### On the order of appearance of product of consecutive Fibonacci numbers

#### Abstract

Let $F_{n}$ be the $n$th Fibonacci number. For each positive integer $m$, the order of appearance of $m$, denoted by $z(m)$, is the smallest positive integer $k$ such that $m$ divides $F_k$. Recently, D. Marques has obtained a formula for $z(F_{n}F_{n+1})$, $z(F_{n}F_{n+1}F_{n+2})$, and $z(F_{n}F_{n+1}F_{n+2}F_{n+3})$. In this paper, we extend Marques' result to the case $z(F_{n}F_{n+1}\cdots F_{n+k})$, for every $4\leq k \leq 6$. For instance, we prove that, for $n\geq1$,

\[z(F_{n}F_{n+1}F_{n+2}F_{n+3}F_{n+4}) =\begin{cases}a, & \text{if $n\equiv1,2,3,4,5,6,7,10 \pmod {12}$, or $n\equiv8,60 \pmod {72}$};\\2a, & \text{if $n\equiv9,11\pmod {12}$, or $n\equiv24,44 \pmod {72}$};\\3a, & \text{if $n\equiv12,32,36,56 \pmod {72}$}; \\6a, & \text{if $n\equiv0,20,48,68 \pmod {72}$}\end{cases}\] where $a=[n,n+1,n+2,n+3,n+4]$.

\[z(F_{n}F_{n+1}F_{n+2}F_{n+3}F_{n+4}) =\begin{cases}a, & \text{if $n\equiv1,2,3,4,5,6,7,10 \pmod {12}$, or $n\equiv8,60 \pmod {72}$};\\2a, & \text{if $n\equiv9,11\pmod {12}$, or $n\equiv24,44 \pmod {72}$};\\3a, & \text{if $n\equiv12,32,36,56 \pmod {72}$}; \\6a, & \text{if $n\equiv0,20,48,68 \pmod {72}$}\end{cases}\] where $a=[n,n+1,n+2,n+3,n+4]$.

#### Keywords

Fibonacci number; least common multiple; the order of appearance

#### Full Text:

PDFDOI: https://doi.org/10.11575/cdm.v13i2.62729

DOI (PDF): https://doi.org/10.11575/cdm.v13i2.62729.g46826

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Contributions to Discrete Mathematics. ISSN: 1715-0868