Approximating the density of $S\cap[a,b]$ using a Folner Sequence of $A$, where $S \subseteq A$?

Edit: I redid my code but it's still not working.

How do we use Mathematica to approximate

$$D(S\cap[a,b])=\lim_{n\to\infty}\frac{\left|S\cap{F_n\cap[a,b]}\right|}{\left|F_n\cap[a,b]\right|}$$

where $D$ is the density of $S\cap[a,b]$ (in $A\cap[a,b]$), $[a,b]$ is an interval for $a,b\in\mathbb{R}$, $F_n$ is the Folner Sequence of $A$, and $S\subseteq A$. For more information, click here (replace $G,X,i,g$ with $A,S,n,a$) and here.

$A$ is countable and dense in $\mathbb{R}$ and can be written as the operations of finite or infinite one-variable functions that, when defined on countable domains dense in $\mathbb{R}$, intersect with the integers.

Example:

$$A=\left\{\frac{m+\ln(w)}{2^{(p/q)}+\sqrt{z}}:m,p,q,z,w\in\mathbb{Z},2^{p/q}+\sqrt{z}\neq0,q\neq0,z>0,w>0\right\}$$

There are many Folner Sequence of $A$. In general, the most natural, "intuitive" sequence is calculated by restricting the whole set, and the variables of each function in the set, by $n$.

$$F_n=\left\{\frac{m+\ln(w)}{2^{(p/q)}+\sqrt{z}}:m,p,q,z\in\mathbb{Z},2^{p/q}+\sqrt{z}\neq0,q\neq0,z>0, w>0,\left|\frac{m+\ln(w)}{2^{(p/q)}+\sqrt{z}}\right|\le n,|m|\le n,|p|\le n,|q|\le n, |z| \le n, |w| \le n\right\}$$

$S$ can be written similarly to $A$ except it must be a subset.

Example:

$$S=\left\{\frac{m_1}{2^{(p_1/q_1)}+\sqrt{z_1}}:m_1,q_1,z_1\in\mathbb{Z},2^{(p_1/q_1)}+\sqrt{z_1}\neq0,q_1\neq0,z_1>0\right\}$$

I attempted to approximate $D(S\cap[0,1])$ when $S=\left\{\frac{m^2}{n^2}:m,n\in\mathbb{Z},n\neq 0\right\}$, $A=\mathbb{Q}$, $F_n=\left\{\frac{p}{2^k(2q+1)}:p,k,q\in\mathbb{Z},2^k \le n, |2q+1|\le n, \left|\frac{p}{2^k(2q+1)}\right|\le n\right\}$.

(In my code I replaced $A$ with A[x_,y_,z_,...] and $F_n$ with F[x_,y_,z_,...], $S$ with S[x_,y_,z_,...], and $D$ with d. I also set $a=0$ and $b=1$.)

I tried to list all elements of $F_n\cap[0,1]$ depending on $n$ and determine which elements in $S\cap[0,1]$ exist in $F_n\cap[0,1]$. Then I counted all elements where this holds and divided it by the total elements in $F_n\cap[0,1]$.

Unprotect[d] Remove[d] A[p_, k_, q_] := p/((2^k)*(2*q + 1)) F[p_, n_] := Table[A[p, k, q], {k, 0, Floor[Log[2, n]]}, {q, 0, Floor[(n - 1)/2]}] f[n_, a_, b_] := p /. Table[ Solve[a <= A[p, k, q] <= b, p, Integers], {k, 0, Floor[Log[2, n]]}, {q, 0, Floor[(n - 1)/2]}] Ff[n_, a_, b_] := DeleteDuplicates[ Flatten[Table[ F[f[n, a, b][[v]][[u]], n][[v]][[u]], {v, 1, Floor[Log[2, n]] + 1}, {u, 1, Floor[(n - 1)/2] + 1}]]] S[j_, k_] := j^2/k^2 X[a_, b_, n_] := Count[Boole[Resolve[Exists[{j, k}, S[j, k] == Ff[n, a, b]]]], 1] Y[a_, b_, n_] := Count[Ff[n, a, b]] d[n_, a_, b_] := N[(Y[a, b, S] - X[a, b, S])/Y[a, b, S]] Ff[4, 1, 2] X[1, 2, 4] Y[1, 2, 4] d[1, 2, 4] 

Instead, I get

 During evaluation of In[629]:= Table::iterb: Iterator {v,1,1+Floor[Log[S]/Log[2]]} does not have appropriate bounds. During evaluation of In[629]:= Table::iterb: Iterator {v,1,1+Floor[Log[S]/Log[2]]} does not have appropriate bounds. During evaluation of In[629]:= Table::iterb: Iterator {v,1,1+Floor[Log[S]/Log[2]]} does not have appropriate bounds. During evaluation of In[629]:= General::stop: Further output of Table::iterb will be suppressed during this calculation. During evaluation of In[629]:= Exists::msgs: Evaluation of S[j,k]==Ff[S,2,4] generated message(s) {General::stop,Table::iterb}. During evaluation of In[629]:= Exists::msgs: Evaluation of j^2/k^2==Table[F[f[<<3>>][[v]][[u]],S][[v]][[u]],{v,1,1+Floor[Log[S]/Log[<<1>>]]},{u,1,Floor[(S-1)/2]+1}] generated message(s) {Table::iterb}. During evaluation of In[629]:= Exists::msgs: Evaluation of j^2/k^2==Table[F[f[<<3>>][[v]][[u]],S][[v]][[u]],{v,1,1+Floor[Log[S]/Log[<<1>>]]},{u,1,Floor[(S-1)/2]+1}] generated message(s) {Table::iterb}. During evaluation of In[629]:= General::stop: Further output of Exists::msgs will be suppressed during this calculation. Out[642]= 1. 

Is there a better and faster method to solving my example? How do we generalize this for any $S\cap[a,b]$ and $A\cap[a,b]$?