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\begin{document}
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\begin{center}
\vskip 1cm{\LARGE\bf Congruences Involving Sums of Ratios of \\
\vskip .1in
Lucas Sequences
}
\vskip 1cm
\large
Evis Ieronymou \\
Department of Mathematics and Statistics\\
University of Cyprus \\
1687 Nicosia \\
Cyprus \\
\href{mailto:ieronymou.evis@ucy.ac.cy}{\tt ieronymou.evis@ucy.ac.cy}
\end{center}
\vskip .2 in
\begin{abstract} Given a pair $(U_t)$ and $(V_t)$ of Lucas sequences,
we establish various congruences involving sums of ratios $\frac{V_t}{U_t}$. More precisely, let $p$ be a prime divisor of the positive integer $m$. We establish congruences, modulo powers of $p$, for the sum $\sum \frac{V_t}{U_t}$, where $t$ runs from $1$ to $r(m)$, the rank of $m$, and $r(q) \nmid t$ for all prime factors $q$ of $m$.
\end{abstract}
\section{Introduction}
Wolstenholme's classic congruence dating back to $1862$ is the following:
$$
1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{p-1} \equiv 0 \pmod {p^2},
$$
where $p \geq 5$ is a prime number. Kimball and Webb \cite{Kim} found a generalization of the above congruence using Lucas sequences. In order to state it we need to recall some basic definitions and terminology:
The pair of Lucas sequences $(U_t)$ and $(V_t)$ associated with a pair of coprime integers $P$ and $Q$ is given by the second-order linear recurrence
$$
X_{n+2}=PX_{n+1}-Q X_n,
$$
and initial conditions $U_0=0$, $U_1=1$, $V_0=2$, $V_1=P$. We set $D=P^2-4Q$. Given an integer $m$ let $r(m)$ denote, if it exists, the least positive integer such that $m$ divides $U_{r(m)}$ and call it the rank of appearance (or apparition) of $m$. For a prime $p$ with $\gcd(p,2Q)=1$ we know that $r(p)$ divides $p- \left( \frac{D}{p} \right)$, where $\left(\frac{D}{p} \right)$ is the Legendre symbol.
A prime is called of maximal rank if $r(p)=p-\left( \frac{D}{p} \right)$.
Kimball and Webb's congruence is the following:
$$
v_p\left(\sum_{t=1}^{r(p)-1}\frac{V_t}{U_t} \right)\geq 2,
$$
when $p\geq 5$ is a prime of maximal rank. In the above inequality and in the rest of this note $v_p(x)$ denotes the standard $p$-adic valuation of the rational number $x$, where $p$ is a prime number.
Note that for $P=2, \ Q=1$ we recover Wolstenholme's congruence.
In turn, Ballot \cite{Ball} generalized Kimball and Webb's congruence in the following:
\begin{proposition}
Let $(U_t)$ and $(V_t)$ be a pair of Lucas sequences.
If $m$ is of maximal rank and $\gcd(m,6Q)$=1 then
$$v_q\left(\sum_{t\in I_m}\frac{V_t}{U_t}\right)\geq 2v_q(m),$$
where $q$ is a prime divisor of $m$ and
$$
I_m=\{n: 1\leq n \leq r(m)\}-\bigcup_{\substack{p \ \textrm{is prime}\\ p \, | \, m}}\Z \cdot r(p).
$$
\end{proposition}
\begin{remark} Ballot \cite{Ball} proves more, most notably the condition that $\gcd(m,6)=1$ is not there and a complete list of what happens in those cases is given.
The reason we will assume that $\gcd(m,6)=1$ in this note will be explained later in the introduction.
\end{remark}
The notion of maximal rank for composite $m$ was introduced in \cite{Ball} and is the following: We say that $m$ has maximal rank if for any two prime divisors $p,q$ of $m$ with $a=v_p(m)$ and $b=v_q(m)$ we have that
(i) $p$ is of maximal rank, (ii) $r(p^{a})=p^{a-1}r(p)$ and (iii) $\gcd(r(p^{a}),r(q^{b}))=1$ for $p\neq q$.
