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\begin{center}
\vskip 1cm{\LARGE\bf On the Reciprocal Sums of Products\\[5pt] of Fibonacci Numbers}
\vskip 1cm
{\large  Younseok Choo\\
Department of Electronic and Electrical Engineering \\
Hongik University\\
Sejong 30016\\
Republic of Korea\\
\href{mailto:yschoo@hongik.ac.kr}{\tt yschoo@hongik.ac.kr}\\
}
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\vskip .2 in

\begin{abstract}
In this paper we study the reciprocal sums of products of two different Fibonacci numbers. We obtain some identities related to the numbers $\lfloor (\sum_{k=n}^{\infty}1/F_{k}F_{k+m})^{-1} \rfloor$, $m \geq 1$, where $\lfloor \cdot \rfloor$ indicates the floor function. 
\end{abstract}

\section{Introduction}
As is well known, the Fibonacci numbers $F_{n}$ are generated from the recurrence relation
\[
F_{n}=F_{n-1}+F_{n-2}\;\; (n \geq 2),
\]
with initial condition $F_{0}=0$ and $F_{1}=1$.

Recently Ohtsuka and Nakamura \cite{ON} found interesting properties of the Fibonacci numbers and proved Theorem \ref{THM1} below.

\begin{theorem}\label{THM1}
For the Fibonacci numbers, the following identities hold:
\begin{equation}\label{eqn1}
\Bigg\lfloor\Bigg(\sum_{k=n}^{\infty}\frac{1}{F_{k}}\Bigg)^{-1}\Bigg\rfloor = 
\begin{cases} F_{n}-F_{n-1}, & \text{ if $n \geq 2$ and $n$ is even};\\
F_{n}-F_{n-1}-1, &\text{ if $n \geq 3$ and $n$ is odd},
\end{cases}\\
\end{equation}
\begin{equation}\label{eqn2}
\Bigg\lfloor\Bigg(\sum_{k=n}^{\infty}\frac{1}{F_{k}^{2}}\Bigg)^{-1}\Bigg\rfloor = 
\begin{cases} F_{n-1}F_{n}-1, &\text{ if $n \geq 2$ and $n$ is even};\\
F_{n-1}F_{n}, &\text{ if $n \geq 3$ and $n$ is odd}.
\end{cases} 
\end{equation}
\end{theorem} 

Following the paper of Ohtsuka and Nakamura \cite{ON}, diverse results in the same direction have been reported in the literature \cite{CH1, CH2, HK}, \cite{LW}, \cite{WW, WY, WZ1, WZ2, WZ3, ZW}. Among them, Liu and Wang \cite{LW} considered the product of two reciprocal Fibonacci numbers, and obtained several interesting results. For example, they proved Theorem \ref{THM2} below for the products of two consecutive Fibonacci numbers.
\newline \null

\begin{theorem}\label{THM2}
\it Let $m \geq 2$. Then
\begin{equation}\label{eqn3}
\Bigg\lfloor\Bigg(\sum_{k=n}^{mn}\frac{1}{F_{k}F_{k+1}}\Bigg)^{-1}\Bigg\rfloor = 
\begin{cases}
F_{n}^{2}, &\text{ if $n \geq 2$ and $n$ is even;}\\
F_{n}^{2}-1,&\text{ if $n \geq 3$ and $n$ is odd.}
\end{cases}
\end{equation}
\end{theorem}

Motivated by Theorem \ref{THM2}, we study the reciprocal sums of products of two different Fibonacci numbers in this paper. We obtain some identities related to the
numbers 
\[
\Bigg\lfloor\Bigg(\sum_{k=n}^{\infty}\frac{1}{F_{k}F_{k+m}}\Bigg)^{-1}\Bigg\rfloor,\;\;m \geq 1.
\]

\begin{remark}
The following identity was conjectured by Ohtsuka and proved by Bruckman \cite{OHT}: 
\[ 
\Bigg(\sum_{k=n}^{\infty}\frac{1}{F_{k}F_{k+m}}\Bigg)^{-1}
=\sum_{k=1}^{n-1}F_{k}F_{k+m}-\frac{1}{3}F_{m-2(-1)^{n}}+
O\Bigg( \frac{1}{F_{n}^{2}}\Bigg),\;\;m \geq 0.
\] 
For the case where $m=0$ and $n$ is large, (\ref{eqn2}) also can be derived from the above result.  
\end{remark}

