\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics,amsmath,amssymb} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \usepackage{enumerate} \DeclareMathOperator{\lcm}{lcm} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.1in} \setlength{\textheight}{8.4in} \newcommand{\seqnum}[1]{\href{http://oeis.org/#1}{\underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \begin{center} \vskip 1cm{\LARGE\bf Some Arithmetic Properties of \\ \vskip .12in Certain Sequences } \vskip 1cm \large E.~L.~Roettger\\ Department of General Education \\ Mount Royal University\\ 4825 Mount Royal Gate SW\\ Calgary, AB T3E 6K6\\ Canada \\ \href{mailto:eroettger@mtroyal.ca}{\tt eroettger@mtroyal.ca} \\ \ \\ H.~C.~Williams\footnote{The second author is supported by NSERC of Canada.} \\ Department of Mathematics and Statistics \\ University of Calgary\\ 2500 University Drive NW\\ Calgary, AB T2N 1N4 \\ Canada \\ \href{mailto:williams@math.ucalgary.ca}{\tt williams@math.ucalgary.ca} \end{center} \vskip .2 in \begin{abstract} In an earlier paper it was argued that two sequences, denoted by $\{U_n\}$ and $\{W_n\}$, constitute the sextic analogues of the well-known Lucas sequences $\{u_n\}$ and $\{v_n\}$. While a number of the properties of $\{U_n\}$ and $\{W_n\}$ were presented previously, several arithmetic properties of these sequences were only mentioned in passing. In this paper we discuss the derived sequences $\{D_n\}$ and $\{E_n\}$, where $D_n=\gcd(W_n-6R^n,U_n)$ and $E_n=\gcd(W_n,U_n)$, in greater detail and show that they possess many number-theoretic properties analogous to those of $\{u_n\}$ and $\{v_n\}$, respectively. \end{abstract} \section{Introduction} Let $p$, $q\in \mathbb{Z}$ be relatively prime and $\alpha$, $\beta$ be the zeros of \[x^2-px+q\] with discriminant $\delta=(\alpha-\beta)^2=p^2-4q$. The well-known Lucas sequences $\{u_n\}$ and $\{v_n\}$ are defined by \begin{align*} u_n=u_n(p,q)=\frac{\alpha^n-\beta^n}{\alpha-\beta}, && v_n=v_n(p,q)=\alpha^n+\beta^n. \end{align*} These sequences possess many interesting properties and have found applications in primality testing, integer factorization, solution of quadratic and cubic congruences, and cryptography (see \cite{williams}). We note here that both sequences are linear recurrence sequences of order 2 and that $u_n$, $v_n \in \mathbb{Z}$ whenever $n\geq 0$. Lucas' problem of extending or generalizing his sequences has been well studied and we refer the reader to \cite[Chapter 1]{Roe09} and \cite[Section 1]{RWG13} for further information on this topic. One possible extension of the Lucas sequences, which involves cubic instead of quadratic irrationalities, was investigated in \cite{Roe09} (also see M{\"u}ller, Roettger and Williams \cite{MRW09}). In this case we let $P$, $Q$, $R \in \mathbb{Z}$ be integers such that $\gcd (P,Q,R)=1$ and let $\alpha$, $\beta$, $\gamma$ be the zeros of \begin{equation} h(x)=x^3-Px^2+Qx-R, \label{h(x)} \end{equation} with discriminant \[ \Delta=(\alpha-\beta)^2(\beta-\gamma)^2(\gamma-\alpha)^2=Q^2P^2-4Q^3-4RP^3+18PQR-27R^2\not=0. \] Roettger's sequences $\{c_n\}$ and $\{w_n\}$ are defined as \[ c_n=c_n(P,Q,R)=(\alpha^n-\beta^n)(\beta^n-\gamma^n)(\gamma^n-\alpha^n)/((\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)) \] and \[ w_n=w_n(P,Q,R)=(\alpha^n+\beta^n)(\beta^n+\gamma^n)(\gamma^n+\alpha^n)-2R^n. \] Note here that if $n\geq0$, we have $c_n$, $w_n \in \mathbb{Z}$ and $\{c_n\}$, $\{w_n\}$ are linear recurrence sequences of order 6. In \cite{Roe09}, it is pointed out that the sequences $\{c_n\}$ and $\{w_n\}$ have many properties analogous to those of $\{u_n\}$ and $\{v_n\}$, respectively. Recently, these sequences were extended further by Roettger, Williams and Guy \cite{RWG13}. If we put $\gamma_1=\alpha/\beta$, $\gamma_2=\beta/\gamma$, $\gamma_3=\gamma/\alpha$, $\lambda=R$, then we can write \begin{eqnarray*} c_n &=&\lambda^{n-1}(1-\gamma_1^n)(1-\gamma_2^n)(1-\gamma_3^n)/((1-\gamma_1)(1-\gamma_2)(1-\gamma_3)) \ \mbox{ and}\\ w_n &=& v_n-2R^n, \quad \mbox{where}\\ v_n &=& \lambda^n(1+\gamma_1^n)(1+\gamma_2^n)(1+\gamma_3^n). \end{eqnarray*} One of the most important properties of the Lucas sequence $\{u_n\}$ when $n\geq0$ is that it is a divisibility sequence. An integer sequence $\{A_n\}$ is said to be a \emph{divisibility sequence} if $A_n \mid A_m$ whenever $n\mid m$ and $A_n \not=0$. For example, Roettger's sequence $\{c_n\}$ $(n\geq 0)$ is a divisibility sequence. Suppose we define \begin{eqnarray} && U_n=\displaystyle\frac{\lambda^{n-1}(1-\gamma_1^n)(1-\gamma_2^n)(1-\gamma_3^n)}{(1-\gamma_1)(1-\gamma_2)(1-\gamma_3)}, \label{Un}\\ && V_n=\lambda^n(1+\gamma_1^n)(1+\gamma_2^n)(1+\gamma_3^n),\label{Vn} \end{eqnarray} where $\lambda$, $\gamma_1$, $\gamma_2$, $\gamma_3 \in \bar{\mathbb{Q}}$; $\gamma_1$, $\gamma_2$, $\gamma_3\not=1$; $\gamma_i\not=\gamma_j$ when $i\not=j$ and $\gamma_1\gamma_2\gamma_3=1$. In \cite{RWG13}, it is shown that if $U_n$, $V_n\in \mathbb{Z}$ whenever $n\geq 0$, $\{U_n\}$ is a linear recurrence sequence and $\{U_n\}$ is also a divisibility sequence, then we must have $\lambda=R \in \mathbb{Z}$ and $\rho_i =R(\gamma_i+1/\gamma_i)$ $(i=1,2,3)$ must be the zeros of a cubic polynomial \begin{equation} g(x)=x^3-S_1x^2+S_2x-S_3, \label{g(x)} \end{equation} where \begin{equation} S_3=RS_1^2-2RS_2-4R^3 \label{S3} \end{equation} and $S_1$, $S_2 \in \mathbb{Z}$. The six zeros of \begin{align*} G(x)&= (x^2-\rho_1x+R^2)(x^2-\rho_2x+R^2)(x^2-\rho_3x+R^2)\\ &= x^6-S_1x^5+(S_2+3R^2)x^4-(S_3+2R^2S_1)x^3 + R^2(S_2+3R^2)x^2-R^4S_1x+R^6 \end{align*} are $R\gamma_i$, $R/\gamma_i$ $(i=1,2,3)$. If we define $W_n=V_n-2R^n$, then both $\{U_n\}$ and $\{W_n\}$ are linear recurrence sequences with characteristic polynomial $G(x)$. Also, $U_0=0$, $U_1=1$, $U_2=S_1+2R$, $U_3=S_1^2+RS_1-S_2-3R^2$, $W_0=6$, $W_1=S_1$, $W_2=S_1^2-2S_2-6R^2$, $W_3=S_1^3-3S_1S_2+3RS_1^2 -6RS_2-3R^2S_1-12R^3$. Furthermore, we have $U_{-n}=-U_n/R^{2n}$, $W_{-n}=W_n/R^{2n}$; hence, $U_n$, $W_n \in \mathbb{Z}$ when $n\geq 0$. It is also the case that $\{U_n\}$ is a divisibility sequence. It is shown in \cite{RWG13} that if \begin{align} S_1=PQ-3R, && S_2=P^3R+Q^3-5PQR+3R^2, \label{relationship} \end{align} then $U_n(S_1,S_2,R)=c_n(P,Q,R)$, $W_n(S_1,S_2,R)=w_n(P,Q,R)$. If, in the expression (\ref{Un}), we define \begin{eqnarray} \Delta&=&\lambda^2(1-\gamma_1)^2(1-\gamma_2)^2(1-\gamma_3)^2 \nonumber \\ &=&R^2(\gamma_1+\gamma_2+\gamma_3-1/\gamma_1-1/\gamma_2-1/\gamma_3)^2, \label{Delta1} \end{eqnarray} we find that \begin{equation} \Delta=S_1^2-4S_2+4RS_1-12R^2, \label{Delta2} \end{equation} but this is the same as $Q^2P^2-4Q^3-4RP^3+18PQR-27R^2$, the discriminant of $h(x)$, when $S_1$ and $S_2$ are given by (\ref{relationship}). If $d$ denotes the discriminant of $g(x)$, then, as shown in \cite{RWG13}, we have $d=\Delta \Gamma$, where \begin{eqnarray} \Gamma&=&R^4(\gamma_1-\gamma_2)^2(\gamma_2-\gamma_3)^2(\gamma_3-\gamma_1)^2\\ &=&S_2^2+10RS_1S_2-4RS_1^3-11R^2S_1^2+12R^3S_1 +24R^2S_2+36R^4.