This notion has the obvious nuisance that it excludes many $m$. For example, if $m$ is of maximal rank then it can have at most one prime divisor outside the prime divisors of $2D$ and moreover all its prime divisors must themselves be of maximal rank. A natural thing to ask is whether we can say anything definite with a weaker notion. The aim of this note is to clarify the situation.
Our first main result tells us what happens for prime powers.
\begin{theorem}\label{bth1}
Let $(U_t)$ and $(V_t)$ be a pair of Lucas sequences and let $p$ be a prime with $\gcd(p,6Q)=1$ and $v_p(U_{r(p)})=1$.
If $p$ has maximal rank then
$$v_p\left(\sum_{t\in I_{p^{a}}}\frac{V_t}{U_t}\right)\geq 2a.$$
If $p$ does not have maximal rank then
$$ v_p\left(\sum_{t\in I_{p^{a}} }\frac{V_t}{U_t}\right)=2a-1.$$
\end{theorem}
The new thing in the above theorem is the inclusion of the case when $p$ does not have maximal rank. It is crucial in the consideration of integers with more than one prime divisor.
\begin{example}
We consider the Fibonacci and Lucas numbers, i.e., the pair of Lucas sequences associated to $P=1$ and $Q=-1$.
$\bullet$ For $m=13^3$ we have $r(13)=7$ and so
$$S\equiv 0 \pmod {13^5} ,\ \ \ S\not\equiv 0 \pmod {13^6},$$
where
$$S=\sum_{\substack{t=1\\ \gcd(t,7)=1}}^{7\cdot 13^2} \frac{L_t}{F_t}.$$
\end{example}
The second main result is concerned with integers with more than one prime divisor.
\begin{theorem}\label{bth2}
Let $m=p^{a}\prod_{i=1}^{\tau} q_i^{a_i}$ be the prime factorization of
$m$.
Suppose that $\gcd(p,6Q)$=1 and $v_p(U_{r(p)})=1$.
Let $r=r(p)$ and $r_i=r(q_i)$.
\begin{enumerate}
\item We have that
$$
v_p\left(\sum_{t\in I_m}\frac{V_t}{U_t} \right)\geq \max(a,\mu + \epsilon),
$$
where $\mu= \max \{ v_p(r_i)\}$ and $\epsilon=\left(\frac{D}{p} \right)^2$.
\item
Suppose that either $\max \{ v_p(r_i) : r \nmid r_i\}=0$ or $\{ r_i : r \nmid r_i\}=\emptyset$. Then
$$
v_p\left(\sum_{t\in I_m}\frac{V_t}{U_t}\right)\geq 2a-1.
$$
If moreover $p$ has maximal rank and $\gcd(r,r_i)=1$ for all $r_i$ with $r\nmid r_i$ the above inequality is strict.
\end{enumerate}
\end{theorem}
Note that we recover Ballot's result stated at the beginning as a special case.
Below we give examples not covered by that.
\begin{example}
We continue with the Fibonacci and Lucas numbers. In each case $S=\sum_{I_m} \frac{L_t}{F_t}$.
