关于二阶线性合数序列
On second order linear sequences of composite numbers
摘要：
本文给出了Dubickas，Novikas，and Siurys在2010年的一个结果的新证明：设$(a,b)\in \mathbb{Z}^2$和$(x_n)_{n\ge 0}$是由初始值$x_0$和$x_1$定义的序列，以及$n\ge 1$的二阶线性递归\begin{equation*} x_{n+1}=ax_n+bx_{n-1} \end{equation*}.假设$b\neq 0$和$(a,b)\neq (2,-1), (-2, -1)$ .则存在两个互质正整数$x_0$，$x_1$使得$|x_n|$是所有$n\in \mathbb{N}$的合数.上述定理推广了Graham在求解$(a,b)=(1,1)$时的一个结果.
英文摘要：
In this paper we present a new proof of the following 2010 result of Dubickas, Novikas, and Siurys: Let $(a,b)\in \mathbb{Z}^2$ and let $(x_n)_{n\ge 0}$ be the sequence defined by some initial values $x_0$ and $x_1$ and the second order linear recurrence \begin{equation*} x_{n+1}=ax_n+bx_{n-1} \end{equation*} for $n\ge 1$. Suppose that $b\neq 0$ and $(a,b)\neq (2,-1), (-2, -1)$. Then there exist two relatively prime positive integers $x_0$, $x_1$ such that $|x_n|$ is a composite integer for all $n\in \mathbb{N}$. The above theorem extends a result of Graham who solved the problem when $(a,b)=(1,1)$.
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