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\begin{center}
\vskip 1cm{\LARGE\bf
Powers of Two as Sums of \\
\vskip .1in
Two Lucas Numbers
}
\vskip 1cm
\large
Jhon J.~Bravo\\
Mathematics Department \\
University of Cauca \\
Street 5 No.~4-70 \\
Popay\'an, Cauca \\
Colombia \\
\href{mailto:jbravo@unicauca.edu.co}{\tt jbravo@unicauca.edu.co} \\
\ \\
Florian Luca\\
School of Mathematics\\
University of the Witwatersrand\\
P.~O.~Box Wits 2050 \\
South Africa\\
and \\
Mathematical Institute \\
UNAM Juriquilla \\
Santiago de Quer\'etaro 76230\\
Quer\'etaro de Arteaga\\
Mexico \\
\href{mailto:fluca@matmor.unam.mx}{\tt fluca@matmor.unam.mx} \\
\end{center}
\vskip .2 in
\begin{abstract}
Let $(L_n)_{n\geq 0}$ be the Lucas sequence given by $L_0=2$, $L_1=1$
and $L_{n+2}=L_{n+1}+L_{n}$ for $n\geq 0$. In this paper, we are
interested in finding all powers of two which are sums of two Lucas
numbers, i.e., we study the Diophantine equation $L_{n}+L_{m}=2^a$ in
nonnegative integers $n$, $m$, and $a$. The proof of our main theorem uses
lower bounds for linear forms in logarithms, properties of continued
fractions, and a version of the Baker-Davenport reduction method in
diophantine approximation. This paper continues our previous work where
we obtained a similar result for the Fibonacci numbers.
\end{abstract}
\section{Introduction}
Let $(F_n)_{n\geq 0}$ be the \emph{Fibonacci sequence} given by $F_0=0$, $F_1=1$ and $F_{n+2}=F_{n+1}+F_{n}$ for all $n\geq 0$. The Fibonacci numbers are famous for possessing wonderful and amazing properties. They are accompanied by the sequence of \emph{Lucas numbers}, which is as important as the Fibonacci sequence. The Lucas sequence $(L_n)_{n\geq 0}$ follows the same recursive pattern as the Fibonacci numbers, but with initial conditions $L_0=2$ and $L_1=1$.
The study of properties of the terms of such sequences, or more
generally, linear recurrence sequences, has a very long history and has
generated a huge literature. For the beauty and rich applications of
these numbers and their relatives, one can see Koshy's book \cite{Kos01}.
For example, a remarkable property of the Fibonacci sequence is that 1,
2 and 8 are the only Fibonacci numbers which are powers of 2. One proof
of this fact follows from Carmichael's primitive divisor theorem
\cite{Car}, which states that for $n$ greater than 12, the $n$th
Fibonacci number $F_n$ has at least one prime factor that is not a
factor of any previous Fibonacci number (see the paper of Bilu, Hanrot,
and Voutier \cite{Bilu} for the most general version of the above
statement). Similarly, it is well known that 1, 2 and 4 are the only
powers of 2 that appear in the Lucas sequence.
The problem of finding all perfect powers in the Fibonacci sequence and the Lucas sequence was a famous open problem finally solved in 2006 in a paper in {\it Annals of Mathematics} by Bugeaud, Mignotte, and Siksek \cite{Bug}. In their work, they applied a combination of Baker's method, the modular approach and some classical techniques to show that the only perfect powers in the Fibonacci sequence are 0, 1, 8 and 144, and the only perfect powers in the Lucas sequence are 1 and 4. A detailed account of this problem can be found in \cite[Section~10]{Bug}.
In our recent paper \cite{BL13}, we found all powers of 2 which are the sums of at most two Fibonacci numbers. Specifically, we proved the following.
\begin{theorem} \label{teo-caseFib}
The only solutions of the Diophantine equation $F_{n}+F_{m}=2^a$ in positive integers $n,m$ and $a$ with $n\geq m$ are given by
\[
2F_1=2, \quad 2F_2=2, \quad 2F_3=4, \quad 2F_6=16,
\]
and
\[
F_2+F_1=2, \quad F_4+F_1=4, \quad F_4+F_2=4, \quad F_5+F_4=8, \quad F_7+F_4=16.
