smc

Note that $x_0=0$ gives the constant sequence $0,0,...$, since $f(0)=4\cdot 0-0^2=0$. Because $f(4)=0,x_0=4$ gives the sequence $4,0,0,...$ with two different values. Similarly, $f(2)=4$, so $x_0=2$ gives the sequence $2,4,0,0,...$ with three values. In general, if $x_0=a_n$ gives the sequence $a_n,a_{n-1},...,a_2,a_1,...$ with $n$ different values, and $f(a_{n+1})=a_n$, then $x_0=a_{n+1}$ gives a sequence with $n+1$ different values. (It is easy to see that we could not have $a_{n+1}=a_i$ for some $i<n+1$.) Thus, it follows by induction that there is a sequence with $n$ distinct values for every positive integer $n$, as long as we can verify that there is always a real number $a_{n+1}$ such that $f(a_{n+1})=a_n$. This makes the answer $\boxed{\infty }$. The verification alluded to above, which completes the proof, follows from the quadratic formula: the solutions to $f(a_{n+1})=4a_{n+1}-a_{n+1}^2=a_n$ are $a_{n+1}=2\pm \sqrt{4-a_n}$. Hence if $0\le a_n\le 4$, then $a_{n+1}$ is real, since the part under the square root is non-negative, and in fact $0\le a_{n+1}\le 4$, since $4-a_n$ will be between $0$ and $4$, so the square root will be between $0$ and $2$, and $2\pm$ something between $0$ and $2$ gives something between $0$ and $4$. Finally, since $a_1=0\le 4$, it follows by induction that all terms satisfy $0\le a_n\le 4$; in particular, they are all real.