Closed form expression for a formal power series

Question

Is there a closed form expression for the following formal power series $$\large\sum_{k\ge 0}\dfrac{z^{2^k}}{1+z^{2^k}}$$ Till now I have tried in vain finding any progress to simplify this expression. Please help.

Answer 1

If we set: $$f(z)=\sum_{k\geq 0}\frac{z^{2^k}}{1+z^{2^k}}\tag{1}$$ we have: $$ f(z^2)=\sum_{k\geq 0}\frac{z^{2\cdot 2^k}}{1+z^{2\cdot 2^k}}=\sum_{k\geq 0}\frac{z^{2^{k+1}}}{1+z^{2^{k+1}}}=f(z)-\frac{z}{1+z}.\tag{2}$$ If we further assume: $$ f(z)=\sum_{m\geq 1} a_m\,z^{m}\tag{3} $$ $(2)$ gives: $$ a_k = a_{2k}+1,\qquad 0 = a_{2k+1}-1\tag{4} $$ hence:

$$ f(z) = \sum_{m\geq 1}\left(1-\nu_2(m)\right) z^m\tag{5} $$

where $$\nu_2(m)=\max\{r\in\mathbb{N}:2^r\mid m\}.$$