\section{Main results}
We will use Lemma \ref{LEM1}  below  to prove our main results.

\begin{lemma}[Koshy \cite{KOS}]  \label{LEM1}
For the Fibonacci numbers, we have
\[
F_{m}F_{n}-F_{m+k}F_{n-k}=(-1)^{n-k}F_{m-n+k}F_{k}.
\]
\end{lemma}

Our main results are stated in the following theorem.

\begin{theorem}\label{THM3}
For the Fibonacci numbers, (a), (b) and (c)  below hold:

(a) Let $m \geq 1$. If
\[\frac{2F_{m}-F_{m+1}}{3} \notin \mathbb{Z},\] 
then there exist positive integers $n_{0}$ and $n_{1}$ such that
\begin{equation}\label{eqn4}
\Bigg\lfloor\Bigg(\sum_{k=n}^{\infty}\frac{1}{F_{k}F_{k+m}}\Bigg)^{-1}\Bigg\rfloor = 
\begin{cases}
 F_{n+m-1}F_{n}+g_{m}-1, 
&\text{ if $n \geq n_{0}$ and $n$ is even;}\\
F_{n+m-1}F_{n}-g_{m},&\text{ if  $n \geq n_{1}$ and $n$ is odd,}
\end{cases}
\end{equation}
where
\[g_{m}=\Bigg\lfloor \frac{2F_{m}-F_{m+1}}{3}\Bigg\rfloor+1.\]

(b) For $m=2$,
\begin{equation}\label{eqn5}
\Bigg\lfloor\Bigg(\sum_{k=n}^{\infty}\frac{1}{F_{k}F_{k+m}}\Bigg)^{-1}\Bigg\rfloor = F_{n+m-1}F_{n}, \text{ for $n \geq 1$}.
\end{equation}