\]
\end{theorem}
In this paper, we prove an analogue of Theorem \ref{teo-caseFib} when the sequence of Fibonacci numbers is replaced by the sequence of the Lucas numbers, i.e., we extend our previous work \cite{BL13} and determine all the solutions of the Diophantine equation
\begin{equation}\label{eqn1}
L_{n}+L_{m}=2^a
\end{equation}
in nonnegative integers $n\ge m$ and $a$.
Similar problems have recently been investigated. For example, repdigits which are sums of at most three Fibonacci numbers were found by Luca \cite{FL12}; Fibonacci numbers which are sums of two repdigits were obtained by D\'iaz and Luca \cite{SL11}, while factorials which are sums of at most three Fibonacci numbers were found by Luca and Siksek \cite{LS10}.
We prove the following result.
\begin{theorem} \label{teo1}
All solutions of the Diophantine equation \eqref{eqn1} in nonnegative integers $n\ge m$ and $a$, are
\[
2L_0=4, \quad 2L_1=2, \quad 2L_3=8, \quad L_2+L_1=4, \quad L_4+L_1=8 \quad \text{and} \quad L_7+L_2=32.
\]
\end{theorem}
\vspace{0.3cm}
Let us give a brief overview of our strategy for proving Theorem \ref{teo1}. First, we rewrite equation \eqref{eqn1} in suitable ways in order to obtain two different linear forms in logarithms which are both nonzero and small. Next, we use a lower bound on such nonzero linear forms in two logarithms due to Laurent, Mignotte, and Nesterenko as well as a general lower bound due to Matveev to find an absolute upper bound for $n$; hence, an absolute upper bound for $m$ and $a$, which we then reduce using standard facts concerning continued fractions.
In this paper, we follow the approach and the presentation described in \cite{BL13}.
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\section{Auxiliary results}
Before proceeding further, we recall that the Binet formula
\[
L_n=\alpha^n+\beta^n \quad \text{holds for all} \quad n\geq 0,
\]
where
\[
\alpha:=\frac{1+\sqrt{5}}{2} \quad \text{and} \quad \beta:=\frac{1-\sqrt{5}}{2}
\]
are the roots of the characteristic equation $x^2-x-1=0$ of $(L_n)_{n\geq 0}$. This will be an important ingredient in what follows. In particular, the inequality
\begin{equation} \label{desfib}
\alpha^{n-1}\leq L_n \leq 2\alpha^{n}
\end{equation}
holds for all $n\geq 0$.
In order to prove Theorem \ref{teo1}, we need a result of Laurent, Mignotte, and Nesterenko \cite{LMN95} about linear forms in two logarithms. But first, some notation.
Let $\eta$ be an algebraic number of degree $d$ with minimal polynomial
\[
a_0x^d+a_1x^{d-1}+\cdots+a_d=a_0\prod_{i=1}^{d}(X-\eta^{(i)}),
\]
where the $a_i$'s are relatively prime integers with $a_0>0$ and the $\eta^{(i)}$'s are conjugates of $\eta$. Then
\[
h(\eta)=\frac{1}{d}\left(\log a_0+\sum_{i=1}^{d}\log\left(\max\{|\eta^{(i)}|,1\}\right)\right)
\]
is called the \emph{logarithmic height} of $\eta$. In particular, if $\eta=p/q$ is a rational number with $\gcd(p,q)=1$ and $q>0$, then $h(\eta)=\log \max \{|p|,q\}$.
The following properties of the logarithmic height, which will be used in the next section without special reference, are also known:
\begin{itemize}
\item $h(\eta\pm\gamma)\leq h(\eta) + h(\gamma)+\log 2$.
\item $h(\eta\gamma^{\pm 1})\leq h(\eta)+h(\gamma)$.
\item $h(\eta^{s})=|s|h(\eta)$.
