I take liberty to start sequence from 0 index: \[ a_{0}=1, \; a_{1}=2, \; a_{2}=24, \; a_{n}=\frac{ 6a_{n-1}^{2}a_{n-3}-8a_{n-1}a_{n-2}^{2}}{a_{n-2}a_{n-3}}.\] Rewriting the recurrent relation, we have \[ 4 a_{n-1}b_{n-1}= a_{n-3}b_{n},\quad\mbox{where}\ b_{n}= a_{n}a_{n-2}-2 a_{n-1}^{2}.\] It follows from the recurrent relation for $ b_{n}$ that $ b_{n}= 2^{2n-1}a_{n-1}a_{n-2}$ and, thus: \[ 2^{2n-1}a_{n-1}a_{n-2}= a_{n}a_{n-2}-2 a_{n-1}^{2}\] or \[ a_{n-2}c_{n}= 2 a_{n-1}c_{n-1}\quad\mbox{where}\ c_{n}= 2^{2n}a_{n-1}-a_{n}.\] It follows from the recurrent relation for $ c_{n}$ that $ c_{n}= 2^{n}a_{n-1}$ and, thus: \[ 2^{n}a_{n-1}= 2^{2n}a_{n-1}-a_{n}\] or \[ a_{n}= (2^{2n}-2^{n}) a_{n-1},\] implying that \[ a_{n}= \prod_{k=1}^{n}(2^{2k}-2^{k}) = 2^{\frac{n(n+1)}{2}}\prod_{k=1}^{n}(2^{k}-1),\] which is obviously integer. [modedit: the $ n|a_{n}$ part has been added since this post, someone please solve it ]

Peter wrote: The sequence $ \{a_{n}\}_{n \ge 1}$ is defined by \[ a_{1} = 1, \; a_{2} = 2, \; a_{3} = 24, \; a_{n} = \frac { 6a_{n - 1}^{2}a_{n - 3} - 8a_{n - 1}a_{n - 2}^{2}}{a_{n - 2}a_{n - 3}}\ \ \ \ (n\ge4). \] Show that $ a_{n}$ is an integer for all $ n$, and show that $ n|a_{n}$ for every $ n\in\mathbb{N}$. $ v_n = \frac {a_n}{a_{n - 1}}$ $ v_1 = 2,v_2 = 12$ $ v_n = 6v_{n - 1} - 8v_{n - 2}$ Then we has $ v_n = 4^n - 2^n$ $ a_n = (4^n - 2^n)a_{n - 1}$ So $ a_n = 2^{\frac {n(n + 1)}{2}}\prod _{k = 1}^n(2^k - 1)$ Result $ n|a_n$ follow from Euler theorem.