Let
$$ a_{0}=1,\; a_{1}=2,\; a_{2}=24,\; a_{n}=\frac{ 6a_{n-1}^{2}a_{n-3}-8a_{n-1}a_{n-2}^{2}}{a_{n-2}a_{n-3}}. $$
Show that every element of this sequence is an integer.

Question

Let
$$ a_{0}=1,\; a_{1}=2,\; a_{2}=24,\; a_{n}=\frac{ 6a_{n-1}^{2}a_{n-3}-8a_{n-1}a_{n-2}^{2}}{a_{n-2}a_{n-3}}. $$
Show that every element of this sequence is an integer.

asnaseer · Answer

$$a_{n}=\frac{ 6a_{n-1}^{2}a_{n-3}-8a_{n-1}a_{n-2}^{2}}{a_{n-2}a_{n-3}}$$$$\qquad=2a_{n-1}(\frac{ 3a_{n-1}a_{n-3}-4a_{n-2}^{2}}{a_{n-2}a_{n-3}})$$$$\qquad=2a_{n-1}(\frac{3a_{n-1}}{a_{n-2}}-\frac{4a_{n-2}}{a_{n-3}})$$therefore:$$\frac{a_n}{a_{n-1}}=2(\frac{3a_{n-1}}{a_{n-2}}-\frac{4a_{n-2}}{a_{n-3}})$$now let:$$r_n=\frac{a_n}{a_{n-1}}$$therefore:$$r_n=2(3r_{n-1}-4r_{n-2})$$and I'm sure we can show that all $r_n$ are also integers using induction as we know from the given values that:$$r_1=\frac{2}{1}=2$$$$r_2=\frac{24}{2}=12$$this would then prove that all $a_n$ are also integers (I think).