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\begin{center}
\vskip 1cm{\LARGE\bf Some Divisibility Properties Concerning  \\
\vskip .1in
Lucas and Elliptic Divisibility Sequences} 
\vskip 1cm
\large
Chatchawan Panraksa and Aram Tangboonduangjit\\
Mahidol University International College\\
Mahidol University\\
Salaya, Nakhon Pathom 73170\\
Thailand \\
\href{mailto:chatchwan.pan@mahidol.edu}{\tt chatchawan.pan@mahidol.edu} \\
\href{mailto:aram.tan@mahidol.edu}{\tt aram.tan@mahidol.edu}\\
\end{center}

\vskip .2 in

\begin{abstract}
We consider sequences of integers formed by a quotient of the Lucas
sequences or elliptic divisibility sequences. We then investigate some
divisibility properties of these quotient sequences. Additionally,
we prove that elliptic divisibility sequences possess a divisibility
property that is analogous to a generalization of Matijasevich's lemma
involving the Fibonacci numbers, which contributed to the solution to
Hilbert's tenth problem.
\end{abstract}


\section{Introduction}
\label{sec:introduction}

Let $P$ and $Q$ be relatively prime integers. The Lucas sequence $\big(U_n(P,Q)\big)_{n\ge 0}$ is defined by $U_0(P,Q) = 0, \, U_1(P,Q) = 1$, and $$U_n(P,Q) = P\cdot U_{n-1}(P,Q) - Q\cdot U_{n-2}(P,Q) \quad \text{for $n\ge 2$.}$$ For example, the Fibonacci sequence $(F_n)_{n\ge 0}$ corresponds to $U_n(1,-1)$ and the sequence $(M_n)_{n\ge 0}$ of Mersenne numbers, where $M_n = 2^n-1$, corresponds to $U_n(3,2)$.  Each Lucas sequence $(U_n(P,Q))_{n\ge 0}$ is associated with a characteristic polynomial of the form $x^2-Px+Q$. In this work, if not stated otherwise, $(U_n)_{n\ge 0}$ denotes a sequence $(U_n(P,Q))_{n\ge 0}$ for some relatively prime integers $P$ and $Q$. We also assume that the sequence $(U_n)_{n\ge 0}$ is nondegenerate in the sense that $Q\neq 0$ and the ratio of the two roots of the associated characteristic polynomial is not a root of unity. Therefore, the discriminant $D=P^2-4Q$ satisfies $D\neq 0$, and the two roots $\alpha, \beta$ of the characteristic polynomial are distinct so that we can express $U_n$ explicitly as 
$$U_n = \frac{\alpha^n-\beta^n}{\alpha-\beta} = \frac{\alpha^n-\beta^n}{\sqrt{D}}.$$

Another sequence of integers that we consider in this work is called the {\it elliptic divisibility sequence} (EDS). It is defined as a sequence of integers $(h_n)_{n\ge 0}$ satisfying the recurrence 
\begin{equation}\label{EDS}
h_{m+n}h_{m-n} = h_{m+1}h_{m-1}h_{n}^2 - h_{n+1}h_{n-1}h_m^2
\end{equation}
for all integers $m \ge n \ge 0$ and being a divisibility sequence; that is, $h_m \mid h_n$ whenever $m\mid n$. Examples \cite{ward} include the sequence of nonnegative integers $(n)_{n\ge 0}$ and the sequence $((-1)^{(n-1)(n-2)/2}F_n)_{n\ge 0}$. We can see the latter by realizing the following identity of the Fibonacci numbers:
\begin{equation}\label{Fib-EDS}
F_{m+n}F_{m-n} = (-1)^{n+1}(F_{m+1}F_{m-1}F_n^2 - F_{n+1}F_{n-1}F_m^2)
\end{equation}
for all $m \ge n \ge 0$. (We can readily verify the identity \eqref{Fib-EDS} by appealing to the well-known Catalan identity $F_{n+r}F_{n-r} = F_n^2+ (-1)^{n+r+1}F_r^2$ for all integers $n$ and $r$.) This means EDS gives us a natural way to generalize the Fibonacci sequence. It also gives examples of divisibility sequences that do not satisfy linear recurrences. A sequence $(h_n)_{n\ge 0}$ with $h_0 = 0$, $h_1=1$, and $h_2h_3 \neq 0$ is said to be {\it proper}. Ward \cite{ward} was the first to study arithmetic properties of proper elliptic divisibility sequences. One of his results states that a proper sequence $(h_n)_{n\ge 0}$ satisfying \eqref{EDS} is an elliptic divisibility sequence if and only if $h_2$, $h_3$, $h_4$ are integers and $h_2\mid h_4$.

However, we can give an alternative definition of EDS via elliptic curves. We recall some theories of elliptic curves. In particular, we consider a curve with the Weierstrass equation given by
\begin{equation}\label{Weierstrass}
y^2+a_1xy + a_3 y = x^3 + a_2 x^2 + a_4 x + a_6
\end{equation}
with integer coefficients. Non-singular rational points on the projective closure of this curve form a group $E(\mathbb{Q})$; see, for instance, \cite{silverman-book}. Let $P$ be a non-identity element in this group. For a positive integer $n$, we denote the addition of point $P$ to itself $P+P+\cdots+P$ up to $n$ terms by $nP$. Then the corresponding coordinate point $(x(nP),y(nP))$ on the curve can be expressed by
$$(x(nP),y(nP)) = \Big(\frac{A_{nP}}{B_{nP}^2},\frac{C_{nP}}{B_{nP}^3}\Big)$$ where $A_{nP}$ and $C_{nP}$ are integers, $B_{nP}$ is a positive integer, and the fractions in the coordinates are in reduced forms. 