(c) Let $m \geq 3$. If 
\[\frac{2F_{m}-F_{m+1}}{3} \in \mathbb{Z},\] 
then there exist positive integers $n_{2}$ and $n_{3}$ such that
\begin{equation}\label{eqn6}
\Bigg\lfloor\Bigg(\sum_{k=n}^{\infty}\frac{1}{F_{k}F_{k+m}}\Bigg)^{-1}\Bigg\rfloor = 
\begin{cases}
 F_{n+m-1}F_{n}+\hat{g}_{m}-1, 
&\text{ if  $n \geq n_{2}$ and $n$ is even;}\\
F_{n+m-1}F_{n}-\hat{g}_{m}-1,&\text{ if $n \geq n_{3}$ and $n$ is odd,}
\end{cases}
\end{equation}
where
\[\hat{g}_{m}= \frac{2F_{m}-F_{m+1}}{3}.\]
\end{theorem}
\begin{proof}
(a) To prove (\ref{eqn4}), consider 
\begin{eqnarray*}
X_{1}&=&\frac{1}{F_{n+m-1}F_{n}+(-1)^{n}g_{m}}-\frac{1}{F_{n+m+1}F_{n+2}+(-1)^{n}g_{m}}-\frac{1}{F_{n}F_{n+m}}-\frac{1}{F_{n+1}F_{n+m+1}}\\
&=&\frac{\hat{X}_{1}}{\{F_{n+m-1}F_{n}+(-1)^{n}g_{m}\}\{F_{n+m+1}F_{n+2}+(-1)^{n}g_{m}\}F_{n}F_{n+m}F_{n+1}F_{n+m+1}},
\end{eqnarray*}
where, by the identity $F_{n+m+1}F_{n+2}-F_{n+m-1}F_{n}=F_{n}F_{n+m}+F_{n+1}F_{n+m+1}$
\[
\hat{X}_{1}=(F_{n}F_{n+m}+F_{n+1}F_{n+m+1})\tilde{X}_{1},\]
with
\begin{eqnarray*}
\tilde{X}_{1}&=&F_{n}F_{n+1}F_{n+m}F_{n+m+1}-
F_{n+m-1}F_{n+m+1}F_{n}F_{n+2}\\
&&-(-1)^{n}g_{m}(F_{n+m-1}F_{n}+F_{n+m+1}F_{n+2})-g_{m}^{2}.
\end{eqnarray*}
From Lemma \ref{LEM1}, we have
\begin{eqnarray*}
F_{n+1}F_{n+m}-F_{n+m+1}F_{n}&=&(-1)^{n}F_{m},\\
F_{n+m+1}F_{n}-F_{n+m-1}F_{n+2}&=&(-1)^{n}(F_{m}-F_{m+1}),\\
F_{n+m+1}F_{n-1}-F_{n+m}F_{n}&=&(-1)^{n}F_{m+1}.
\end{eqnarray*}
Then
\begin{eqnarray*}
&&F_{n}F_{n+1}F_{n+m}F_{n+m+1}-F_{n+m-1}F_{n+m+1}F_{n}F_{n+2}\\
&=&F_{n+m+1}F_{n}\Big\{F_{n+m+1}F_{n}+(-1)^{n}F_{m}\Big\}\\
&&-F_{n+m+1}F_{n}\Big\{F_{n+m+1}F_{n}+(-1)^{n}(F_{m+1}-F_{m})\Big \} \\
&=&(-1)^{n}F_{n+m+1}F_{n}(2F_{m}-F_{m+1}),
\end{eqnarray*}
and
\begin{eqnarray*}
F_{n+m-1}F_{n}+F_{n+m+1}F_{n+2}&=&3F_{n+m+1}F_{n}+F_{n+m+1}F_{n-1}-F_{n+m}F_{n}\\
&=&3F_{n+m+1}F_{n}+(-1)^{n}F_{m+1}.
\end{eqnarray*}
Hence
\begin{eqnarray*}
\tilde{X}_{1}&=&(-1)^{n}F_{n+m+1}F_{n} (2F_{m}-F_{m+1}-3g_{m})
-g_{m}F_{m+1}-g_{m}^{2}.
\end{eqnarray*}
Assume that $n$ is even.
Since $g_{m}>0$ and $2F_{m}-F_{m+1}-3g_{m}<0,$
 then $X_{1}<0$ and
\[
\frac{1}{F_{n+m-1}F_{n}+g_{m}}-\frac{1}{F_{n+m+1}F_{n+2}+g_{m}}<
\frac{1}{F_{n}F_{n+m}}+\frac{1}{F_{n+1}F_{n+m+1}}.
\]
Repeatedly applying the above inequality, we have
\begin{equation}\label{eqn7}
\frac{1}{F_{n+m-1}F_{n}+g_{m}}<\sum_{k=n}^{\infty}\frac{1}{F_{k}F_{k+m}}, \text{ if $n \geq 2$ and $n$ is even.}
\end{equation}
Similarly, if $n$ is odd, then there exists a positive integer $m_{1}$ such that, for $n \geq m_{1}$, $X_{1}>0$  and
\[
\frac{1}{F_{n}F_{n+m}}+\frac{1}{F_{n+1}F_{n+m+1}}<
\frac{1}{F_{n+m-1}F_{n}-g_{m}}-\frac{1}{F_{n+m+1}F_{n+2}-g_{m}},
\]
from which we obtain 
\begin{equation}\label{eqn8}
\sum_{k=n}^{\infty}\frac{1}{F_{k}F_{k+m}}<
\frac{1}{F_{n+m-1}F_{n}-g_{m}}, \text{ if $n \geq m_{1}$ and $n$ is odd.}
\end{equation}


