Solving $a_{n}a_{n-1}=1,\, a_1=2$

Question

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.

Answer 1

Let

$$b_n:=\frac{3\log_2a_n+1}2$$

and the equation

$$a_na_{n-1}=1$$ can be written

$$b_n+b_{n-1}=1$$ with $b_1=2$.

Hence the solution

$$b_n=\frac{1-3(-1)^n}2$$

or

$$a_n=2^{-(-1)^n}.$$

Answer 2

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 $a_{2m+1}=\cdots=?$

Answer 3

Not that I can see. But if you want to follow through with your first steps, we hope all $a_n$ are greater than $0$. So $ln(a_n) + ln(a_{n-1}) = 0$ implies $$ ln(a_n) = ln\left(\frac{1}{a_{n-1}}\right) $$ and so $a_n = 1 / a_{n-1}$. Given initial condition, you get $$ 2,1/2,2,1/2,... $$

Answer 4

Let $$A_n=\frac{k+r^n}{k+r^{n-1}}~~~~(1)$$ Then $$A_n A_{n-1}=1 \implies r=\pm 1. $$ $r=1$ gives trivial result $A_n=1$ which contradicts $A_2=2$. So from (1), we $k=3$. Hence $$A_n=\frac{3+(-1)^n}{3+(-1)^{n-1}}$$