\end{itemize}
With the above notation, Laurent, Mignotte, and Nesterenko \cite[Corollary~1]{LMN95} proved the following theorem.
\begin{theorem}\label{teoLMN}
Let $\gamma_1, \gamma_2$ be two non-zero algebraic numbers, and let $\log \gamma_1$ and $\log \gamma_2$ be any determinations of their logarithms. Put $D=[\mathbb{Q}(\gamma_1,\gamma_2):\mathbb{Q}]/[\mathbb{R}(\gamma_1,\gamma_2):\mathbb{R}]$, and
\[
\Gamma:=b_2\log \gamma_2-b_1\log \gamma_1,
\]
where $b_1$ and $b_2$ are positive integers. Further, let $A_1, A_2$ be real numbers $>1$ such that
\[
\log A_i\geq \max\left\{h(\gamma_i),\frac{|\log \gamma_i|}{D}, \frac{1}{D}\right\}, \quad \text{$i=1,2$}.
\]
Then, assuming that $\gamma_1$ and $\gamma_2$ are multiplicatively independent, we have
\[
\log|\Gamma|>-30.9 \cdot D^4\left(\max\left\{\log b^{\prime},\frac{21}{D},\frac{1}{2} \right\} \right)^2 \log A_1 \cdot \log A_2,
\]
where
\[
b^{\prime}=\frac{b_1}{D\log A_2}+\frac{b_2}{D\log A_1}.
\]
\end{theorem}
We shall also need the following general lower bound for linear forms in logarithms due to Matveev \cite{Matveev} (see also the paper of Bugeaud, Mignotte, and Siksek \cite[Theorem~9.4]{Bug}).
\begin{theorem}[Matveev's theorem]\label{teoMatveev}
Assume that $\gamma_1, \ldots, \gamma_t$ are positive real algebraic numbers in a real algebraic number field $\mathbb{K}$ of degree $D$, $b_1,\ldots,b_t$ are rational integers, and
\[
\Lambda:=\gamma_1^{b_1}\cdots\gamma_t^{b_t}-1,
\]
is not zero. Then
\[
|\Lambda|>\exp\left(-1.4\cdot 30^{t+3}\cdot t^{4.5}\cdot D^2(1+\log D)(1+\log B)A_1\cdots A_t\right),
\]
where
\[
B\geq \max\{|b_1|,\ldots,|b_t|\},
\]
and
\[
A_i\geq \max\{Dh(\gamma_i),|\log \gamma_i|, 0.16\}, \quad \text{for all} \quad i=1,\ldots,t.
\]
\end{theorem}
In 1998, Dujella and Peth\H o in \cite[Lemma~5 (a)]{DP} gave a version of the reduction method based on the Baker-Davenport lemma \cite{Baker-Davenport}. To conclude this section of auxiliary results, we present the following lemma from \cite{BL1}, which is an immediate variation of the result due to Dujella and Peth\H o from \cite{DP}, and will be one of the key tools used in this paper to reduce the upper bounds on the variables of the equation \eqref{eqn1}.
\begin{lemma} \label{reduce}
Let $M$ be a positive integer, let $p/q$ be a convergent of the continued fraction of the irrational $\gamma$ such that $q>6M$, and let $A,B,\mu$ be some real numbers with $A>0$ and $B>1$. Let $\epsilon:=||\mu q||-M||\gamma q||$, where $||\cdot||$ denotes the distance from the nearest integer. If $\epsilon >0$, then there is no solution to the inequality
\[
0__m$.
If $n\leq 200$, then a brute force search with \emph{Mathematica} in the range $0\leq m 200$.
Let us now get a relation between $n$ and $a$. Combining \eqref{eqn1} with the right inequality of \eqref{desfib}, one gets that
\[
2^{a} \leq 2\alpha^{n}+2\alpha^{m}<2^{n+1}+2^{m+1}=2^{n+1}(1+2^{m-n})\leq 2^{n+1}(1+2^{-1})<2^{n+2},
\]
which leads to $a\leq n+1$. This estimate is essential for our purpose.
On the other hand, we rewrite equation \eqref{eqn1} as
\[
\alpha^n-2^a=-\beta^n-L_m.
\]
We now take absolute values in the above relation obtaining
\[
\left|\alpha^n-2^a\right|\leq |\beta|^n+L_m<\frac{1}{2}+2\alpha^m.
