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\begin{center}
\vskip 1cm{\LARGE\bf{On the Euler Function of Fibonacci Numbers}}
\vskip 1cm
\normalsize

Florian~Luca\\
Instituto de Matem{\'a}ticas\\
Universidad Nacional Autonoma de M{\'e}xico \\
C.P. 58089, Morelia, Michoac{\'a}n\\
M{\'e}xico\\
\href{mailto:fluca@matmor.unam.mx}{\tt fluca@matmor.unam.mx}\\
\ \\
V. Janitzio Mej\'{\i}a Huguet \\
Departamento de Ciencias B\'asicas\\
Universidad Aut\'onoma Metropolitana-Azcapotzalco \\
Av. San Pablo \#180 \\
Col. Reynosa Tamaulipas\\
C. P. 02200, Azcapotzalco DF\\
M{\'e}xico\\
\href{mailto:vjanitzio@gmail.com}{\tt vjanitzio@gmail.com}\\
\ \\
Florin Nicolae\\
Institut f\"ur Mathematik\\
Technische Universit\"at Berlin \\
MA 8-1, Strasse des 17 Juni 136 \\
D-10623 Berlin\\
Germany\\
\href{mailto:nicolae@math.tu-berlin.de}{\tt nicolae@math.tu-berlin.de}\\
and \\
Institute of Mathematics\\
Romanian Academy\\
P.O. Box 1-764, RO-014700, Bucharest\\
Romania\\
\end{center}

\vskip .2 in

\begin{abstract}
In this paper, we show that for any positive integer $k$, the set
$$
\left\{\left(\frac{\phi(F_{n+1})}{\phi(F_n)},\frac{\phi(F_{n+2})}{\phi(F_n)},\ldots,\frac{\phi(F_{n+k})}{\phi(F_n)}\right):n\ge
1\right\}
$$
is dense in $\R^{k}_{\ge 0}$, where $\phi(m)$ is the Euler function of
the positive integer $m$ and $F_n$ is the $n$th Fibonacci number.
\end{abstract}

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\section{Introduction}

Let $(F_n)_{n\ge 1}$ be the Fibonacci sequence given by $F_1=F_2=1$
and $F_{n+2}=F_{n+1}+F_n$ for all $n\ge 1$. Recently, there has been
some interest in investigating values of classical arithmetic
functions with Fibonacci numbers. For a positive integer $m$, we put
$\phi(m)$ for the Euler function of $m$ which counts the number of
positive integers less than or equal to $m$  and coprime to $m$, and
$\sigma(m)$ for the sum of divisors of $m$.

\medskip

Luca \cite{Luperf} proved that there is no perfect Fibonacci
number; that is, there is no positive integer $n$ such that
$\sigma(F_n)=2F_n$. Luca and Nicolae \cite{LN} proved that if
$\phi(F_n)=F_m$, then $n\in \{1,2,3,4\}$. Luca \cite{LuLeh} proved
that if $\phi(F_n)\mid F_n-1$, then $F_n$ is prime, confirming
in this way that a well-known conjecture of Lehmer holds for the
Fibonacci numbers.

\medskip

It follows from the results from Luca and Shparlinski \cite{LS} that the asymptotic
$$
\frac{1}{x}\sum_{n\le
x}\left(\frac{\phi(F_n)}{F_n}\right)^k=\Gamma_k+O_k\left(\frac{(\log
x)^k}{x}\right)
$$
holds for all positive integers $k$ as $x\to\infty$ with some
positive constant $\Gamma_k$. In particular, the function
$\phi(F_n)/F_n$ has a distribution function. That is, for every real
number $z$, the asymptotic density of the set of positive integers
$n$ such that $\phi(F_n)/F_n&lt;z$ exists. Luca \cite{LuS} proved
that if we put $M_n=2^n-1$, then the set $\{\phi(M_n)/M_n:n\ge 1\}$
is dense in $[0,1]$. In the same paper, it is said that the same
property holds true if one replaces the sequence of Mersenne numbers $(M_n)_{n\ge 0}$ by the sequence of Fibonacci numbers $(F_n)_{n\ge 0}$.

\medskip

In this paper, we revisit values of the Euler function with
Fibonacci numbers and prove the following result.

\begin{thm}
\label{thm:main} For every positive integer $k$, the set
\begin{equation}
\label{eq:1}
\left\{\left(\frac{\phi(F_{n+1})}{\phi(F_n)},\frac{\phi(F_{n+2})}{\phi(F_n)},
\ldots,\frac{\phi(F_{n+k})}{\phi(F_n)}\right):n\ge
1\right\}
\end{equation}
is dense in $\R^k_{\ge 0}$.
\end{thm}

Theorem \ref{thm:main} remains true when the Euler function $\phi$ is replaced by the sum of divisors function $\sigma$. It also remains true if the sequence of Fibonacci numbers $(F_n)_{n\ge 0}$ is replaced by other Lucas sequences such as the sequence of
Mersenne numbers $(M_n)_{n\ge 0}$. If the sequence of Fibonacci numbers $(F_n)_{n\ge 0}$ is replaced by the sequence of all natural numbers, then the statement 
that for every positive integer $k$, the set
$$
\left\{\left(\frac{\phi(n+1)}{\phi(n)},\frac{\phi(n+2)}{\phi(n)},
\ldots,\frac{\phi(n+k)}{\phi(n)}\right):n\ge
1\right\}
$$
is dense in $\R^k_{\ge 0}$ is a result of Schinzel of 1955 \cite{Sch}. More general forms of this result appear in a paper by Erd\H os and Schinzel of 1960 \cite{ES}, and in the 
recent work of Wong \cite{Wo}.

