I would start by trial and error. If you do this for the first $12$ positive integers you may be able to see a more deterministic way of doing it.
You need to write your number as a sum of numbers chosen from
$$1,\,2,\,3,\,5,\,8,\,13,\,21,\,\ldots$$
in such a way that you never choose two adjacent numbers. For example you cannot choose both $5$ and $8$.
An example: if you want to make up $16$ then you could start with $13$, then you need another $3$. You can't take $1$ and $2$ because they are adjacent, but you can simply take $3$ itself. So you have $16=13+3$, and you write this as $100100$, where the $1$s indicate the numbers $13$ and $3$ which are used, and the $0$s indicate the numbers $8,\,5,\,2,\,1$ which are not used.
I would start by trial and error. If you do this for the first $12$ positive integers you may be able to see a more deterministic way of doing it.
You need to write your number as a sum of numbers chosen from $$1,\,2,\,3,\,5,\,8,\,13,\,21,\,\ldots$$ in such a way that you never choose two adjacent numbers. For example you cannot choose both $5$ and $8$.
An example: if you want to make up $16$ then you could start with $13$, then you need another $3$. You can't take $1$ and $2$ because they are adjacent, but you can simply take $3$ itself. So you have $16=13+3$, and you write this as $100100$, where the $1$s indicate the numbers $13$ and $3$ which are used, and the $0$s indicate the numbers $8,\,5,\,2,\,1$ which are not used.
See if you can do $1$ to $12$ for yourself.