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Let all functions be integer functions herein. I.e. $\Bbb{Z}\to\Bbb{Z}$ or $\Bbb{N}\to\Bbb{Z}$ where appropriate.

I found this jewel of floor functions.

So that made me wonder whether, we can solve for the simultaneous solution to:

$$ f(x) = g(y) $$

where

$$f(x) = x +2 \left\lfloor\frac{x}{3}\right\rfloor + 1 - [3 \mid x]\\ g(y) = y + 2 \left\lfloor\frac{y}{5}\right\rfloor + 1 - [5\mid y] $$

where $x_{k} = x \mod k$ the least non-negative residue. And $[3\mid x]=$ Iverson bracket for whether $3$ divides $x$.

But more specifically, can we describe a single-parameter function $h(t) = (x(t), y(t))$ such that:

$$ f(x(t)) = g(y(t)) $$

?

Attempt.

$$ x + 2\frac{ x - x_{3}}{3} + 1 - [3\mid x] = y + 2\frac{ y - y_{5}}{5} + 1 - [5\mid y] \\ \iff \\ 25x +10x_{3} - 15[3\mid x ] = 21y + 6y_{5} - 15[5\mid y] $$

Well, I've managed to reduce it for us in terms of variables modulo $3,5$, Iverson brackets and naked variables/coefficients, all "linearly" combined.


In other words how do we go about actually computing an equalizer (category theory term)?

Alternative Question. Clearly $h(t)$ is necessarily also injective and increasing as long as $x(t), y(t)$ are. Does this mean that in some way that the function $$k(z) = \# \{ 0 \leq y \leq z: y = f(x(t))[=g(y(t))] \text{ for some } t \in \Bbb{N}\} $$ has something to do with a possible parameterization?

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    $\begingroup$ What does $~[ ~3 ~| ~x ~] ~$ represent? $\endgroup$ Commented Jun 23 at 8:34
  • $\begingroup$ @user2661923 $[3\mid x] = \begin{cases}1 \text{ if } 3 \text{ divides } x\\ 0 \text{ otherwise} \end{cases}$ a function $\Bbb{Z} \to \{0,1\} \subset \Bbb{Z}$. $\endgroup$ Commented Jun 23 at 8:35
  • $\begingroup$ My first try, which might not succeed, would be to assume that $~x = 3P + Q + r ~: ~P \in \Bbb{Z}, ~Q \in \{0,1,2\}, ~0 \leq r < 1. ~$ Also, if I have understood your question correctly, it looks like you are assuming that $~x \in \Bbb{Z} \implies r = 0.$ $\endgroup$ Commented Jun 23 at 8:38
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    $\begingroup$ You can rewrite $~[ ~k ~| ~x ~] ~$ as $\left\lfloor\frac{x}{k}\right\rfloor - \left\lfloor\frac{x-1}{k}\right\rfloor$ if it helps $\endgroup$
    – Henry
    Commented Jun 23 at 8:46
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    $\begingroup$ I don't know if this helps: These functions $(x(t), y(t))$ look like a line as you zoom out. I know how to compute the exact line's "slope" $\frac{y(t)}{x(t)}$. Simply drop all floors, and combine the coefficients of $x$ i.e. you could definitely start from @Henry 's representation, which is all floor (no Iverson bracket). Then the deviation of other solutions from this trendline is something small. $\endgroup$ Commented Jun 23 at 8:49

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