How to calculate the density of any subset of $\mathbb{Q}$ using a particular Folner Sequence of $\mathbb{Q}$?

Question

Suppose we choose a particular Folner Sequence of $\mathbb{Q}$ such as $F_n=\left\{\frac{p}{2^k(2q+1)}:p,q\in\mathbb{Z},k\in\mathbb{N},\gcd\left(p,2^k(2q+1) \right)=1,2^k\le n, |2q+1| \le n, \left|\frac{p}{2^k(2q+1)}\right|\le n\right\}$? Is this a Folner Sequence of $\mathbb{Q}$?

How do we calculate the density of $A$, where $A\subseteq{\mathbb{Q}}$.

(Informally we define the density of $A$ as)

$$D(A)=\lim_{n\to\infty}\frac{\left|A\cap F_n\right|}{\left|F_n\right|}$$

(Formally we use Ultrafilters to replace $D(A)$ as a probability measure.)

In order to find $|A\cap F_n|$, we must present the intersection as a set with fully reduced elements. I will attempt an example but I don't know if I am correct. If I am not how do we correctly calculate the answer?

If $A=\left\{\frac{f^3}{g^6}:f,g\in\mathbb{Z},g\neq 0\right\}$ and $F_n=\left\{\frac{p}{2^k(2q+1)}:p,q\in\mathbb{Z},k\in\mathbb{N},\gcd\left(p,2^k(2q+1) \right)=1,2^k\le n, |2q+1| \le n, \left|\frac{p}{2^k(2q+1)}\right|\le n\right\}$, the cardinality of the interesection of $F_n$ and interval $[0,1]$ is $|F_n\cap[0,1]|=\sum\limits_{|k|\le\lfloor\log_2(n) \rfloor}\sum\limits_{|q|\le\lfloor (n-1)/2\rfloor}\phi(2^k(2q+1))=\sum\limits_{|k|\le \lfloor \log_{2}(n) \rfloor}\phi(2^k)\sum\limits_{|q|\le \lfloor (n-1)/2\rfloor}\phi(2q+1)\approx\frac{2}{\pi^2}n^2$ for $n\in\mathbb{N}$ using this and $|F_n|\lessapprox(2n)\sum\limits_{q=1}^{n}\phi(q)\approx(2n)\cdot\frac{2}{\pi^2}n^2$, where $\sum\limits_{q=1}^{n}\phi(q)$ is Euler's Summatory function. We must present $|A\cap F_n|$ in a similar way.

For example, take $A=\left\{\frac{f^3}{g^6}:f,g\in\mathbb{Z},g \neq 0\right\}$. Using the identity I found here, $\left\{\frac{f^3}{g^6}:f,g\in\mathbb{Z},g\neq 0\right\}=\left\{\frac{c^2}{d^2}:c,d\in\mathbb{Z}, d\neq 0\right\}=\left\{\frac{c^2}{2^{2k}(2d+1)^2}:c,d\in\mathbb{Z}, 2d+1 \le n, 2^{2k}\le n,\gcd(c^2,d^2)=1, d\neq 0\right\}$.

Using the answer to this, we can create an asymptotic series of $\left|A\cap F_n\right|=\left|\left\{\frac{c^2}{d^2}:c,d\in\mathbb{Z},\gcd(c^2,d^2)=1,d\neq 0,d^2\le n,\left|\frac{c^2}{d^2}\right|\le n\right\}\right|=F(\sqrt{n})$ where $F(n)$ is the asymptotic series of $(2n)\sum\limits_{d=1}^{\lfloor n \rfloor}\sqrt{\phi(d^2)}$. I do not understand @Reuns answer so I do not know what the asymptotic series would equal (I will call this $m(n)$). Our answer should be

$$D(A)\lessapprox\lim_{n\to\infty}\frac{|A\cap F_n|}{|F_n|}=\lim_{n\to\infty}\frac{(2n)m(n)}{(2n)\frac{3}{\pi}n^3}$$

I don't what the lower bound is but am I correct so far? If not what is the lower and upper bound?

(Using Generalized Euler's Phi Function can be helpful)

Finally, if we generalized $A$ into $\left\{\frac{s_1}{s_2}\right\}$, where $s_1:D_1\cap\mathbb{Z}\to R_1$, $s_2:D_2\cap\mathbb{Z}\to R_2$ such that $D_1,D_2$ is the domain of $s_1,s_2$ and $R_1,R_2$ is the range of $s_1,s_2$; then what is the asymptotic series of $\left|A\cap F_n \right|$? What would be the density of $A$ if we took a Folner sequence of $\left\{\sqrt[b]{\frac{p}{q}}:p,q\in\mathbb{N},q\neq 0\right\}$, where $b\in\mathbb{Q}$ is a constant variable?

Answer 1

Update

Is this a Folner Sequence of $\mathbb{Q}$?

It seems so, but this claim needs to be checked.

