\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} \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 A New Kind of Fibonacci-Like Sequence \\ \vskip .11in of Composite Numbers } \vskip 1cm \large Dan Ismailescu\\ Department of Mathematics \\ Hofstra University\\ 103 Hofstra University \\ Hempstead, NY 11549 \\ USA\\ \href{mailto:dan.p.ismailescu@hofstra.edu}{\tt dan.p.ismailescu@hofstra.edu}\\ \ \\ Jaesung Son\\ Ridgewood High School\\ 627 East Ridgewood Ave\\ Ridgewood, NJ 07450 \\ USA\\ \href{mailto:sonjaesung@gmail.com}{\tt sonjaesung@gmail.com }\\ \end{center} \vskip .2 in \begin{abstract} An integer sequence $(x_n)_{n\ge 0}$ is said to be {\it Fibonacci-like} if it satisfies the binary recurrence relation \begin{equation*} x_n=x_{n-1}+x_{n-2}, \quad n\geq 2. \end{equation*} We construct a new type of Fibonacci-like sequence of composite numbers. \end{abstract} \section{The problem and previous results}\label{fibo} In this paper we consider {\it Fibonacci-like} sequences, that is, sequences $({x_n})_{n=0}^\infty$ satisfying the binary recurrence relation \begin{equation}\label{recrel} x_n=x_{n-1}+x_{n-2}, n\geq 2. \end{equation} If $x_0=0$ and $x_1=1$ then $x_n=F_n$, the classical Fibonacci sequence. Similarly, when $x_0=2$ and $x_1=1$ then $x_n=L_n$, the Lucas sequence. Graham \cite{Graham} proved that there exist relatively prime positive integers $x_0$ and $x_1$ such that the sequence $({x_n})_{n=0}^\infty$ defined by the recurrence above contains no prime numbers; $x_0$ and $x_1$ have 33 and 34 digits, respectively. Knuth \cite{Knuth} improved on Graham's method and found a 17-digit pair. Soon after, Wilf \cite{Wilf} discovered a smaller 17-digit pair. Nicol \cite{Nicol} refined Knuth's idea and found a 12-digit pair. Finally, Vsemirnov \cite{Vsemirnov} found a smaller pair of 12 and 11 digits: $x_0=106276436867$, $x_1=35256392432$. Let us describe the common idea used in proving these results. Start by looking for a finite set of quadruples $(p_i, m_i, r_i,c_i)$, $1\le i\le t$ with the following properties: \begin{itemize} \item[(a)]{each $p_i$ is a prime;} \item[(b)]{every $p_i$ divides $F_{m_i}$, the $m_i$-th Fibonacci number;} \item[(c)]{every positive integer $n$ satisfies a congruence $n\equiv r_i$ (mod $m_i$) for some $i=1,\,2,\ldots,\,t$. In other words, $\{(r_i,m_i)\}_{i=0}^{i=t}$ is a {\it covering system of the integers.}} \end{itemize} Next, define $x_0$ and $x_1$ as follows: \begin{equation}\label{define} x_0\equiv c_iF_{m_i-r_i}\pmod{p_i} \qquad \text {and} \quad x_1\equiv c_iF_{m_i-r_i+1}\pmod{p_i} \quad \text{for}\,\, i=1,\,2,\,\ldots,\, t. \end{equation} From the recurrence relation \eqref{recrel} it follows that in general, $x_n\equiv c_iF_{n+m_i-r_i}$ (mod $p_i$). The divisibility property $F_m\,|\,F_{sm}$ and condition (b) imply that $p_i\,|\,x_n$ if $n\equiv r_i$ (mod $m_i$). Since $x_n$ is an increasing sequence and all primes $p_i$ are relatively small, condition (c) guarantees that $({x_n})_{n=0}^\infty$ contains only composite numbers. The role of the parameters $c_i$ is to minimize the solution corresponding to a given covering system. As mentioned earlier, the current record is due to Vsemirnov whose construction is based on the following set of $t=17$ quadruples $(p_i, m_i, r_i, c_i)$: \medskip \begin{table}[ht] \begin{tabular} {|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline $p_i$&3&2&5&7&17&11&47&19&61&23&107&31&1103&181&41&541&2521\\ \hline $m_i$&4&3&5&8&9&10&16&18&15&24&36&30&48&90&20&90&60\\ \hline $r_i$&3&1&4&5&2&6&9&14&12&17&8&0&33&80&18&62&48\\ \hline $c_i$&2&1&2&3&5&6&34&14&29&6&19&21&9&58&11&185&306\\ \hline \end{tabular} \bigskip \caption{Vsemirnov's quadruples} \end{table} Graham, Knuth and Wilf used similar covering systems except with primes $2207$, $1087$, $4481$, $53$, $109$ and $5779$ instead of $23$, $1103$, $107$, $181$ and $541$. Nicol used primes $53$, $109$, $5779$ instead of $107$, $181$, $541$. A major factor in deciding the size of a solution is the product of the primes in the covering system: $P=\prod_{i=1}^t p_i$. The smaller the value of $P$, the greater the chance to find a smaller solution. Of all constructions mentioned above, Vsemirnov's attains the smallest $P$. It is not known whether a better covering system can be found. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{A new construction} We note that Izotov \cite{Izotov} was the first to propose an alternative approach to a different problem, namely that of construction of Sierpi\'{n}ski numbers, for which the only known solutions involved the use of covering systems. In fact, Erd\H{o}s \cite[Section F13]{Guy} conjectured that Sierpi\'{n}ski numbers could only be constructed by the use of covering systems. Similarly, all known examples of Fibonacci-like sequences of composite numbers are based on the existence of a finite covering set of primes $\{p_1,p_2,\ldots,p_t\}$. In other words, all examples mentioned in the previous section have the property that for every positive integer $n$, there exists an $i\in\{1,2,,\ldots,t\}$ with $x_n\equiv 0$ (mod $p_i$). In this paper we construct a Fibonacci-like sequence of composite numbers for which such a covering set does not appear to exist. Our approach can be summarized as follows: On one hand, we are going to choose two relatively prime positive integers $x_0$ and $x_1$, such that for every nonnegative integer $n$, $x_{2n+1}$ is equal to the product of two integers greater than $1$, both of which can be written explicitly in terms of $n$, $x_0$ and $x_1$. On the other hand, we are going to find a finite set of prime numbers $\{p_1,p_2,\ldots,p_t\}$ such that for every nonnegative integer $n$, $x_{2n}\equiv 0$ (mod $p_i$) for some $i\in\{1,2,\ldots,t\}$. We will thus obtain the desired Fibonacci-like sequence of composite numbers. Notice that there are different reasons why $x_n$ is composite depending on the parity of $n$. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{theorem}\label{odd} Consider the sequence given by $x_0=p^2+q^2$, $x_1=2pq+q^2$, $x_n=x_{n-1}+x_{n-2}$ for all $n\ge 2$, where $p$ and $q$ are integers. Then for every $n\ge 0$ we have \begin{equation}\label{eqodd} x_{2n+1}=(pF_n+qF_{n+1})(pL_n+qL_{n+1}). \end{equation} In particular, if $p\ge 1$ and $q\ge 2$, then $x_{2n+1}$ is composite for all $n\ge 0$. \end{theorem} \begin{proof} It is known that the Fibonacci numbers can be written in matrix form as below \begin{equation*} \begin{bmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{bmatrix} = {\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}}^n \quad {\rm for\,\, all\,\,} n\ge 1. \end{equation*} Using the fact that $A^{m+n}=A^mA^n$ with $A=\bigl[\begin{smallmatrix}1&1\\ 1&0 \end{smallmatrix} \bigr]$ it follows that \begin{equation*} \begin{bmatrix} F_{m+n+1} & F_{m+n} \\ F_{m+n} & F_{m+n-1} \end{bmatrix}= \begin{bmatrix} F_{m+1} & F_m \\ F_m & F_{m-1} \end{bmatrix}\cdot \begin{bmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{bmatrix} \quad {\rm from\,\, which} \end{equation*} \begin{equation*}\label{mn} F_{m+n+1}=F_{m+1}F_{n+1}+F_mF_n. \end{equation*} In particular, taking $m=n$ and $m=n-1$ respectively, we obtain \begin{align} &F_{2n+1}=F^2_{n+1}+F_n^2,\label{2n+1}\\ &F_{2n}=2F_nF_{n+1}-F^2_n\label{2n}. \end{align} It is easy to prove by induction that $x_n=x_0F_{n-1}+x_1F_n$\label{x_n}. It follows that \begin{equation*} x_{2n+1}=x_0F_{2n}+x_1F_{2n+1} \end{equation*} which after using equalities \eqref{2n+1} and \eqref{2n} gives \begin{equation}\label{xodd} x_{2n+1}=x_0(2F_nF_{n+1}-F_n^2)+x_1(F_{n+1}^2+F_n^2)=(x_1-x_0)F_n^2+2x_0F_nF_{n+1}+x_1F_{n+1}^2. \end{equation} Regard the right hand term in the above equation as a quadratic form in the variables $F_n$ and $F_{n+1}$. We want this quadratic form to be reducible over the integers for all $n$, which is equivalent to requiring the discriminant $\Delta=x_0^2-(x_1-x_0)x_1$ to be a perfect square. In other words we want to choose $x_0$ and $x_1$ as solutions of the diophantine equation \begin{equation}\label{perfectsquare} x_0^2+x_0x_1-x_1^2=k^2. \end{equation} It is straightforward to check that $x_0=p^2+q^2$, $x_1=2pq+q^2$ is a solution of the above equation. Indeed, with the above choices for $x_0$ and $x_1$ we have \begin{align*} x_0^2+x_0x_1-x_1^2&=(p^2+q^2)^2+(p^2+q^2)(2pq+q^2)-(2pq+q^2)^2=\\ &=p^4+2p^3q-p^2q^2-2pq^3+q^4=(p^2+pq-q^2)^2. \end{align*} This solution can be obtained by using the techniques for solving the general equation $ax^2+bxy+cy^2=k^2$; the interested reader may consult \cite[Chapter XIII, pp.\ 404--409]{Dickson}. This explains our choices for $x_0$ and $x_1$. Substituting now $x_0=p^2+q^2$ and $x_1=2pq+q^2$ into \eqref{xodd} we obtain \begin{align*} x_{2n+1}&=(2pq-p^2)F_n^2+2(p^2+q^2)F_nF_{n+1}+(2pq+q^2)F^2_{n+1}=\\ &=p^2(2F_nF_{n+1}-F_n^2) + pq(2F_n^2+2F_{n+1}^2) + q^2(2F_nF_{n+1} + F_{n+1}^2)=\\ &=p^2F_n(2F_{n+1}-F_n) + 2pq(F_n^2+F_{n+1}^2) + q^2F_{n+1}(2F_n + F_{n+1}). \end{align*} We use now some basic identities relating the Fibonacci and the Lucas numbers. \begin{align*} 2F_{n+1} - F_n &= F_{n+1} + (F_{n+1} - F_n) = F_{n+1} + F_{n-1} = L_n.\\ 2F_n + F_{n+1} &= F_n + (F_n + F_{n+1}) = F_n + F_{n+2} = L_{n+1}.\\ 2(F_n^2+F_{n+1}^2)&=F_n(2F_n+F_{n+1})+F_{n+1}(2F_{n+1}-F_n)=F_nL_{n+1}+F_{n+1}L_n. \end{align*} Substituting these into the last equality, we obtain \begin{equation*} x_{2n+1}=p^2F_nL_n+pq(F_nL_{n+1}+F_{n+1}L_n)+q^2F_{n+1}L_{n+1}= (pF_n + qF_{n+1})\cdot(pL_n + qL_{n+1}). \end{equation*} Thus, equation \eqref{eqodd} is satisfied. If $p\ge 1$ and $q\ge 2$ it follows that for all nonnegative integers $n$ we have that $pF_n + qF_{n+1}\ge q\ge 2$ and $pL_n +qL_{n+1}>p+q\ge 3$. It follows that $x_{2n+1}$ is always composite. The proof is complete. \end{proof} From Theorem \ref{odd} it follows that if one chooses $x_0=p^2+q^2$ and $x_1=2pq+q^2$, then for every $n\ge 0$ we have that $x_{2n+1}$ is composite. It remains to ensure that for every $n\ge 0$, $x_{2n}$ is also composite. In order to achieve this, we construct a finite partial covering set as described below. We are looking for a collection of quadruples $\{(p_i,m_i,r_i,c_i)\}^{i=t}_{i=1}$ such that \begin{itemize} \item[(a)] {each $p_i$ is a prime;} \item[(b)] {every $p_i$ divides $F_{m_i}$, the $m_i$-th Fibonacci number;} \item[(c)] {every even positive integer $2n$ satisfies at least a congruence $2n\equiv r_i$ (mod $m_i$) for some $i=1,\,2,\ldots,\,t$. In other words, $\{(r_i,m_i)\}_{i=0}^{i=t}$ is a partial covering system, as it covers all even integer values.