\label{Gamma1} \end{eqnarray} The discriminant $D$ of $G(x)$ is given by $D=Ed^2R^{12}$, where \[ E=R^2\Delta(S_1+2R)^2=(\rho_1-4R^2)(\rho_2-4R^2)(\rho_3-4R^2). \] If $S_1$ and $S_2$ are given by (\ref{relationship}), then \begin{equation} \Gamma=(RP^3-Q^3)^2.\label{Gamma2} \end{equation} We remark that the condition analogous to $\gcd(P,Q,R)=1$ for Roettger's sequences is $\gcd(S_1,S_2,R)=1$ for the more general $\{W_n\}$ and $\{U_n\}$ sequences. The duplication formulas are \begin{align} 2W_{2n}=W_n^2+\Delta U_n^2-4R^nW_n, && U_{2n}=U_n(W_n+2R^n) \label{duplication} \end{align} and the triplication formulas are \begin{eqnarray} && 4W_{3n}= 3\Delta U_n^2 (W_n+2R^n)+W_n^2(W_n-6R^n)+24R^{2n},\label{tripWn}\\ && 4U_{3n}=U_n(3W_n^2+\Delta U_n^2).\label{tripUn} \end{eqnarray} Since $\{U_n\}$ is a divisibility sequence, we must have $U_{3n}/U_n \in \mathbb{Z}$ $(n\geq0)$ and by (\ref{tripUn}), this means that $4\mid W_n^2-\Delta U_n^2$. Thus, if $2\mid U_n$, then $2\mid W_n$ and we have proved Proposition \ref{div2}. \begin{proposition}\label{div2} If $n\geq 0$, then $2\mid \gcd(W_n, U_n)$ if and only if $2\mid U_n$. \end{proposition} \noindent The general multiplication formulas for $\{W_n\}$ and $\{U_n\}$ are given as \cite[(7.7) and (7.8)]{RWG13}. We observe here that in general for a given $S_1$, $S_2$ $R \in \mathbb{Z}$ there do not always exist, $P$, $Q \in \mathbb{Z}$ such that (\ref{relationship}) holds. As a simple example consider $S_1=-1$, $S_2=-4$, and $R=1$; it is not possible to find integers $P$, $Q$ such that $PQ=2$ and $P^3+Q^3=3$. Thus, the sequences $\{W_n(S_1,S_2,R)\}$, $\{U_n(S_1,S_2,R)\}$ represent a non-trivial extension of Roettger's sequences $\{w_n\}$ and $\{c_n\}$. In \cite{RWG13} it is mentioned that if we define \[ D_n=\gcd(W_n-6R^n,U_n) \quad \mbox{and} \quad E_n=\gcd(W_n,U_n), \] then the sequences $\{D_n\}$ and $\{E_n\}$ possess many number-theoretic properties in common with $\{u_n\}$ and $\{v_n\}$, respectively. Indeed, some of these properties were presented in \cite{RWG13} without proof. The purpose of this paper is to supply these proofs or sketches thereof and to develop some new results concerning $\{D_n\}$ and $\{E_n\}$. \section{Some properties of $\{D_n\}$} In this section we will produce some results concerning $\{D_n\}$ that are similar to those possessed by $\{u_n\}$. We begin with two simple propositions that easily follow from Lemma 8.1 of \cite{RWG13} and results immediately following that lemma. \begin{proposition}\label{gcdDnR1} If $\gcd(S_1,S_2,R)=1$, then for $n\geq 0$ we have \[ \gcd(D_n,R) \mid 2. \] \end{proposition} \begin{proposition}\label{gcdDnR2} If $\gcd(S_1,S_2,R)=1$, then for any $n\geq 0$, we must have $4\nmid D_n$ whenever $2\mid R$. \end{proposition} \noindent In the sequel we will assume that $S_1$, $S_2$, $R$ have been selected such that $\gcd(S_1,S_2,R)=1$. If we define \begin{displaymath} F_n=\begin{cases} \Delta U_n^2, & \text{when $2\nmid \Delta U_n$;}\\ \Delta U_n^2/4, & \text{when $2\mid \Delta U_n$} \end{cases} \end{displaymath} we see that since $4\mid W_n^2-\Delta U_n^2$, $F_n$ must be an integer. If $M$ is any divisor of $F_n$ and $(M,R)=1$, then we can use \cite[(7.7) and (7.8)]{RWG13} to show that \begin{eqnarray} && U_{mn}/U_n\equiv R^{n(m-1)}K_m(W_n/2R^n)\pmod{M},\label{UmnUnmodM}\\ &&W_{mn}\equiv 2R^{mn}L_m(W_n/2R^n)\pmod{M},\label{WmnmodM} \end{eqnarray} where the polynomials $K_m(x)$ and $L_m(x)$ are respectively defined in \cite[\S 4.3 and \S 5.1]{Roe09}. Also, from results in \cite{Roe09} it is easy to show that $L_m(3)=3$ and $K_m(3)=m^3$. We next establish that like $\{u_n\}$, $\{D_n\}$ is a divisibility sequence. \begin{theorem}\label{Dndivis} If $n$, $m \geq 1$, then $D_n\mid D_{mn}$. \end{theorem} \begin{proof} Since $\{U_n\}$ is a divisibility sequence it suffices to show\\ $D_n \mid W_{mn}-6R^{mn}$. We let $2^\lambda \mid\mid D_n$. If $\lambda=0$ or $\lambda\geq1$ and $2\nmid R$, then $D_n\mid F_n$. By Proposition \ref{gcdDnR1}, we have $\gcd(D_n,R)=1$ and by (\ref{WmnmodM}) we get \[ W_{mn}\equiv 2R^{mn}L_m(W_n/2R^n)\equiv 2R^{mn} L_m(3) \equiv 6R^{mn}\pmod{D_n}. \] If $\lambda=1$, then $\gcd(D_n/2,R)=1$ and $D_n/2 \mid F_n$; hence, \[ W_{mn}\equiv 6R^{mn} \pmod{D_n/2}. \] Also, since $2\mid U_n$, we have $2 \mid U_{mn}$ and $2\mid W_{mn}$ (Proposition \ref{div2}). It follows that $W_{mn}\equiv 6R^{mn}\pmod{2}$ and since $\gcd(2, D_n/2)=1$ we get \[ W_{mn}\equiv 6R^{mn}\pmod{D_n}. \] There remains the case of $\lambda>1$ and $2\mid R$, but this is impossible by Proposition \ref{gcdDnR2}. \end{proof} Let $p$ be any prime. We are next able to present the \emph{law of repetition} for $p$ in $\{D_n\}$. We denote by $\nu_p(x)$ $(x\in \mathbb{Z})$ that value of $\lambda$ such that $p^\lambda \mid\mid x$. \begin{theorem}\label{lawrepDn} Let $p$ be any prime such that $p>3$ and suppose that $\nu_p(D_n)\geq 1$. \begin{enumerate} \item If $\nu_p(U_n)>\nu_p(W_n-6R^n)$, then $\nu_p(D_{pn})=\nu_p(D_n)+2$ and\\ $\nu_p(W_{pn}-6R^{pn})<\nu_p(U_{pn})$. \item If $\nu_p(U_n)=\nu_p(W_n-6R^n)$ and $\nu_p(U_n)>1$, then\\ $\nu_p(D_{pn})=\nu_p(D_n)+2$ and $\nu_p(W_{pn}-6R^{pn})<\nu_p(U_{pn})$. \item If $\nu_p(U_n)<\nu_p(W_n-6R^n)$, then if $\nu_p(U_n)>1$, \\ $\nu_p(D_{pn})=\nu_p(D_n)+3$. \item If $\lambda=1$, then $\nu_p(D_{pn})\geq 2$. \end{enumerate} \end{theorem} \begin{proof} These results can be established by making use of the techniques of \cite[\S 5.2]{Roe09}, together with the polynomial congruence \begin{multline*} L_p(x)\equiv 3+p^2(x-3)+(p^2(p^2-1)/12)(x-3)^2\\ +(p^2(p^2-1)(p^2-4)/360)(x-3)^3\pmod{(x-3)^4}, \end{multline*} which holds for all primes $p\geq 5$. \end{proof} When $p=3$, the law of repetition for $3$ in $\{D_n\}$ is given below. \begin{theorem} Let $\nu_3(D_n)\geq 1$. \begin{enumerate} \item If $\nu_3(U_n)\geq \nu_3(W_n-6R^n)>1$, then $\nu_3(D_{3n})=\nu_3(D_n)+2$. \item If $\nu_3(U_n)\geq \nu_3(W_n-6R^n)=1$, then $\nu_3(D_{3n})\geq \nu_3(D_n)+2$. \item If $\nu_3(U_n)< \nu_3(W_n-6R^n)$, then \[ \nu_3(D_{3n})=\nu_3(D_n)+3 \quad \mbox{when} \quad \nu_3(D_n)>1 \] or \[ \nu_3(D_{3n})\geq \nu_3(D_n)+3 \quad \mbox{when} \quad \nu_3(D_n)=1. \] \end{enumerate} \end{theorem} \begin{proof} These results can be easily proved by making use of the the triplication formulas (\ref{tripWn}) and (\ref{tripUn}). \end{proof} In the case of $p=2$, there exists a rather complicated law of repetition for $p$ in $\{D_n\}$. We will not provide the complete law here, but we remark that if $\nu_2(D_n)>1$, then the duplication formulas (\ref{duplication}) can be used to show that $\nu_2(D_{2n})\geq \nu_2(D_n)+1$. The case of $\nu_2(D_n)=1$, however, is more problematical. Certainly, if $2\mid R$, there is no law of repetition for $2$ in $\{D_n\}$ by Proposition \ref{gcdDnR2}. Thus, we need only consider the case of $2\mid \mid D_n$ and $2\nmid R$. In this case, we can use the duplication and triplication formulas to find that if \begin{enumerate}[i)] \item $4\mid U_n$, $2\mid\mid W_n-6R^n$; \item $2\mid\mid U_n$, $2\mid\mid W_n-6R^n$, $2\mid\Delta$; \item $2\mid\mid U_n$, $4\mid W_n-6R^n$, $2\nmid\Delta$; \end{enumerate} then $4\mid D_{3n}$ and $4\nmid D_{2n}$. In all other cases we have $4\mid D_{2n}$. We also have the following companion result to the law of repetition for any odd prime in $\{D_n\}$. \begin{theorem} \label{companion} If $p$ is odd and $\nu_p(D_n)\geq 1$, then $\nu_p(D_{mn})=\nu_p(D_n)$ whenever $p\nmid m$. \end{theorem} \begin{proof} Since $p\not=2$, we have $p^{2\lambda}\mid F_n$ when $\lambda=\nu_p(D_n)$, $\gcd(p,R)=1$ and $W_n\equiv 6R^n \pmod{p^\lambda}$. It follows from (\ref{WmnmodM}) that \[ W_{mn}\equiv 2R^{mn}L_m(W_n/2R^n)\equiv 2R^{mn}L_m(3)\equiv 6R^{mn}\pmod{p^\lambda} \] and by (\ref{UmnUnmodM}) that \[ U_{mn}/U_n\equiv R^{n(m-1)}K_m(3)\equiv m^3 R^{n(m-1)}\pmod{p^\lambda}. \] Since $p\nmid m$, it follows that $p^\lambda\mid\mid U_{mn}$ and $p^\lambda\mid W_{mn}-6R^{mn}$; hence $p^\lambda \mid\mid D_{mn}$. \end{proof} \noindent In the case of $p=2$, Theorem \ref{companion} is not in general true when $\lambda=1$ and $2\nmid R$, as we have seen in the above remarks. Of course, we could eliminate this problem if we could impose additional restrictions on $S_1$, $S_2$, $R$ such that none of i), ii) or iii) could occur. If $2\mid \mid D_n$ and $2\nmid R$, it is easy to show that cases i), ii) or iii) can occur if and only if $2\mid\tilde{Q}_n$, where $\tilde{Q}_n=(W_n^2-\Delta U_n^2)/4$. In a later section we will discuss the parity of $\tilde{Q}_n$ when $2\mid D_n$. Note that if $4\mid D_n$, then $2\nmid R$ and \\ $\tilde{Q}_n\equiv 1 \pmod{2}$. If $\lambda>1$, then we certainly have $2^\lambda \mid D_{mn}$ by Theorem \ref{Dndivis} and since $W_n/2R^n\equiv 3 \pmod{2^{\lambda-1}}$, we get \[ U_{mn}/U_n\equiv m^3R^{n(m-1)}\pmod{2^{\lambda-1}}. \] Thus, if $m$ is odd, then $2\nmid U_{mn}/U_n$ and $2^\lambda\mid\mid D_{mn}$. Hence Theorem \ref{companion} is true when $p=2$ and $\nu_2(D_n)>1$. We conclude this section with a result that is often useful. \begin{theorem}\label{useful} If $m$, $n\geq 1$, then $\gcd(U_{mn}/U_n,D_n)\mid 2m^3$. \end{theorem} \begin{proof} It is easy to show this when $2\nmid D_n$ because $D_n \mid F_n$ and $\gcd(D_n, R)=1$. Suppose $2\mid D_n$; then because $U_n/2\mid F_n$, we have $D_n/2\mid F_n$. Also, $\gcd(D_n/2,R)=1$ by Propositions \ref{gcdDnR1} and \ref{gcdDnR2}. Hence, by (\ref{UmnUnmodM}) \[ U_{mn}/U_n\equiv m^3R^{n(m-1)}\pmod{D_n/2}. \] It follows that \[ \gcd(U_{mn}/U_n,D_n/2)\mid m^3 \] and \[ \gcd(U_{mn}/U_n,D_n)\mid 2m^3. \] \end{proof} \section{The law of apparition for $m$ in $\{D_n\}$} In this section we deal with the problem of when $m\mid D_n$, when $m>1$. We note that if $p$ is an odd prime and $p\mid R$, then $p\nmid D_n$ $(n\geq 0)$ by Proposition \ref{gcdDnR1}. Thus, we may assume that if $m$ is odd, then $\gcd(m,R)=1$. We define $\omega=\omega(m)$, if it exists, to be the least positive value of $n$ such that $m\mid D_n$. We call $\omega$ the \emph{rank of apparition} of $m$ in $\{D_n\}$. We begin by examining the case where $m$ is a prime $p$ where $p\mid d$ and $p\nmid 2R$. \begin{theorem} Let $p$ by any prime such that $p\nmid 2R$ and $p\mid d$. There exists a rank of apparition $\omega$ of $p$ in $\{D_n\}$ and if $p\mid D_n$ for some $n\geq 0$, then $\omega \mid n$. Also, $\omega=p$ or $\omega\mid p \pm 1$. \end{theorem} \begin{proof} By results in the early part of \cite[\S 9]{RWG13}, we know that if \\ $p\mid S_1^2-3S_2$, then $p$ has a simple rank of apparition $r_1$ in $\{U_n\}$. It is not difficult to show that $p\mid D_n$ if and only if $r_1\mid n$; hence, $\omega=r_1$. If $p\nmid S_1^2-3S_2$, then $p$ can have two ranks of apparition in $\{U_n\}$ when $p\mid \Delta$ and only one when $p \nmid \Delta$. In either case, it is a simple matter to show that there is a rank of apparition $\omega$ of $p$ in $\{D_n\}$, that $\omega\not=p$ and that if $p\mid D_n$, then $\omega \mid n$. \end{proof} We next consider the case of $p=3$ and $3\nmid d$. \begin{lemma} If $p=3$ and $3\nmid dR$, then $\omega=\omega(3)$ always exists in $\{D_n\}$ and if $3\mid D_n$, then $\omega\mid n$. \end{lemma} \begin{proof} We see from \cite[Table 2]{RWG13} that there is single rank of apparition $r$ of $3$ in $\{U_n\}$. From the duplication formulas we see that if $3\mid U_n$ and $3\nmid W_n$, then $3\mid W_{2n}$ if and only if $W_n\equiv R^n \pmod{3}$ and $3\mid W_{4n}$ if and only if $W_n \equiv -R^n \pmod{3}$. Thus, $\omega(3)$ always exists and $\omega=r$, $2r$ or $4r$. Furthermore, if $3\mid D_n$, then $\omega \mid n$. \end{proof} There remains the case of odd $p$ where $p\nmid 3dR$. We first need to establish a simple lemma in this case. Here and in the sequel we will denote by $\mathbb{K}_p$ the splitting field of $G(x) \in \mathbb{F}_p[x]$. We can denote the zeros of $G(x) \in \mathbb{F}_p[x]$ by $R\gamma_i$ and $R/\gamma_i$ $(i=1,2,3)$. \begin{lemma} \label{gammaK} If $p\nmid 2\Delta R$, then $p\mid D_n$ if and only if $\gamma_1^n=\gamma_2^n=\gamma_3^n=1$ in $\mathbb{K}_p$. \end{lemma} \begin{proof} Certainly, if $\gamma_1^n=\gamma_2^n=\gamma_3^n=1$ in $\mathbb{K}_p$, then $p\mid W_n-6R^n$ and $p\mid U_n$ by (\ref{Un}) and (\ref{Vn}); hence, $p\mid D_n$. If $p\mid D_n$, then since $p\mid U_n$ and $p\nmid \Delta$, we may assume without loss of generality