Fibonacci but with ratio of sum and difference of the the previous two terms

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I have discovered a pretty weird sequence $$a_0=1$$ $$a_1= 2$$ $$a_n=\frac{a_{n-1}+a_{n-2}}{a_{n-1}-a_{n-2}}$$

The first few terms are: $$1,\ 2,\ 3,\ 5,\,\ 4,\ -9,\ \frac{5}{13},\ -\frac{56}{61},\ \frac{423}{1033},...$$

The sequence is very unpredictable and looks very random. I wonder if there is a closed-form formula, perhaps analogous to Binet's formula, that calculates the $n$-th element of this sequence.

To simplify the problem, i tried splitting the sequence to its numerator and denominator $$a_n = \frac{A_n}{B_n}$$ Combining with the recursive equation yields, $$\frac{A_n}{B_n} = \frac{\frac{A_{n-1}}{B_{n-1}} +\frac{A_{n-2}}{B_{n-2}}}{\frac{A_{n-1}}{B_{n-1}} -\frac{A_{n-2}}{B_{n-2}} } = \frac{A_{n-1} B_{n-2}+A_{n-2}B_{n-1}}{A_{n-1} B_{n-2}-A_{n-2}B_{n-1}} $$

Then we set, $$A_n = A_{n-1}B_{n-2} + A_{n-2}B_{n-1}$$ $$B_n = A_{n-1}B_{n-2} - A_{n-2}B_{n-1}$$ $$A_0 = B_0 = B_1 = 1$$ $$A_1 = 2$$

While i fell short of my goal of finding a closed-form formula, i think this is a great start. Can anyone find a closed-form formula?