} \end{itemize} For every $1\le i\le t$, we have $1\le c_i \le p_i-1$. These values are going to come into play later on, as we will require a certain system of congruences to be compatible. Suppose we found such a set of quadruples. Choose $x_0$ and $x_1$ such that \begin{equation}\label{choicex0x1} x_0 \equiv c_iF_{m_i-r_i}\pmod{p_i},\quad x_1 \equiv c_iF_{m_i-r_i+1} \pmod{p_i}\quad \text{for all}\,\, 1\le i\le t. \end{equation} Let $n\ge 0$ and let $1\le i\le t$ be such that $2n\equiv r_i$ (mod $m_i$). The existence of such $i$ is guaranteed by condition (b). From \eqref{x_n}, we have \begin{align*} x_{2n} = x_0F_{2n-1} + x_1F_{2n} &\equiv c_i(F_{m_i-r_i}\cdot F_{2n-1} + F_{m_i-r_i+1}\cdot F_{2n})\pmod{p_i}\\ &\equiv c_i\, F_{2n+m_i-r_i} \pmod{p_i} \\ &\equiv c_i \, F_{sm_i} \pmod{p_i} \\ & \equiv 0 \pmod{p_i}. \end{align*} Thus, $x_{2n}\equiv 0$ (mod $p_i$) and therefore composite. There is however one major difficulty in finding a partial covering system with the properties (a), (b) and (c) listed above, as every odd prime $p_i$ has to satisfy the congruence $p_i\equiv 1$ (mod $4$). Let us explain why that is the case. First, we need the following simple result. \begin{lemma}\label{1mod4} For any positive odd integer $m$, the Fibonacci number $F_m$ has no prime factors of the form $4l+3$. \end{lemma} \begin{proof} Let $m$ be odd and let $p\,|\,F_m$. From Cassini's identity, we have $F_{m+1}^2-F_mF_{m+2} = (-1)^m$, which after reducing modulo $p$ gives $F_{m+1}^2 \equiv -1$ (mod $p$). This implies that $-1$ is a quadratic residue modulo $p$. By the quadratic reciprocity law this is true only when $p \equiv 1$ (mod 4). \end{proof} Recall that in Theorem \ref{odd} we chose $x_0$ and $x_1$ such that $x_0^2 + x_0x_1 - x_1^2 = k^2$. On the other hand, we selected $x_0$ and $x_1$ as in \eqref{choicex0x1}. It follows that for every $1\le i\le t$ we have that \begin{equation} c_i^2(F_{m_i-r_i}^2 + F_{m_i-r_i}\cdot F_{m_i-r_i+1} - F_{m_i-r_i+1}^2) \equiv k^2 \pmod{p_i} \end{equation} Using Cassini's identity again, the above congruence can be simplified to $c_i^2\cdot (-1)^{m_i-r_i+1} \equiv k^2 \pmod {p_i}$. In particular, $(-1)^{m_i-r_i+1}$ is a quadratic residue modulo $p_i$ for every $1 \le i \le t$. We want that for every $n\ge 0$, the congruence $2n\equiv r_i\pmod{m_i}$ holds for some $1\le i\le t$. If $m_i$ is even, then $r_i$ is even as well, and therefore $(-1)^{m_i-r_i+1}=-1$, which is a quadratic residue modulo $p_i$ if and only if $p_i\equiv 1\pmod{4}$. If $m_i$ is odd, then condition (b) states that $p_i\,|\,F_{m_i}$, and hence $p_i\equiv 1$ (mod $4$) follows from Lemma \ref{1mod4}. Hence, none of the primes $p_i$ in the partial covering system with properties (a), (b) and (c) can be of the form $4l+3$. Consider the set $\{(p_i, m_i, r_i, c_i)\}_{i=1}^{i=30}$ given in Table \ref{table2}. It is straightforward to check that this system of quadruples has the desired properties. \medskip \begin{table}[ht]\label{table2} \begin{centering} \begin{tabular} {|c|c|c|c|} \hline $p_i$ & $m_i$ & $r_i$ & $c_i$\\ \hline 2&3&1&1\\ \hline 5&5&1&2\\ \hline 13&7&1&5\\ \hline 17&9&3&11\\ \hline 29&14&2&5\\ \hline 41&20&4&3\\ \hline 61&15&2&41\\ \hline 181&90&8&46\\ \hline 241&120&14&109\\ \hline 281&28&4&207\\ \hline 421&21&3&171\\ \hline 541&90&38&243\\ \hline 