that $\gamma_1^n=1$. By \cite[(8.4)]{RWG13}, we have $\gamma_2^n-1=0$ and therefore $\gamma_3^n=1/(\gamma_1^n \gamma_2^n)=1$. \end{proof} \begin{corollary} If $p\nmid 2 \Delta R$ and $\omega=\omega(p)$ exists for $p$ in $\{D_n\}$, then $p\mid D_n$ if and only if $\omega \mid n$. \end{corollary} \begin{proof} Certainly $p\mid D_n$ when $\omega\mid n$ because $\{D_n\}$ is a divisibility sequence. Suppose next that $\omega \nmid n$ and $p\mid D_n$. In this case we have $n=qw+r$, where $01$. By the law of repetition we know that $p^\alpha\mid D_n$ for some $n>0$; hence $\omega(p^\alpha)$ must exist. If we put $\omega=\omega(p)$, then since $p\mid D_{\omega(p^\alpha)}$, we must have $\omega\mid \omega(p^\alpha)$ by Theorem \ref{rankDn}. Put $s=\omega(p^\alpha)/\omega$ and let $p^\nu\mid\mid s$, then $s=p^\nu t$, where $p\nmid t$. If $p^\lambda\mid \mid D_{p^\nu \omega}$ and $\lambda<\alpha$, then $p^\lambda \mid \mid D_{p^\nu \omega t}$ by Theorem \ref{companion}, which is a contradiction; thus $\omega(p^\alpha)=p^\nu \omega$. Notice that $\nu$ is the least positive integer such that $p^\alpha\mid D_{p^\nu \omega}$. Next, suppose that $2\nmid m$ and the prime power decomposition of $m$ is \[ m=\prod_{i=1}^k p_i^{\alpha_i}; \] we must have \begin{equation} \label{omegam} \omega(m)= \lcm (\omega(p_i^{\alpha_i}):i=1,2,\dots,k). \end{equation} Thus, if $(m,2R)=1$, then $\omega(m)$ always exists and is given by (\ref{omegam}). \section{The auxiliary sequences $\{U_n^{*}\}$ and $\{W_n^{*}\}$} In order to prove some results concerning $\{U_n\}$ and $\{W_n\}$, it is often useful to make use of the auxiliary sequences $\{U_n^{*}\}$ and $\{W_n^{*}\}$. We put $\gamma_1^{*}=\gamma_2/\gamma_1$, $\gamma_2^{*}=\gamma_3/\gamma_2$, $\gamma_3^{*}=\gamma_1/\gamma_3$, $R^{*}=R^2$ and define \begin{eqnarray*} && V_n^{*}=R^{*n}(1+\gamma_1^{*n})(1+\gamma_2^{*n})(1+\gamma_3^{*n}),\\ && U_n^{*}=R^{*n-1}(1-\gamma_1^{*n})(1-\gamma_2^{*n})(1-\gamma_3^{*n})/((1-\gamma_1^{*})(1-\gamma_2^{*})(1-\gamma_3^{*})),\\ && W_n^{*}=V_n^{*}-2R^{*n}, \end{eqnarray*} where \begin{equation}\label{Delta*} \Delta^{*}=R^{*2}(1-\gamma_1^{*})^2(1-\gamma_2^{*})^2(1-\gamma_3^{*})^2=\Gamma \not=0. \end{equation} Notice also that \begin{eqnarray*} \Gamma^{*} &=& R^{*4}(\gamma_1^{*}-\gamma_2^{*})^2(\gamma_2^{*}-\gamma_3^{*})^2(\gamma_3^{*}-\gamma_1^{*})^2\\ &=& \Delta R^2U_3^2. \end{eqnarray*} If we put $\gamma_1^{**}=\gamma_2^{*}/\gamma_1^{*}=1/\gamma_2^3$, then $\gamma_1^{**}=1/\gamma_2^3$. We also have \\$\gamma_2^{**}= \gamma_3^{*}/\gamma_2^{*}=1/\gamma_3^3$, $\gamma_3^{**}=\gamma_1^{*}/\gamma_3^{*}=1/\gamma_1^3$; hence, \begin{align} W_n^{**}=R^nW_{3n}, && U_n^{**}=R^{n-1}U_{3n}/U_3. \label{relate} \end{align} If we put $\rho_i^{*}=R^{*}(\gamma_i^{*}+1/\gamma_i^{*})$ $(i=1,2,3)$, we get \begin{equation} S_1^{*}=\rho_1^{*}+\rho_2^{*}+\rho_3^{*}=S_2-RS_1\label{S1star} \end{equation} and \begin{eqnarray}\nonumber S_2^{*}&=& \rho_1^{*}\rho_2^{*}+\rho_2^{*}\rho_3^{*}+\rho_3^{*}\rho_1^{*}=RW_3+R^2S_1^{*}\\ &=&RS_1^3-3RS_1S_2+3R^2S_1-5R^2S_2-4R^3S_1-12R^4.\label{S2star} \end{eqnarray} Also, \begin{eqnarray*} S_3^{*}&=&\rho_1^{*}\rho_2^{*}\rho_2^{*}\\ &=&R^{*}S_1^{*2}-2R^{*}S_2^{*}-4R^{*3}. \end{eqnarray*} It follows, then, from the results mentioned in \S 1, that if we compute the initial values of $U_n^{*}$ and $W_n^{*}(=V_n^*-2R^{*n})$ by replacing $R, S_1, S_2$ by $R^{*}, S_1^{*}, S_2^{*}$, respectively, then we have both $\{U_n^{*}\}$ and $\{W_n^{*}\}$ to be linear recurrence sequences of order 6 with characteristic polynomial $G^{*}(x)$ and $\{U_n^{*}\}$ is a divisibility sequence. It is easy to show as well that $W_{-n}^{*}=W_n^{*}/R^{*2n}$ and $U_{-n}^{*}=-U_n^{*}/R^{*2n}$. We observe further that $\gcd(S_1^{*}, S_2^{*}, S_3^{*})=1$ if and only if $\gcd(S_1, S_2, S_3)=1$. Thus, the sequences $\{U_n^{*}\}$ and $\{W_n^{*}\}$ have the same properties as $\{U_n\}$ and $\{W_n\}$ with $R, S_1, S_2$, replaced by $R^{*}, S_1^{*}, S_2^{*}$, respectively. We have shown how to relate the $\{U_n^{**}\}$ and $\{W_n^{**}\}$ sequences to $\{U_n\}$ and $\{W_n\}$ in (\ref{relate}); we can also relate the $\{U_n^{*}\}$ and $\{W_n^{*}\}$ sequences to $\{U_n\}$ and $\{W_n\}$. We define $\rho_i^{(n)}=R^n(\gamma_i^n+1/\gamma_i^n)$ $(i=1,2,3)$ and find that \begin{equation} S_1^{(n)}=\rho_1^{(n)}+\rho_2^{(n)}+\rho_3^{(n)}=W_n \label{S1relate} \end{equation} and \begin{equation} S_2^{(n)}=\rho_1^{(n)}\rho_2^{(n)}+\rho_2^{(n)}\rho_3^{(n)}+\rho_1^{(n)}\rho_3^{(n)}=W_n^{*}+R^nW_n.\label{S2relate} \end{equation} Since \begin{eqnarray*} \Delta U_n^2&=& R^{2n}(1-\gamma_1^n)^2(1-\gamma_2^n)^2(1-\gamma_3^n)^2\\ &=&S_1^{(n)2}-4S_2^{(n)}+4R^nS_1^{(n)}-12R^{2n}, \end{eqnarray*} we get \begin{equation} \Delta U_n^2=W_n^2-4W_n^{*}-12R^{2n} \label{generalDelta} \end{equation} using (\ref{S1relate}) and (\ref{S2relate}). This formula, which generalizes (\ref{Delta2}), is similar to the well-known Lucas function identity \[ v_n^2-\delta u_n^2=4q^n. \] Note also that we get \begin{equation} \tilde{Q}_n=W_n^{*}+3R^n \label{tilQ} \end{equation} from (\ref{generalDelta}) and \[ 4W_n^{*}=W_n^2 -\Delta U_n^2-12R^{2n}, \] the relation connecting $W_n^{*}$ to $W_n$ and $U_n$. To relate $U_n^{*}$ to $W_n$ and $U_n$ is somewhat more complicated. From (\ref{generalDelta}), we have \[ \Delta^{*} U_n^{*2}=W_n^{*2}-4W_n^{**}-12R^{*2n}. \] Hence, from (\ref{Delta*}), (\ref{relate}), and (\ref{generalDelta}), we get \[ \Gamma U_n^{*2}=((W_n^2-\Delta U_n^2)/4-3R^{2n})^2-4R^nW_{3n}-12R^{4n}. \] From (\ref{tripWn}), we find that \begin{multline} 16 \Gamma U_n^{*2}= W_n^4-16R^nW_n^3-48R^n\Delta W_n U_n^2+72R^{2n}W_n^2-72R^{2n}\Delta U_n^2\\ -2\Delta W_n^2U_n^2+\Delta^2U_n^4-432R^{4n}, \end{multline} a formula that generalizes (\ref{Gamma1}). As promised in \S 2 we will now investigate the parity of $\tilde{Q}_n$ when $2\nmid R$ and $2\mid D_n$. If $2\nmid S_1$ and $2\mid S_2$, then by (\ref{S1star}) and (\ref{S2star}), we have $2\nmid S_1^{*}$ and $2\mid S_2^{*}$. It follows that $2\mid U_n^{*}$ if and only if $7\mid n$ and $2\mid W_n^{*}$ when $2\mid D_n$. In this case we find from (\ref{tilQ}) that $2\nmid \tilde{Q}_n$ whenever $2\mid D_n$. If $2\mid S_1$ and $2\mid S_2$, then $2\mid S_1^{*}$ and $2\mid S_2^{*}$; hence, $2\mid U_n^{*}$ if and only if $2\mid n$ and we get $2\mid W_n^{*}$, $\tilde{Q}_n\equiv 1\pmod{2}$ whenever $2\mid D_n$. If $2\nmid S_1$ and $2\nmid S_2$, then $\Delta^{*}=\Gamma \equiv (S_2+RS_1)^2\equiv 0 \pmod{4}$ from (\ref{Gamma1}). Since $4\mid W_n^{*2}-\Delta^{*}U_n^{*2}$, we get $2\mid W_n^{*}$ and $\tilde{Q}_n \equiv 1 \pmod{2}$. The only remaining case is $2\mid S_1$ and $2\nmid S_2$. In this case $4\mid \Delta$ and case (iii) can never occur. We get $U_2\equiv S_1+2\pmod{4}$ and $W_2-6R^2 \equiv 2 \pmod{4}$; thus, we see that cases (i) and (ii) can always occur, depending on the parity of $S_1/ 2$. In either of these cases, we get $4 \mid D_6$. It follows that if we eliminate the case of $2\mid S_1$ and $2\nmid S_2R$, then Thereom \ref{companion}, will be true for all primes $p$. Also, we have already seen in \S 3 that if we eliminate this case, then we have a rank of apparition $\omega$ of $2$ in $\{D_n\}$ and $2\mid D_n$ if and only if $\omega \mid n$; indeed, if $\gcd(m,R)=1$, there always exists a rank of apparition $\omega$ of $m$ in $\{D_n\}$ given by (\ref{omegam}) such that $m\mid D_n$ if and only if $\omega \mid n$. We remark here that if $S_1$ and $S_2$ are given by (\ref{relationship}), then if $2\nmid R$ and $2\mid S_1$, we must have $2\mid S_2$. Thus, for the sequences $\{c_n\}$ and $\{w_n\}$ we cannot have the case of $2\mid S_1$ and $2\nmid S_2R$. If $p$ is an I-prime and $p \equiv \epsilon = (\Delta/p) \pmod{3}$, then $3\mid p^2+\epsilon p+1$. Since we know in this case that $p\mid D_{p^2+\epsilon p+1}$, it is of some interest to determine a criterion for deciding whether or not $p\mid D_{(p^2+\epsilon p+1)/3}$. Roettger showed for the case of the $\{c_n\}$ and $\{w_n\}$ sequences that $p\mid D_{(p^2+ p+1)/3}$ ($\epsilon=1$ in this case if $p$ is an I-prime) if and only if $R^{(p-1)/3}\equiv 1 \pmod{p}$ in \cite[Theorem 5.14]{Roe09}. In what follows we will extend this result to the $\{U_n\}$ and $\{W_n\}$ sequences. We begin with three preliminary lemmas. \begin{lemma}\label{Iprime1} If $3W_1^2\equiv -\Delta \pmod{p}$, then $p$ cannot be an I-prime. \end{lemma} \begin{proof} We have $W_1=S_1$ and by (\ref{Delta2}) we find that \[ S_2 \equiv RS_1^2-2RS_1-4R^3 \pmod{p} \] and by (\ref{S3}) \[ S_3\equiv -RS_1^2-2R^2S_1+2R^3 \pmod{p}. \] Hence \[ g(x)\equiv (x+R)(x^2-(S_1+R)x+S_1^2+2RS_1-2R^2)\pmod{p}. \] Since $g(x)$ is reducible modulo $p$, $p$ cannot be an I-prime. \end{proof} \begin{lemma}\label{Isplit} Let $p$ be an I-prime and let $\mathbb{K}_p$ be the splitting field of $G(x) \in \mathbb{F}[x]$. If $\zeta$ is a primitive cube root of unity in $\mathbb{K}_p$, then in $\mathbb{K}_p$ we can have \begin{equation}\label{deltas} \zeta^k(\gamma_1+\gamma_2+\gamma_3)+\zeta^{-k}(\gamma_1^{-1}+\gamma_2^{-1}+\gamma_3^{-1})=\gamma_1+\gamma_2+\gamma_3+\gamma_1^{-1}+\gamma_2^{-1}+\gamma_3^{-1} \end{equation} if and only if $3\mid k$. \end{lemma} \begin{proof} If $3\mid k$ it is trivial that (\ref{deltas}) must hold. If $3\nmid k$, we first observe that $\zeta^k+\zeta^{-k}=-1$ and we have \begin{align*} \zeta^k+1/2=(\zeta^k-\zeta^{-k})/2, && \zeta^{-k}+1/2=(\zeta^{-k}-\zeta^{k})/2. \end{align*} Thus (\ref{deltas}) can hold only if \[ \frac{\zeta^k-\zeta^{-k}}{2}(\gamma_1+\gamma_2+\gamma_3-\gamma_1^{-1}-\gamma_2^{-1}-\gamma_3^{-1})=\frac{3}{2}(\gamma_1+\gamma_2+\gamma_3+\gamma_1^{-1}+\gamma_2^{-1}+\gamma_3^{-1}). \] On multiplying both sides by $2R$ and squaring we find that \[ 3W_1^2\equiv-\Delta \pmod{p}, \] which by the previous lemma is impossible. \end{proof} \begin{lemma} \label{IprimeDn} If $p$ is an I-prime and $p\mid U_n$, then $p\mid D_n$. \end{lemma} \begin{proof} Since $p\mid U_n$, we must have $\gamma_i^n=1$ in $\mathbb{K}_p$ for some $i \in \{1,2,3\}$ by (\ref{Un}). We may assume that $\gamma_1^n=1$. From the proof of \cite[Theorem 9.8]{RWG13}, we have $1=\gamma_1^{pn}=\gamma_2^{\pm n}$; hence, $\gamma_2^n=1$ and $\gamma_3^n=1/(\gamma_1^n\gamma_2^n)=1$. The result now follows by Lemma \ref{gammaK}. \end{proof} We are now able to derive our criterion for when $p\mid D_{(p^2+\epsilon p+1)/3}$. \begin{theorem}\label{CritD} If $p$ is an I-prime and $p\equiv \epsilon \pmod{3}$, then $p\mid D_{(p^2+\epsilon p+1)/3}$ if and only if \[ W_{(p-\epsilon)/3}^{*}\equiv R^{2(p-\epsilon)/3-1}W_1 \pmod{p}. \] \end{theorem} \begin{proof} We first note by Lemma \ref{IprimeDn} and \ref{gammaK} that $p\mid U_{(p^2+\epsilon p+1)/3}$ if and only if $\gamma_i^{(p^2+\epsilon p+1)/3}=1$ in $\mathbb{K}_p$ for all $i \in \{1,2,3\}$. Since $\gamma_1^{p^2+\epsilon p+1}=1$ in $\mathbb{K}_p$, we must have \[ \gamma_1^{\frac{p^2+\epsilon p+1}{3}}=\zeta^k, \] where $\zeta$ is a primitive cube root of unity in $\mathbb{K}_p$. It follows that\\ $p\mid D_{(p^2+\epsilon p+1)/3}$ if and only if $3\mid k$. Now \[ (p^2+\epsilon p+1)/3=(p-\epsilon)(p+2\epsilon)/3+1. \] Hence, \[ \zeta^k=\gamma_1^{(p^2+\epsilon p+1)/3}=(\gamma_1^{p+2\epsilon})^{(p-\epsilon)/3}\gamma_1. \] Since $\gamma_1^p=\gamma_2^\epsilon$ (see the proof of \cite[Theorem 9.8]{RWG13}), we get \[ \zeta^k=(\gamma_2\gamma_1^2)^{\epsilon(p-\epsilon)/3}\gamma_1=\gamma_3^{*\epsilon(p-\epsilon)/3}\gamma_1 \] and \[ \gamma_3^{*(p-\epsilon)/3}=(\zeta^k/\gamma_1)^\epsilon. \] Since $\gamma_3^{*p}=\gamma_1^p/\gamma_3^p=\gamma_2^\epsilon/\gamma_1^\epsilon=\gamma_1^{*\epsilon}$, we get \[ \gamma_1^{*\epsilon(p-\epsilon)/3}=(\zeta^{kp}/\gamma_1^p)^\epsilon=\zeta^k/\gamma_2 \] and\[ \gamma_1^{*(p-\epsilon)/3}=(\zeta^k/\gamma_2)^\epsilon. \] Similarly $\gamma_2^{*(p-\epsilon)/3}=(\zeta^k/\gamma_3)^\epsilon$. It follows that \[ W_{(p-\epsilon)/3}^{*}=R^{*(p-\epsilon)/3}[\zeta^{-k\epsilon}(\gamma_1^\epsilon+\gamma_2^\epsilon+\gamma_3^\epsilon)+\zeta^{k\epsilon}(\gamma_1^{-\epsilon} +\gamma_2^{-\epsilon}+\gamma_3^{-\epsilon})]. \] By Lemma \ref{Isplit}, we see that $3\mid k$ if an only if \[ W_{(p-\epsilon)/3}^{*}\equiv R^{2(p-\epsilon)/3-1}W_1 \pmod{p}. \] \end{proof} This criterion can easily be converted to one that involves only the $\{U_n\}$ and $\{W_n\}$ sequences by using (\ref{generalDelta}). At first glance, the criterion of Theorem \ref{CritD} does not resemble the more elegant rule for $p\mid D_{(p^2+\epsilon p+1)/3}$ when dealing with Roettger's sequences. In this case we have $\gamma_1=\alpha/\beta$, $\gamma_2=\beta/\gamma$, $\gamma_3=\gamma/\alpha$ and $R=\alpha \beta \gamma$. We can deduce Roettger's rule in the following corollary of Theorem \ref{CritD}. \begin{corollary} Suppose $D_n=\gcd(w_n-6R^n,c_n)$ and $p$ is an I-prime with respect to $h(x)\in \mathbb{F}_p[x]$, then if $p\equiv 1 \pmod{3}$, we have \[ p\mid D_{(p^2+\epsilon p+1)/3} \Leftrightarrow R^{(p-1)/3}\equiv 