Solving $a_{n}a_{n-1}=1,\, a_1=2$ The solution of the recurrence
$$a_{n}+a_{n-1}=1,\, a_1=2$$
is
$$a_n=\frac{1-3(-1)^n}{2}.$$
Could this be somehow used to solve
$$a_{n}a_{n-1}=1,\, a_1=2?$$
Logarithms would turn this to
$$\ln a_n+\ln a_{n-1}=0$$
but that doesn't seem to help.

Hint:
$$a_na_{n-1}=a_{n-1}a_{n-2}\implies a_n=a_{n-2}$$
So, $a_{2m}=\cdots=a_2=a_0$
and $a_{2m+1}=\cdots=?$