1009&126&90&294\\ \hline 1601&80&34&1259\\ \hline 2161&40&10&1706\\ \hline \end{tabular} \qquad\qquad \begin{tabular} {|c|c|c|c|} \hline $p_i$ & $m_i$ & $r_i$ & $c_i$\\ \hline 2521&60&20&636\\ \hline 3041&80&74&790\\ \hline 8641&360&18&4664\\ \hline 20641&120&110&1405\\ \hline 31249&126&42&901\\ \hline 103681&72&54&80856\\ \hline 109441&45&23&16635\\ \hline 141961&35&12&12156\\ \hline 721561&420&180&529617\\ \hline 1461601&252&186&970625\\ \hline 35239681&63&6&25860534\\ \hline 764940961&252&0&562105967\\ \hline 8288823481&105&33&83463210\\ \hline 10783342081&180&162&7785411056\\ \hline 571385160581761&504&222&49367403415248\\ \hline \end{tabular} \bigskip \caption{A finite partial covering system with no primes $\equiv 3\pmod{4}$} \end{centering} \end{table} \medskip \noindent Let us notice that in order to check condition (c), it suffices to test the even numbers $\le 5040$. This is indeed the case since $5040$ is the least common multiple of all the $m_i$, $1\le i\le 30$. We are now in position to state the main result of this paper. \begin{theorem}\label{mainthm} Let $p=1$ and $q=12951150255508108245872399074061259209531943793351-\\ 2025195406541068394745828231264515958532145970461367703231950382110924410768870$. Define a sequence $(x_n)_{n\ge 0}$ by $x_0=p^2+q^2$, $x_1=2pq+q^2$, $x_n=x_{n-1}+x_{n-2}$, for all $n\ge 2$. Then $\gcd(x_0, x_1)=1$ and $x_n$ is composite for all $n\ge 0$. \end{theorem} \begin{proof} By our choice of $x_0$ and $x_1$, Theorem \ref{odd} immediately implies that $x_{2n+1}$ is composite for every integer $n\ge 0$. We claim that for every $n\ge 1$, $x_{2n}$ has a factor in the set $\{p_1, p_2, \ldots, p_{30}\}$, where the primes are those in Table \ref{table2}. Indeed, let us first choose $x_0$ and $x_1$ according to \eqref{choicex0x1} \begin{equation*} x_0 \equiv c_iF_{m_i-r_i}\pmod{p_i},\quad x_1 \equiv c_iF_{m_i-r_i+1} \pmod{p_i}\quad \text{for all}\,\, 1\le i\le 30, \end{equation*} where $(p_i, m_i, r_i, c_i)$ are those given in Table \ref{table2}. Since $p=1$, $x_0=p^2+q^2$ and $x_1=2pq+q^2$, these congruences can be written as \begin{equation}\label{q} 1+q^2 \equiv c_iF_{m_i-r_i}\pmod{p_i},\quad 2q+q^2 \equiv c_iF_{m_i-r_i+1} \pmod{p_i}\quad \text{for all}\,\, 1\le i\le 30. \end{equation} \begin{table}[ht] \begin{centering} \begin{tabular} {|c|c|} \hline $p_i$ & $ q \bmod{p_i}$ \\ \hline 2&0\\ \hline 5&0\\ \hline 13&0\\ \hline 17&11\\ \hline 29&20\\ \hline 41&34\\ \hline 61&55\\ \hline 181&149\\ \hline 241&134\\ \hline 281&45\\ \hline 421&140\\ \hline 541&307\\ \hline 1009&818\\ \hline 1601&1347\\ \hline 2161&799\\ \hline \end{tabular} \qquad\qquad \begin{tabular} {|c|c|} \hline $p_i$ & $q\bmod{p_i}$\\ \hline 2521&1934\\ \hline 3041&455\\ \hline 8641&1277\\ \hline 20641&13565\\ \hline 31249&24574\\ \hline 103681&22094\\ \hline 109441&43164\\ \hline 141961&112001\\ \hline 721561&170379\\ \hline 1461601&442479\\ \hline 35239681&5419606\\ \hline 764940961&483887978\\ \hline 8288823481&6095337569\\ \hline 10783342081&54018520\\ \hline 571385160581761&504780818763137\\ \hline \end{tabular} \bigskip \caption{Set of congruences satisfied by the solution of \eqref{q}} \label{table3} \end{centering} \end{table} The choices of $c_i$ ensure that this system has solutions: in particular, a solution is given by the set of congruences listed in Table \ref{table3}. The Chinese remainder theorem guarantees that there exists a $q$ that satisfies all the above congruences. The smallest