why this sequence is period $a_{n+5}=a_{n}$ Let $a_{0}=a>0,a_{1}=b>0$,and such 
$$a_{n+1}a_{n-1}=\max\{(a_{n},1)\},\forall n\in N^{+}$$
show that
$$a_{n+5}=a_{n}$$
Even $a,b$ with the 1 uncertainty,so we can't $$a_{2}=\dfrac{\max{(a_{1},1})}{a_{0}}=\begin{cases}\dfrac{a_{1}}{a_{0}}&a_{1}\ge 1\\
\dfrac{1}{a_{0}},&a_{1}<1
\end{cases}$$, But This sequnece always is period $5$. it looks very interesting.following when I determine$a_{3}$,I can't.Thanks so much for any suggestion.

Do 5 different cases,


*

*$b\geq a\geq 1$

*$b\geq 1> a$

*$a\geq b \geq1$

*$b<1,a<1$

*$b<1, a\geq 1$
Case 1. $a_2=b/a, a_3=1/a, a_4=a/b, a_5=a, a_6=b, ...$
Case 2. $a_2=b/a, a_3=1/a, a_4=1/b, a_5=a, a_6=b, ...$
Case 3. $a_2=b/a, a_3=1/b, a_4=a/b, a_5=a, a_6=b, ...$
Case 4. $a_2=1/a, a_3=1/ab, a_4=1/b, a_5=a, a_6=b, ...$
Case 5. $a_2=1/a, a_3=1/b, a_4=a/b, a_5=a, a_6=b, ...$