$\bullet$ For $m=7^4\cdot 11^3$ we have $r(7)=8$, $r(11)=10$ and so
$$S\equiv 0 \pmod {7^7\cdot 11^5}.$$
$\bullet$ For $m=7 \cdot 17^2$ we have $r(7)=8$, $r(17)=9$ and so
$$S\equiv 0 \pmod {7^2 \cdot 17^3}.$$
$\bullet$ For $m=7\cdot 97$ we have $r(7)=8$, $r(97)=7^2$ and so
$$S\equiv 0 \pmod {7^3\cdot 97}.$$
$\bullet$ For $m=5\cdot 11^2$ we have $r(5)=5$, $r(11)=10$ and so
$$S\equiv 0 \pmod {5^2\cdot 11^3}. $$
$\bullet$ For $m=79\cdot 859^2$ we have $r(79)=r(859)=78$ and so
$$S\equiv 0 \pmod {79^2\cdot 859^3}.$$
$\bullet$ For $m=37\cdot 73$ we have $r(37)=19$, $r(73)=37$ and so
$$S\equiv 0 \pmod {37^2\cdot 73}.$$
$\bullet$ For $m=19\cdot 73^2 \cdot 89^3\cdot 97^4$ we have $r(19)=18$, $r(73)=37$, $r(89)=11$, $r(97)=49$ and so
$$S\equiv 0 \pmod {19^2\cdot73^3 \cdot 89^5 \cdot 97^7 }.$$
\end{example}
\begin{remark} Note that there is no claim for optimality in the congruences. It can be checked that when $m\in\{7\cdot 17^2, 5\cdot 11^2, 79\cdot 859^2, 37\cdot 73\}$ the congruences above are indeed optimal for each prime dividing $m$. However when $m=7\cdot 97$ the congruence is optimal modulo $97$ but not modulo $7$ as we actually have that $v_7(S)=4$.
\end{remark}
A few words about our assumptions and definitions. If we seek a nice generalization of Kimball and Webb's result summing up to $r(m)$ is more or less forced (cf.\ \cite[pg.\ 7]{Ball}). Now, if we want to say something about the sum modulo powers of prime divisors of $m$ we want the individual terms to make sense modulo $p$ for any prime $p$ dividing $m$.
That is why we exclude the multiples of $r(p)$ in the definition of $I_m$. This, however, has hidden ramifications since integers we excluded because of $q$ might alter the sum modulo powers of $p$ in an unpredictable way and we have to keep track of the various interactions as we add more primes.
In this note we only impose the condition that $v_p(U_{r(p)})=1$ and see what happens. This is a reasonable condition: from considering prime powers we see that $I_m$ does not see the difference between $p$ and $p^{v_p(U_{r(p)})}$. Moreover, when $v_p(U_{r(p)})>1$ we can still use Lemma \ref{vo} to draw conclusions. With regards to our other assumptions, the exclusion of $3$ is mostly a matter of aesthetics: the same arguments would go through but we would have to subtract $1$ in various results when working modulo powers of $3$ thus complicating the statement of the results. The exclusion of $2$ is qualitatively different: not only the results would be more tedious to write but they would require different arguments to establish. Finally, we take $\gcd(p,Q)=1$ since otherwise $r(p)$ does not exist.
This note is organized as follows. In Section 2 we recall some basic facts about Lucas sequences and set up two auxiliary lemmata.
In Section 3 we take care of the case when $m$ is a prime power by proving our first main result. In Section 4 we move on to the case when $m$ has more than one prime divisor, and we prove our second main result.
The assertions about the ranks of appearance in the examples were checked using the computer algebra system {\tt magma}.
\section{Preliminaries}
Lucas sequences are well-studied and satisfy many identities. For the convenience of the reader we collect some well-known properties of Lucas sequences we will be using in the following lemma. The interested reader can look at \cite{Ballnr}, \cite{Rib} or \cite{Roet} and references therein for details of the proofs and much more about Lucas sequences. We remind the reader of our assumption that $\gcd(P,Q)=1$.
\begin{lemma}\label{bas}
\begin{enumerate}
\item If $s,t$ are positive integers and $s$ divides $t$ then $U_s$ divides $U_t$.
\item $m$ divides $U_n$ if and only if $r(m)$ divides $n$.
\item
$$r(\lcm(a,b))=\lcm(r(a),r(b)).$$
\item If $\gcd(n,2Q)=1$ and $r(n)$ does not divide either $s$ or $t$ then
$$\frac{V_s}{U_s}\equiv \frac{V_t}{U_t} \pmod n \ \Leftrightarrow s \equiv t \pmod{ r(n)}.$$
\item (Law of repetition) If $p$ is an odd prime and $p$ divides $U_n$ then $$v_p(U_{kn})=v_p(U_{n})+v_p(k).$$
\item If $\ell=v_p(U_{r(p)})$ then $r(p^{a})=p^{\max(a,\ell)-\ell}r(p)$.