1\pmod{p}. \] \end{corollary} \begin{proof} Suppose first that $p\nmid \Gamma$. In this case $p$ is an I-prime with respect to $g(x) \in \mathbb{F}_p[x]$ and $1=(d/p)=(\Gamma\Delta/p)=(\Delta/p)=\epsilon$. By Theorem \ref{CritD} we have $p\mid D_{(p^2+\epsilon p+1)/3}$ if and only if $W_{(p-\epsilon)/3}^{*}\equiv R^{2(p-\epsilon)/3-1}W_1 \pmod{p}$. But in $\mathbb{K}_p$, we have $\gamma_1^{*}=\gamma_2/\gamma_1=\beta^2/(\alpha \gamma)=\beta^3/R$; hence, \[ \gamma_1^{*\frac{p-1}{3}}=\beta^{p-1}/R^{(p-1)/3}=(\alpha/\beta)/R^{(p-1)/3} =\gamma_2^{-1}/R^{(p-1)/3}. \] Similarly, $\gamma_2^{*\frac{p-1}{3}} = \gamma_3^{-1}/R^{(p-1)/3}$, $\gamma_3^{*\frac{p-1}{3}} =\gamma_1^{-1}/R^{(p-1)/3}$. It follows that \[ W_{\frac{p-1}{3}}^{*}=R^{*(p-1)/3}(R^{(p-1)/3}(\gamma_1+\gamma_2+\gamma_3)+R^{-(p-1)/3}(\gamma_1^{-1}+\gamma_2^{-1}+\gamma_3^{-1})) \] and by Lemma \ref{Isplit} $W_{\frac{p-1}{3}}^{*}\equiv R^{2(p-1)/3-1} W_1\pmod{p}$, if and only if $R^{(p-1)/3}=1$ in $\mathbb{K}_p$. Suppose next that $p\mid \Gamma$. In this case, $p$ cannot be an I-prime with respect to $g(x)$. If $p\nmid P$, then by (\ref{Gamma2}) we have $R\equiv (Q/P)^3 \pmod{p}$ and $h(Q/P)\equiv 0 \pmod{p}$. In this case $p$ is not an I-prime with respect to $h(x)$, a contradiction. If $p\mid P$, then $p\mid Q$ and $\alpha^3=\beta^3=\gamma^3=R$ in $\mathbb{L}_1$. We have $\alpha^{p-1}=\beta^{p-1}=\gamma^{p-1}=R^{(p-1)/3}$ and if $R^{(p-1)/3}\equiv 1\pmod{p}$, we get $\alpha^p=\alpha$, and $p$ is not an I-prime with respect to $h(x) \in \mathbb{F}_p[x]$, a contradiction. Now $p\mid D_3$ and since $3\nmid (p^2+\epsilon p+1)/3$, we have $p\nmid D_{(p^2+\epsilon p+1)/3}$. Thus, if $p$ is an I-prime with respect to $h(x) \in \mathbb{F}_p[x]$, then $R^{(p-1)/3}\not\equiv 1\pmod{p}$ and $p\nmid D_{(p^2+\epsilon p+1)/3}$. \end{proof} We conclude this section with the following result concerning \[ D_n^{*}=\gcd(W_n^{*}-6R^{*n},U_n). \] \begin{theorem} If $p$ is an I-prime and $p\equiv \epsilon \pmod{3}$, then $p\mid D_{(p^2+\epsilon p+1)/3}^{*}$. \end{theorem} \begin{proof} We observe as above that $\gamma_1^{*}=\gamma_2/\gamma_1$ and \[ (p^2+\epsilon p+1)/3=(p-\epsilon)(p+2\epsilon)/3+1. \] Hence \[ \gamma_1^{*(p^2+\epsilon p+1)/3}=(\gamma_2/\gamma_1)((\gamma_2/\gamma_1)^{p+2\epsilon})^{(p-\epsilon)/3} \] in $\mathbb{K}_p$. Now $\gamma_2^p=\gamma_3^\epsilon$, $\gamma_1^p=\gamma_2^\epsilon$; hence, \[ (\gamma_2/\gamma_1)^{p+2\epsilon}=(\gamma_2\gamma_3/\gamma_1^2)^\epsilon=\gamma_1^{-3\epsilon}. \] It follows that \[ ((\gamma_2/\gamma_1)^{p+2\epsilon})^{(p-\epsilon)/3}=\gamma_1^{-\epsilon(p-\epsilon}=\gamma_1/\gamma_2 \] and \[ \gamma_1^{*(p^2+\epsilon p+1)/3}=1. \] Hence, $p\mid D_{(p^2+\epsilon p+1)/3}^{*}$. \end{proof} \section{Some properties of $\{E_n\}$} We will devote the major portion of this section to the proof that if $p$ $(>3)$ is a prime and $p\mid E_n$, then $p\equiv (\Gamma/p)\pmod{3}$. This generalizes \cite[Theorem 6.2]{Roe09}. We observe that by Proposition \ref{gcdDnR1} we have $\gcd(E_n,R)=2$. We now need some preliminary results. \begin{lemma}\label{pEn} Let $p$ be any prime such that $p>3$. If $p\mid E_n$, then in $\mathbb{K}_p$ we must have \begin{align*} \gamma_i^n=1, && \gamma_j^{2n}+\gamma_j^n+1=0, \end{align*} where $i \in \{1,2,3\}$ and all $j \in \{1,2,3\}$ such that $j\not=i$. \end{lemma} \begin{proof} If $p\nmid \Delta$ and $p \mid U_n$, we may assume with no loss of generality that $\gamma_1^n=1$ in $\mathbb{K}_p$. If $p\mid \Delta$ we may assume with no loss of generality that $\gamma_1=1$ (and $\gamma_1^n=1$) in $\mathbb{K}_p$. Now \begin{eqnarray*} W_n=V_n-2R^n&=& R^n(1+\gamma_1^n)(1+\gamma_2^n)(1+\gamma_3^n)-2R^n\\ &=& 2R^n(\gamma_2^n\gamma_3^n+\gamma_2^n+\gamma_3^n)\\ &=& 2R^n(1+\gamma_2^n+1/\gamma_2^n)\\ &=& 2R^n(1+1/\gamma_3^n+\gamma_3^n), \end{eqnarray*} the latter results following from $\gamma_1^n=1$ and $\gamma_1^n\gamma_2^n\gamma_3^n=1$. Since $W_n=0$ in $\mathbb{K}_p$, we have $\gamma_2^{2n}+\gamma_2^n+1=\gamma_3^{2n}+\gamma_3^n+1=0$. \end{proof} \begin{lemma}\label{GamEn} If $p$ $ (>3)$ is a prime, then $p\nmid (E_n, \Gamma)$. \end{lemma} \begin{proof} If $p\mid \Gamma$, then $\gamma_1=\gamma_2$, $\gamma_2=\gamma_3$ or $\gamma_3=\gamma_1$ in $\mathbb{K}_p$ by (\ref{Gamma1}). If $p\mid E_n$, then we may assume that $\gamma_1^n=1$ and $\gamma_2^{2n}+\gamma_2^n+1=0$ in $\mathbb{K}_p$ by Lemma \ref{pEn}. If $\gamma_1=\gamma_2$, then $\gamma_2^n=1$, which is impossible because $p>3$. The same is true if $\gamma_2=\gamma_3$ or $\gamma_3=\gamma_1$. \end{proof} \begin{lemma} If $p$ $ (>3)$ is a prime, $p\mid \Delta$ and $p\mid E_n$, then \[ p\equiv (\Gamma/p)\pmod{3}. \] \end{lemma} \begin{proof} Since $p\mid \Delta$, we may assume with no loss of generality that $\gamma_1=1$ and therefore $\gamma_2\gamma_3=1$ in $\mathbb{K}_p=\mathbb{F}_{p^2}$. Also, by Lemma \ref{pEn} we may assume that if $p\mid E_n$, then \[ \gamma_2^{2n}+\gamma_2^n+1=0 \] in $\mathbb{K}_p$. Hence, $\gamma_2^{3n}=1$ and $\gamma_2^{n} \not=1$ in $\mathbb{K}_p$. By Lemma \ref{GamEn}, $p\nmid \Gamma$ and \begin{eqnarray}\nonumber \Gamma^{\frac{p-1}{2}}&=&(\gamma_1-\gamma_2)^{p-1}(\gamma_2-\gamma_3)^{p-1}(\gamma_3-\gamma_1)^{p-1}\\ \label{Gammastuff} &=&\frac{(1-\gamma_2^p)(\gamma_2^p-\gamma_3^p)(\gamma_3^p-1)}{(1-\gamma_2)(\gamma_2-\gamma_3)(\gamma_3-1)}. \end{eqnarray} If $\gamma_2\in \mathbb{F}_p$, then $\Gamma^{\frac{p-1}{2}}=1$. Also, from $\gamma_2^{pn}=\gamma_2^n$, we get $\gamma_2^{(p-1)n}=1$, which, since $\gamma_2^n\not=1$ means that $3\mid p-1$ and $p\equiv (\Gamma/p)\pmod{3}$. If $\gamma_2 \in \mathbb{F}_{p^2} \backslash\mathbb{F}_p$, then $\gamma_2^p=\gamma_3$ and $\gamma_2^{(p-1)n}=-1$ by (\ref{Gammastuff}). Since $\gamma_2^{pn}=\gamma_3^n=1/\gamma_2^n$ and $\gamma_2^{(p+1)n}=1$, we see that $3\mid p+1$ and $p\equiv (\Gamma/p)\pmod{3}$. \end{proof} We now show that if $p$ is an I-prime, then $p\nmid E_n$. \begin{theorem}\label{IprimeEnmidn} If $p$ is an I-prime, then $p\nmid E_n$. \end{theorem} \begin{proof} As noted above we know that if $p$ is an I-prime, then $\gamma_1^p=\gamma_2^\epsilon$, $\gamma_2^p=\gamma_3^\epsilon$, $\gamma_3^p=\gamma_1^\epsilon$ in $\mathbb{K}_p$. If $p\mid E_n$, then by Lemma \ref{pEn}, we have $\gamma_1^n=1$ and $\gamma_2^{2n}+\gamma_2^n+1=0$. Now $\gamma_2^{p^2}=\gamma_3^{\epsilon p}=\gamma_1^{\epsilon^2}=\gamma_1$ and $\gamma_2^{p^2n}=\gamma_1^n$. Hence, \[ 0=(\gamma_2^{2n}+\gamma_2^n+1)^{p^2}=3, \] which is a contradiction. \end{proof} We next deal with the case where $p\mid S_1+2R$. \begin{lemma} If $p$ $ (>3)$ is a prime, $p\nmid d$, $p\mid S_1+2R$ and $p\mid E_n$, then \[ p\equiv (\Gamma/p)\pmod{3}. \] \end{lemma} \begin{proof} Since $p\mid S_1+2R$ and $S_1+2R=R(\gamma_1+1)(\gamma_2+1)(\gamma_3+1)$, we may assume in $\mathbb{K}_p$ that $\gamma_1=-1$ and $\gamma_2\gamma_3=-1$. We get \[ (\gamma_1+\gamma_2)(\gamma_2+\gamma_3)(\gamma_3+\gamma_1)=-(\gamma_2^2+1/\gamma_2^2-2). \] Since $S_1\equiv -2R \pmod{p}$, we get $S_3\equiv -2RS_2 \pmod{p}$ from (\ref{S3}) and \[ g(x)=(x+2R)(x^2+S_2) \in \mathbb{F}_p[x]. \] Since $\rho_1=R(\gamma_1+1/\gamma_1)=-2R$, we get $\rho_2^2=\rho_3^2=-S_2$ and $\gamma_2^2+1/\gamma_2^2=\rho_2^2/R^2-2=-S_2/R^2-2\in \mathbb{F}_p$. It follows that $(\gamma_1+\gamma_2)(\gamma_2+\gamma_3)(\gamma_3+\gamma_1) \in \mathbb{F}_p$ and \begin{eqnarray}\nonumber ((\gamma_1^2-\gamma_2^2)(\gamma_2^2-\gamma_3^2)(\gamma_3^2-\gamma_1^2))^{p-1} \!