It is now conventional to define an elliptic divisibility sequence as follows. 
\begin{definition}\label{EDS-def}
Let $E/\mathbb{Q}$ be an elliptic curve with the Weierstrass equation given by \eqref{Weierstrass}. Let $P \in E(\mathbb{Q})$ be a non-torsion point. For each positive integer $n$, denote the integer $B_{nP}$ by $B_n$. Then the sequence $(B_n)_{n\ge 1}$ is said to be an {\it elliptic divisibility sequence (EDS)}. An EDS $(B_n)_{n\ge 1}$ is said to be {\it normalized} if $B_1=1$. We can normalize an EDS by a change of variables in the defining Weierstrass equation. The obtained sequence $(B_n/B_1)_{n\ge 1}$ is then an EDS. 
\end{definition}

In this work, we follow Definition \ref{EDS-def} for the definition of the elliptic divisibility sequence. If not stated otherwise, all elliptic divisibility sequences are assumed to be normalized. See, for example, \cite{verzobio, silverman} for the connection between the two definitions of  the elliptic divisibility sequence. Elliptic divisibility sequences have gained popularity recently due to their application in cryptography; see, for example, \cite{stange-1}. In Section \ref{Lucas sequences}, we define a new sequence as a ratio of Lucas sequences and study specific arithmetic properties this sequence possesses. In Section \ref{Elliptic divisibility sequences}, we do the same for elliptic divisibility sequences. In 1970, Matijasevich \cite{matijasevich} completed the theorem (based on combined work by Davis, Putnam, and Robinson) that solved Hilbert's tenth problem. Part of the proof of this theorem used specific arithmetic properties of the Fibonacci numbers. Onphaeng and Pongsriiam \cite{onphaeng} generalized one of these properties to the Lucas sequences. The final result of this paper (Theorem \ref{Generalization of Matijasevich's lemma}) generalizes the exact property to elliptic divisibility sequences.   

\section{Lucas sequences}
\label{Lucas sequences}
A sequence $(u_n)$ of integers is said to be a \emph{strong divisibility sequence} if $\gcd(u_m,u_n) = u_{\gcd(m,n)}$ for all $m,n$. Let $N$ be a positive integer. A sequence $(u_n)$ of rational numbers is said to be an  \emph{$N$-almost strong divisibility sequence} if  for all $m$ and $n$ where $u_m$ and $u_n$ are integers we have $\gcd(u_m,u_n) = u_{\gcd(m,n)}$ whenever $\gcd(mn, N) = 1$. Let the sequence $(T_n)_{n\ge 1}$ be defined by 
\begin{equation}\label{Tn-defn}
T_n = \abs{\frac{U_{n\Delta}}{U_nU_\Delta}},	
\end{equation}
 where $\Delta = \abs{D}$ and $D$ is the discriminant of the characteristic polynomial associated with the Lucas sequence $(U_n)_{n\ge 0}$. In general, some of the terms of this sequence might not be integers. For example, for the sequence $U(4,-7)$, we have $\Delta =12$ and the first five terms of the sequence $$T_n = \abs{\frac{U_{12n}}{U_{12}U_n}}$$ are $$1, \, \frac{76751}{2},\, \frac{3240525601}{3}, \, \frac{158095946378449}{2},\, 7471977820027132645$$ while for the Fibonacci sequence $F_n = U_n(1,-1),$ we have $\Delta = 5$ and the first five terms of the sequence
\begin{equation}\label{Tn-as-Fibonacci}
	T_n = \frac{F_{5n}}{5F_n} 
\end{equation}
are 
$$1,\, 11,\, 61,\, 451,\, 3001.$$
We can show that each term of the sequence \eqref{Tn-as-Fibonacci} is an integer \cite{aram-chatchawan-2}. Filipponi and Freitag \cite{filipponi} gave the Zeckendorf decomposition of the sequence \eqref{Tn-as-Fibonacci}. The present authors \cite{aram-chatchawan-2} also studied and gave certain curious arithmetic properties of the sequence \eqref{Tn-as-Fibonacci} . In particular, we proved that $T^k_n$ divides $T(nT(nT(\cdots (nT(n)))))$ exactly, where $n$ appears $k$ times in this composition of indices. For example, for $k=4$, we have $$T_n^4 \edv T(nT(nT(nT(n)))),$$ where $a^n \edv b$ means $a^n$ divides $b$ exactly, that is, $a^n$ divides $b$ but $a^{n+1}$ does not. In this work, we generalize this result to the sequence $(T_n)_{n\ge 1}$ defined in terms of the Lucas sequences in \eqref{Tn-defn} (Corollary \ref{nested-divisibility-Lucas}). We also prove that this sequence $(T_n)_{n\ge 1}$ is a $\Delta$-almost strong divisibility sequence (Theorem \ref{Main_Thm}). 

\subsection{$p$-adic valuations of Lucas sequences}
\label{sec:p-adic valuations}

If $p$ is a prime such that $p\nmid Q$, then the {\it rank of apparition of $p$} in the sequence $(U_n)_{n\ge 0}$, denoted $\tau(p)$, is defined to be the least positive integer such that $p \mid U_{\tau(p)}$. The following basic facts about $\tau(p)$ are well-known: $\tau(p)$ exists for each $p$, and $p\mid U_n$ if and only if $\tau(p) \mid n$. Sanna~\cite{sanna} gave an explicit formula for the $p$-adic valuation $\nu_p(U_n)$ of nondegenerate Lucas sequences in terms of the prime $p$, the index $n$, and the rank of apparition $\tau(p)$, generalizing the result by Lengyel~\cite{lengyel} who gave the same formula but only for the Fibonacci sequence. We quote this result by Sanna \cite[Thm.\ 1.5, Cor.\ 1.6]{sanna} as Theorem \ref{Sanna_theorem} below. 

