= Solution
Take $n=2^k$. Every proper divisor of $n$ divides $n/2=2^{k-1}$, so the points fixed by $F^n$ that do not have exact period $n$ are precisely those fixed by $F^{n/2}$. Part (ii) therefore gives
$$
\begin{aligned}
\#\{x:\text{exact period }2^k\}
&=(2^{2^k}-1)-(2^{2^{k-1}}-1)\\
&=\boxed{2^{2^k}-2^{2^{k-1}}}.
\end{aligned}
$$
Because different starting points are counted separately, no division by the period is made. This is the <exact power-of-two periods of the doubling map>.
Back to article page
