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Factorizations of some periodic linear recurrence systems

Colloquium Mathematicum MSC: Primary 11B39; Secondary 15A18, 11B37. DOI: 10.4064/cm8362-1-2021 Published online: 25 February 2021

Abstract

Let $P$ and $Q$ be relatively prime integers. The Lucas sequences are defined by $U_0 = 0$, $U_1 = 1$, $V_0 = 2$, $V_1 = P$, and $$U_n = P U_{n-1} - Q U_{n-2} \quad \text {and} \quad V_n = P V_{n-1} - Q V_{n-2} ,$$ where $n \ge 2$. We show that $$\begin {array}{rll} U_n &= \prod _{k=1}^{n-1} \biggl ( P - 2 \sqrt {Q} \cos \frac {k\pi }{n} \biggr ) , \quad \ &n &\ge 2,\\ V_n &= \prod _{k=1}^n \biggl ( P - 2 \sqrt {Q} \cos \frac {(k- {1}/{2})\pi }{n} \biggr ) ,\quad \ &n&\ge 1. \end {array}$$ The proofs depend on finding the eigenvalues and eigenvectors of certain tridiagonal matrices.

Next, let $a_1$, $a_2$, $b_1$, and $b_2$ be real numbers. A period two second order linear recurrence system is defined to be the sequence $f_0 = 1$, $f_1 = a_1$, and $$f_{2n} = a_2 f_{2n-1} + b_1 f_{2n-2} \quad \text {and}\quad f_{2n+1}= a_1 f_{2n} + b_2 f_{2n-1}$$ for $n \ge 1$. Also, let $D = a_1 a_2 + b_1 + b_2$ and assume $D^2 - 4b_1 b_2 \ne 0$. We show that $$f_{2n+1} = a_1 \prod _{k=1}^n \biggl ( \frac {a_1 + a_2}{2} \pm \sqrt { \biggl ( \frac {a_1 - a_2}{2} \biggr ) ^2 - b_1 - b_2 + 2 \sqrt {b_1 b_2} \cos \frac {k\pi }{n+1}} \biggr )$$ for $n \ge 0$. The proof also depends on finding the eigenvalues and eigenvectors of a certain tridiagonal matrix.

Authors

• Curtis CooperSchool of Computer Science and Mathematics
University of Central Missouri
Warrensburg, MO 64093, U.S.A.
e-mail
• Richard Parry Jr.School of Computer Science and Mathematics
University of Central Missouri
Warrensburg, MO 64093, U.S.A.

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