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.

I know frankooo has already posted an answer, but I started my post beforehand and I want to go through the whole process of solving one case (as frankooo's answer doesn't show a full proof). I hope this helps in formulating the remaining proofs! Lets start by assuming $a_1 \geq a_0 \geq 1$. We then plug it into the definition
$$a_n = \frac{\max(a_{n-1},1)}{a_{n-2}}\qquad$$
To get
$$a_2 = \frac{\max(a_1,1)}{a_0} = \frac{a_1}{a_0}$$
Since $a_1\geq a_0$ the quantity $a_2 \geq 1$ and $a_2 \geq a_1$
$$a_3 = \frac{\max(a_2,1)}{a_1} = \frac{a_2}{a_1} = \frac{1}{a_0}$$
Note that we get the last equality above by substituting in the definition of $a_2$. Since we assumed $a_0 \geq 1$ we have that $a_3 \leq 1$
$$a_4 = \frac{\max(a_3,1)}{a_2} = \frac{1}{a_2} = \frac{a_0}{a_1}$$
Note that since $a_1 \geq a_0$ the quantity $a_4 \leq 1$
$$a_5 = \frac{\max(a_4,1)}{a_3} = \frac{1}{a_3} = a_0$$
Finally, since $a_0 \geq 1$ we get
$$a_6 = \frac{\max(a_5,1)}{a_4} = \frac{a_5}{a_4} = \frac{a_0}{a_4} = a_1$$
From this we can see that the cycle would continue with the desired cyclic behavior. To preserve my sanity I will leave the rest of the proofs up to you!  
Note: there may be a faster way to prove this, but I haven't done any number theory prior to this... I made this up as I went along using Avika's comment to check my answers