\begin{theorem}\label{Sanna_theorem} 
Let $p$ be a prime such that $p\nmid Q$. Then, for each positive integer $n$,
$$\nu_p(U_n) = \begin{cases} \nu_p(n) + \nu_p(U_p)-1, &amp; \text{$p\mid D$ and $p\mid n$;}\\ 
	0, &amp; \text{$p\mid D$ and  $p\nmid n$;}\\
	\nu_p(n) + \nu_p(U_{p\tau(p)})-1, &amp; \text{$p\nmid D,\,\tau(p)\mid n$, and $p\mid n$;}\\
	\nu_p(U_{\tau(p)}), &amp;  \text{$p\nmid D,\, \tau(p)\mid n$, and $p\nmid n$;} \\
	0, &amp;  \text{$p\nmid D$ and $\tau(p)\nmid n$.}
\end{cases}$$
In particular, if $p$ is an odd prime such that $p \nmid Q$, then, for each positive integer $n$,
$$\nu_p(U_n) = \begin{cases} \nu_p(n) + \nu_p(U_p)-1, &amp; \text{$p\mid D$ and $p\mid n$;}\\ 
	0, &amp; \text{$p\mid D$ and  $p\nmid n$;}\\
	\nu_p(n) + \nu_p(U_{\tau(p)}), &amp; \text{$p\nmid D$ and $\tau(p)\mid n$;}\\
	0, &amp;  \text{$p\nmid D$ and $\tau(p)\nmid n$.}
\end{cases}	
$$
\end{theorem}

\begin{remark}\label{when-p-divides-Q} If $p$ is a prime and $p\mid Q$ then since $\gcd(P,Q)=1$, we can prove by induction that $\nu_p(U_n) = 0$ for all $n\ge 1$. In fact, since $U_1 = 1$, we have $p\ndv U_1$. Assume that $p\ndv U_{n-1}$ for some $n\ge 2$. If $p\dv U_n$, then since $PU_{n-1} = U_n+QU_{n-2}$,  $p \dv Q$, and $p\ndv U_{n-1}$, it follows that $p\dv P$ contradicting the assumption that $\gcd(P, Q) = 1$. Hence $p\ndv U_n$.
\end{remark}

We also need the following result by the present authors \cite[Lem.\ 2.3]{aram-chatchawan-1} for $p$-adic valuation of Lucas sequence with integer-multiple index. 
\begin{lemma}\label{Riasat_lemma}
Let $n,k \ge 1$ and $p$ a prime factor of $U_k$ such that $p\nmid Q$. Then
\begin{enumerate}
	\item if (i)\, $p$ is odd, or (ii)\, $p=2$ and $k$ is even, or (iii)\, $p=2$ and $n$ is odd, we have 
$$\nu_p(U_{kn}) = \nu_p(n) + \nu_p(U_k);$$
	\item if $k$ and $D$ are odd and $n$ is even, we have    
	$$\nu_2(U_{kn}) =  \nu_2 (n)+\nu_2(U_k) +\left(\nu_2(U_{2\tau(2)})-\nu_2(U_{\tau(2)})-1 \right) \ge \nu_2 (n)+\nu_2(U_k).$$

\end{enumerate}
\end{lemma}

We now state some properties of the sequence $(T_n)_{n\ge 1}$.

\begin{lemma}\label{lemma-B}
Suppose $\gcd(n,\Delta) = 1$. Then $T_n$ is an integer. 
\end{lemma}

\begin{proof}
	Since $\gcd(P,Q) = 1$, it follows that $(U_n)_{n\ge 0}$ is a strong divisibility sequence. From the definition of $T_n$, it suffices to show that $\gcd(U_n,U_\Delta) = 1$. Indeed, we have $\gcd(U_n, U_\Delta) = U_{\gcd(n,\Delta)} = U_1 = 1$. 
\end{proof}


\begin{lemma}\label{lemma-A}
Suppose $\gcd(n, \Delta)=1$. Then $\gcd(T_n, \Delta) = 1 = \gcd(T_n, U_n)$.
\end{lemma}

\begin{proof}
To prove $\gcd(T_n,\Delta)=1$, suppose $p$ is a prime factor of $\Delta$. Then by Theorem \ref{Sanna_theorem}, since $\gcd(n,\Delta) = 1$ and $p\mid \Delta$,  we have
\begin{align*}
    \nu_p(U_{n\Delta}) &amp;= \nu_p(n\Delta) + \nu_p(U_p) - 1 = \nu_p(n) +\nu_p(\Delta) + \nu_p(U_p) - 1 \\
    &amp;= \nu_p(\Delta) + \nu_p(U_p) - 1
    \end{align*}
and
\begin{align*}
\nu_p(U_nU_\Delta) &amp;= \nu_p(U_n) + \nu_p(U_\Delta) =0+\nu_p(\Delta) + \nu_p(U_p) - 1 \\
&amp;= \nu_p(\Delta) + \nu_p(U_p) - 1.
\end{align*}
 We see that $\nu_p(U_{n\Delta}) = \nu_p(U_nU_\Delta)$. Therefore, $\gcd(T_n,\Delta) = 1$.

To prove $\gcd(T_n, U_n) = 1$, suppose $p$ is a prime divisor of $U_n$. If $p\mid \Delta$, then since $\gcd(n,\Delta) = 1$, we have $p\nmid n$. By Theorem \ref{Sanna_theorem}, it follows that $\nu_p(U_n) =0$, contradicting the fact that $p$ is a prime divisor of $U_n$. Thus $p\nmid\Delta$.  By case (1) of Lemma \ref{Riasat_lemma}, we have 
$$\nu_p(U_{n\Delta}) = \nu_p(\Delta) + \nu_p(U_n) = 0 + \nu_p(U_n) = \nu_p(U_n).$$ This implies $\nu_p(T_n) = 0$ and therefore $\gcd(T_n,U_n) = 1$. 
\end{proof}