such value is the $129$-digit number mentioned in the statement of Theorem~\ref{mainthm}. It remains to argue why we believe this particular sequence $(x_n)_{n\ge 0}$ does not have a finite covering set of primes. Computer verifications show that there are $803$ values of $0\le n\le 200000$ such that $x_n$ has no prime factor $\le 2\times 10^6$ or a prime factor among the primes $p_1,\,p_2,\ldots,\,p_{30}$ given in table \ref{table2}. Moreover, any two such terms are mutually prime. Also, it can be checked that for these choices for $p$ and $q$, the numbers $pF_{913}+qF_{914}$, $pL_{913}+qL_{914}$, $pF_{943}+qF_{944}$ and $pL_{943}+qL_{944}$ are all primes of lengths $319$, $320$, $326$ and $326$, respectively. Since \eqref{eqodd} gives that $x_{1827}=(pF_{913}+qF_{914})(pL_{913}+qL_{914})$ and $x_{1887}=(pF_{943}+qF_{944})(pL_{943}+qL_{944})$, it follows that $x_{1827}$ and $x_{1887}$ are both products of exactly two primes, with the least prime factor having lengths 319 and 326, respectively. As a result, if a finite covering with primes would exist, it has to have at least $803$ primes larger than $2\times 10^6$, and among these, at least two primes of length $319$ or greater. It seems difficult to prove that the least prime factor of $x_n$ is unbounded as $n$ tends to infinity. As already noticed by Filaseta, Finch and Kozek \cite{Filaseta}, this type of question is open even for much simpler sequences such as $F_n$, $5\cdot 2^n+1$ or $11\cdot 5^n-1$. \end{proof} \begin{thebibliography}{9} \bibitem{Dickson} L. E. Dickson, \emph{History of the Theory of Numbers}, Vol.~II: Diophantine Analysis, Chelsea Publishing Co., 1966. \bibitem{Filaseta} M. Filaseta, C. Finch and M. Kozek, On powers associated with Sierpi\'{n}ski numbers, Riesel numbers and Polignac's conjecture, {\it J. Number Theory} {\bf 128} (2008), 1916--1940. \bibitem{Graham} R. L. Graham, A Fibonacci-like sequence of composite numbers, {\it Math. Mag.} {\bf 37} (1964), 322--324. \bibitem{Guy} R. K. Guy, {\it Unsolved Problems in Number Theory} (3rd ed.), Problem Books in Mathematics, Springer-Verlag, 2004. \bibitem{Izotov} A. S. Izotov, A note on Sierpi\'{n}ski numbers, {\it Fibonacci Quart.} {\bf 33} (1995), 206--207. \bibitem{Knuth} D. E. Knuth, A Fibonacci-like sequence of composite numbers, {\it Math. Mag.} {\bf 63} (1990), 21--25. \bibitem{Nicol} J. W. Nicol, A Fibonacci-like sequence of composite numbers, {\it Electron. J. Combin.} {\bf 6} (1999), Research Paper \#R44. \bibitem{Vsemirnov} M. Vsemirnov, A new Fibonacci-like sequence of composite numbers, {\it J. Integer Seq.} {\bf 7} (2004), \href{https://cs.uwaterloo.ca/journals/JIS/VOL7/Vsemirnov/vsem5.html}{Article 04.3.7}. \bibitem{Wilf} H. S. Wilf, Letters to the Editor, {\it Math. Mag.} {\bf 63} (1990), 284. \end{thebibliography} \bigskip \hrule \bigskip \noindent 2010 {\it Mathematics Subject Classification}: Primary 11B39; Secondary 11A51, 11B37. \noindent \emph{Keywords: } Fibonacci-like recurrence, composite numbers, system of covering congruences. \bigskip \hrule \bigskip \noindent (Concerned with sequences \seqnum{A082411}, \seqnum{A083103}, \seqnum{A083104}, \seqnum{A083105}, \seqnum{A083216}, and \seqnum{A221286}.) \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received March 24 2014; revised version received July 8 2014; July 21 2014. Published in {\it Journal of Integer Sequences}, July 25 2014. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document}