We point out that the similar looking statement that the set
$$
\left\{\left(\frac{\gamma(n+1)}{\gamma(n)},\ldots,\frac{\gamma(n+k)}{\gamma(n)}\right):n\ge
1\right\}
$$
is dense in $\R^k_{\ge 0}$, where $\gamma(m)=\prod_{p\mid m} p$ is
the square-free kernel of $m$, or the product of all the distinct
primes dividing $m$, was proved by Luca and Shparlinski \cite{LSShi}. 

We have the following immediate corollary of Theorem \ref{thm:main}.

\begin{cor}
\label{cor:cor}
For any positive integer $k$ and every permutation
$(i_1,\ldots,i_k)$ of $(1,\ldots,k)$ there exist infinitely many
positive integers $n$ such that
\begin{equation}
\label{eq:2} \phi(F_{n+i_1})&lt;\phi(F_{n+i_2})&lt;\cdots
&lt;\phi(F_{n+i_k}).
\end{equation}
\end{cor}

P.~Erd\H os, K.~Gy\H ory and Z.~Papp \cite{EGP} call two functions $f(n)$ and $g(n)$
defined on the set of positive integers to be {\it independent} if
for every $k\ge 1$ and permutations $(i_1,\ldots,i_k)$ and
$(j_1,\ldots,j_k)$ of $(1,\ldots,k)$, there exist infinitely many
$n$ such that
$$
f(n+i_1)&lt;f(n+i_2)&lt;\cdots&lt;f(n+i_k),
$$
while
$$
g(n+j_1)&lt;g(n+j_2)&lt;\cdots&lt;g(n+j_k).
$$
N.~ Doyon and  F. ~Luca \cite{DoLu} proved that the Euler function $\phi(n)$ and
the Carmichael function $\lambda(n)$ (also known as the universal
exponent modulo $n$) are independent, while Hernane and Luca \cite{LuMo} proved
that the two functions $\sigma(\phi(n))$ and $\phi(\sigma(n))$
are independent. In light of \eqref{eq:2}, it makes sense to ask if
we can pair the function $\phi(F_n)$ independently with some other
interesting multiplicative functions. Here are some questions which
are left for the reader.

\begin{prob}
\begin{itemize}
\ \\
\item[(i)] Are the functions $\phi(F_n)$ and $F_{\phi(n)}$
independent?
\item[(ii)] Are the functions $\phi(F_n)$ and $\phi(M_n)$
independent?
\end{itemize}
\end{prob}

\section{Notation}

We put $\alpha=(1+{\sqrt{5}})/2$ and $\beta=(1-{\sqrt{5}})/2$. We
let $(L_n)_{n\ge 1}$ be the Lucas companion of the Fibonacci
sequence given by $L_1=1,~L_2=3$ and $L_{n+2}=L_{n+1}+L_n$ for all
$n\ge 1$. It is well-known that $F_n$ and $L_n$ can be expressed in
terms of $\alpha$ and $\beta$ as
\begin{equation}
\label{eq:genterms}
F_n=\frac{\alpha^n-\beta^n}{\alpha-\beta}\qquad
{\text{\rm and}}\qquad L_n=\alpha^n+\beta^n\qquad {\text{\rm
for}}~n=1,2,\ldots.
\end{equation}
It is also well-known that $F_n$ and $L_n$ satisfy various relations
such as
\begin{equation}
\label{eq:relations} F_{2n}=F_nL_n\qquad {\text{\rm and }}\qquad
L_n^2-5F_n^2=4(-1)^n,
\end{equation}
which will be used in the sequel.

\medskip

Throughout this paper, we use the letters $p$ and $q$ with or
without subscripts for prime numbers and $\ell$, $m$ and $n$ for
positive integers. We write $\omega(n)$ for the number of distinct
prime factors of the positive integer $n$ and $p(n)$ for the
smallest prime factor of $n$. We write $z(n)$ for the order of
apparition of $n$ in the Fibonacci sequence, which is the smallest
positive integer $k$ such that $n\mid F_k$. It is well-known that
$z(n)$ exists and that $n\mid F_m$ if and only if $z(n)\mid m$.
Furthermore, if $p$ is an odd prime, then $p\mid F_{p-(5|p)}$, where
we write $(a|p)$ for the Legendre symbol of $a$ with respect to $p$.

\medskip

We write $\log x$ for the natural logarithm, and for an integer
$k\ge 1$ we write $\log_k x$ for the recursively defined function
given by $\log_1 x=\max\{1,\log x\}$ and $\log_k
x=\max\{1,\log(\log_{k-1} x)\}$ for $k\ge 2$. When $k=1$ we omit the
index of the logarithm. Hence, all logarithms which will appear are
$\ge 1$. We use the Vinogradov symbols $\ll,~\gg$ and the Landau
symbols $O$ and $o$ with their usual meanings. Recall that the
statements $A\ll B,~B\gg A$ and $A=O(B)$ are all equivalent to the
fact that the inequality $|A|&lt;cB$ holds with some positive constant
$c$, while $A=o(B)$ means that $A/B\to 0$. The constants implied by
these symbols may be absolute or may depend on some other parameters
such as $k$, or a fixed point ${\bf
\gamma}=(\gamma_1,\ldots,\gamma_k)\in \R_{&gt;0}^k$, etc. We use $x_0$
for a large positive real number. The meaning of ``large" might
change from a line to the next.