\subsection{Almost strong divisibility property} 
\label{sec:main1}

\begin{theorem}\label{Main_Thm}
 The sequence $(T_n)_{n\ge 1}$ is a $\Delta$-almost strong divisibility sequence. 
\end{theorem}

\begin{proof}
Suppose $m$ and $n$ satisfy $\gcd(mn, \Delta)=1$. Let $p$ be a prime such that $p\mid T_m$ and $p\mid T_n$. By Lemma \ref{lemma-A}, $p\nmid U_m$, $p\nmid U_n$, and $p\nmid \Delta$.
If $p$ is odd, by Theorem \ref{Sanna_theorem} we have
\begin{align*}
\nu_p(T_m) &amp;= \nu_p(U_{m\Delta}) - \nu_p(U_\Delta) = \nu_p(m\Delta) + \nu_p(U_{\tau(p)})  - \nu_p(U_\Delta)\\
&amp;= \nu_p(m) + \nu_p(U_{\tau(p)})  - \nu_p(U_\Delta).
\end{align*}
Similarly, we have  $\nu_p(T_n) = \nu_p(n) + \nu_p(U_{\tau(p)}) - \nu_p(U_\Delta)$. Let $\ell = \gcd(m,n)$. Then 
\begin{align*}
\nu_p(T_\ell) &amp;= \nu_p(U_{\ell \Delta})  - \nu_p(U_\Delta) = \nu_p(\ell) + \nu_p(U_{\tau(p)}) - \nu_p(U_\Delta) \\
&amp;= \min\{\nu_p(m),\nu_p(n)\} + \nu_p(U_{\tau(p)}) - \nu_p(U_\Delta).
\end{align*}
For $p = 2$, we consider four cases.

\noindent\textbf{Case 1:} Suppose $p\mid m$ and $p\mid n$. We have
\begin{align*}
\nu_p(T_m) &amp;= \nu_p(U_{m\Delta}) - \nu_p(U_\Delta) = \nu_p(m\Delta) + \nu_p(U_{p\tau(p)}) - 1 - \nu_p(U_\Delta) \\
&amp;= \nu_p(m) + \nu_p(U_{p\tau(p)}) - 1 - \nu_p(U_\Delta).
\end{align*}
Similarly, we have $\nu_p(T_n) =  \nu_p(n) + \nu_p(U_{p\tau(p)}) - 1 - \nu_p(U_\Delta)$. Let $\ell = \gcd(m,n)$ so that $p\mid \ell$. Then 
\begin{align*}\nu_p(T_\ell) &amp;= \nu_p(U_{\ell \Delta})  - \nu_p(U_\Delta) = \nu_p(\ell) + \nu_p(U_{p\tau(p)}) - 1 - \nu_p(U_\Delta) \\ &amp;= \min\{\nu_p(m),\nu_p(n)\} + \nu_p(U_{p\tau(p)})-1 - \nu_p(U_\Delta). 
\end{align*}

\noindent\textbf{Case 2:} Suppose $p\nmid m$ and $p\mid n$. We have
$$\nu_p(T_m) = \nu_p(U_{m\Delta}) - \nu_p(U_\Delta) = \nu_p(U_{\tau(p)}) - \nu_p(U_\Delta)$$ and 
\begin{align*}
\nu_p(T_n) &amp;= \nu_p(n) + \nu_p(U_{p\tau(p)}) -1  - \nu_p(U_\Delta)  \ge \nu_p(U_{p\tau(p)})  - \nu_p(U_\Delta) \\
&amp;\ge \nu_p(U_{\tau(p)}) - \nu_p(U_\Delta).
\end{align*} 
Let $\ell = \gcd(m,n)$ so that $p\nmid \ell$. Then
$$\nu_p(T_\ell) = \nu_p(U_{\ell \Delta}) - \nu_p(U_\Delta) = \nu_p(U_{\tau(p)}) - \nu_p(U_\Delta) = \min\{\nu_p(T_m), \nu_p(T_n)\}.$$

\noindent\textbf{Case 3:} Suppose $p\mid m$ and $p\nmid n$. This is analogous to Case 2.

\noindent\textbf{Case 4:} Suppose $p\nmid m$ and $p\nmid n$. This is also analogous to Case 2. 


We have shown that $\nu_p(\gcd(T_m,T_n)) = \nu_p(T_{\gcd(m,n)})$ for all primes $p$. Hence $$\gcd(T_m,T_n) = T_{\gcd(m,n)},$$
as we wish to show. 
\end{proof}

\subsection{Nested exact divisibility property}
\label{sec:main2}

Let $n$ be a positive integer. Define the sequence $(H_k(n))_{k\ge 1}$ by $H_1(n) = T_n$ and $H_k(n) = T_{nH_{k-1}(n)}$ for $k\ge 2$. The first few terms of the sequence $(H_k(n))_{k\ge 1}$ are 
$$T(n), \quad T(nT(n)), \quad T(nT(nT(n))), \quad T(nT(nT(nT(n)))).$$ We see that this forms a subsequence of $(T_m)_{m\ge 1}$ for each $n$, and the sequence of nested indices is increasing very rapidly. We first observe the following property of the sequence $(H_k(n))_{k\ge 1}$.

\begin{lemma}\label{lemma-C}
	Suppose $\gcd(n,\Delta) = 1$. Then $\gcd(U_{nH_k(n)}, T_n) = 1$ for all positive integers $k$.
\end{lemma}

\begin{proof} We prove that $\gcd(H_k(n),\Delta) = 1$ for all positive integers $k$ by induction. For $k=1$, by Lemma \ref{lemma-A}, we have 
$$\gcd(H_k(n),\Delta) = \gcd(H_1(n),\Delta) = \gcd(T_n,\Delta) = 1.$$ Assume that $\gcd(H_{k}(n),\Delta) = 1$ for some $k \ge 1$. Then $\gcd(nH_{k}(n), \Delta) = 1$ and by Lemma \ref{lemma-A}, we have
$$\gcd(H_{k+1}(n),\Delta) = \gcd(T_{nH_{k}(n)}, \Delta) = 1.$$
Now since $(U_n)_{n\ge 0}$ is a strong divisibility sequence, we have $$\gcd(U_{nH_k(n)},U_{n\Delta}) = U_{\gcd(nH_k(n),n\Delta)} = U_{n\cdot\gcd(H_k(n),\Delta)}=U_n.$$
	Hence $$\gcd(U_{nH_k(n)}, T_n)  = \gcd\left(U_{nH_k(n)}, \frac{U_{n\Delta}}{U_nU_\Delta}\right) = 1.$$ 
\end{proof}

