An Elegant function with elegant conditions

$\begin {cases} f(1) =1 \\f(3) =3 \\f(2n) = f(n)\\ f(4n+1) =2f(2n+1)-f(n)\\ f(4n+3)=3f(2n+1)-2f(n) \end {cases}$ The function $f$ is defined on the set of all positive integers as above. Find the number of $n$ with $f(n)=n$ for $1\leq n\leq 1988$ .