$|F_n\cap[0,1]|=\sum\limits_{|k|\le\lfloor\log_2(n) \rfloor}\sum\limits_{|q|\le\lfloor (n-1)/2\rfloor}\phi(2^k(2q+1))$

I assume that the sets $F_n$ in the second definition are the same as these from the first definition (in the second definition we still have a condition $2q+1\le n$)

If $|2q+1|\le n$ then the bounds for $q$ should be $\lceil (–n-1)/2\rceil \le q \le \lfloor (n-1)/2\rfloor$.

Also there is a double counting: if $p$ is an odd number with $\operatorname{gcd}(p,2q+1)$ then a fraction $\frac {p}{2^k(2q+1)}$ equals to a fraction $\frac {-p}{2^k(2(-q-1)+1)}$.

$\sum\limits_{|k|\le \lfloor \log_{2}(n) \rfloor}\phi(2^k)\sum\limits_{|q|\le \lfloor (n-1)/2\rfloor}\phi(2q+1)\approx\frac{2}{\pi^2}n^2$

I didn’t find at your links explicit formulae implying this equality. Assuming that $\phi(-x)=\phi(x)$ for each natural $x$, we have $\phi(2^k)=2^{k-1}$, so

$$\sum\limits_{|k|\le \lfloor \log_{2}(n) \rfloor}\phi(2^k)\approx\sum\limits_{|k|\le \lfloor \log_{2}(n) \rfloor} 2^{k-1}\approx 2\cdot 2^{\lfloor \log_{2}(n) \rfloor}$$ $$\sum\limits_{|q|\le \lfloor (n-1)/2\rfloor}\phi(2q+1)\approx 2\sum\limits_{1\le q\le \lfloor (n-1)/2\rfloor}\phi(2q+1)\approx $$ $$2\sum\limits_{1\le q\le \lfloor (n-1)/2\rfloor}\phi(q)- 2\sum\limits_{1\le q\le \lfloor (n-1)/2\rfloor}\phi(2q).$$

The first sum should be about $2\Phi(\lfloor (n-1)/2\rfloor)$, but the members of the second corresponding to even $q$’s should be estimated.

If I am not correct, then how would you solve my examples? If I am incorrect how would we solve the density of my examples of $A$ using my edited Folner Sequence of $\Bbb Q$.

If I would have such a task, which looks rather technical and long, I’d accurately do it step by step, trying to use, if needed, some number theory results, like properties of Totient Summatory Function, or even computational guesses, but I’m still not sure that I would be able to finish this task due to possible number theoretical complications.

End of the update

Too long for a comment.

How to calculate the density of any subset of $\mathbb{Q}$ using a particular Folner Sequence of $\mathbb{Q}$?

In general it depends on given a subset of $\mathbb{Q}$, a Følner sequence, and (possible) an (ultra)filter $\mathcal F$ with respect to which we take the limit (in order to assure its existence).

I tried to answer your particular question, but its formulation is so inaccurate that I failed to extract an exact and correct question :-(, so I listed my remarks below instead in order to help you to formulate such a question.

1) You regularly miss condition $q\ne 0$, when you divide by $q$.

2) The notation $F_n$ is used for sets with different definitions.

3) It cannot be $|F_n\cap[0,1]|=\frac{3}{\pi^2}n^2$, because the first quantity is integer. The same remark concerns an equality $|F_n|=(2n)\cdot\frac{3}{\pi^2}n^2$.

4) Even taking into account Remark 1, the sets $A=\left\{\frac{f^3}{g^6}:f,g\in\mathbb{Z}, g\ne 0\right\}$ and $B=\left\{\frac{f^2}{g^2}:f,g\in\mathbb{Z}, g\ne 0\right\}$ are not equal. For instance, a number $8=\frac {2^3}{1^6}\in A$, but it is not a square of a rational number, so it does not belong to $B$.

5) I guess that a formula $$\left|A\cap F_n\right|=\left|\left\{\frac{f^2}{g^2}:f,g\in\mathbb{Z},\gcd(f^2,g^2)=1,g^2\le n,\left|\frac{f^2}{g^2}\right|\le n\right\}\right|$$ fails for boths definitions of $F_n$, because the first of them misses a condition $\gcd(f,g)=1$ and in the second misses a bound $g^2\le n$.

6) If we generalize $A$ into $\left\{\frac{s_1}{s_2}\right\}$, where $s_1:D_1\to\mathbb{R}$, $s_2:D_2\to\mathbb{R}\setminus{\left\{0 \right\}}$ and $D_1,D_2\subseteq \mathbb{Z}$ then $s_1$ and $s_2$ are not numbers, but functions, so elements of $A$, being quotients of functions with possible distinct domains, should be not rational numbers, but can be even undefined.

7) It is not clear why the generalized Euler Phi function for Dedekind abstract number rings or number fields pointed by a link to MathOverflow question is needed for the previous.