SAMC

The recursive relation is equivalent to $$3a_{n+1}-1=\frac{1}{2}\left(9a_n^2-6a_n+1\right),n=1,2,\dots$$ Let $x_n=3a_n-1$, $n=1,2,\dots$, and get $x_1=8=2^3$, $$x_{n+1}=\frac{1}{2}{x}_n^2,n=1,2,\dots$$ Let $x_n=2^{y_n}$, $y_1=3$, $n=1,2,\dots$, and obtain $$y_{n+1}=2y_n-1,n=1,2,\dots$$ It follows $y_n=2^n+1$, $n=1,2,\dots$, hence $x_n=2^{2^n+1}$, and we get $$a_n=\frac{1}{3}\left(2^{2^n}+1\right),n=1,2,\dots$$ If $n=3^k$, then $$2^{2\cdot 3^k}=4^{3^k}=(3+1{)}^{3^k}\equiv 1(\mathrm{mod}{3}^{k+1})$$ hence $$\left(2^{3^k}-1\right)\left(2^{3^k}+1\right)\equiv 0(\mathrm{mod}{3}^{k+1})$$ But, we have $$2^{3^k}-1=(3-1{)}^{3^k}-1\equiv -2(\mathrm{mod}{3}^{k+1})$$ and we obtain $$2^{3^k}+1\equiv 0(\mathrm{mod}{3}^{k+1})$$ hence $$2^{3^k}\equiv -1(\mathrm{mod}{3}^{k+1}).$$ It follows $2^{3^k}=3^{k+1}m-1$, for some odd integer $m$, hence $$\begin{array}{c}{2}^{2^{3^k+1}+1}=2^{3^{k+1}m}+1=(3-1{)}^{3^{k+1}m}+1\\ \equiv (-1{)}^{3^{k+1}m}+1(\mathrm{mod}{3}^{k+1})\equiv 0(\mathrm{mod}{3}^{k+1})\end{array}$$ We obtain $3^{k+1}\mid 2^{2^n}+1$, hence $3^k\mid a_n$, that is $n\mid a_n$.