Ratio occurrences of ($x \operatorname{AND} x^{-1} = 0$) to ($x \operatorname{AND} x^{-1} > 0$) for some range $[0...n]$, $x \in \mathbb{N}$

Just a start, looking at the case when $n=2^{k+1}-1.$ Specifically, if the values $A_n,B_n$ Correspond to values $x<n,$ the we have a subsequence of $A_n/B_n$ which converges to zero.

I have the feeling some details might be slightly off, but the approach is sound.

In the numbers between $2^k$ and $2^{k+1}-1,$ inclusive, there are $2^k$ values. Of those, there are $$\sum_{j=0}^{\lfloor (k-1)/2\rfloor}2^j\binom{\lfloor (k-1)/2\rfloor}{j}\tag1$$ with the value $$x\operatorname{AND} x^{-1}=0.\tag2$$

This is because, if we think of the digits as a subset of $f(x)\subset\{0,1,2,\dots,k\},$ with $k\in f(x).$ $x$ satisfies $(2)$ if $\forall a,b\in f(x),$ $a+b\neq k.$ This means we can think of $f(x)$ as a subset $T\subseteq \left\{1,\dots,\lfloor (k-1)/2\rfloor\right\}$, plus a choice of exactly one element $\{a,k-a\}$ for each $a\in T.$ There are $2^{|T|}$ such choices. (Note, it is never true for such $x$ that $a=k-a.$)

This means the number of $T$ with this property of size $j$ is $\binom{\lfloor (k-1)/2\rfloor}{j},$ so you get the sum formula $(1).$

So for $x\leq 2^{k+1}-1,$ you get:

$$\begin{align} A_{2^{k+1}}&=\sum_{i=1}^{k}\sum_{j=0}^{\lfloor (i -1)/2\rfloor}2^j\binom{\lfloor (i -1)/2\rfloor}j\\ &=\sum_{j=0}^{\lfloor(k-1)/2\rfloor}2^j\sum_{i=2j+1}^{k} \binom{\lfloor (i -1)/2\rfloor}j \end{align} $$

When $k=2q$ is even:

$$A_{2^{k+1}}= \sum_{j=0}^{q-1}2^{j+1}\binom{q+1}{j+1}=3^{q+1}-1-2^{q+1}$$

And with $B_{2^{2q+1}}=2^{2q+1}-1-A_{2^{2q+1}}$ you get $\dfrac{A_{2^{2q+1}}}{B_{2^{2q+1}}}\to0$ as $q\to\infty.$

But this only proves for a subsequence of $A_n/B_n.$