Hi I have this question,

What is the least number of points of prime period n that the logistic map can have in each of the cases: (i) n is a prime number, (ii) n is 2^k where k is a natural number.

I know that the logistic map is Q_{\mu}=\mu x(1-x) and that x is a prime period n point if Q^n_{\mu}(x)=x where n is the lowest integer for which this holds.

I'm not sure how to go about this because it seems to me, the number of period points depends on \mu and not on n.

Because if \mu = 1 then there is only one period point of prime period 1 which is x=0.

And if \mu=4 then there are 2^n period n points, and 2^{n-1} of these are prime period n points. Is this not right?

Please help me understand this question,