-(n+1)}&lt;0),$$
and 
$$\frac{1}{T_n-T_{n-1}}&lt;\sum_{k=n}^\infty\frac{1}{T_n}&lt;\frac{1}{T_n-T_{n-1}-1}\ \ \ \ \ (T_{-(n+1)}&gt;0).$$ 
Hence, 
$${\left\lfloor \left(\sum_{k=n}^\infty \frac{1}{T_k} \right)^{-1}\right\rfloor}=\begin{cases}
                            T_n-T_{n-1},
              &amp;\text{if}~T_{-(n+1)}&lt;0;\\
                            T_n-T_{n-1}-1,\ \ &amp;\text{if}~T_{-(n+1)}&gt; 0.
                           \end{cases}$$
\end{proof}
We now calculate the analogous values to Theorem \ref{main} for $n=3$ (as $T_{-4}=0$ is not included). Since 
$$\sum_{k=3}^\infty\frac{1}{T_k}&gt;\sum_{k=1}^\infty\frac{1}{2^k}=1,$$                                                
we have
$$\left\lfloor \left(\displaystyle\sum_{k=3}^\infty \frac{1}{T_k} \right)^{-1}\right\rfloor = 0.$$ 


Observe for $1\le n \le19$ that, if  $3|n$, the value of $T_{-n}$ is negative. We have the following corollary. 

\begin{corollary}
Let $n&lt;19$ be a positive integer with $n\not=16$. Then 
$${\left\lfloor \left(\sum_{k=n}^\infty \frac{1}{T_k} \right)^{-1}\right\rfloor}=\begin{cases}
                            T_n-T_{n-1},
              &amp;\text{if}~3|(n+1);\\
                            T_n-T_{n-1}-1,\ \ &amp;\text{if}~3\not|(n+1).
                           \end{cases}$$
\end{corollary}

\begin{theorem}    \label{main1}
Let $n&gt;1$ be a positive integer. Then 
$$\left\lfloor\left(\sum_{k=n}^\infty \frac{(-1)^k}{T_k} \right)^{-1}\right\rfloor =\begin{cases}
                       (-1)^n(T_n+T_{n-1}),
              &amp;\text{if}~(-1)^nT_{-(n+1)}&gt;0;\\
                       (-1)^n(T_n+T_{n-1})-1,\ \ &amp;\text{if}~(-1)^nT_{-(n+1)}&lt;0.
                           \end{cases}$$                            
\end{theorem}
\begin{proof}
Observe that
$$\frac{1}{(-1)^n(T_n+T_{n-1})}-\frac{(-1)^n}{T_n}-\frac{1}{(-1)^{n+1}(T_{n+1}+T_n)}
=\frac{(-1)^nT_{-(n+1)}}{T_n(T_n+T_{n-1})(T_{n+1}+T_n)}.$$
If $(-1)^nT_{-(n+1)}&gt;0$, then 
$$\frac{1}{(-1)^n(T_n+T_{n-1})}&gt;\frac{(-1)^n}{T_n}+\frac{1}{(-1)^{n+1}(T_{n+1}+T_n)},$$
we obtain
$$\sum_{k=n}^\infty\frac{(-1)^k}{T_k}&lt;\frac{1}{(-1)^n(T_n+T_{n-1})}.$$
Similarly, if $(-1)^nT_{-(n+1)}&lt;0$, then 
$$\sum_{k=n}^\infty\frac{(-1)^k}{T_k}&gt;\frac{1}{(-1)^n(T_n+T_{n-1})}.$$
Combining the inequality of Lemma \ref{recip1} (ii) and the above inequalities, we obtain the desired result.
\end{proof}

\begin{theorem}    \label{main2}
Let $n\ge4$ be a positive integer. Then 
$${\left\lfloor \left(\sum_{k=n}^\infty \frac{1}{\sum_{i=1}^k T_i} \right)^{-1}\right\rfloor}=
                            T_n-1.$$                           
\end{theorem}
\begin{proof}
We shall prove our theorem directly, using Lemma \ref{ident} (ii), it is equivalent to
$$\frac{1}{T_n}&lt;\sum_{k=n}^\infty\frac{2}{T_{k+2}+T_k-1}&lt;\frac{1}{T_n-1}\ \ \ \ \ \ (n\ge 4).$$
Consider
\begin{align*}
\frac{1}{T_n-1}-\frac{2}{T_{n+2}+T_n-1}-\frac{1}{T_{n+1}+1}&amp;=\frac{T_{n+2}T_{n+1}-T_nT_{n+1}+T_{n+1}+3T_n-T_nT_{n+2}-T_n^2-2}{(T_n-1)(T_{n+2}+T_n-1)(T_{n+1}-1)}\\
&amp;=\frac{T^2
_{n+1}+T_{n-1}T_{n+1}+T_{n+1}+3T_n-T_nT_{n+2}-T_n^2-2}{(T_n-1)(T_{n+2}+T_n-1)(T_{n+1}-1)}\\
&amp;=\frac{T_{-(n+2)}-T_{-(n+1)}+T_{n+1}+3T_n-2}{(T_n-1)(T_{n+2}+T_n-1)(T_{n+1}-1)}\\&amp;&gt;\ 0,
\end{align*}
we have
$$\frac{2}{T_{n+2}+T_n-1}&lt;\frac{1}{T_n-1}-\frac{1}{T_{n+1}}.$$
Thus,
$$\sum_{k=n}^\infty\frac{2}{T_{k+2}+T_k-1}&lt;\frac{1}{T_n-1}.$$
In a similar way, we have 
\begin{align*}
\frac{1}{T_n}-\frac{2}{T_{n+2}+T_n-1}-\frac{1}{T_{n+1}}&amp;=\frac{T_{n+2}T_{n+1}-T_nT_{n+1}-T_{n+1}+T_n-T_nT_{n+2}-T_n^2}{T_n(T_{n+2}+T_n-1)T_{n+1}}\\
&amp;=\frac{T_{n+1}^2-T_nT_{n+2}-T_{n+1}+T_{n-1}T_{n+1}-T_n^2+T_n}{T_n(T_{n+2}+T_n-1)T_{n+1}}\\
&amp;=\frac{T_{-(n+2)}-T_{n+1}-T_{-{(n+1)}}+T_n}{T_n(T_{n+2}+T_n-1)T_{n+1}}\\
&amp;=\frac{T_{-(n+2)}-(T_n+T_{n-1}+T_{n-2})-T_{-{(n+1)}}+T_n}{T_n(T_{n+2}+T_n-1)T_{n+1}}\\
&amp;=\frac{T_{-(n+2)}-T_{n-1}-T_{n-2}-T_{-{(n+1)}}}{T_n(T_{n+2}+T_n-1)T_{n+1}}\\
&amp;&lt;\ 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \   (n\ge4).
\end{align*}
Therefore,
$$\sum_{k=n}^\infty\frac{2}{T_{k+2}+T_k-1}&gt;\frac{1}{T_n}\ \ \ \ \ \ \ \ \\ \ \ (n\ge4).$$
\end{proof}

\section{Acknowledgments}

The authors thank the anonymous referee for his/her careful reading and helpful comments. This work is supported by the Kasetsart University Research and Development Institute (KURDI), Thailand.

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\end{thebibliography}

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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11B39; Secondary 11B37.

\noindent \emph{Keywords: }
tribonacci number, reciprocal sum.

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\noindent (Concerned with sequences
\seqnum{A000045},
\seqnum{A000073}, and
\seqnum{A136431}.)

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\noindent
Received September 16 2015;
revised version received January 8 2016.
Published in {\it Journal of Integer Sequences}, January 9 2016.

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\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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