Compute $\sum_{l=1}^\infty N(l)z^l$ (find a simple expression for this sum)

Question

Consider the doubling map $T =T_2 :\Bbb T→\Bbb T$ (where $\Bbb T = [0, 1]$ with $0$ and $1$ identified) defined as$T(θ)=2θ$ (mod $1$). What is the number $N(l)$ of points $\theta ∈\Bbb T$ are there such that $T^l(θ) = θ$, for fixed $l ≥ 1$. Compute $\sum_{l=1}^\infty N(l)z^l$ (find a simple expression for this sum).

Here for fixed $l ≥ 1$ we have $\frac{p}{2^l-1}$ is fixed where $p=0,\ldots,2^l-2$ so, $N(l)=2^l-1$ and $\sum_{l=1}^\infty N(l)z^l=\sum_{l=1}^\infty (2z)^l-\sum_{l=1}^\infty z^l$. Now how to simply it further?

Answer 1

Since $$ \sum_{l=1}^\infty w^l=\frac w{1-w}\quad \text{for} \ |w|<1, $$ we get $$ \sum_{l=1}^\infty N(l)z^l=\frac {2z}{1-2z}-\frac z{1-z}=\frac{z}{(1-2z)(1-z)} $$ for $|z|<1/2$.

Now that you are at it, you can try to compute the corresponding zeta function $$ \zeta_T(z)=\exp\sum_{l=1}^\infty \frac{N(l)}l z^l. $$ Hint: note that $\zeta_T'(z)=\zeta_T(z)\sum_{l=1}^\infty N(l)z^{l-1}$.