\section{Preliminary Results}

\begin{lem}
\label{lem:1}
Let $p\equiv 13\pmod {20}$ be a prime. Then  $p\mid
F_{(p+1)/2}$.
\end{lem}

\begin{proof}
This should be well-known but we supply a quick proof of it for
completeness. Since $p\equiv 3\pmod 5$, it follows that $(5|p)=-1$,
therefore $p\mid F_{p+1}=F_{(p+1)/2}L_{(p+1)/2}$. Suppose for a
contradiction that $p\nmid F_{(p+1)/2}$. Then $p\mid L_{(p+1)/2}$.
Reducing the relation
$$
L_{(p+1)/2}^2-5F_{(p+1)/2}^2=4(-1)^{(p+1)/2}=-4
$$
(see \eqref{eq:relations}) modulo $p$, we get $5F_{(p+1)/2}^2\equiv
4\pmod p$, leading to the conclusion that $(5|p)=1$, which is a
contradiction.
\end{proof}

For every positive integer $m$ let ${\mathcal P}_m=\{p:z(p)=m\}$.
The following lemma turns out to be useful.

\begin{lem}
\label{lem:2}
We have the estimate
\begin{equation}
\label{eq:primitiveprimes}
\sum_{p\in {\mathcal P}_m}\frac{1}{p}\ll
\frac{\log m}{m}.
\end{equation}
\end{lem}

\begin{proof}
We may assume that $m&gt;5$. Since $p\mid F_{p\pm 1}$, we get that
$p\equiv \pm 1\pmod m$ for all $p\in {\mathcal P}_m$. Thus,
$$
\alpha^m&gt;F_m\ge \prod_{p\in {\mathcal P}_m} p\ge (m-1)^{\#{\mathcal
P}_m},
$$
leading to $\#{\mathcal P}_m\ll  m/\log m$.

We split the desired sum at $m^2$. We have
\begin{equation}
\label{eq:sum}
\sum_{p\in {\mathcal P}_m}\frac{1}{p}\le
\sum_{\substack{p\in {\mathcal P}_m\\
p&lt;m^2}}\frac{1}{p}+\sum_{\substack{p\in {\mathcal P}_m\\
p&gt;m^2}}\frac{1}{p}:=S_1+S_2.
\end{equation}
Observe that if $p\in {\mathcal P}_m\cap (1,m^2)$, then $p=\pm
1+m\ell$ for some integer $\ell \in [1,m]$. Thus
\begin{equation}
\label{eq:S1}
 S_1\le \sum_{1\le \ell\le m}\left(\frac{1}{-1+m\ell}+\frac{1}{1+m\ell}\right)\ll
\frac{1}{m}\sum_{1\le \ell \le m}\frac{1}{\ell}\ll \frac{\log m}{m},
\end{equation}
As for $S_2$, we have that
\begin{equation}
\label{eq:S2} S_2\le \frac{\#{\mathcal P}_m}{m^2}\ll \frac{1}{m\log
m}.
\end{equation}
The desired conclusion now follows at once from estimates
\eqref{eq:sum}, \eqref{eq:S1} and \eqref{eq:S2}.
\end{proof}

\begin{lem}
\label{lem:3} We have the estimate
\begin{equation}
\label{eq:logs} \sum_{d|m}\frac{\log d}{d}\ll (\log_2 m)^2.
\end{equation}
\end{lem}

\begin{proof}
This is Lemma 2 in Luca \cite{LuS}.
\end{proof}

\begin{lem}
\label{lem:4} Let ${\mathcal P}$ be the set of all primes $p$ which
fulfill the following conditions:
\begin{itemize}
\item[(i)] $p\equiv 13\pmod {20}$;
\item[(ii)] $p((p+1)/2)&gt;\log_2 p$; (Here, we recall that $p(m)$ stands for the smallest prime factor of $m$.)
\item[(iii)] $p+1$ is square-free.
\end{itemize}
Let ${\mathcal P}(t)={\mathcal P}\cap (1,t)$. Then the estimate
$$
\#{\mathcal P}(t)\gg \frac{t}{\log t\log_3 t}
$$
holds.
\end{lem}

\begin{proof}
This follows by typographic changes from the proof of Lemma 1 in
Luca \cite{LuS}.
\end{proof}

\section{$\phi(F_n)/F_n$ is dense in $[0,1]$ revisited}

As we have mentioned in the Introduction, the fact that
$\{\phi(F_n)/F_n:n\ge 1\}$ is dense in $[0,1]$ alluded to in the
title of this section was already proved in Luca \cite{LuS}. Here, we
revisit that proof especially since that proof yields a bit more. We
follow the arguments from Luca \cite{LuS} although we slightly change
some of the parameters.

We let $x$ be a large positive real number, put $y=\log_2 x$,
$z=\log x$, and ${\mathcal P}(y,z)={\mathcal P}\cap (y,z)$. We let
$$
S=\sum_{p\in {\mathcal P}(y,z)}\frac{1}{p}.
$$
We first show that
\begin{equation}
\label{eq:S}
S&gt;2\log_4 x
\end{equation}
holds for $x&gt;x_0$. Indeed, observe that, by Abel's summation formula,
we have
\begin{equation*}
S=\sum_{\substack{p\in {\mathcal P}\\ y&lt;p&lt;z}}\frac{1}{p}=\int_y^z
\frac{d(\#{\mathcal P}(t))}{t}=\frac{\#{\mathcal
P}(t)}{t}\Big|_{t=y}^{t=z}+\int_y^z\frac{\#{\mathcal P}(t)}{t^2}.
\end{equation*}
Clearly,
$$
\Big|\frac{\#{\mathcal P}(t)}{t}\Big|_{t=y}^{t=z}\Big|\le
\frac{\pi(y)}{y}=O\left(\frac{1}{\log_3 x}\right).
$$
For large $x$, by Lemma \ref{lem:4}, we have that
$$
\#{\mathcal P}(t)\ge \frac{3 t}{\log t\log_2 t}.
$$
Thus, we get that
\begin{eqnarray*}
S &amp; \ge &amp; 3\int_y^z\frac{dt}{t\log t\log_2 t}+O\left(\frac{1}{\log_3
x}\right)
 =  3(\log_3 t)|_{t=y}^{t=z}+o(1)\\
&amp; = &amp; 3\log_4x -3\log_5 x+o(1) &gt;  2\log_4 x
\end{eqnarray*}
for $x&gt;x_0$.