\end{enumerate}
\end{lemma}
\begin{remark} For the sake of clarity, in order to make sense of the last two statements of the lemma when some $U_n=0$ we use the conventions that
$v_p(0)=\infty$ and $\infty- \infty=0$. This has no relevance to the next two sections.
\end{remark}
We now establish two auxiliary lemmata which we will be using in the next sections.
\begin{lemma}\label{vo}
Let $p$ be an odd prime, $s=r(p^{a})$ and $\nu=v_p(U_{s})$.
Let $J\subset \Z$ be a finite set such that
(i) none of its elements is divisible by $r(p)$, and
(ii) there is a $k$ such that sending $x$ to $ks-x$ maps $J$ to $J$.
Let $\ell=v_p(U_{ks})$ and denote $ks-x$ by $\hat{x}$.
Then
$$v_p\left( \sum_{t\in J}\frac{V_t}{U_t} \right)=\ell +v_p \left(\sum_{t\in J}\frac{1}{U_tU_{\hat{t}}} \right),$$
$$2 V_{ks}\sum_{t\in J}\frac{1}{U_tU_{\hat{t}}} \equiv D|J|-\sum_{t\in J}
\left(\frac{V_t}{U_t} \right)^2 \pmod {p ^{2\ell}}.$$
Moreover, if $J'$ is the translation of $J$ by $s$, i.e., $J'=J+s$, then $J'$ satisfies the two assumptions (with the new k being k+2) and the value of
$\sum\frac{1}{U_tU_{\hat{t}}}$ is multiplied by $\frac{4}{V_s^{2}}$ $\pmod {p^{\nu}}$.
\end{lemma}
\begin{proof}
The easy algebraic manipulations (which are based on well known-formulae for Lucas sequences) that establish the result can be found in \cite{Ball} or \cite{Kim}.
For completeness and the convenience of the reader, we repeat the main points here.
The first equality can be easily established by using the equality
$$
\frac{V_t}{U_t}+\frac{V_{\hat{t}}}{U_{\hat{t}}}=2\frac{U_{ks}}{U_tU_{\hat{t}}}.
$$
The congruence is established by noting that its left hand side equals
$$
\sum_{t\in J}\frac{V_tV_{\hat{t}}}{U_tU_{\hat{t}}}+ D|J|,
$$
and that we also have
$$
\sum_{t\in J}\frac{V_tV_{\hat{t}}}{U_tU_{\hat{t}}}=\frac{1}{2}\sum_{t\in J}\frac{(2U_{ks})^2}{(U_tU_{\hat{t}})^2}-\sum_{t\in J} \left(\frac{V_t}{U_t} \right)^2.
$$
The only non-trivial part of the last sentence is the final assertion, which follows immediately from the fact that
$$
2U_{s+t}\equiv U_tV_s \pmod {U_s}.
$$
\end{proof}
\begin{lemma}\label{mainvo}
Let $p\geq 5$ be a prime and $r=r(p)$.
If $p$ has maximal rank then
$$
\sum_{t=1}^{r-1}\left(\frac{V_t}{U_t}\right)^2 \equiv D(r-1) \pmod p.
$$
If $p$ does not have maximal rank then
$$
\sum_{t=1}^{r-1}\left(\frac{V_t}{U_t}\right)^2 \not\equiv D(r-1) \pmod p.
$$
\end{lemma}
\begin{proof}
In the first case it is easy to calculate explicitly the set $\{ \frac{V_t}{U_t}: 1\leq t \leq r-1 \}$ modulo $p$ and establish the result
(cf.\ \cite[proof of Theorem]{Kim}).
