Closed Form of Recursion $a_n = \frac{6}{a_{n-1}-1}$ Given that $a_0=2$ and $a_n = \frac{6}{a_{n-1}-1}$, find a closed form for $a_n$.
I tried listing out the first few values of $a_n: 2, 6, 6/5, 30, 6/29$, but no pattern came out. 

The function $f(x) = \dfrac{6}{x-1}$ is a fractional linear transformation.
Composition of these corresponds to matrix multiplication:
 $$\text{if}\ T(x) = \dfrac{a_{11} x + a_{12 }}{a_{21} x + a_{22}} \ \text{and} \ S(x) = \dfrac{b_{11} x + b_{12}}{b_{21} x + b_{22}}$$
$$\text{then}\ T(S(x)) =  \dfrac{c_{11} x + c_{12}}{c_{21} x + c_{22}}\ \text{where}\ 
\pmatrix{c_{11} & c_{12}\cr c_{21} & c_{22}\cr} = \pmatrix{a_{11} & a_{12}\cr a_{21} & a_{22}\cr} \pmatrix{b_{11} & b_{12}\cr b_{21} & b_{22}\cr}$$
In your case the matrix is
$$M = \pmatrix{0 & 6\cr 1 & -1\cr}$$
and we'd like to find a closed form for its powers.  We diagonalize the matrix as 
$ M = E D E^{-1}$ where
$$ E = \pmatrix{-2 & 3\cr 1 & 1\cr}, \ D = \pmatrix{-3 & 0\cr 0 & 2\cr},\
E^{-1} = \dfrac{1}{5}\pmatrix{-1 & 3\cr 1 & 2\cr}$$ 
so $$M^n = E D^n E^{-1} =\dfrac{1}{5} \left( \begin {array}{cc} 2\, \left( -3 \right) ^{n}+3\cdot{2}^{n}&-6\,
 \left( -3 \right) ^{n}+6\cdot{2}^{n}\\ - \left( -3
 \right) ^{n}+{2}^{n}&3\, \left( -3 \right) ^{n}+2\cdot{2}^{n}
\end {array} \right)
$$
i.e. $$a_n =  \left(\dfrac{ (2 \left( -3 \right) ^{n}+3\cdot{2}^{n}) \cdot 2 -6
 \left( -3 \right) ^{n}+6\cdot{2}^{n}}{ \left(- \left( -3
 \right) ^{n}+{2}^{n}\right) \cdot 2  + 3 \left( -3 \right) ^{n}+2\cdot{2}^{n}
} \right) = \dfrac{-2(-3)^n + 12 \cdot 2^n}{(-3)^n + 4 \cdot 2^n}$$