Next let
$$
N=\lcm[(p+1)/2:p\in {\mathcal P}(y,z)].
$$
Observe that $N$ is square-free, so we can write it as
$N=q_1q_2\cdots q_T$, where $q_1&lt;q_2&lt;\cdots &lt;q_T$. Since
$q_T&lt;(z+1)/2$, we have, by the Prime Number Theorem, that $N\le
\prod_{q&lt;(z+1)/2} q&lt;\exp((0.5+o(1))z)$ as $x\to\infty$. In
particular, $N&lt;x$ for $x&gt;x_0$.


We next put $n_0=1$ and $n_{j}=\prod_{i\le j} q_i$ for
$j=1,\ldots,T$. Let
$$
s_j=\frac{\phi(F_{n_j})}{F_{n_j}}.
$$
Since $n_0\mid n_1\mid n_2\mid \cdots \mid n_T$, we have that
$$
s_0\ge s_1\ge \cdots \ge s_T.
$$
Obviously, $s_0=1$. We now look at $s_T$. Observe that, by Lemma
\ref{lem:1} and the fact that $F_a\mid F_b$ whenever $a\mid b$, we
get that
$$
p\mid F_{(p+1)/2}\mid F_N=F_{n_T}\qquad {\text{\rm for~all}}~p\in
{\mathcal P}(y,z).
$$
Thus,
\begin{eqnarray*}
s_T &amp; = &amp; \frac{\phi(F_{n_T})}{F_{n_T}}=\prod_{p\mid
F_{n_T}}\left(1-\frac{1}{p}\right)\le \prod_{p\in {\mathcal
P}(y,z)}\left(1-\frac{1}{p}\right)\\
&amp; = &amp; \exp\left(-\sum_{p\in {\mathcal
P}(y,z)}\frac{1}{p}+O\left(\sum_{p}\frac{1}{p^2}\right)\right)\\
&amp; = &amp; \exp(-S+O(1))&lt; \exp(-2\log_4 x+O(1))  &lt;  \frac{1}{\log_3 x}
\end{eqnarray*}
for $x&gt;x_0$, where the last inequality follows from \eqref{eq:S}.

Next we show that
\begin{equation}
\label{eq:ratios} \frac{s_{j+1}}{s_j}=1+O\left(\frac{(\log_5
x)^2}{\log_4 x}\right)\qquad {\text{\rm
holds~for~all}}~j=0,1,\ldots,T-1.
\end{equation}
First of all notice that every prime factor $p$ of $F_{n_j}$ for
some $j\ge 1$ has the property that $z(p)\mid n_j$. In particular,
$z(p)&gt;\log_2 y=\log_3 z$. Since also $p\equiv \pm 1 \pmod {z(p)}$, we get
that $p\ge (\log_3 z)-1$.

Thus,
\begin{eqnarray}
\label{eq:cons}
\frac{s_{j+1}}{s_j} &amp; = &amp; \prod_{\substack{p\mid F_{n_{j+1}}\\
p\nmid F_{n_j}}}\left(1-\frac{1}{p}\right)=\prod_{\substack{z(p)\mid
n_{j+1}\\ z(p)\nmid n_j}}\left(1-\frac{1}{p}\right)
 =  \prod_{d\mid n_j}\prod_{p\in {\mathcal
P}_{dq_{j+1}}}\left(1-\frac{1}{p}\right)\nonumber\\
&amp; = &amp; \exp\left(-\sum_{d\mid n_j}\sum_{p\in {\mathcal
P}_{dq_{j+1}}}\frac{1}{p}+O\left(\sum_{p\ge
(\log_3 z)-1}\frac{1}{p^2}\right)\right)\nonumber\\
&amp; = &amp; \exp\left(O\left(\sum_{d\mid n_j}
\frac{\log(dq_{j+1})}{dq_{j+1}}+\frac{1}{\log_3 z}\right)\right),
\end{eqnarray}
where in the last estimate above we used Lemma \ref{lem:2}. Now note
that
\begin{eqnarray}
\label{eq:interm} \sum_{d\mid n_j} \frac{\log (dq_{j+1})}{dq_{j+1}}
&amp; = &amp; \frac{\log q_{j+1}}{q_{j+1}}\sum_{d\mid
n_j}\frac{1}{d}+\frac{1}{q_{j+1}}\sum_{d\mid n_j}\frac{\log
d}{d}\nonumber\\
&amp; \ll &amp; \frac{1}{q_{j+1}}\left(\log q_{j+1}\log_2 n_j+(\log_2
n_j)^2\right),
\end{eqnarray}
where here we used Lemma \ref{lem:3} together with the well-known
estimate
$$
\sum_{d\mid m}\frac{1}{d}=\frac{\sigma(m)}{m}\ll \log_2 m
$$
with $m=n_j$. Since the primes $q_1,\ldots,q_{j+1}$ are arranged
increasingly, we have that
$$
n_j\le \prod_{q&lt;q_{j+1}} q&lt;\exp(2q_{j+1})
$$
for $x&gt;x_0$, therefore $\log_2 n_j\ll \log q_{j+1}$. This last
inequality together with inequality \eqref{eq:interm} yields
$$
\sum_{d\mid n_j}\frac{\log(dq_{j+1})}{dq_{j+1}}\ll \frac{(\log
q_{j+1})^2}{q_{j+1}}\ll \frac{(\log_3 y)^2}{\log_2 y}=\frac{(\log_4 z)^2}{\log_3 z}.
$$
Inserting the above inequality into inequality \eqref{eq:cons}, we
get that
$$
\frac{s_{j+1}}{s_j}=\exp\left(O\left(\frac{(\log_4 z)^2}{\log_3
z}\right)\right)=1+O\left(\frac{(\log_4 z)^2}{\log_3
z}\right)=1+O\left(\frac{(\log_5 x)^2}{\log_4 x}\right),
$$
which is precisely inequality \eqref{eq:ratios}.

