Limit associated with two Beatty sequences that are not a Beatty pair Suppose that $r>1$ and $s>1$ are irrational numbers, and let $a_n=\lfloor nr \rfloor$ and $b_n=\lfloor ns \rfloor$.  Assume that $r$ and $s$ are numbers for which $\{a_n\}\cap\{b_n\}$ is infinite, and let $(c_n)$ be the increasing sequence of numbers in $\{a_n\}\cap\{b_n\}$.  It appears that the number
$$t = \lim_{n \to \infty}\frac{c_n}{n}$$
exists.  Can someone determine its value?
Here are a few estimates:
$$\begin{array}{c|c|c|}
r & s & t \\ \hline
\sqrt{3} & \sqrt{3}+1 & 4.098 \\ \hline
\sqrt{3} & \sqrt{5} & 3.864 \\ \hline
\sqrt{5} & \sqrt{5}+1 & 6.792 \\ \hline
\sqrt{5} & \sqrt{5}-1 & 6.696 \\ \hline
\pi & e & 8.539 \\ \hline
\end{array}$$
 A: I will give a complete answer when $1,1/r$ and $1/s$ are linearly independent over $\mathbb Q$ and a recipe to compute your $t$ otherwise.
First of all, notice that $n$ lies in a Beatty sequence $\lfloor m\alpha\rfloor$, where $\alpha>1$ if and only if $\{n/\alpha\}>1-\frac{1}{\alpha}$. Indeed, if $n=\lfloor m\alpha \rfloor$, then
$$
\frac{n}{\alpha}=\frac{m\alpha-\{m\alpha\}}{\alpha}=m-\frac{\{m\alpha\}}{\alpha}, \text{ so }\{n/\alpha\}>1-\frac{1}{\alpha}.
$$
On the other hand, if $\{n/\alpha\}>1-\frac{1}{\alpha}$, then one can set $m=\lfloor n/\alpha \rfloor+1=n/\alpha+1-\{n/\alpha\}$ and get
$$
m\alpha=n+\alpha\left(1-\{n/\alpha\}\right),
$$
so $\lfloor m\alpha \rfloor=n$, as needed.
Now, Case 1: $1,1/r$ and $1/s$ are linearly independent over $\mathbb Q$. In this case, it is known (say, by Weyl's criterion), that the points
$$
t_n=(n/r \mod 1, n/s \mod 1)\in \mathbb R^2 \diagup \mathbb Z^2
$$
are uniformly distributed on the torus. This means that for $N\to +\infty$ the proportion of $n\leq N$ with
$$
\{n/r\}>1-\frac{1}{r}\text{ and }\{n/s\}>1-\frac{1}{s}
$$
is equal to the measure of the set $(x,y) \in \mathbb R^2\diagup \mathbb Z^2$ with $x>1-\frac{1}{r}, y>1-\frac{1}{s}$, hence the inverse $t$ of this limiting proportion is equal to $rs$. For example, for $\pi$ and $e$ we get $\pi e\approx 8.5397$ and for $\sqrt{3}$ and $\sqrt{5}$ we get $\sqrt{15}\approx 3.873$.
Case 1.5 is when both $r$ and $s$ are rational. In this case, your problem reduces to enumeration of certain congruence classes $\mod \mathrm{num}(r)\mathrm{num}(s)$.
Finally, Case 2 is when $1/r$ is irrational, but for some $a,p,q\in \mathbb Z$ we have $1/s=p/q\cdot1/r+a/q$. Without loss of generality one can assume that $p,q>0$, $a\geq 0$. Next, let us divide $\mathbb N$ into congruence classes $\mod q$ and evaluate the (relative) density $\delta_b$ of all $n\equiv b\pmod q$ such that
$$
\{n/r\}>1-\frac{1}{r}\text{ and }\{n/s\}>1-\frac{1}{s}.
$$
In this case, the answer would be
$$
t=\frac{q}{\delta_0+\delta_1+\ldots+\delta_{q-1}}.
$$
Fix $b$. The congruence $n\equiv b\pmod q$ means that $n=qm+b$. Now, we need
$$
\{qm/r+b/r\}>1-\frac{1}{r}\text{ and }\{pm/r+b/s\}>1-1/s.
$$
Since $1/r$ is irrational, $\{m/r\}$ is equidistributed $\mod 1$. This means that our relative density $\delta_b$ is equal to the measure of all $x\in [0,1]$ such that
$$
\{qx+b/r\}>1-\frac{1}{r}\text{ and }\{px+b/s\}>1-1/s.
$$
(By relative density I mean the density of corresponding $n$ in the residue class, i.e. the actual density of your $n$ with $n\equiv b\pmod q$ is $\delta_b/q$)