Next, consider
\begin{eqnarray*}
X_{2}&=&\frac{1}{F_{n+m-1}F_{n}+(-1)^{n}g_{m}-1}-\frac{1}{F_{n+m}F_{n+1}+(-1)^{n+1}g_{m}-1}-\frac{1}{F_{n}F_{n+m}}\\
&=&\frac{\hat{X}_{2}}{\{F_{n+m-1}F_{n}+(-1)^{n}g_{m}-1\}\{F_{n+m}F_{n+1}+(-1)^{n+1}g_{m}-1\}F_{n}F_{n+m}},
\end{eqnarray*}
where
\begin{eqnarray*}
\hat{X}_{2}&=&F_{n}F_{n+m}^{2}F_{n+1}-F_{n+m}F_{n+m-1}F_{n}F_{n+1}-F_{n}^{2}F_{n+m-1}F_{n+m}\\
&&-(-1)^{n}g_{m}(2F_{n}F_{n+m}-F_{n+m-1}F_{n}+F_{n+m}F_{n+1})\\
&&+F_{n+m-1}F_{n}+F_{n+m}F_{n+1}+g_{m}^{2}-1.
\end{eqnarray*}
From Lemma \ref{LEM1}, we have
\[F_{n+m-1}F_{n}-F_{n+m-2}F_{n+1}=(-1)^{n+1}F_{m-2}=(-1)^{n}(F_{m+1}-2F_{m}).\]
Then
\begin{eqnarray*}
&&F_{n}F_{n+m}F_{n+1}F_{n+m}-F_{n+m}F_{n+m-1}F_{n}F_{n+1}-F_{n}^{2}F_{n+m-1}F_{n+m}\\
&=&F_{n}F_{n+m}(F_{n+1}F_{n+m-2}-F_{n}F_{n+m-1})\\
&=&(-1)^{n}F_{n}F_{n+m}(2F_{m}-F_{m+1}),
\end{eqnarray*}
and
\begin{eqnarray*}
&&2F_{n}F_{n+m}+F_{n+m}F_{n+1}-F_{n+m-1}F_{n}\\
&=&3F_{n}F_{n+m}+F_{n+m}F_{n-1}-F_{n+m-1}F_{n}\\
&=&3F_{n}F_{n+m}+(-1)^{n}(2F_{m+2}-F_{m+3}).
\end{eqnarray*}
Hence
\begin{eqnarray*}
\hat{X}_{2}&=&(-1)^{n}F_{n}F_{n+m}(2F_{m}-F_{m+1}-3g_{m} )+F_{n+m-1}F_{n}
+F_{n+m}F_{n+1}\\
&&-g_{m}(2F_{m+2}-F_{m+3})+g_{m}^{2}-1.
\end{eqnarray*}
Suppose that $n$ is even. 
Since
\[-2 \leq 2F_{m}-F_{m+1}-3g_{m} \leq -1,\]
then
\begin{eqnarray*}
&&F_{n}F_{n+m}(2F_{m}-F_{m+1}-3g_{m})+(F_{n+m-1}F_{n}+F_{n+m}F_{n+1})\\
& \geq & -2F_{n}F_{n+m}+F_{n}F_{n+m-1}+F_{n+1}F_{n+m}\\
&=&(F_{n-1}-F_{n})(F_{n+m-1}+F_{n-m-2})+F_{n}F_{n+m-1}\\
&=&F_{n-1}F_{n+m-1}-F_{n-2}F_{n+m-2}\\
&>&0,
\end{eqnarray*}
and there exists a positive integer $m_{2}$ such that, for $n \geq m_{2}$, $X_{2}>0$ and
\[
\frac{1}{F_{n}F_{n+m}}<\frac{1}{F_{n+m-1}F_{n}+(-1)^{n}g_{m}-1}-\frac{1}{F_{n+m}F_{n+1}+(-1)^{n+1}g_{m}-1}.
\] 
Repeatedly applying the above inequality, we have
\begin{equation}\label{eqn9}
\sum_{k=n}^{\infty}\frac{1}{F_{k}F_{k+m}}<\frac{1}{F_{n+m-1}F_{n}+g_{m}-1}, \text{ if $n \geq m_{2}$ and $n$ is even.}
\end{equation} 