\]
Dividing both sides of the above expression by $\alpha^{n}$ and taking into account that $n>m$, we get
\begin{equation} \label{deflambda1}
\left|1-2^{a}\cdot\alpha^{-n}\right|<\frac{3}{\alpha^{n-m}}.
\end{equation}
In order to apply Theorem \ref{teoLMN}, we take $\gamma_1:=\alpha$, $\gamma_2:=2$, $b_1:=n$ and $b_2:=a$. So,
\[
\Gamma:=b_2\log\gamma_2-b_1\log\gamma_1,
\]
and therefore estimation \eqref{deflambda1} can be rewritten as
\begin{equation} \label{deflambda1B}
\left|1-e^{\Gamma}\right|<\frac{3}{\alpha^{n-m}}.
\end{equation}
The algebraic number field containing $\gamma_1,\gamma_2$ is $\mathbb{Q}(\sqrt{5})$, so we can take $D:=2$. By using \eqref{eqn1} and the Binet formula for the Lucas sequence, we have
\[
\alpha^n=L_n-\beta^n0$. This, together with \eqref{deflambda1B}, gives
\begin{equation} \label{des-gamma}
0<\Gamma <\frac{3}{\alpha^{n-m}},
\end{equation}
where we have also used the fact that $x\leq e^x-1$ for all $x\in\mathbb{R}$. Hence,
\begin{equation} \label{exp-gamma}
\log\Gamma <\log 3-(n-m)\log\alpha.
\end{equation}
Note further that $h(\gamma_1)=(\log\alpha)/2=0.2406\cdots$ and $h(\gamma_2)=\log 2=0.6931\cdots$; thus, we can choose $\log A_1:=0.5$ and $\log A_2:=0.7$. Finally, by recalling that $a\leq n+1$, we get
\[
b^{\prime}=\frac{n}{1.4}+a < 1.71429\,n+1<2n.
\]
Since $\alpha$ and 2 are multiplicatively independent, we have, by Theorem \ref{teoLMN}, that
\begin{align}
\log \Gamma\geq & \,-30.9\cdot 2^4\cdot \left(\max\left\{\log(2n),21/2,1/2\right\} \right)^2\cdot 0.5\cdot 0.7 \notag\\
>& -174\cdot \left(\max\left\{\log(2n),21/2,1/2\right\} \right)^2. \label{aplicLMN}
\end{align}
We now combine \eqref{exp-gamma} and \eqref{aplicLMN} to obtain
\begin{equation} \label{res-aplic-LMN}
(n-m)\log\alpha<180\cdot \left(\max\left\{\log(2n),21/2\right\} \right)^2.
\end{equation}
Let us now get a second linear form in logarithms. To this end, we now rewrite \eqref{eqn1} as follows:
\[
\alpha^n(1+\alpha^{m-n})-2^a=-\beta^n-\beta^m.
\]
Taking absolute values in the above relation and using the fact that $\beta=(1-\sqrt{5})/2$, we get
\[
|\alpha^n(1+\alpha^{m-n})-2^a|=|\beta|^n+|\beta|^m<2
\]
for all $n>200$ and $m\geq 0$. Dividing both sides of the above inequality by the first term of the left-hand side, we obtain
\begin{equation} \label{deflambda2}
\left|1-2^{a}\cdot \alpha^{-n}\cdot (1+\alpha^{m-n})^{-1}\right|<\frac{2}{\alpha^{n}}.
\end{equation}
We are now ready to apply Matveev's result Theorem \ref{teoMatveev}. To do this, we take the parameters $t:=3$ and
\[
\gamma_1:=2, \quad \gamma_2:=\alpha, \quad \gamma_3:=1+\alpha^{m-n}.
\]
We take $b_1:=a$, $b_2:=-n$ and $b_3:=-1$. As before, $\mathbb{K}:=\mathbb{Q}(\sqrt{5})$ contains $\gamma_1,\gamma_2,\gamma_3$ and has $D:=[\mathbb{K}:\mathbb{Q}]=2$. To see why the left-hand side of \eqref{deflambda2} is not zero, note that otherwise, we would get the relation
\begin{equation} \label{exp1}
2^a=\alpha^n+\alpha^m.
\end{equation}
Conjugating the above relation in $\mathbb{Q}(\sqrt{5})$, we get
\begin{equation} \label{exp2}
2^a=\beta^n+\beta^m.
