How can I find the closed form of the recurrence relation: $a_{n+1}=\frac{1}{1-a_n}$ ?
I found the following values:
$$a_1=3$$
$$n=1: a_2=\frac{1}{1-3}=-\frac{1}{2}$$
$$n=2: a_3=\frac{1}{1+\frac{1}{2}}=\frac{1}{\frac{3}{2}}=\frac{2}{3}$$
$$n=3: a_4=\frac{1}{1-a_3}=\frac{1}{1-\frac{2}{3}}=\frac{1}{\frac{1}{3}}=3$$
$$n=4: a_5=\frac{1}{1-a_4}=\frac{1}{1-\frac{1}{3}}=\frac{1}{\frac{2}{3}}=\frac{3}{2}$$
$$n=5: a_6=\frac{1}{1-a_5}=\frac{1}{1-\frac{3}{2}}=\frac{1}{\frac{-1}{2}}=-2$$