Now it is easy to see that $\{\phi(F_n)/F_n:n\ge 1\}$ is dense in
$[0,1]$ in the following way. Let $\gamma\in (0,1)$ and
$\varepsilon&gt;0$ arbitrary. Let $j\in \{0,\ldots,T\}$ be maximal such
that $s_j&gt;\gamma$. Since $s_T=o(1)$ for large $x$, it follows that
$j\ne T$. Let $j$ be maximal with this property. Then $s_{j+1}\le
\gamma$. However,
$$
\gamma\ge s_{j+1}=s_j\left(1+O\left(\frac{(\log_5 x)^2}{\log_4
x}\right)\right)&gt; s_j-\frac{c(\log_5 x)^2}{\log_4 x},
$$
where $c&gt;0$ is some absolute constant. This shows that
$$
\frac{\phi(F_{n_j})}{F_{n_j}}=s_j\in
\left(\gamma,\gamma+c\frac{(\log_5 x)^2}{\log_4 x}\right).
$$
Choosing $x$ large enough such that
$$
\frac{c(\log_5 x)^2}{\log_4
x}&lt;\varepsilon,
$$
we get that $\phi(F_{n_j})/F_{n_j}\in
(\gamma,\gamma+\varepsilon)$.

This certainly implies that the sequence $(\phi(F_n)/F_n)_{n\ge 1}$
is dense in $(0,1)$.

However, notice that we have proved the following stronger
statement.

\begin{lem}
\label{lem:6} Let $\gamma\in (0,1)$. Then for $x&gt;x_0$, there exists
a square-free $n&lt;x$ such that $p(n)&gt;\log_2 x$ and such that
furthermore
$$
\frac{\phi(F_n)}{F_n}\in \left(\gamma,\gamma+\frac{(\log_5
x)^3}{\log_4 x}\right).
$$
\end{lem}

Now let $i\ge 1$ be fixed and write $\alpha_i=\alpha^i$ and
$\beta_i=\beta^i$. Look at the sequence
$$
F_n^{(i)}=\frac{\alpha_i^n-\beta_i^n}{\alpha_i-\beta_i}\qquad
{\text{\rm for~all}}~n=1,2,\ldots.
$$
Observe that
$$
F_{n}^{(i)}=\frac{\alpha^{in}-\beta^{in}}{\alpha^i-\beta^i}=\frac{F_{ni}}{F_i}.
$$
The sequence $(F_{n}^{(i)})_{n\ge 1}$ shares the same divisibility
properties of the sequence of Fibonacci numbers. We record two of
them.

\begin{lem}
\label{lem:7} Let $i\ge 1$ be fixed.
\begin{itemize}
\item[(i)] If $a\mid b$, then $F_a^{(i)}\mid F_b^{(i)}$.
\item[(ii)] If $p$ is a prime dividing both $F_i$ and $F_a^{(i)}$, then $p$ divides $a$.
\item [(ii)] If $p\equiv 13\pmod {20}$ and $i$ and
$(p+1)/2$ are coprime, then $p\mid F_{(p+1)/2}^{(i)}$.
\end{itemize}
\end{lem}

\begin{proof}
The first one is obvious and the second one is well-known and easy to prove. To see the third one, note that since
$p\equiv 13\pmod {20}$, it follows, by Lemma \ref{lem:1}, that
$p\mid F_{(p+1)/2}\mid F_{i(p+1)/2}$. Since $i$ and $(p+1)/2$ are
coprime and $z(p)\mid (p+1)/2$, it follows, in particular, that
$z(p)$ does not divide $i$. Thus, $F_i$ is not a multiple of $p$, so
$$
p\mid \frac{F_{i(p+1)/2}}{F_i}=F_{(p+1)/2}^{(i)}.
$$
\end{proof}

Now the arguments that proved Lemma \ref{lem:6} also prove the
following result.

\begin{lem}
\label{lem:8} Let $i\ge 1$ be fixed and $\gamma\in (0,1)$. There
exists $x_0(i)$ depending on $i$ such that for $x&gt;x_0(i)$ there
exists a square-free $n&lt;x$ with $p(n)&gt;\log_2 x$ and
$$
\frac{\phi(F_n^{(i)})}{F_n^{(i)}}\in
\left(\gamma,\gamma+\frac{(\log_5 x)^3}{\log_4 x}\right).
$$
\end{lem}

\section{The Proof of Theorem \ref{thm:main}}

We may assume that $k\ge 5$. In particular, $F_{K}\ge K$ for all
$K\ge k$.

We write $f(n)=\phi(n)/n$. First we make some reductions. Observe
that
$$
\frac{\phi(F_{n+i})}{\phi(F_n)}=\frac{f(F_{n+i})}{f(F_n)}\cdot\frac{F_{n+i}}{F_n},
$$
and
$$
\frac{F_{n+i}}{F_n}=\alpha^i(1+o(1))\qquad {\text{\rm
as}}~n\to\infty.
$$
Thus, in order to prove that the set of vectors shown in the
statement of Theorem \ref{thm:main} is dense in $\R_{\ge 0}^k$, it
suffices to show that the set of vectors
\begin{equation}
\label{eq:newset}
\left\{\left(\frac{f(F_{n+1})}{f(F_{n})},\frac{f(F_{n+2})}{f(F_n)},\cdots,\frac{f(F_{n+k})}{f(F_n)}\right):n\ge
1\right\}
\end{equation}
is dense in $\R_{\ge 0}^k$.

