Six seats are arranged in a circular table. Each seat is to be painted in red, blue or green such that any two adjacent seats have different colours. How many ways are there to paint the seats?
I had thought that if you choose a specific seat to start with, say, $a_1$, and WLOG it is red. Then there are two cases: $a_2$ and $a_6$ (chairs adjacent to $a_1$) are both the same colour or are two different colours.
Case 1: $a_2$ is blue and $a_6$ is green. Then if both $a_3$ and $a_5$ are red, $a_4$ has two choices: green or blue. So from this example there are $3 \cdot 2 \cdot 2 = 12$ possible ways.
Case 2: Again, $a_2$ is blue and $a_6$ is green. But if $a_3$ is green and $a_5$ is blue, then $a_4$ must be red. There are $3 \cdot 2$ choices.
Case 3: As before, $a_2$ is blue and $a_6$ is green. But if $a_3$ is red and $a_5$ is blue, then $a_4$ must be green. $3 \cdot 2 \cdot 2 = 12$
So in total I had $12 + 6 + 12 = 30$ ways but the answer was $66$. I do think my reasoning might have some flaws here, but I need help.