\begin{lemma}\label{lemma-D}
Suppose $\gcd(n,\Delta) = 1$. If $2\mid T_n$, then $2\mid n\Delta$.
\end{lemma}

\begin{proof}
We proceed by contradiction. Assume that $n\Delta$ is odd. Then both $n$ and $\Delta$ are odd. Since $2\mid T_n$, we have $2\mid U_{n\Delta}$ so that $\nu_2(U_{n\Delta})\ge 1$. Consequently, Theorem \ref{Sanna_theorem} implies that $\nu_2(U_{n\Delta}) = \nu_2(U_{\tau(2)})$ and $\tau(2) \mid n\Delta$. Since $\Delta$ is odd, we have that $P$ is odd and thus, by Remark \ref{when-p-divides-Q}, $\tau(2)=3$. Therefore $\nu_2(U_{n\Delta}) = \nu_2(U_3)$ and $3\mid n\Delta$. This implies $3\mid \Delta$ or $3\mid n$. If $3\mid \Delta$, by the fact that $(U_n)_{n\ge 0}$ is a divisibility sequence, we have $$\nu_2(U_3) = \nu_2(U_{n\Delta}) \ge \nu_2(U_{\Delta}) \ge \nu_2(U_3).$$ This yields $\nu_2(U_\Delta) =\nu_2(U_3)$. Similarly, if $3\mid n$, then $\nu_2(U_n) = \nu_2(U_3)$. Consequently, $\max\{\nu_2(U_\Delta),\nu_2(U_n)\} = \nu_2(U_3)$, and we have
\begin{align*}
0 &amp;\le \nu_2(T_n) = \nu_2\Big(\frac{U_{n\Delta}}{U_nU_\Delta}\Big) \\ 
&amp;\le \min\Big\{\nu_2\Big(\frac{U_{n\Delta}}{U_n} \Big),\nu_2\Big(\frac{U_{n\Delta}}{U_\Delta} \Big)\Big\} \\
&amp;= \nu_2(U_{n\Delta}) - \nu_2(U_3) = \nu_2(U_3) - \nu_2(U_3)=0.
\end{align*}
Hence, $\nu_2(T_n) = 0$, contradicting the assumption that $2\mid T_n$. 
\end{proof}


We have the following theorem about the exact divisibility property of the sequence $(H_k(n))_{k\ge 1}$. 

\begin{theorem}\label{nested-valuation-Lucas}
	Suppose $\gcd(n,\Delta) = 1$. Let $p$ be a prime such that $\nu_p(T_n) &gt; 0$. Then $\nu_p(H_k(n)) = k\nu_p(T_n)$ for all positive integers $k$.
\end{theorem}


\begin{proof}
	We prove by induction on $k$. The case when $k=1$ is obvious. Assume that the statement holds for some positive integer $k$.  Let $p$ be a prime such that $p\dv T_n$. Put $a= \nu_p(T_n) \ge 1$. Then $$\nu_p(T_n^k) = ka = \nu_p(H_k(n)).$$ Since $p\dv T_n$, it follows from Lemma~\ref{lemma-A} that  $p\ndv U_n$ and from Lemma~\ref{lemma-C} that $p\ndv U_{nH_k(n)}$. Thus,
	$$\nu_p(H_{k+1}(n))=\nu_p(T_{nH_k(n)}) = \nu_p\left(\frac{U_{nH_k(n)\cdot \Delta}}{U_{nH_k(n)}U_\Delta}\right) = \nu_p(U_{nH_k(n)\cdot \Delta})-\nu_p(U_\Delta).$$
If $p$ is odd, then, by Theorem \ref{Sanna_theorem} and Lemma \ref{Riasat_lemma}, we have 
\begin{align*}
\nu_p(U_{nH_k(n)\cdot \Delta})-\nu_p(U_\Delta) &amp;= \nu_p(H_k(n))+\nu_p(U_{n\Delta})-\nu_p(U_\Delta) \\&amp;= ka + \nu_p\left(\frac{U_{n\Delta}}{U_\Delta}\right) = ka + \nu_p\left(\frac{U_{n\Delta}}{U_nU_\Delta}\right)\\ &amp;= ka+\nu_p(T_n) = ka + a = (k+1)a.
\end{align*}
If $p = 2$, then by Lemma \ref{lemma-D}, we have that $n\Delta$ is even and by case 1 (ii) of Lemma \ref{Riasat_lemma}, we have the same result as above. 

\noindent Hence, for all primes $p$, we have
$$\nu_p(H_{k+1}(n)) = (k+1)a = (k+1)\nu_p(T_n).$$
\end{proof}

We conclude this section with an immediate corollary.
\begin{corollary}\label{nested-divisibility-Lucas}
Suppose $\gcd(n,\Delta) = 1$ and $T_n\neq 1$. Then, for each positive integer $k$,
	$$T_n^k \edv H_k(n).$$
\end{corollary}


\section{Elliptic divisibility sequences}
\label{Elliptic divisibility sequences}

We have the following result summarized by Reynolds \cite[Lem.\  2.1]{reynolds} for specific basic arithmetic properties of elliptic divisibility sequences.
\begin{lemma}\label{edsv}
Let $(B_n)_{n\ge 1}$ be an elliptic divisibility sequence corresponding to an elliptic curve $E$ with the Weierstrass equation \eqref{Weierstrass} and a non-torsion point $P$ in $E(\mathbb{Q})$.
\begin{itemize}
    \item[(i)] Let $p$ be a prime. There exists a smallest positive integer $n_0$ such that $p\mid B_{n_0}.$ Moreover, for every positive integer $n$, 
    $$p\mid B_n \quad \text{if and only if}\quad  n_0\mid n.$$
    \item[(ii)] Let $p$ be an odd prime. For every pair of positive integers $m,n$, if $\nu_p (B_n)&gt;0$ then
    $$\nu_p (B_{mn})=\nu_p (B_n)+\nu_p (m).$$
    \item[(iii)] For every pair of positive integers $m,n$, if $\nu_2(B_n)&gt;0$ then
    $$\nu_2 (B_{mn})=\nu_2 (B_n)+\nu_2 (m)$$
    if the coefficient $a_1$ is even and
    $$\abs{\nu_2(B_{mn})-(\nu_2 (B_n)+\nu_2 (m))}\leq \epsilon$$
    otherwise, where the constant $\epsilon$ depends only on $E$ and $P$.
    \item[(iv)] For all positive integers $m,n$, 
    $$\gcd(B_m,B_n)=B_{\gcd(m,n)}.$$
\end{itemize}
\end{lemma} 