Next let $K=((k+1)!)^2$ and choose $n=Kn_0$ with some positive integer
$n_0$. Then
$$
n+i=Kn_0+i=i\left(\frac{K}{i}n_0+1\right):=in_i\qquad {\text{\rm
for~all}}~i=1,\ldots,k.
$$
Furthermore, note that the numbers $n_i$ are coprime with all primes
$p\le k+1$ for $i=1,\ldots,k$. That is, $p(n_i)&gt;k+1$ for $i=1,\ldots,k$.
Assume in fact more, namely that $p(n_i)&gt;F_k$ for all
$i=1,\ldots,k$. Then
$$
F_{n+i}=F_{in_i}=F_i\left(\frac{F_{in_i}}{F_i}\right)=F_iF_{n_i}^{(i)}
$$
for all $i=1,\ldots,k$. By (ii) of Lemma \ref{lem:7}, the prime factors of
$\gcd(F_i,F_{n_i}^{(i)})$ divide $n_i$. Since $p(n_i)&gt;F_k$ for all
$i=1,\ldots,k$, it follows that $F_i$ and $F_{n_i}^{(i)}$ are
coprime. Thus, by the multiplicativity of $f(n)$, we get that
$$
f(F_{n+i})=f(F_i)f(F_{n_i}^{(i)})\qquad {\text{\rm
for~all}}~i=1,\ldots,k.
$$
Similarly, assuming that $p(n_0)&gt;F_K$, we get that
$F_{n}=F_KF_{n_0}^{(K)}$ and
$$
f(F_n)=f(F_K)f(F_{n_0}^{(K)}).
$$
With these conventions, the set shown at \eqref{eq:newset} will be
dense in $\R_{\ge 0}^k$ provided that the set of vectors
\begin{equation}
\label{eq:set3}
\left\{\left(\frac{f(F_{n_1})}{f(F_{n_0}^{(K)})},\frac{f(F_{n_2}^{(2)})}{f(F_{n_0}^{(K)})},
\ldots,\frac{f(F_{n_k}^{(k)})}{f(F_{n_0}^{(K)})}\right)~:~p(n_0\cdots
n_k)&gt;F_K\right\}
\end{equation}
is dense in $\R_{\ge 0}^k$.

Now let ${\bf \gamma}=(\gamma_1,\ldots,\gamma_k)\in {\bf R}_{&gt;0}^k$.
Write ${\bf \gamma}=(\delta_1/\delta_0,\ldots,\delta_k/\delta_0)$,
where $\delta_i\in (0,1)$ for $i=1,\ldots,k$. It suffices to take
$\delta_0=1/(2\Gamma)$, where
$\Gamma=\max\{1,\gamma_1,\ldots,\gamma_k\}$ and
$\delta_i=\gamma_i\delta_0$ for $i=1,\ldots,k$. In order to prove
that the set shown at \eqref{eq:set3} is dense in $\R_{&gt;0}^k$, it
suffices to exhibit an infinite sequence ${\mathcal N}$ of $n$'s
such that
$$
\left(f(F_{n_0}^{(K)}),f(F_{n_1}),f(F_{n_2}^{(2)}),\ldots,f(F_{n_k}^{(k)})\right)\to
(\delta_0,\delta_1,\ldots,\delta_k)
$$
as $n\to\infty$ in the sequence ${\mathcal N}$.

We shall now explain how to do this. We next fix some arbitrary
$\varepsilon&gt;0$, and recursively choose positive square-free
integers $m_0,m_1\ldots,m_k$ satisfying the following properties:
\begin{itemize}
\item[(i)] $p(m_0\cdots m_k)&gt;F_K$;
\item[(ii)]
$$
f(F_{m_0}^{(K)})\in (\delta_0,\delta_0+\varepsilon),~~ f(F_{m_1})\in
(\delta_1,\delta_1+\varepsilon),~~\ldots~~ f(F_{m_k}^{(k)})\in
(\delta_k,\delta_k+\varepsilon);
$$
\item[(iii)] $\gcd(m_i,m_j)=1$ for all $0\le i&lt;j\le k$.
\end{itemize}
To do this, we proceed as follows. We take a large $x&gt;x_0(K)$, such that
both inequalities $\log_2 x&gt;F_K$ and $(\log_5 x)^3/(\log_4 x)&lt;\varepsilon$ hold, and choose
a square-free positive integer $m_0&lt;x$ as in Lemma \ref{lem:8} with
$i=K$ and $\gamma=\delta_0$. Such $m_0$ will satisfy $p(m_0)&gt;\log_2
x$ and
$$
f(F_{m_0}^{(K)})\in \left(\delta_0,\delta_0+\frac{(\log_5
x)^3}{\log_4 x}\right)\subset (\delta_0,\delta_0+\varepsilon).
$$
After having chosen $m_0$, we fix it and take a new $x$ which is
larger and such that $x&gt;x_0(1)$ and $\log_2 x&gt;\max\{F_K,m_0\}$.
Applying now Lemma \ref{lem:8} with $i=1$ and $\gamma=\delta_1$, we
get some square-free positive integer $m_1&lt;x$ with $p(m_1)&gt;\log_2
x&gt;\max\{F_K,m_0\}$ such that
$$
f(F_{m_1}^{(1)})\in (\delta_1,\delta_1+\varepsilon).
$$
Note that $m_1$ is coprime to $m_0$ because $p(m_1)&gt;m_0$. Now we
take a new $x$, perhaps larger, such that $x&gt;x_0(2)$ and $\log_2
x&gt;\max\{F_K,m_0,m_1\}$, and apply Lemma \ref{lem:8} with $i=2$ and
$\delta=\delta_2$ to get some square-free positive integer $m_2&lt;x$
such that $p(m_2)&gt;\log_2 x&gt;\max\{F_K,m_0,m_1\}$ and
$$
f(F_{m_2}^{(2)})\in (\delta_2,\delta_2+\varepsilon).
$$
Observe that $m_2$ is coprime to both $m_0$ and $m_1$ since
$p(m_2)&gt;\max\{m_0,m_1\}$. Proceeding in this way $k$ times, we end
up with some square-free positive integers $m_0,\ldots,m_k$
satisfying the above three properties.