For the second case, the result is established by hand for the cases $r=2,3,4$. When $r\geq 5$ we will use \cite[Theorem 1.1]{Pan} which states that if $n\geq5$ then
$$
\sum_{t=1}^{n-1}\frac{V_t}{U_t} \equiv \frac{(n^2-1)D}{6}\cdot \frac{U_n}{V_n} \pmod {w_n^2},
$$
where $w_n$ is the largest divisor of $U_n$ relatively prime to $U_1,\ldots,U_{n-1}$. Taking $n=r$ in the above, and noting that from our assumptions $\frac{(r^2-1)D}{6V_r}$ is not divisible by $p$ we deduce that
$$ v_p\left(\sum_{t=1}^{r-1}\frac{V_t}{U_t}\right)=v_p(U_r).$$
The result now follows from the two displayed formulas of Lemma \ref{vo} (taking $s=r$, $k=1$ and $J=\{1,\ldots,r-1\}$).
\end{proof}
\section{The case of prime powers}
In this section $p$ is a prime with $\gcd(p,6Q)=1$ and $v_p(U_{r(p)})=1$.
Note that the second assumption implies that $$r(p^{a})=p^{a-1}r(p) \, , \ v_p(U_{r(p^{a})})=a.$$
Recall the definition
$$
I_m=
\{x: 1\leq x \leq r(m)\}-
\bigcup_{\substack{q \ \textrm{is prime}\\ q \, | \, m}}\Z \cdot r(q).
$$
\begin{lemma}\label{kro}
Let $r=r(p)$. Then
$$v_p\left( \sum_{t\in I_{p^{a}}} \left(\frac{V_t}{U_t} \right)^2 - D|I_{p^{a}}| \right) \ \ \text{is} \ \
\begin{cases}
\geq a, &\text{if $p$ has maximal rank;} \\=a-1,&\text{if not.}\end{cases}
$$
\end{lemma}
\begin{proof}
Let $K:=\ker(\Z/p^{a}\to \Z/p)$.
Note that the elements of $\{\frac{V_t}{U_t} : t \in I_p\}$ are pairwise distinct modulo $p$, the elements of $\{\frac{V_t}{U_t} : t \in I_{p^{a}}\}$ are pairwise distinct modulo $p^{a}$, $\frac{V_t}{U_t}\equiv \frac{V_{t+r}}{U_{t+r}}$ (mod $p$) and $|I_{p^{a}}|=p^{a-1}|I_p|$. Hence, working in $\Z/p^{a}$ the set $\{\frac{V_t}{U_t} : t \in I_{p^{a}}\}$ is the disjoint union of the cosets $K+\frac{V_t}{U_t}$ with $t\in I_p$. Since $\sum_{x\in K}x=\sum_{x\in K}x^2=0 \in \Z/p^{a}$, we therefore have that
$$
\sum_{t\in I_{p^{a}}} \left(\frac{V_t}{U_t} \right)^2\equiv
p^{a-1} \sum_{t\in I_{p}} \left(\frac{V_t}{U_t} \right)^2 \pmod {p^{a}}.
$$
The result now follows from Lemma \ref{mainvo}.
\end{proof}
\begin{remark} It is clear that the same result holds if we replace $I_{p^{a}}$ by $I_{p^{a}}+l \cdot r(p^{a})$, where $l$ is a positive integer.
\end{remark}
{\em Proof of Theorem \ref{bth1}:}
Combine Lemma \ref{vo} and Lemma \ref{kro}. \qed
\begin{remark}\label{lre} Looking at Lemma \ref{vo} and Lemma \ref{kro} it is easy to see what happens if we replace $I_{p^{a}}$ by $I_{p^{a}}+l\cdot r(p^{a})$ in Theorem \ref{bth1}: If $p$ has maximal rank the result remains the same.
If $p$ does not have maximal rank then the result is the same when $p$ does not divide $2l+1$ whereas the $p$-adic valuation is increased by at least one when $p$ divides $2l+1$.