Now we define a sequence similar in the form to sequence \eqref{Tn-as-Fibonacci} but in the context of elliptic divisibility sequences. Let $\T$ be a positive integer and  $(B_n)_{n\ge 1}$ an elliptic divisibility sequence corresponding to an elliptic curve with the Weierstrass equation \eqref{Weierstrass} and a non-torsion point $P$. We define the sequence $(K_n)_{n\ge 1}$ as follows. For each positive integer $n$, let $$K_n = \frac{B_{\T n}}{B_\T B_n}.$$ Then we have the following result. 
\begin{theorem}\label{main-thm-2}
Let $(K_n)_{n\ge 1}$ be a sequence defined as above with the coefficient $a_1$ in the Weierstrass equation \eqref{Weierstrass} being even and $\T\mid B_\T$. Then $(K_n)_{n\ge 1}$ is a $\T$-almost strong divisibility sequence. That is, for all positive integers $m$, $n$, if $\gcd(mn,\T)=1,$ then $$\gcd(K_m, K_n)=K_{\gcd(m,n)}.$$
\end{theorem}
\begin{proof}
Let $m$ and $n$ be positive integers that are relatively prime to $\tau$. By Lemma \ref{edsv}(iv), we have $B_\T\mid B_{\T n}$ and $B_n\mid B_{\T n}$. Since  $\gcd(n,\T)=1$, it follows that $K_n$ is an integer. Similarly, we have that $K_m$ is an integer. Let $p$ be a prime. We consider two cases.

\bigskip
\noindent{\bf Case 1:} Suppose $\nu_p(B_\T)&gt;0$.
Since $\gcd(\T,mn)=1$, by Lemma \ref{edsv}(iv), we have $\nu_p(B_n)=0=\nu_p(B_m)$. By Lemma \ref{edsv}(ii, iii), we have $\nu_p(B_{\T n})=\nu_p(B_\T)+\nu_p(n)$ and $\nu_p(B_{\T m})=\nu_p(B_\T)+\nu_p(m)$. This implies that
\begin{align*}
\nu_p(K_n) &amp;= \nu_p(B_{\T n})-\nu_p(B_\T)-\nu_p(B_n)\\
&amp;=\nu_p(B_\T)+\nu_p(n)-\nu_p(B_\T)-0 \\
&amp;=\nu_p(n).
\end{align*}
Similarly, we have $\nu_p(K_m) = \nu_p(m)$. Consequently,
$$\nu_p(\gcd(K_m,K_n)) = \min\{\nu_p(K_m), \nu_p(K_n)\} = \min\{\nu_p(m), \nu_p(n)\}.$$
Now, we consider $\nu_p(K_{\gcd(m,n)})$. By Lemma \ref{edsv}(iv), we observe that $$\nu_p(B_{\gcd(m,n)})= \nu_p(\gcd(B_m,B_n)) = \min\{\nu_p(B_m),\nu_p(B_n)\} =0.$$ 
By Lemma \ref{edsv}(ii, iii), we obtain that
\begin{align*}
    \nu_p(K_{\gcd(m,n)}) &amp;= \nu_p(B_{\T\cdot\gcd(m,n)})-\nu_p(B_\T)-\nu_p(B_{\gcd(m,n)})\\
    &amp;=\nu_p(B_\T)+\nu_p(\gcd(m,n))-\nu_p(B_\T)-\nu_p(B_{\gcd(m,n)})\\
    &amp;=\nu_p(\gcd(m,n)) = \min\{\nu_p(m),\nu_p(n)\}.
\end{align*}
Comparing the values, we conclude that 
$$\nu_p(\gcd(K_m,K_n)) = \nu_p(K_{\gcd(m,n)}).$$

\bigskip
\noindent{\bf Case 2:} Suppose $\nu_p(B_\T)=0$. We consider three subcases.\\
\noindent{\bf Case 2.1:} Suppose $\nu_p(B_m)=0$ and $\nu_p(B_n)=0$.
We have, $$\nu_p(K_n)=\nu_p(B_{\T n})-\nu_p(B_\T)-\nu_p(B_n)=\nu_p(B_{\T n}).$$ Similarly, we have $\nu_p(K_m)=\nu_p(B_{\T m})$. Consequently, by Lemma \ref{edsv}(iv), 
\begin{align*}
    \nu_p(\gcd(K_m,K_n))&amp;=\nu_p(\gcd(B_{\T m},B_{\T n}))
    =\nu_p(B_{\gcd(\T m,\T n)})
    =\nu_p(B_{\T\cdot \gcd(m,n)}).
\end{align*}
On the other hand, since  $\min\{\nu_p(B_m),\nu_p(B_n)\}=0$, we have, by Lemma \ref{edsv}(iv), $$\nu_p(B_{\gcd(m,n)})=\nu_p(\gcd(B_m,B_n))=\min\{\nu_p(B_m),\nu_p(B_n)\}=0.$$ Hence,
$$\nu_p(K_{\gcd(m,n)}) = \nu_p(B_{\T\cdot \gcd(m,n)})-\nu_p(B_\T)-\nu_p(B_{\gcd(m,n)}) = \nu_p(B_{\T\cdot \gcd(m,n)}).$$
Comparing the values, we conclude that 
$$\nu_p(\gcd(K_m,K_n)) = \nu_p(K_{\gcd(m,n)}).$$


