Pythagorean Fibonacci

Algebra Level 3

A recursive relation $(a_n)$ satisfies

$\left\{\begin{array}{c}aa_{2n+1}=\sqrt{(a_{2n})^{2}+(a_{2n-1})^{2}}\\a_{2n}=a_{2n-1}\end{array}\right.$

for positive integer $n\geq1$ .

If $a_{1}=1, a_{2}=1$ ,

find $a_{2018}$ .

If the value of $a_{2018}=2^{p}$ such that $p$ is a positive integer, compute $p$ .

This is part of the set Fun With Problem-Solving .

1 solution

Donglin Loo
Jan 25, 2018

For $n\geq1$ ,

$a_{1}=1$

$a_{2}=1$

$a_{3}=\sqrt{2}$

$a_{4}=\sqrt{2}$

$a_{5}=2$

$a_{6}=2$

$a_{7}=2 \sqrt{2}$

$a_{8}=2 \sqrt{2}$

Notice that the recurrence comes in a group of four with corresponding terms in each subsequent group being multiplied by $2$ .

$2018\equiv2\pmod4$

Hence, $a_{2018}$ corresponds with $a_{2}=1$ .

Let $a_{2018}=1\cdot 2^{x}$

$2018=2+4x$

$4x=2016$

$x=504$

So, $a_{2018}=1\cdot 2^{504}$