The series $(a_n)_{n\in\mathbb{N}}$ is given through $$a_1=1,\quad a_2=\frac{1}{2},\quad a_{n+2}=a_na_{n+1} \quad\text{ for } n\geq1.$$ I want to show that $$a_n = 2^{-f_{n-1}}$$ whereas $\,f_n$ is the $n$-th Fibonacci number.
I did a proof with induction, but now I came up with this. Out of the initial figures, the definition of $f_n$, and the exponent and logarithm rules I think it is
\begin{align} &\quad a_{n} = 2^{\,\log_2 a_n} = 2^{\,\log_2 (a_{n-2} \,\cdot\, a_{n-1})}\\ \Longleftrightarrow &\quad a_n = 2^{\,\log_2 (a_{n-2}) \,+\, \log_2 (a_{n-1})}\\ \Longleftrightarrow &\quad \log_2 a_n = \log_2 2^{\,\log_2 (a_{n-2}) \,+\, \log_2 (a_{n-1})}\\ \Longleftrightarrow &\quad \log_2 a_n = \log_2 a_{n-2} \,+\, \log_2 a_{n-1}\\ \Longleftrightarrow &\quad (-\log_2 a_n) = (-\log_2 a_{n-2}) + (-\log_2 a_{n-1}) \quad(*) \end{align}
and with the initial figures $-\log_2 a_1 = 0$ and $-\log_2 a_2 = 1$ the negative expononents form the Fibonacci sequence $2^{-f_{n-1}}$ with $(*)$.