\noindent{\bf Case 2.2:} Suppose $\nu_p(B_m)&gt;0$ and $\nu_p(B_n)&gt;0$. By Lemma \ref{edsv}(ii, iii), we have $\nu_p(B_{\T n})=\nu_p(B_n)+\nu_p(\T)$. Therefore,
\begin{align*}
\nu_p(K_n)&amp;=\nu_p(B_{\T n})-\nu_p(B_\T)-\nu_p(B_n)\\
&amp;=\nu_p(B_n)+\nu_p(\T)-\nu_p(B_\T)-\nu_p(B_n)\\
&amp;=\nu_p(\T).
\end{align*}
Similarly, we have $\nu_p(K_m)=\nu_p(\T)$. Thus,  $$\nu_p(\gcd(K_m,K_n))=\min\{\nu_p(K_m),\nu_p(K_n)\} = \nu_p(\T).$$ 

On the other hand, since $\nu_p(B_m)&gt;0$ and $\nu_p(B_n)&gt;0$, by Lemma \ref{edsv}(iv), it follows that $\nu_p(B_{\gcd(m,n)}) &gt;0$. By Lemma \ref{edsv}(ii), we have
\begin{align*}
    \nu_p(K_{\gcd(m,n)})&amp;=\nu_p(B_{\T\cdot\gcd(m,n)})-\nu_p(B_\T)-\nu_p(B_{\gcd(m,n)})\\
    &amp;=\nu_p(B_{\gcd(m,n)})+\nu_p(\tau)-\nu_p(B_\T)-\nu_p(B_{\gcd(m,n)})\\
    &amp;=\nu_p(\T).
\end{align*}
Again, for this case, we have 
$$\nu_p(\gcd(K_m,K_n)) = \nu_p(K_{\gcd(m,n)}).$$

\noindent{\bf Case 2.3:} Suppose $\nu_p(B_m)&gt;0$ and $\nu_p(B_n)=0$. 
By following the same argument as in Case 2.1 and Case 2.2, we have $\nu_p (K_m) = \nu_p(\T)$ and $\nu_p (K_n)=\nu_p(B_{\T n})$.
Since $\T\mid B_\T$, by Lemma \ref{edsv}(iv), we have
$$\nu_p(\gcd(K_m, K_n)) = \nu_p(\gcd(\T, B_{\T n})) \le \nu_p(\gcd(B_\T, B_{\T n})) = \nu_p(B_\T) = 0.$$ Hence, $\nu_p(\gcd(K_m, K_n)) = 0$. Now we claim that $\nu_p(K_{\gcd(m,n)}) = 0$. Since, by Lemma \ref{edsv}(iv), $\nu_p(B_{\gcd(m,n)}) = \nu_p(\gcd(B_m,B_n)) = \min\{\nu_p(B_m),\nu_p(B_n)\} = 0$, it follows that
$$\nu_p(K_{\gcd(m,n)}) = \nu_p(B_{\T\cdot \gcd(m,n)})-\nu_p(B_\T)-\nu_p(B_{\gcd(m,n)}) = \nu_p(B_{\T\cdot \gcd(m,n)}).$$ Assume that $\nu_p(B_{\T\cdot \gcd(m,n)}) &gt; 0$. Since $\nu_p(B_m) &gt; 0$, by Lemma \ref{edsv}(i), there exists a smallest positive integer $m_0$ such that $\nu_p(B_{m_0}) &gt; 0$ and $m_0\mid m$. Since $p\mid B_{\T\cdot \gcd(m,n)}$, by Lemma \ref{edsv}(i), $m_0\mid \T\cdot\gcd(m,n)$. Since $\gcd(mn,\T)=1$ and $m_0\mid m$, we have $\gcd(m_0,\T) = 1$. Consequently, $m_0 \mid \gcd(m,n)$. This implies $m_0 \mid n$, so that, by Lemma \ref{edsv}(i), $\nu_p(B_n) &gt; 0$, which is a contradiction. Hence, $\nu_p(B_{\T\cdot \gcd(m,n)}) = 0 = \nu_p(K_{\gcd(m,n)})$. We conclude for this case that $$\nu_p(\gcd(K_m,K_n)) = \nu_p(K_{\gcd(m,n)}).$$

Combining all cases, we have proved that for every prime $p$, we have $$\nu_p(\gcd(K_m,K_n)) = \nu_p(K_{\gcd(m,n)}).$$ Hence $$\gcd(K_m, K_n)=K_{\gcd(m,n)},$$ as desired.
\end{proof}
We point out that the condition in Theorem \ref{main-thm-2} on integer $\T$ dividing $B_\T$ is not particularly restrictive. This condition is investigated extensively in \cite{gottschlich, silverman}. 

\begin{example}\label{ex:1}
\cite{silverman} The elliptic divisibility sequence $(B_n)_{n\ge 1}$ corresponding to elliptic curve $E: y^2+y = x^3-x$ and non-torsion point $P = (0,0)$ is 
$$(1, 1, 1, 1, 2, 1, 3, 5, 7, 4, 23, 29, 59, 129,
314, 65, 1529, 3689, 8209, 16264, 83313, \dots ).$$
(See \cite[\seqnum{A006769}]{Slo}.)
The indices $\T$ with the property $\T\mid B_\T$ are $$1, 40, 53, 63, 80, 127, 160, 189, 200, 320, 400, 441, 443, \dots .$$ According to Theorem \ref{main-thm-2} with $\T=40$, the sequence $(K_n)_{n\ge 1}$ defined by $$K_n = \frac{B_{40n}}{B_{40}B_n} = \frac{B_{40n}}{(40\cdot 13 526 278 251 270 010)B_n}$$ for all $n\ge 1$ satisfies $\gcd(K_m,K_n) = K_{\gcd(m,n)}$ whenever $\gcd(mn, 40) = 1$.
\end{example}


