Sequence of floor functions If you write out the multiples of π and take the floor of each term, you get the following sequence: 3, 6, 9, 12, 15, 18, 21, 25, 28, etc.
Now, let’s consider the integers that aren’t in this sequence: 1, 2, 4, 5, 7, 8, …
Surprisingly, there also exists a real number t such that the n-th term of this sequence is the floor of  t.n. Find t to the nearest thousandth (3 decimals).
How can I proceed.Thanks in advance.
 A: Hint:
The equation $\lfloor nt\rfloor=d$ gives you a possible range for $t$ (by $d\le nt<d+1$). The intersection of the ranges for various $n$ will narrow down the interval.
A: As pointed out by others in comment, sequences of the form $a_n = \lfloor \alpha n \rfloor$ is known as Beatty sequence.
If you are given two irrational numbers $\alpha, \beta > 1$ such that $\frac{1}{\alpha} + \frac{1}{\beta} = 1$, the two Beatty sequences $\lfloor \alpha n\rfloor$ and $\lfloor \beta n \rfloor$ will be complementary to each other.
The set of positive integers will be a disjoint union of them:
$$\mathbb{Z}_{+} = \{ \lfloor \alpha n\rfloor : n \in \mathbb{Z}_{+} \}
\sqcup \{ \lfloor \beta n\rfloor : n \in \mathbb{Z}_{+} \}$$
For the question at hand, $t$ do exists and equals to  $\frac{1}{1 - \frac{1}{\pi}} = \frac{\pi}{\pi - 1}$.
It is actually not that hard to prove this from first principle.
Let $A = \{ \lfloor \pi n \rfloor : n \in \mathbb{Z}_{+} \}$ and $B = \mathbb{Z}_{+} \setminus A$. Let $\beta_n$ be the $n^{th}$ member of $B$.
Being the $n^{th}$ positive integer not belongs to $A$, $\beta_n$ satisfies:
$$\#\{ x \in B : x \le \beta_n \} = n \iff \left\lfloor \frac{\beta_n}{\pi}\right\rfloor = \#\{ x \in A : x \le \beta_n \}  = \beta_n - n$$
This leads to
$$\left(1 - \frac{1}{\pi}\right)\beta_n + \left\{\frac{\beta_n}{\pi}\right\} = n
\iff
\beta_n + \frac{\pi}{\pi - 1}\left\{\frac{\beta_n}{\pi}\right\}
= \frac{\pi}{\pi - 1}n\\
\implies \beta_n + \left\lfloor\frac{\pi}{\pi - 1}\left\{\frac{\beta_n}{\pi}\right\}\right\rfloor = \left\lfloor \frac{\pi}{\pi - 1}n \right\rfloor\tag{*1}
$$
Notice $\beta_n \notin A$. For any integer multiple $\pi k$ of $\pi$, whenever
$\pi k \ge \beta_n$, we need to have $\pi k \ge \beta_n+1$. Since $\pi$ is irrational, the inequality is strict. This implies
$$\pi\left(\left\lfloor\frac{\beta_n}{\pi}\right\rfloor + 1 \right) > \beta_n + 1
\iff \pi - 1 > \pi \left\{\frac{\beta_n}{\pi}\right\} 
\implies
 \left\lfloor\frac{\pi}{\pi - 1 }\left\{\frac{\beta_n}{\pi}\right\}\right\rfloor = 0\tag{*2}$$
Combine $(*1)$ and $(*2)$, we obtain
$\displaystyle\;\beta_n = \left\lfloor \frac{\pi}{\pi - 1} n \right\rfloor$ and
$$t = \frac{\pi}{\pi - 1} \approx 1.46694220692426...$$
The desired answer is $1.467$.
