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 $\displaystyle 2^k$ where k is a natural number.

I know that the logistic map is $\displaystyle Q_{\mu}=\mu x(1-x)$ and that x is a prime period n point if $\displaystyle Q^n_{\mu}(x)=x$ 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 $\displaystyle \mu$ and not on n.

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

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

Please help me understand this question,

Thanks

Katy