\end{remark}
\section{More than one prime divisor}
In this section $p$ is a prime with $\gcd(p,6Q)=1$ and $v_p(U_{r(p)})=1$.
First we recall the following well-known fact, easily established by Binet's formulae:
Let $U'$, $V'$ denote the pair of Lucas sequences associated with $P'=V_n$, $Q'=Q^n$.
Then
\begin{equation}\label{mulfor}
U_{nk}=U_nU'_k \ \text{and} \ \ V_{nk}=V'_k.
\end{equation}
It is also straightforward to see that: (i) If $r(p)$ divides $n$ then $r'(p)=p$ and $v_p(U'_p)=1$,
(ii) If $r(p)$ does not divide $n$ then $r'(p)=\frac{r(p)}{d}$ and $v_p(U'_{r'(p)})=v_p(U_{r(p)})+v_p(\frac{n}{d})$ where $d=\gcd(r(p),n)$.
Next, we fix some notation. For a finite set $J\subset \Z$ denote
$$
S(J):=\sum_{t\in J}\frac{V_t}{U_t}.
$$
Trivially
\begin{equation*}
S(X\cup Y)=S(X)+S(Y)-S(X\cap Y),
\end{equation*}
and so if three of the above terms have $p$-adic valuation at least $\nu$ then so does the fourth.
We will use this with no further mention.
Recall that if $X$ is a subset of $\Z$ we write
$$X+n=\{x+n: x \in X\} \ \ \text{and} \ \ n\cdot X=\{nx : x\in X\}.$$
We also introduce the following:
$$
L_{n:b_1,..,b_k}:=\{x: 1 \leq x \leq n\}-\bigcup_{1\leq i \leq k} \Z \cdot b_i.
$$
%
%$$
%L_{n:b_1,..,b_k:c_1,..,c_t}:=L_{n:b_1,..,b_k,\lcm(c_1,\ldots,c_t)}
%$$
Whenever we use the notation above,
we will always have that $n$ is divisible by all the $b_i$'s.
We list some identities which are elementary to establish
\begin{equation}\label{du}
L_{nsb:b}=\bigsqcup_{0\leq t \leq n-1} L_{sb:b}+ tsb,
\end{equation}
\begin{equation}\label{aug}
L_{n:b_1,\ldots,b_k}\cap L_{n:b_{k+1},\ldots,b_{k+t}}=L_{n:b_1,\ldots,b_{k+t}},
\end{equation}
\begin{equation}\label{aug1}
L_{n:b_1,\ldots,b_k,c_1}\cup L_{n:b_1,\ldots,b_k,c_2}=L_{n:b_1,\ldots,b_k,\lcm(c_1,c_2)},
\end{equation}
\begin{equation}\label{find}
L_{n:b,c}\cup (L_{n:b}\cap L_{n:c}^*)=L_{n:b},
\end{equation}
\begin{equation}\label{mlt}
L_{n:b}\cap L_{n:c}^*=c\cdot L_{\frac{n}{c}:\frac{b}{\gcd(b,c)}},
\end{equation}
where ``$\bigsqcup$'' stands for disjoint union and $L_{n:c}^*$ is the complement of $L_{n:c}$ in $L_{n:n}$.
We are now ready to prove our main result.
\bigskip
{\em Proof of Theorem \ref{bth2}:} Let
$$
N=v_p\left(\sum_{t\in I_m}\frac{V_t}{U_t} \right)=v_p(S(I_m)).
$$
\begin{enumerate}
\item
By Lemma \ref{vo} we have that $N\geq v_p(U_{r(m)})=
a+v_p \left(\frac{r(m)}{r(p^{a})} \right)$. Since
\begin{equation}\label{rp}
r(m)=\lcm(p^{a-1}r,r(q_i^{a_i})),
\end{equation}
the result then follows from the fact that $v_p(r)=1$ when $p$ divides $D$ and $v_p(r)=0$ when $p$ does not divide $D$.