\!\! &=& \!\!\!((\gamma_1-\gamma_2)^2(\gamma_2-\gamma_3)^2(\gamma_3-\gamma_1)^2)^{\frac{p-1}{2}}\nonumber \\ \!\!\!&=& \!\!\!(\Gamma/p).\label{Gamp} \end{eqnarray} As $\gamma_2^2+1/\gamma_2^2 \in \mathbb{F}_p$, we must have $\gamma_2^2$, $1/\gamma_2^2 \in \mathbb{F}_{p^2}$ and $\gamma_2^{2p}=\gamma_2^2$ or $\gamma_2^{2p}=\gamma_3^2$. Since $p\nmid d$, we see from (\ref{Gamp}), that $(\Gamma/p)=1$, when $\gamma_2^{2p}=\gamma_2^2$ and $(\Gamma/p)=-1$, when $\gamma_2^{2p}=\gamma_3^2$. If $p\mid E_n$, then by Lemma \ref{pEn}, we have $\gamma_i^n=1$ for some $i \in \{1,2,3\}$ and $\gamma_j^{2n}+\gamma_j^n+1=0$ $(i\not=j)$. Since $\gamma_1=-1$, we see that $i=1$ and $2\mid n$. If $(\Gamma/p)=1$, then $\gamma_2^{np}=\gamma_2^n$ and $\gamma_2^{n(p-1)}=1$. Since $\gamma_2^{3n}=1$ and $\gamma_2^n\not=1$, we see that $3\mid p-1$ and $p\equiv (\Gamma/p)\pmod{3}$. If $(\Gamma/p)=-1$, then $\gamma_2^{np}=\gamma_3^n=1/\gamma_2^n$ and $\gamma_2^{n(p+1)}=1$; hence $3\mid p+1$ and $p\equiv (\Gamma/p)\pmod{3}$. \end{proof} We are now ready to prove our main result. \begin{theorem}\label{mainresult} If $p$ $ (>3)$ is a prime divisor of $E_n$, then $p\equiv (\Gamma/p)\pmod{3}$. \end{theorem} \begin{proof} We have already proved this result when $p\mid d$ and when $p\nmid d$ and $p\mid S_1+2R$. We may assume, then, that $p\nmid d$ and $p\nmid S_1+2R$. Since $p\mid E_n$, $p$ can only be an S-prime or a Q-prime by Theorem \ref{IprimeEnmidn}. If $p$ is an S-prime, then $1=(d/p)=(\Delta/p)(\Gamma/p)$ and $(\Gamma/p)=\epsilon$; if $p$ is an Q-prime, then $-1=(d/p)=(\Delta/p)(\Gamma/p)$ and $(\Gamma/p)=-\epsilon$. Suppose $p$ is an S-prime. By results in the proof of \cite[Theorem 9.4]{RWG13}, we have $\gamma_i^p=\gamma_i^\epsilon$ $(i=1,2,3)$ in $\mathbb{K}_p$. By Lemma~\ref{pEn}, we get $\gamma_2^{3n}=1$, $\gamma_2^n\not=1$; also, $\gamma_2^{np}=\gamma_2^{n\epsilon}$ means that $\gamma_2^{(p-\epsilon)n}=1$ and $3\mid p-\epsilon$. Similarly, if $p$ is a Q-prime, then by the results in the proof of \cite[Theorem 9.6]{RWG13}, we have \begin{align*} \gamma_2^p=\gamma_3^\epsilon, && \gamma_3^p=\gamma_2^\epsilon, && \gamma_3^p=\gamma_1^\epsilon \end{align*} in $\mathbb{K}_p$. In this case we get $\gamma_2^{pn}=\gamma_3^{\epsilon n}= (1/\gamma_2)^{\epsilon n}$ and $\gamma_2^{n(p+\epsilon)}=1$, $\gamma_2^{3n}=1$ and $\gamma_2^n\not=1$. Hence $3\mid p+\epsilon$ and in either case $p\equiv (\Gamma/p)\pmod{3}$. \end{proof} In order to extend Theorem \ref{mainresult}, we need to prove the following result. \begin{theorem} \label{extend} For any $n>0$, we have $E_n\mid D_{3n}$. \end{theorem} \begin{proof} We can rewrite (\ref{tripWn}) as \begin{equation}\label{modifiedtripWn} W_{3n}-6R^{3n}=(W_n-6R^n)\tilde{Q}_n+\Delta W_n U_n^2, \end{equation} where $\tilde{Q}_n=(W_n^2-\Delta U_n)/4$. Suppose $p$ is any odd prime and $p^\lambda\mid\mid E_n$, where $\lambda \geq 1$. Since $p^\lambda \mid U_n$, we must have $p^\lambda \mid U_{3n}$. Also, $p^{2\lambda}\mid \tilde{Q}_n$ and $p^\lambda \mid W_{3n}-6R^{3n}$ by (\ref{modifiedtripWn}). Next, suppose that $2^\lambda\mid \mid E_n$ and $\lambda \geq 1$. We have $2\mid W_n-6R^n$ and $2^{2\lambda-2}\mid \tilde{Q}_n$, $2^\lambda \mid U_n$. By (\ref{modifiedtripWn}) we see that $2^{2\lambda-1}\mid W_{3n}-6R^{3n}$ and since $\lambda \geq 1$, we have $2\lambda-1\geq \lambda$ and $2^\lambda \mid D_{3n}$. Hence, $E_n\mid D_{3n}$. \end{proof} We next prove a result which is analogous to the theorem that states that if $p$ is an odd prime and $p\mid v_n$, then $p\equiv \pm 1 \pmod{2^{\nu+1}}$, where $2^\nu\mid\mid n$. (See \cite[Theorem 2.20]{Roe09}). \begin{theorem} \label{Ennu} If $p$ $(>3)$ is a prime and $p\mid E_n$, then $p\equiv (\Gamma/p) \pmod{3^{\nu+1}}$, where $3^\nu\mid\mid n$. \end{theorem} \begin{proof} Since $p\mid E_n$ and $p>3$, we have $p\nmid D_n$, as $p\nmid 6R$. But, by Theorem \ref{extend}, we know that $p\mid D_{3n}$. Thus, if $\omega$ is the rank of apparition of $p$ in $\{D_n\}$, we have $\omega \mid 3n$ and $\omega\nmid n$. It follows that $3^{\nu+1}\mid \omega$. Also, since $p$ is not an I-prime and $p\nmid 6R$, we must have $\omega=p$ or $\omega \mid p^2-1$ by results in \S 3. Since $3\mid \omega$ we cannot have $\omega=p$ and therefore $\omega\mid p^2-1$ and $3^{\nu+1}\mid p^2-1$. Since $p\nmid \Gamma$, we have $p^2-1=(p-(\Gamma/p))(p+(\Gamma/p))$ and $3\mid p-(\Gamma/p)$. Hence $3^{\nu+1}\mid p-(\Gamma/p)$. \end{proof} \section{Primality tests} In Williams \cite{williams}, it is shown how Lucas used the properties of $\{u_n\}$ and $\{v_n\}$ to develop primality tests for certain families of integers. In this section we will indicate how the properties of $\{U_n\}$ and $\{W_n\}$ can be used to produce some primality tests. We begin with a simple result concerning integers of the form $A3^n+\eta$, where $\eta^2=1$. \begin{theorem}\label{primetest} Let $N=A3^n+\eta$, where $2\mid A$, $n\geq 2$, $3\nmid A$, $\eta\in \{1,-1\}$ and $A<3^n$. If \[ N\mid U_{N-\eta}/U_{(N-\eta)/3}, \] then $N$ is a prime. \end{theorem} \begin{proof} Let $p$ be any prime divisor of $N$ and put $m=(N-\eta)/3$. We note that $p\not=2,3$ and by (\ref{tripUn}) \[ 4U_{3m}/U_m=3W_m^2+\Delta U_m^2. \] Since $p\mid U_{3m}$, there must exist some rank of apparition $r$ of $p$ in $\{U_n\}$ such that $r\mid 3m$. If $p\mid U_m$ and $p\mid W_m$, then $p\mid E_m$ and $p\equiv (\Gamma/p)\pmod{3^n}$ by Theorem \ref{Ennu}. If $p\nmid U_m$, then $r\nmid m$ and $r\mid 3m$ means that $3^n\mid r$. Suppose $p\nmid dR$. If $p$ is an S-prime or a Q-prime, then by \cite[Corollary 9.5 and Theorem 9.7]{RWG13} we must have $r\mid p-\epsilon$, where $\epsilon =(\Delta/p)$; hence $p\equiv (\Delta/p)\pmod{3^n}$. If $p$ is an I-prime, then $r\mid p^2+\epsilon p+1$ by Theorem 9.9 of \cite{RWG13}. Since $9\mid r$, this is impossible. If $p\mid dR$, then $r=3, p$ or divides $p\pm 1$. Since $9\mid r$, $r\not=3$ and since $p\nmid N-\eta$, we cannot have $r=p$. Thus, in all possible cases, we find that $p\equiv \pm 1\pmod{3^n}$ and since $p$ is odd, we have $p\geq 2 \cdot 3^n-1$. Since $( 2 \cdot 3^n-1)^2>N$, $N$ can only be a prime. \end{proof} \noindent We also note that if $N$ obeys the conditions in the first line of Theorem \ref{primetest} and $N\mid E_{(N-\eta)/3}$, then $N$ must be a prime. By extending the results in \cite[Chapter 7]{Roe09} it is possible to select the parameters of $S_1$, $S_2$ to make Theorem \ref{primetest} both a necessary and sufficient test for the primality of $N$, but this test is much less efficient than one based on the Lucas Functions. In \cite[\S 9]{RWG13} several primality tests for $N$ are presented. These tests can be easily proved by using the techniques in \cite[Chapter 7]{Roe09}, but to be usable they require that we know the complete factorization of \[ N^2+N+1 \quad \mbox{or} \quad N^2-N+1. \] Of course, such a circumstance is very unlikely, but we might have a partial factorization of $N^2\pm N+1$. In what follows we will devise a test for the primality of $N$ in this case. We first require a simple lemma. \begin{lemma}\label{simplelemma} If $p$ and $q$ are distinct primes, $p>3$ and $p\mid D_{qn}$ and $p\mid U_{qn}/U_n$, then $q^{\lambda+1}\mid \omega$, where $\omega$ is the rank of apparition of $p$ in $\{D_n\}$ and $q^\lambda\mid\mid n$. \end{lemma} \begin{proof} Suppose $p\mid D_n$. If $p\mid U_{qn}/U_n$, then by Theorem \ref{useful}, we get $p \mid 2q^3$, which is impossible. Hence, $p\nmid D_n$. It follows that since $p\mid D_{qn}$ ($\{D_n\}$ is a divisibility sequence), we get $\omega\mid qn$ and $\omega \nmid n$, which means that $q^{\lambda+1}\mid \omega$. \end{proof} We will also need the easily established technical lemma below. \begin{lemma}\label{tech} If $x\geq 5$, then \[ (x^2+x+1)^2<2(x^4-x^2+1). \] \end{lemma} \begin{theorem} Let $N$ be a positive integer such that $\gcd(N,6)=1$ and put $\eta=1$ or $-1$. Let $T=N^2+\eta N+1$ and suppose that $T^{\prime}\mid T$, where $\gcd(T^{\prime}, T/T^{\prime})=1$ and $T^{\prime 2}>2T$. If $N\mid D_T$ and $N\mid U_T/U_{T/q}$ for all distinct primes $q$ such that $q\mid T^{\prime}$, then $N$ is a prime. \end{theorem} \begin{proof} Let $p$ be any prime divisor of $N$ and $q$ be any prime divisor of $T^{\prime}$; then $p\geq 5$ and by Lemma~\ref{simplelemma} we have $q^\lambda\mid \omega(p)$, where $\omega(p)$ is the rank of apparition of $p$ in $\{D_n\}$ and $q^\lambda\mid \mid T$. Since $\gcd(T^{\prime}, T/T^{\prime})=1$, we have $q^\lambda\mid \mid T^{\prime}$; hence, $T^{\prime}\mid \omega(p)$. Let $\omega$ denote the rank of apparition of $T$ in $\{D_n\}$. We have $\omega\mid T$ and $\omega/q\nmid T$; hence, $q^\lambda\mid \omega$, where $q^\lambda \mid \mid T$ and therefore $T^{\prime}\mid \omega$. By (\ref{omegam}), we have \[ \omega= \lcm (\omega(p_i^{\alpha_i}):i=1,2,\dots,j), \] where \[ N=\prod_{i=1}^j p_i^{\alpha_i} \] is the prime power factorization of $N$. Since $\omega(p_i^{\alpha_i})=p_i^{\nu_i}\omega(p_i)$, we must have $\nu_i=1$ because $p_i\nmid T$. We get \[ \omega= \lcm (\omega(p_i):i=1,2,\dots,j_\mid T^{\prime} \prod_{i=1}^j\frac{\omega(p_i)}{T^{\prime}}. \] If we put $T=k\omega$, then \[ T\leq kT^{\prime} \prod_{i=1}^j\frac{\omega(p_i)}{T^{\prime}}\leq k T^{\prime} \prod_{i=1}^j\frac{p_i^2+p_i+1}{T^{\prime}} \] by Theorem \ref{SQIDn}. Also, since \[ T=N^2+\eta N+1 > 2 \prod_{i=1}^j\frac{p_i^2+p_i+1}{2}, \] (\cite[Lemma 9.11]{RWG13}, cf.~\cite[Lemma 7.1]{Roe09}) we get \[ k T^{\prime} \prod_{i=1}^j\frac{p_i^2+p_i+1}{T^{\prime}}>2 \prod_{i=1}^j\frac{p_i^2+p_i+1}{2} \] and \[ k T^{\prime} 2^j > 2(T^{\prime})^j. \] Hence, \[ k>(T^{\prime}/2)^{j-1} \geq T^{\prime}/2 \quad \mbox{(when $j\geq 2$)}. \] But since $T/T^{\prime}=k\omega/T^{\prime}$, we have $k\leq T/T^{\prime}2T$, we have for $\alpha \geq 2$ \[ (p^2+p+1)^2>2(p^{2\alpha}+\eta p^\alpha+1)\geq 2(p^{2\alpha}- p^\alpha+1) \geq2(p^4-p^2+1) \] which is impossible by Lemma~\ref{tech}. Hence we can only have $N=p$. \end{proof} Many other primality tests can be devised by making use of the ideas in \cite[Chapter 7]{Roe09}, but the above should suffice to illustrate the kind of results that can be established. \section{Conclusions} In \cite{RWG13} we showed that the $\{U_n\}$ and $\{W_n\}$ sequences can be considered, respectively, as the sextic analogues of Lucas' $\{u_n\}$ and $\{v_n\}$ sequences. In this paper we have produced a number of results that are the number-theoretic analogues of well-known properties of the Lucas functions. Of course, there are many other properties of $\{D_n\}$ and $\{E_n\}$ that are similar to those of the $\{D_n\}$ and $\{E_n\}$ sequences discussed at some length in \cite{Roe09}, and these can be proved by using the results presented here and the techniques of \cite{Roe09}. \begin{thebibliography}{1} \bibitem{MRW09} S.~M{\"u}ller, H.~C. Williams, and E.~Roettger, A cubic extension of the {{L}}ucas functions, {\em Ann. Sci. Math. Qu{\'{e}}bec} {\bf 33} (2009), 185--224. \bibitem{Roe09} E.~Roettger, {\em A Cubic Extension of the {L}ucas Functions}, PhD thesis, University of Calgary, 2009. \newblock Available online at \url{http://people.ucalgary.ca/~williams/}. \bibitem{RWG13} E.~L. Roettger, H.~C. Williams, and R.~K. Guy, Some extensions of the {L}ucas functions. \newblock In {\em Number {T}heory and {R}elated {F}ields}, Springer, 2013, pp.~279--319. \bibitem{williams} H.~C. Williams, {\em {{\'{E}}}douard {L}ucas and Primality Testing}, Wiley-Interscience, 1998. \end{thebibliography} \bigskip \hrule \bigskip \noindent 2010 {\it Mathematics Subject Classification}: Primary 11B37; Secondary 11Y11, 11B50. \noindent \emph{Keywords: } linear recurrence, Lucas function, primality testing. \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received February 5 2015; revised version received May 11 2015; May 29 2015. Published in {\it Journal of Integer Sequences}, May 30 2015. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document}