The following result is an analog to Theorem \ref{nested-valuation-Lucas} for $p$-adic valuation of sequence with nested indices. Here again, to describe the $n$th term of a sequence $(a_n)$, we use the function notation $a(n)$ interchangeably with the subscript notation $a_n$. 

\begin{theorem} Let $(B_n)_{n\ge 1}$ be an elliptic divisibility sequence corresponding to an elliptic curve whose Weierstrass equation \eqref{Weierstrass} has coefficient $a_1$ even. Let $n$ be a positive integer. Define a sequence $(G_k(n))_{k\ge 1}$ as follows: $G_1(n) = B_n$ and $G_{k+1}(n) = B(nG_k(n))$ for $k\ge 1$. Let $p$ be a prime number. If $\nu_p(B_n) &gt; 0$, then 
$$\nu_p(G_k(n))=k\nu_p(B_n).$$ Therefore, if $B_n\neq 1$, then $B_n^k \edv G_k(n)$ for all positive integers $k$.
\end{theorem}
\begin{proof}
We prove by induction on $k$. For $k=1$, we have $G_1(n) = B_n$. Hence the statement holds. Assume that the statement holds for some positive integer $k$. Then, Lemma \ref{edsv}(ii, iii), we have 
\begin{align*}
\nu_p(G_{k+1}(n)) &amp;= \nu_p(B(n G_k(n))) = \nu_p(G_k(n))+\nu_p(B_n)\\
&amp;=k\nu_p(B_n) + \nu_p(B_n) = (k+1)\nu_p(B_n).
\end{align*}
By mathematical induction, the statement holds for all positive integers $k$. 
\end{proof}

The final result is a generalization of Matijasevich's lemma \cite[Lem.\ 17]{matijasevich} for elliptic divisibility sequences. Matijasevich's lemma states that for $n &gt; 2$ we have $F_n^2 \mid F_m$ if and only if $nF_n \mid m$ where $(F_n)_{n\ge 0}$ is the Fibonacci sequence. 

\begin{theorem}\label{Generalization of Matijasevich's lemma} Let $(B_n)_{n\ge 1}$ be an elliptic divisibility sequence corresponding to an elliptic curve whose Weierstrass equation \eqref{Weierstrass} has coefficient $a_1$ even. Moreover, suppose that there exists a positive integer $N$ such that all terms of the sequence $(B_n)_{n\ge N}$ are distinct and none of the terms $B_1, \dots, B_{N-1}$ appears in $(B_n)_{n\ge N}$. Then, for all integers $n$, $r\ge N$ and for all positive integers $k$, we have $B_n^k\mid B_r$ if and only if $nB_n^{k-1}\mid r$.
\end{theorem}

\begin{proof} Let $n$, $r$ be at least $N$ and $k$ a positive integer. Assume that $B_n^k\mid B_r$.  Since $B_n\mid B_r$, by Lemma \ref{edsv}(iv), we have  
$B_{\gcd(n,r)} = \gcd (B_n, B_r) = B_n$. Since all terms of the sequence $(B_n)_{n\ge N}$ are distinct and none is equal to any $B_1,\dots,B_{N-1}$, it follows that $\gcd(n,r) = n$ and therefore $n\mid r$. Write $r = n s$ for some positive integer $s$ and let $p$ be a prime dividing $nB_n$. We consider two cases.

\noindent\textbf{Case 1:} Suppose $p\mid B_n$. Let $\nu_p(B_n) = \ell \ge 1$. 
Then $k\ell=\nu_p(B_n^{k})\le \nu_p(B_r)$. By Lemma \ref{edsv}(ii, iii), we have
$$k\ell\le \nu_p(B_r)=\nu_p(B_{ns})=\nu_p(B_n)+\nu_p(s) =\ell+\nu_p(s).$$
Thus, $\nu_p(s)\geq (k-1)\ell$. Consequently, 
\begin{align*}
\nu_p(nB_n^{k-1}) &amp;= \nu_p(n)+\nu_p(B_n^{k-1})=\nu_p(n)+(k-1)\ell \\
&amp;\le \nu_p(n)+\nu_p(s) = \nu_p(ns) = \nu_p(r).
\end{align*}

\noindent\textbf{Case 2:} Suppose $p\nmid B_n$. Then $p\mid n$ and we have $\nu_p(nB_n^{k-1}) = \nu_p(n) \le \nu_p(r)$.

Combining all cases, we have $nB_n^{k-1}\mid r$.

For the converse, we assume that $nB_n^{k-1}\mid r$. Let $p$ be a prime dividing $B_n$. Let $\nu_p(B_n)=\ell\ge 1$. Then Lemma \ref{edsv}(ii, iii, iv) implies
\begin{align*}
    \nu_p(B_r) &amp;\ge \nu_p (B_{nB_n^{k-1}})=\nu_p (B_n)+\nu_p (B_n^{k-1})\\
    &amp;=\ell+(k-1)\ell=k\ell = \nu_p(B_n^k).
\end{align*}
Since this is true for all primes $p$ dividing $B_n$, it follows that $B_n^k\mid B_r$.
\end{proof}



\section{Acknowledgments}
The authors would like to thank the anonymous referee who provided constructive comments and valuable suggestions that led to the  improvement of the quality of this paper. We would also like to thank Brian Hopkins for proofreading the first draft of this paper.


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\noindent 2020 {\it Mathematics Subject Classification}: Primary 11B39; Secondary 11B83. \\
\noindent \emph{Keywords:} Fibonacci number, Lucas number, elliptic divisibility sequence, strong divisibility sequence, almost strong divisibility sequence, Matijasevich, Hilbert's tenth problem.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000045}, \seqnum{A000225}, \seqnum{A006769}, and \seqnum{A088545}.)

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\vspace*{+.1in}
\noindent
Received May 16 2022; 
revised versions received  July 31 2022; August 7 2022.
Published in {\it Journal of Integer Sequences}, October 28 2022.

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\noindent
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