Yes. For any $n$ we can take the Zeckendorf representation of $n$, i.e. writing $n = F_{a_1} + F_{a_2} + \cdots + F_{a_k}$ for positive integers $a_i > 1$ which are non-consecutive. Then simply let $f(n)$ be $F_{a_1+1} + F_{a_2 + 1} + \cdots + F_{a_k + 1}.$ This is easily checked to work. $\square$