Let $A=1+\dots\in\mathbb F[[x]]$ be a (multiplicatively) invertible series over the field $\mathbb F_2$ of two elements. Writing $A=\sum_{n\geq 0}\alpha_n x^n$ and $\frac{1}{A}=\sum_{n\geq 0} \beta_n x^n$ with $\alpha_i,\beta_i\in\lbrace 0,1\rbrace$, we consider $$\delta=\delta(A)=\limsup_n\frac{\sum_{j=0}^{n-1}\alpha_j+\beta_j}{2n}\ .$$

Can $\delta$ be arbitrarily close to $1$?

Can $\delta$ be arbitrarily close to $1$ for a rational fraction?

The current record-holder for $\delta$ is $\frac{5}{6}$, achieved by $A=\frac{1+x}{1+x+x^2}=1+\sum_{k\geq 1}x^{3k-1}+x^{3k}$ with inverse $\frac{1}{A}=1+\sum_{k\geq 2} x^k$. (There is no higher $\delta$ for rational $A$ with numerator and denominator of degree $\leq 8$.)

Let $A=1+\dots\in\mathbb F[[x]]$ be a (multiplicatively) invertible series over the field $\mathbb F_2$ of two elements. Writing $A=\sum_{n\geq 0}\alpha_n x^n$ and $\frac{1}{A}=\sum_{n\geq 0} \beta_n x^n$ with $\alpha_i,\beta_i\in\lbrace 0,1\rbrace$, we consider $$\delta=\delta(A)=\limsup_n\frac{\sum_{j=0}^{n-1}\alpha_j+\beta_j}{2n}\ .$$

Can $\delta$ be arbitrarily close to $1$?

Can $\delta$ be arbitrarily close to $1$ for a rational fraction?

The current record-holder for $\delta$ is $\frac{5}{6}$, achieved by $A=\frac{1+x}{1+x+x^2}=1+\sum_{k\geq 1}x^{3k-1}+x^{3k}$ with inverse $\frac{1}{A}=1+\sum_{k\geq 2} x^k$. (There is no higher $\delta$ for rational $A$ with numerator and denominator of degree $\leq 8$.)

The corresponding question concerning matrices is without interest: Given an invertible square matrix $A$ of size $n\times n$ with coefficients in $\mathbb F_2$, the total number of $1$'s in $A$ and $A^{-1}$ can be equal to $2n^2-2n$: Take $n$ even and $A=A^{-1}$ the all $1-$matrix minus the diagonal matrix.