We now fix these numbers $m_0,\ldots,m_k$. Throughout the remaining
of this section, the constants implied by the symbols $\gg,~\ll$ and
$O$ will depend on $k,~{\bf \gamma}$ and $m_0,\ldots,m_k$ (but not
on $\varepsilon$).


We start by searching for $n_0$ satisfying the following congruences
$$
n_0\equiv 0\pmod {m_0}\quad {\text{\rm and}}\quad (K/i)n_0+1\equiv
0\pmod  {m_i}\quad {\text{\rm for}}~i=1,\ldots,k.
$$
Since the largest prime factor of $K$ is $\le k+1\le K\le F_K&lt;p(m_i)$
for all $i=1,\ldots,k$, it follows that $K/i$ is invertible modulo
$m_i$. In particular, the congruence $(K/i)n_0+1\equiv 0\pmod {m_i}$
is solvable for $n_0$ and puts $n_0$ into a certain congruence class
$N_{0,i}\pmod {m_i}$. Since $m_0,\ldots,m_k$ are pairwise coprime,
the above system is solvable by the Chinese Remainder Theorem and
puts $n_0$ into a certain progression $N_0\pmod M$, where
$M=m_0\cdots m_k$. We assume that $N_0$ is the smallest positive
integer in this progression. Write
$$
n_0=N_0+M\ell.
$$
Then
\begin{eqnarray}
\label{eq:forms}
n_0 &amp; = &amp; m_0\left(\frac{N_0}{m_0}+\left(\frac{M}{m_0}\right)\ell\right);\\
(K/i)n_0+1&amp; = &amp;
m_i\left(\frac{(K/i)N_0+1}{m_i}+(K/i)\left(\frac{M}{m_i}\right)\ell\right)\quad
{\text{\rm for}}~i=1,\ldots,k.
\end{eqnarray}
Put $A_0={N_0}/m_0,~B_0=M/m_0$ and for $i=1,\ldots,k$ put
$A_i=((K/i)N_0+1)/m_i$ and $B_i=(K/i)M/m_i$. It is easy to check that
$\gcd(A_i,B_i)=1$ for all $0,\ldots,k$. 

Indeed, say that $p\mid \gcd(A_0,B_0)$ for some prime $p$. Then
$p\mid m_i$ for some $i\in \{1,\ldots,k\}$, so that $p\mid
(K/i)n_0+1$. However, we also have that $p\mid n_0$, and so $p\mid
1=((K/i)n_0+1)-(K/i)n_0$, which is impossible.

Assume now that $p\mid \gcd(A_i,B_i)$ for some $i\in
\{1,\ldots,k\}$. If $p\mid K/i$, then $p\le k+1$, but such primes cannot divide $A_i$.
Thus, $p\mid M/m_i$, and, in particular, $p\mid m_j$ for some $j\ne i$. But
then $p\mid (K/i)n_0+1$ and $p\mid (K/j)n_0+1$, therefore $p\mid
n_0(K/i-K/j)=n_0K(j-i)/ij$. Clearly, $p\nmid n_0$ since $p\mid
(K/i)n_0+1$. Thus, $p\mid K(i-j)/ij$, and so, in particular, $p\le
k+1$. However, $p(m_j)&gt;F_K\ge K&gt;k+1$, and we obtained a contradiction.

We next see that
$$
\Delta=\prod_{0\le i&lt;j\le k} (A_iB_j-A_jB_i)\ne 0.
$$
Indeed, if $\Delta=0$, then $A_iB_j=A_jB_i$ holds for some $0\le
i&lt;j\le k$. Thus, $A_i/B_i=A_j/B_j$, and since both fractions are
reduced and with positive denominators, we must have $B_i=B_j$. Since all prime factors of $m_i$ are $&gt;K$,  this implies $m_i=m_j$, which is impossible.

