Consider the sequence $\{a_n\}$ defined by $\;a_1=1,\; a_2=2,\; a_3=24\;$ and $$a_n=\frac{6a_{n-1}^2a_{n-3}-8a_{n-1}a_{n-2}^2}{a_{n-2}a_{n-3}},\; \forall n\geq 4.$$
Show that $\;\forall n,\; a_n\;$ is an integer that is multiple of $n$.
Solution: we have $\frac{a_n}{a_{n-1}}=6\frac{a_{n-1}}{a_{n-2}}-8\frac{a_{n-2}}{a_{n-3}}$. Set $b_n=\frac{a_n}{a_{n-1}}$. Thus $b_n=2^{n-1}(2^{n-1}-1)$ and $a_n=2^{n(n-1)/2}\prod_{i=1}^{n-1}(2^i-1)$.
Claim: $a_n$ is a multiple of $n$.
Let $n=2^km$ where $m$ is odd. Then $k\leq n \leq n(n-1)/2$ and there exists $i \leq m-1$ such that $m$ is a divisor of $2^i-1$[for $i=\phi(m)$, $\phi$ is euler totient function]. Consequently $a_n$ is a multiple of $n$.
So, I am done right?