\end{equation}
Combining \eqref{exp1} and \eqref{exp2}, we obtain
\[
\alpha ^n<\alpha^n+\alpha^m=|\beta^n+\beta^m|\leq |\beta|^n+|\beta|^m<2,
\]
which is impossible for $n>200$. Hence, indeed the left-hand side of inequality \eqref{deflambda2} is nonzero.
In this application of Matveev's theorem we take $A_1:=1.4$ and $A_2:=0.5$. Since $a\leq n+1$ it follows that we can take $B:=n+1$. Let us now estimate $h(\gamma_3)$. We begin by observing that
\[
\gamma_3=1+\alpha^{m-n}<2 \quad \text{and} \quad \gamma_3^{-1}=\frac{1}{1+\alpha^{m-n}}<1,
\]
so that $|\log \gamma_3|<1$. Next, notice that
\[
h(\gamma_3) \leq |m-n|\left(\frac{\log \alpha}{2}\right)+\log 2=\log 2+(n-m)\left(\frac{\log \alpha}{2}\right).
\]
Hence, we can take
\[
A_3:=2+(n-m)\log \alpha >\max\{2h(\gamma_3),|\log \gamma_3|,0.16\}.
\]
Now Matveev's theorem implies that a lower bound on the left-hand side of \eqref{deflambda2} is
\[
\exp\left(-C\cdot (1+\log(n+1)) \cdot 1.4 \cdot 0.5 \cdot (2+(n-m)\log \alpha \right)
\]
where $C:=1.4\cdot 30^{6}\cdot 3^{4.5}\cdot 2^2 (1+\log 2)<9.7\times 10^{11}$. So, inequality \eqref{deflambda2} yields
\begin{equation} \label{exp3}
n\log \alpha-\log 2 < 1.36\times 10^{12}\log n \cdot (2+(n-m)\log \alpha),
\end{equation}
where we used the inequality $1+\log(n+1)<2\log n$, which holds because $n>200$.
Using now \eqref{res-aplic-LMN} in the right-most term of the above inequality \eqref{exp3} and performing the respective calculations, we arrive at
\begin{equation} \label{exp-fin-n}
n<6\times 10^{14}\log n\cdot \left(\max\left\{\log(2n),21/2\right\} \right)^2.
\end{equation}
If $\max\left\{\log(2n),21/2\right\}=21/2$, it then follows from \eqref{exp-fin-n} that $n< 8\times 10^{16}\log n$ giving $n<3.5\times 10^{18}$. If on the other hand we have that $\max\left\{\log(2n),21/2\right\}=\log(2n)$, then, from \eqref{exp-fin-n}, we get $n< 6\times 10^{14}\log n\log^2(2n)$ and so $n<5.9\times 10^{19}$. In any case, we have that $n<5.9\times 10^{19}$ always holds. We summarize what we have proved so far in the following lemma.
\begin{lemma}\label{cota_an}
If $(n,m,a)$ is a solution in positive integers of equation \eqref{eqn1} with $n>m$ and $n>200$, then inequalities
\[
a\leq n+1<6\times 10^{19}
\]
hold.
\end{lemma}
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\section{Reducing the bound on $n$}
After finding an upper bound on $n$ the next step is to reduce it. To do this, we first use some properties of continued fractions to obtain a suitable upper bound on $n-m$, and secondly we use Lemma \ref{reduce} to conclude that $n$ must be relatively small. Let us see.
Turning back to inequality \eqref{des-gamma}, we obtain
\[
00}
0\frac{1}{(a_M+2)a}.
\end{equation}
Comparing estimates \eqref{Gamma>0} and \eqref{aplic-frac-cont}, we get right away that
\[
\alpha^{n-m}<7\cdot 136\cdot a<6\times 10^{22},
\]
leading to $n-m\leq 110$. Let us now work a little bit on \eqref{deflambda2} in order to find an improved upper bound on $n$. Put
\begin{equation} \label{def-z}
z:=a \log 2-n\log\alpha-\log \varphi(n-m),
\end{equation}
where $\varphi$ is the function given by the formula $\varphi(t):=1+\alpha^{-t}$. Therefore, \eqref{deflambda2} implies that
\begin{equation} \label{des-z}
|1-e^{z}|<\frac{2}{\alpha^{n}}.