On the other hand,
\begin{eqnarray*}
X_{3}&=&\frac{1}{F_{n+m-1}F_{n}+(-1)^{n}g_{m}+1}-\frac{1}{F_{n+m}F_{n+1}+(-1)^{n+1}g_{m}+1}-\frac{1}{F_{n}F_{n+m}}\\
&=&\frac{\hat{X}_{3}}{\{F_{n+m-1}F_{n}+(-1)^{n}g_{m}+1\}\{F_{n+m-1}F_{n+1}+(-1)^{n+1}g_{m}+1\}F_{n}F_{n+m}},
\end{eqnarray*}
where
\begin{eqnarray*}
\hat{X}_{3}&=&\hat{X}_{2}-2(F_{n+m-1}F_{n}+F_{n+m}F_{n+1})\\
&=&(-1)^{n}F_{n}F_{n+m}(2F_{m}-F_{m+1}-3g_{m} )-F_{n+m-1}F_{n}-F_{n+1}F_{n+1}\\
&&-g_{m}(2F_{m+2}-F_{m+3})+g_{m}^{2}-1.
\end{eqnarray*}
Suppose that $n$ is odd. As shown above, we have
\[
-F_{n}F_{n+m}(2F_{m}-F_{m+1}-3g_{m})-F_{n+m-1}F_{n}-F_{n+m}F_{n+1}
<F_{n-2}F_{n+m-2}-F_{n-1}F_{n+m-1}.
\]
Hence 
there exists a positive integer  $m_{3}$ such that, for $n \geq m_{3}$, $X_{3} <0$ and
\[
\frac{1}{F_{n+m-1}F_{n}+(-1)^{n}g_{m}+1}-\frac{1}{F_{n+m}F_{n+1}+(-1)^{n+1}g_{m}+1}<\frac{1}{F_{n}F_{n+m}},
\]
from which we have
\begin{equation}\label{eqn10}
\frac{1}{F_{n+m-1}F_{n}-g_{m}+1}<\sum_{k=n}^{\infty}\frac{1}{F_{k}F_{k+m}}, \text{ if $n \geq m_{3}$ and $n$ is odd.}
\end{equation}
Then, (\ref{eqn4}) follows from (\ref{eqn7}), (\ref{eqn8}), (\ref{eqn9}) and (\ref{eqn10}). 
\newline \null

(b) Since $F_{n+2}F_{n+3}-F_{n}F_{n+1}=F_{n}F_{n+2}+F_{n+1}F_{n+3}$, we have
\[
\frac{1}{F_{n}F_{n\!+\!1}}\!-\!\frac{1}{F_{n\!+\!2}F_{n\!+\!3}}\!-\!\frac{1}{F_{n}F_{n\!+\!2}}\!-\!\frac{1}{F_{n\!+\!1}F_{n+3}}
\!=\!\frac{F_{n\!+\!2}F_{n\!+\!3}\!-\!F_{n}F_{n\!+\!1}\!-\!(F_{n}F_{n\!+\!2}\!+\!F_{n\!+\!1}F_{n\!+\!3})}{F_{n}F_{n\!+\!1}F_{n\!+\!2}F_{n\!+\!3}}\!=\!0,
\] 
i.e.,
\[
\frac{1}{F_{n}F_{n+1}}-\frac{1}{F_{n+2}F_{n+3}}=\frac{1}{F_{n}F_{n+2}}+\frac{1}{F_{n+1}F_{n+3}}.
\]
Repeatedly applying the above equality, we obtain (\ref{eqn5}).
\newline \null

(c) Let $m \geq 3$ and assume that
\[\hat{g}_{m}=\frac{2F_{m}-F_{m+1}}{3} \in \mathbb{Z}.\]
We recall the proof of (a). Replacing $g_{m}$ by $\hat{g}_{m}$ in $\tilde{X}_{1}$, we have
\[
\tilde{X}_{1}=-\hat{g}_{m}F_{m+1}-\hat{g}_{m}^{2}<0.
\]
Then $X_{1}<0$ if $n \geq 2$ and $n$ is even  or if $n \geq m_{4}$ and $n$ is odd for some positive integer $m_{4}$, and we have 
\begin{equation}\label{eqn11}
\frac{1}{F_{n+m-1}F_{n}+(-1)^{n}\hat{g}_{m}}<\sum_{k=n}^{\infty}\frac{1}{F_{k}F_{k+m}}, \text{ if $n \geq 2$ ($n$ is even) 
 or if  $n \geq m_{4}$ ($n$ is odd).}
\end{equation}
Similarly there exist positive integers $m_{5}$ and $m_{6}$ such that $X_{2}>0$ if $n \geq m_{5}$ and $n$ is even, or if $n \geq m_{6}$ and $n$ is odd, from which
we have
\begin{equation}\label{eqn12}
\sum_{k=n}^{\infty}\frac{1}{F_{k}F_{k+m}}<\frac{1}{F_{n+m-1}F_{n}+(-1)^{n}\hat{g}_{m}-1}, \text{ if $n \geq m_{5}$ ($n$ is even) or 
if $n \geq m_{6}$ ($n$ is odd).}
\end{equation}
Then, (\ref{eqn6}) follows from (\ref{eqn11}) and (\ref{eqn12}).
\end{proof}