Thus, we have that
$$
n_0=m_0(A_0+B_0\ell)\quad {\text{\rm and}}\quad
(K/i)n_0+1=m_i(A_i+B_i\ell)\quad {\text{\rm for}}~i=1,\ldots,k,
$$
and the linear forms $A_i+B_i\ell$ are primitive and distinct for
$i=0,\ldots,k$. Next we show that for each prime $p$, there is $\ell$ such that none of the above $k+1$ forms evaluated in $\ell$ is zero modulo $p$. Well, if $p\le k+1$, this follows because $A_i+B_i\ell\equiv A_i\not\equiv 0\pmod p$ for all $i=0,\ldots,k$, while if $p&gt;k+1$, then since the degree $k+1$ polynomial
$$
\prod_{i=0}^{k} (A_i+B_ix)
$$
is not the zero polynomial modulo $p$, there must exists a congruence class $x\equiv \ell \pmod p$ which is not a zero of the above polynomial. Under these conditions, it is known from the Brun sieve (see, for example,
Theorem 2.6 on page 85 in H.~Halberstam and H.~E.~Richert \cite{HR}), that there exists a constant
$c&gt;0$ depending on $k$ and the numbers $m_0,\ldots,m_k$ such that if
$x&gt;x_0$ (depending on $m_0,\ldots,m_k$), then there are $\gg x/(\log
x)^{k+1}$ positive integers $\ell&lt;x$ such that
$$
p(A_i+B_i\ell)&gt;x^{c}\qquad {\text{\rm for~all}}~i=0,\ldots,k.
$$
In particular, $\omega(A_i+B_i\ell)= O(1)$ holds for all
$i=0,\ldots,k$, once $x$ sufficiently large. We now take such an
$\ell$. Observe that
$$
F_{n_0}^{(K)}=\frac{F_{Kn_0}}{F_K}=\frac{F_{Km_0}}{F_K}\cdot
\frac{F_{Km_0(A_0+B_0\ell)}}{F_{Km_0}}.
$$
The first factor on the left is $F_{m_0}^{(K)}$. If there is some
prime number dividing both $F_{Km_0}$ and $F_{Km_0(A_0+B_0\ell)}/F_{Km_0}$,
then this prime must also divide $A_0+B_0\ell$. This is false for large
$x$ since $p(A_0+B_0\ell)&gt;x^c$ exceeds $F_{Km_0}$. Thus,
$$
f(F_{n_0}^{(K)})=f(F_{m_0}^{(K)})f\left(\frac{F_{Km_0(A_0+B_0\ell)}}{F_{Km_0}}\right).
$$
Let us look at the primes $p\mid F_{Km_0(A_0+B_0\ell)}/F_{Km_0}$.
Such primes have an index of apparition $z(p)$ dividing
$Km_0(A_0+B_0\ell)$ but not $Km_0$. In particular, $z(p)\ge x^c$,
therefore $p\ge x^{c}-1$. Now by an argument already used in the
proof of Lemma \ref{lem:6} and based on Lemma \ref{lem:2}, we have
that
\begin{eqnarray*}
f\left(\frac{F_{Km_0(A_0+B_0\ell)}}{F_{Km_0}}\right) &amp; = &amp;
\prod_{\substack{z(p)\mid Km_0(A_0+B_0\ell)\\ z(p)\nmid
Km_0}}\left(1-\frac{1}{p}\right)\\
&amp; = &amp; \exp\left(O\left(\sum_{\substack{d\mid A_0+B_0\ell\\
d&gt;1}}\sum_{d_1\mid
Km_0}\frac{\log(dd_1)}{dd_1}+\sum_{p&gt;x^{c}-1}\frac{1}{p^2}\right)\right).
\end{eqnarray*}
Now $Km_0$ has $O(1)$ divisors. Furthermore, since
$p(A_0+B_0\ell)&gt;x^{c}$, the number $A_0+B_0\ell$ also has $O(1)$
divisors $&gt;1$, the smallest one being $\ge x^{c}$. Putting all this
together, we get that
$$
\sum_{\substack{d\mid A_0+B_0\ell\\ d&gt;1}}\sum_{d_1\mid
Km_0}\frac{\log(dd_1)}{dd_1}\ll \frac{\log x}{x^{c}}=o(1),
$$
while also
$$
\sum_{p&gt;x^{c}-1}\frac{1}{p^2}\ll \frac{1}{x^c}=o(1),
$$
both when $x\to\infty$. Thus, we have just showed that
$$
f(F_{n_0}^{(K)})=f(F_{m_0}^{(K)})(1+o(1))=\delta_0+\varepsilon+o(1)
$$
as $x\to\infty$. In the same way, one proves that
$$
f(F_{n_i}^{(i)})=f(F_{m_i}^{(i)})(1+o(1))=\delta_i+\varepsilon+o(1)\
\qquad {\text{\rm for~all}}~i=1,\ldots,k,
$$
as $x\to\infty$. Thus, choosing a large $x$, we get an $n$ such that
$$
f(F_{n_0}^{(K)})\in
(\delta_0-2\varepsilon,\delta_0+2\varepsilon)\quad {\text{\rm
and}}\quad f(F_{n_i}^{(i)})\in
(\delta_i-2\varepsilon,\delta_i+2\varepsilon)\quad {\text{\rm
for~all}}~i=1,\ldots,k.
$$
Since $\varepsilon&gt;0$ was arbitrary, the result follows.



\section{Comments and Remarks}

As we have already pointed out in the Introduction, typographical changes, such as working with primes $p\equiv 7\pmod
8$ and with $(p-1)/2$ instead of $(p+1)/2$ show that the conclusion
of Theorem \ref{thm:main} remains valid when the Fibonacci numbers
$F_n$ are replaced by the Mersenne numbers $M_n$ (see Luca \cite{LuS}).
Quite likely, it also applies to other Lucas sequences of the first
kind not only to the Fibonacci and Mersenne numbers. Furthermore,
the conclusion of the theorem remains also valid when the Euler
function $\phi(n)$ is replaced by the sum of divisors function
$\sigma(n)$, or, more generally, by any multiplicative function
$g(n)$ such that there exist two constants $c\ne 0$ and $\lambda&gt;1$
with the property that on prime powers $p^a$ we have
$g(p^a)=1+c/p+O(1/p^{\lambda})$. An example of such a function is
the function $\alpha(n)$ which associates to every positive integer
$n$ the average order of the elements in the cyclic group of order
$n$. Such function was studied in Luca \cite{LuR}. On the other hand, it
seems very difficult to prove the analogous Theorem \ref{thm:main}
when the function $\phi(n)$ is replaced by one of the functions
$\omega(n)$, or $\Omega(n)$, which count the number of prime power
divisors of $n$, or $\tau(n)$, which counts the number of divisors
of $n$. We leave such problems as possible research projects for the
interested reader.


\section{Acknowledgments}

We thank the anonymous referee for pointing out various inaccuracies in a previous version of this manuscript. This work was done during a visit of
V.~J.~M.~H.~ and F.~N. at the Mathematical Institute of the UNAM in
Morelia, Mexico. During the preparation of this paper, F.~L. was
supported in part by Grants SEP-CONACyT 79685 and PAPIIT 100508 and
V.~J.~M.~H. was supported by Grant UAM-A 2232508 and a Postdoctoral Position at the IFM of UMSNH. He thanks the people of these institutions
for their hospitality.


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\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11A25; Secondary 11B39.

\noindent \emph{Keywords: } Fibonacci numbers, Euler function,
sieve methods.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequence
\seqnum{A000045}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received May 20 2009;
revised version received  September 22 2009;
Published in {\it Journal of Integer Sequences}, September 22 2009.

\bigskip
\hrule
\bigskip

\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
\vskip .1in


\end{document}

                                                                                


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