Densities of the reciprocals of polynomials in binary power series Define the density $[f(z)]$ of a power series $f(z) = f_0+f_1z+f_2z^2+\cdots$ in the binary power series ring $F_2[[z]]$ as the natural density of the set $E_f := \{i : f_i = 1\}.$
$D := \{[f^{-1}] : 0 \in E_f\}$ is the set of densities of the reciprocals of the polynomials with non-zero constant term. 
I think every arithmetic sequence has the form $E_{f^{-1}}$ where $f(z)$ is a polynomial with $f(0)=1$
1) Any insight into whether $D$ is nowhere dense or not ? What are the accumulation points of $D$ ? 
2) Any insight into upper bounds for $D$ ? $\inf(D)=0$, $\sup(D)=1$ 
I remember some of this brought up in Cooper & Bryant's "Reciprocals Of Binary Power Series" lead by questions on the parity of the partition function and among other things the densities of the reciprocals of polynomials.
 A: I felt this synopsis of Aaron Meyerowitz's post with Pietro Majer's comment and my own might be useful, possibly even insightful, in characterizing $D$ even further.
$A(z)=\sum_{0}^{\infty} a_iz^i\in \Bbb F_2[[z]]$ is called periodic by any of the following equivalent definitions :


*

*For some $n\in\Bbb N,B(z)\in\Bbb F_2[z]$ with $deg(B(z))<n$ we have $$A(z)*(1+z^n)=B(z)$$ (The period $d$ of $A(z)$ is the smallest possible value for $n$ while $k$ is the exact number of terms in $B(z)$)

*$E_A$ is the union of $k$ distinct arithmetic sequences of the same common difference $n$.

*$(a_i)_i^\infty$ is a periodic sequence of period $d$, which divides $n$, and has exactly $k$ 1s up to its first $n$ coefficients.


*

*Therefore, there are exactly $2^n$ periodic power series whose period divides $n$. Each periodic power series whose period divides $n$ has the form $\frac{B(z)}{1+z^n}$ and has density $\frac{k}{n}$ where $k$ is the number of nonzero coefficients in $B(z)$. Of these the polynomial reciprocals are the reciprocals of the non-constant factors of $1+z^n$ and thus each of the form $\frac{b(z)}{1+z^n}$ where $b(z)$ is a proper divisor of $1+z^n$.

*Therefore $D=${$\frac{k}{n}:$ $n\in\Bbb N,$ $b(z)$ has exactly $k$ terms and properly divides $1+z^n$} as each invertible polynomial $f(z)$ divides the characteristic polynomial $\psi_m(z)=z+z^{2^m}$ of its splitting field $\Bbb F_{2^m}$ where $m = deg(f(z))$ when $f$ is irreducible over $\Bbb F_2$. 
Claim (Meyerowitz) : Given $m\in\Bbb N$ there exists a $2^{m-1}$ term polynomial $b(z)$ that divides $1+z^{2^m-1}$.
Assuming this claim then $\frac{2^j}{c(2^k-1)}\in D$ for all $c,j,k\in\Bbb N$ with $j<k$ because the $\Bbb F_2[[z]]$ ring endomorphism $\phi_n : f(z)\rightarrow f(z^n)$ has the obvious property $$[\phi_n(f(z))]=\frac{1}{n}*[f(z)]$$ for all binary power series $f(z)$.
