Let $f_n$ be the Fibonacci sequence, $f_1=1$, $f_2=2$.
Prove that each positive integer admits a unique representation in a form $a_1f_1+a_2f_2+\cdots+a_nf_n+\cdots$ such that
1) each $a_i$ is either $0$ or $1$;
2) there are finitely many numbers $a_i$ equal to $1$;
3) no two consecutive numbers are equal to $1$.
Attempt:
Since every Fibonacci number can be represented as $1$ times itself, obviously satisfying above three properties, we only need to show integers between every two consecutive Fibonacci numbers satisfy three properties.
Since the gap between two consecutive Fibonacci numbers $F(n)$ and $F(n+1)$ is equal to $F(n-1)$, we only need to show that integers from $1$ to $(F(n-1)-1)$ can be represented in the required form, since we can add them to $F(n)$ to get every integer in the gap between $F(n)$ and $F(n+1)$.
I try to use induction.
The first Fibonacci number subtracted by $1$ that is not another Fibonacci number is $F(4)$, since $F(4)-1=4$, $1$ to $4$ obviously satisfy three properties.
Suppose $1$ to $F(n)-1$ can be represented in the form satisfying three properties, we prove $1$ to $F(n+1)-1$ can also do that.
However I found it hard to continue, maybe this is not a good idea in the first place. Is there a better way to do this problem? Another question is that I don't understand what the point of the second property is. Since there are only finitely many Fibonacci numbers that are less than a integer, how could it possible to represent it as a combination of them with infinitely many coefficients that is equal to $1$?
