How to show that $a_n$ : integer $a_{n+1}=\frac{1+a_na_{n-1}}{a_{n-2}}(n\geq3), a_1=a_2=a_3=1$

I would appreciate if somebody could help me with the following problem:

Q: How to show that $a_n$ : integer

$$a_{n+1}=\frac{1+a_na_{n-1}}{a_{n-2}}(n\geq3), a_1=a_2=a_3=1$$

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Prove that $a_n = 4a_{n-2} – a_{n-4}$.

Since the question Prove that $a_{n+1}a_{n-2}-a_{n}a_{n-1}=1$ is always an integer is now (correctly) marked as a duplicate, I’ll post my answer here:

Write the relations for $n$ (given) and $n-1$; subtract and see if you can say anything about the periodicity of the sequence $b_n = (a_{n+1}+a_{n-1})/a_n$. Then conclude that $a_{n+1} = a_nb_n – a_{n-1}$ is indeed an integer.

With an extra bit of work you can get $a_n = 4a_{n-2} – a_{n-4}$, as mentioned in another answer. While that would allow you to find the formula for the general term of the sequence (not pretty … ), you don’t need to get there in order to prove that $a_n$ is always an integer.