\begin{remark}
From Theorem \ref{THM3}, we have
\begin{eqnarray*}
\Bigg\lfloor\Bigg(\sum_{k=n}^{\infty}\frac{1}{F_{k}F_{k+1}}\Bigg)^{-1}\Bigg\rfloor &=& 
\left \{ \begin{array}{ll} F_{n}^{2}, 
&\text{ if $n \geq 2$ and $n$ is even;}\\
F_{n}^{2}-1,&\text{ if $n \geq 1$ and $n$ is odd,}
\end{array} \right.\\
\Bigg\lfloor\Bigg(\sum_{k=n}^{\infty}\frac{1}{F_{k}F_{k+6}}\Bigg)^{-1}\Bigg\rfloor &=& 
\left \{ \begin{array}{ll} F_{n+5}F_{n}, 
&\text{ if $n \geq 2$ and $n$ is even;}\\
F_{n+5}F_{n}-2,&\text{ if $n \geq 1$ and $n$ is odd,}
\end{array} \right.\\
\end{eqnarray*}
etc.
\end{remark}

\section{Acknowledgments}

The author is thankful to the editor-in-chief and to the anonymous referee for their helpful comments which led to the improved presentation of the paper.

\begin{thebibliography}{99}

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1365--1373.

\bibitem{CH2} Y. Choo, On the finite sums of reciprocal Lucas numbers,
{\em Int. J. Math. Anal.} {\bf 11} (2017), 519--529. 

\bibitem{HK} S. Holliday and T. Komatsu, On the  sum of reciprocal generalized Fibonacci numbers, {\em Integers} {\bf 11A} (2011), Article 11.

\bibitem{KOS} T. Koshy, {\em Fibonacci and Lucas numbers with Applications}, Wiley, 2001.

\bibitem{LW} R. Liu and A. Wang, Sums of products of two reciprocal Fibonacci numbers, {\em Adv. Diff. Eqn.} {\bf 16} (2016), Article 1136.

\bibitem{OHT} H. Ohtsuka, Problem H-698 (ii), {\em Fibonacci Quart.} {\bf 50} (2012), 284.
 
\bibitem{ON} H. Ohtsuka and S. Nakamura, On the sum of reciprocal Fibonacci numbers, {\em Fibonacci Quart.} {\bf 46/47} (2008/2009), 153--159.

\bibitem{WW} A. Wang and P. Wen, On the partial finite sums of the reciprocals of the Fibonacci numbers, {\em J. Inequal. Appl.} {\bf 15} (2015), Article 73.

\bibitem{WY} A. Wang and T. Yuan, Alternating sums of the reciprocal Fibonacci numbers, {\em J. Integer Seq.} {\bf 20} (2017), Article 17.1.4.

\bibitem{WZ1} A. Wang and F. Zhang, The reciprocal sums of even and odd terms in the Fibonacci sequence, {\em J. Inequal. Appl.} {\bf 15} (2015), Article 376.
 
\bibitem{WZ2} Z. Wu and W. Zhang, The sums of the reciprocals of Fibonacci polynomials and Lucas polynomials, {\em J. Inequal. Appl.} {\bf 12} (2012), Article 134.

\bibitem{WZ3} A. Wang and F. Zhang, The reciprocal sums of the Fibonacci 3-subsequences, {\em Adv. Diff. Eqn.} {\bf 16} (2016), Article 27.

\bibitem{ZW} W. Zhang and T. Wang, The infinite sum of reciprocal Pell numbers, {\em Appl. Math. Comput.} {\bf 218} (2012), 6164--6167. 
\end{thebibliography}


\bigskip
\hrule
\bigskip

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

\noindent \emph{Keywords: }
Fibonacci number, reciprocal, floor.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequence
\seqnum{A000108}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received July 17 2017;
revised versions received  December 29 2017; January 22 2018.
Published in {\it Journal of Integer Sequences}, March 8 2018.

\bigskip
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\noindent
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