\item
By Theorem \ref{bth1}, Remark \ref{lre} and formula \eqref{du} we have that
\begin{equation}\label{ineq}
v_p(S(L_{nr:r}))\geq 2v_p(n)+1.
\end{equation}
Write $\{1\leq i \leq \tau \}=T\bigsqcup T'$ where $i\in T'$ if and only if $r$ divides $r_i$.
Note that if $T=\emptyset$ then $I_m=L_{r(m):r,r_1,\ldots,r_{\tau}}=L_{r(m):r}$, and we are done by what was said above.
To treat the case when $T\neq \emptyset$ we first establish the following.
{\em Claim:} Let c be the $\lcm$ of some of the $r_i$ with $i\in T$. Then
$$
v_p(S(L_{r(m):r,c}))\geq 2a-1.
$$
{\em Proof of claim:} By equation \eqref{find} it suffices to show that
$$v_p(S(L_{r(m):r}))\geq 2a-1 \ \ \ \ \text{and} \ \ \ \ v_p(S(L_{r(m):r}\cap L_{r(m):c}^*)\geq 2a-1.
$$
Using \eqref{ineq} and \eqref{rp} we immediately get the first inequality. If $r$ divides $c$ then $L_{r(m):r}$ equals $L_{r(m):r,c}$ and we are done.
Hence, we may assume that $r$ does not divide $c$.
By \eqref{mlt} and \eqref{mulfor} we have that
$$
S(L_{r(m):r}\cap L_{r(m):c}^*)=\frac{1}{U_c}\sum_{t\in M}\frac{V'_{t}}{U'_{t}}, \ \ \text{where} \ \ M=L_{\frac{r(m)}{c}:\frac{r}{\gcd(r,c)}}
$$
and $U',V'$ are defined as in the beginning of this section with $n=c$.
Since $v_p(r_i)=0$ when $i\in T$, we have that $v_p(c)=0$. Hence, our assumptions imply that $v_p(U'_{r'(p)})=1$ and $r'(p)=\frac{r}{\gcd(r,c)}$. Therefore, we deduce by \eqref{ineq} that $v_p(S(L_{r(m):r}\cap L_{r(m):c}^*)\geq 2\gamma +1$, where $\gamma=v_p \left( \frac{r(m)}{\lcm(r,c)} \right)$. Since
$$v_p \left(\frac{r(m)}{\lcm(r,c)} \right)=v_p \left(\frac{r(m)}{r} \right)\geq a-1,$$
the proof of the claim is complete.
QED
It is now an easy induction using \eqref{aug} and \eqref{aug1} to augment the claim to
$$
v_p(S(L_{r(m):r,c,r_{i_1},\ldots,r_{i_k}}))\geq 2a-1,
$$
where $i_j\in T$.
Note that if $p$ has maximal rank and $\gcd(r,r_i)=1$ for all $i\in T$ then whenever we invoke Theorem \ref{bth1} in the proof above the corresponding inequality is strict, and so in this case the result is that $v_p(S(L_{r(m):r,c,r_{i_1},\ldots,r_{i_k}}))> 2a-1$.
Finally, to finish the proof we need only note that by definition $I_m=L_{r(m):r,r_1,\ldots,r_{\tau}}$, and that this set is unaltered if we omit
all the $r_i$'s that are multiples of $r$. \qed
\end{enumerate}
\begin{remark}
Looking at the proof, it is not difficult to see that the same results hold if we replace $I_{m}$ by $I_{m}+l\cdot r(m)$, where $l$ is a positive integer.
\end{remark}
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\end{thebibliography}
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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11B39; Secondary 11A07.
\noindent \emph{Keywords: }
Lucas sequence, rank of appearance, congruence.
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\vspace*{+.1in}
\noindent
Received June 5 2014;
revised versions received July 18 2014; August 1 2014; August 8 2014.
Published in {\it Journal of Integer Sequences}, August 12 2014.
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