\end{equation}
Note that $z\neq 0$; thus, we distinguish the following cases. If $z>0$, then, from \eqref{des-z}, we obtain
\[
00}
00} for all choices $n-m\in\{1,\ldots,110\}$ except when $n-m=1,3$ and get that
\[
n<\frac{\log(Aq/\epsilon)}{\log B},
\]
where $q>6M$ is a denominator of a convergent of the continued fraction of $\gamma$ such that $\epsilon=||\mu q||-M||\gamma q||>0$. Indeed, with the help of \emph{Mathematica} we find that if $(n,m,a)$ is a possible solution of the equation \eqref{eqn1} with $z>0$ and $n-m\neq 1,3$, then $n\leq 130$. This is false because our assumption that $n>200$.
Suppose now that $z<0$. First, note that $2/\alpha^{n}<1/2$ since $n>200$. Then, from \eqref{des-z}, we have that $|1-e^{z}|<1/2$ and therefore $e^{|z|}<2$. Since $z<0$, we have
\[
0<|z|\leq e^{|z|}-1=e^{|z|}|e^{z}-1|<\frac{4}{\alpha^{n}}.
\]
Then, by the same arguments used for proving \eqref{z>0}, we obtain
\begin{equation} \label{z<0}
00} or \eqref{z<0} (according to whether $z$ is positive or negative, respectively), the corresponding parameter $\mu$ appearing in Lemma \ref{reduce} is either
\[
-\frac{\log\varphi(t)}{\log\alpha}=
\begin{cases}
-1, & \text{if } t=1; \\ 1-\frac{\log 2}{\log\alpha}, & \text {if } t=3.
\end{cases}
\quad \text{or} \quad
\frac{\log\varphi(t)}{\log 2}=
\begin{cases}
\frac{\log\alpha}{\log 2}, & \text{if } t=1; \\ 1-\frac{\log \alpha}{\log 2}, & \text{if } t=3.
\end{cases}
\]
But, in any case, one can see that the corresponding value of $\epsilon$ from Lemma \ref{reduce} is always negative and therefore the reduction method is not useful for reducing the bound on $n$ in these instances. For this reason we need to treat these cases differently.
All we want to do here is solve the equations
\begin{equation} \label{eq:n-m=1,3(A)}
L_{m+1}+L_m=2^a \quad \text{and} \quad L_{m+3}+L_m=2^a
\end{equation}
in positive integers $m$ and $a$ with $m+1>200$ and $m+3>200$, respectively. But, by definition $L_{m+1}+L_m=L_{m+2}$. Moreover, $L_{m+3}+L_m=2L_{m+2}$, which is easily checked. We see from the above discussion that equations \eqref{eq:n-m=1,3(A)} are transformed into the simpler equations
\begin{equation} \label{eq:n-m=1,3(B)}
L_{m+2}=2^a \quad \text{and} \quad L_{m+2}=2^{a-1}
\end{equation}
to be resolved in positive integers $m$ and $a$ with $m>199$ and $m>197$, respectively. But, we quickly see that the above equations \eqref{eq:n-m=1,3(B)} have no solutions for $m>1$ as mentioned earlier. This completes the analysis of the cases when $n-m=1,3$ and therefore the proof of Theorem \ref{teo1}.
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\section{Acknowledgments}
We thank the referee for suggestions which improved the quality of this
paper. J.~J.~B.~was partially supported by University of Cauca and
F.~L.~was supported in part by Project PAPIIT IN104512, UNAM, Mexico.
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\noindent 2010 {\it Mathematics Subject Classification}: Primary 11B39; Secondary 11J86.
\noindent \emph{Keywords: } Fibonacci number, Lucas number, linear forms in logarithms, continued fraction, reduction method.
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\noindent (Concerned with sequences
\seqnum{A000032} and
\seqnum{A000045}.)
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\noindent
Received March 17 2014;
revised versions received July 22 2014; July 30 2014.
Published in {\it Journal of Integer Sequences}, July 30 2014.
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\noindent
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