I initially wanted to solve the following recurrence relation: $$F_n = c \cdot \left\lfloor \frac{F_{n-1}}{d} \right\rfloor \text{ for } F_0, c, d \in \mathbb{N} \tag{1}$$
Out of interest, I've also approached the problem more generally:
\begin{align*} F_n &= a \cdot \lfloor b \cdot F_{n-1} \rfloor + c \tag{2} \\ C_n &= a \cdot \lceil b \cdot C_{n-1} \rceil + c, \text{ for } a,b,c \in \mathbb{Q}^+ \end{align*}
I wish to solve these generally.
Special case for (1) where $d|c$ and $c|d$
A special case for (1) where $c | d$ and $d|c$ can be found relatively easily.
If $d | c$, then: $$F_n = c \left(\frac{c}{d}\right)^{c-1} \left\lfloor \frac{F_0}{d} \right\rfloor$$ If $c|d$, then:
$$F_n = c \left\lfloor \left(\frac{c}{d}\right)^{n-1} \frac{F_0}{d} \right\rfloor$$
The first is proven by using $d | c \implies \frac{c}{d} \in \mathbb{Z}$, and $\forall n,m \in \mathbb{Z} \left(\lfloor nm \rfloor = nm\right)$ on:
$$F_n = c \cdot \left\lfloor \frac{c}{d} \cdot \left\lfloor \frac{F_{n-2}}{d} \right\rfloor \right\rfloor = c \cdot \frac{c}{d} \cdot \left\lfloor \frac{F_{n-2}}{d} \right\rfloor $$
The second is proven using $c|d \implies \exists n \in \mathbb{Z} \left(\frac{d}{c} = \frac{1}{n}\right)$ and the property $\left\lfloor \frac{\lfloor x / m \rfloor}{n} \right\rfloor = \left\lfloor \frac{x}{mn} \right\rfloor$:
$$F_n = c \cdot \left\lfloor \frac{c}{d} \cdot \left\lfloor \frac{F_{n-2}}{d} \right\rfloor \right\rfloor = c \cdot \left\lfloor \frac{c}{d} \cdot \frac{F_{n-2}}{d} \right\rfloor $$
Example for (1) where $c=3, d=2$
The simplest case with some interesting behaviour is $c = 3$ and $d = 2$. Despite different initial conditions $F_0$, quite a few of sequences become the same sequence when they share the same values:
e.g.:
$F_0 = 4, F_1 = 6, F_2 = 9,\color{red}{F_3 = 12, F_4 = 18,\dots}$
$F_0 = 8 ,\color{red}{F_1 = 12, F_2 = 18,\dots}$
For clarity, I'll refer to the sequence $F_0, F_1, F_2, \dots$ as $\mathscr{F}(F_0)$
(e.g.: So $\mathscr{F}(4)$ is the sequence $4, 6, 9, \dots$)
Some observations that can be made are:
Given sequence $(F_i)$ the following sequences share values:
- $\mathscr{F}(F_i)$
- $\mathscr{F}\left(\frac{2}{3}F_i\right)$ if $\frac{2}{3}F_i \in \mathbb{N}$
- $\mathscr{F}(F_i + 1)$ if $F_i$ is even
It also seems every $\mathscr{F}(4 + 6n)$ is always a "new" sequence?
It seems that if $\mathscr{F}(n)$ where $n < 4 + 6n$
Then $\mathscr{F}(n)$ and $\mathscr{F}(4 + 6n)$ do not share any values.
Keeping track of every time an odd number appears in $\mathscr{F}(4)$ results in A087791.
Special case for (2)
A special case for (2) can be solved using Theorem 3.10 from Concrete Mathematics:
Let $f(x)$ be any continuous, monotonically increasing function with the property that: $$f(x) \in \mathbb{Z} \implies x \in \mathbb{Z}$$ Then: $$\left\lfloor f(x) \right \rfloor = \left\lfloor f(\lfloor x \rfloor) \right\rfloor$$
Using $f(x) = abx + bc$, we can prove by induction that: $$F_n = a\lfloor f^n(bF_0) \rfloor +c \text{ where } f^n(x) = f(\underbrace{f(f(\dots))}_\textrm{n times}$$ We can then prove $f^n(x_0) = (ab)^n x_0 + \frac{1-(ab)^n}{1-ab}cb$, and obtain the special case solution:
If $a,b,c \in \mathbb{R^+}$ satisfies $\left(abx + bc \in \mathbb{Z} \implies x \in \mathbb{Z}\right)$
Then $$F_n = a \left\lfloor (ab)^{n-1} bF_0 + \frac{1-(ab)^{n-1}}{1-ab}cb \right\rfloor + c $$
Admittedly, I'm not quite happy with this result as the conditions are incredibly restrictive.
It also does not solve my original problem (even for cases $d|c$ or $c|d$?)
Are there general solutions to my problem?
For the original problem:
Set $G_n = \frac{F_n}{c}$. Then we have $G_n = \left\lfloor\frac{F_{n-1}}{d}\right\rfloor = \left\lfloor\frac{c}{d}G_{n-1}\right\rfloor$
I think this form of the problem probably has been studied for some values of $\frac{c}{d}$ and some initial values. For example, for $c=3$ and $d=2$, with initial condition $G_1=2$ can be found in the OEIS, while the corresponding $F_n$ sequence cannot.
For that particular case it's known that $G_n = \left\lceil(\tfrac{3}{2})^{n-1}2\ \alpha\right\rceil$ where $\alpha$ is a constant related with the case $k=3$ of the Josephus problem
Sequences of the form $x_{n+1} = \lfloor \alpha\ x_n \rfloor$ for other values of $\alpha$ (not necessarily rational) can also be found on the OEIS, for example, $\alpha = \sqrt{2}$ or $\alpha = \pi$