3
$\begingroup$

During research I came to the following sequence:

Let $\lambda>1$ and define $n_{k+1}=\text{IntergerPart}[\lambda\cdot n_k]$ where we assume that $n_0$ is sufficently large integer, so that the sequence $n_k$ is strictly increasing. Finally let $x_k=\text{FractionalPart}[\lambda\cdot n_k]$.

Question: Are the following claims true or false?

Claim 1: If $\lambda\in \mathbb{R}\backslash\mathbb{Q}$ then the sequence $(x_k)_k$ is dense in $[0,1)$.

Claim 2: If $\lambda\in \mathbb{Q}\backslash\mathbb{N}$ then the sequence $(x_k)_k$ is not periodic.

Since I'm not a specialist in this field I'm looking for some hints how to prove or tackle these two problems. Claims were made after some computer experiments.

P.s. It is clear that in Claim 2 the sequence is generated by finitely many numbers.

$\endgroup$
4
  • 1
    $\begingroup$ We usually denote the integer part by $[x]$ or $\lfloor x\rfloor$, and the fractional part by $\{x\}$. $\endgroup$
    – GH from MO
    Commented Nov 10, 2020 at 14:17
  • 1
    $\begingroup$ @GHfromMO Thank you for your comment. I am familiar with the notation for the interger part. To avoid any confusions, I decided to write it as a Mathematica Functions. $\endgroup$ Commented Nov 10, 2020 at 16:15
  • 1
    $\begingroup$ By "generated by finitely many numbers" do you mean "contains only finitely many distinct values"? $\endgroup$ Commented Nov 10, 2020 at 19:41
  • $\begingroup$ @GregMartin Yes it means "contains only finitely many distinct values". $\endgroup$ Commented Nov 11, 2020 at 6:46

1 Answer 1

3
$\begingroup$

Claim 1 is false. Let $\phi = \frac{1+\sqrt{5}}{2}$, $\overline{\phi} = \frac{1-\sqrt{5}}{2}$, $\lambda = \phi^{2}$ and $n_{0} = 1$.

I claim that $n_{k} = F_{2k+1}$, the $(2k+1)$st Fibonacci number for $k \geq 1$. By strong induction and Binet's formula $F_{k} = \frac{1}{\sqrt{5}}\left(\phi^{k} - \overline{\phi}^{k}\right)$, we have $$ n_{k} = \lfloor \phi^{2} F_{2k-1} \rfloor = \left\lfloor \phi^{2} \cdot \frac{1}{\sqrt{5}} (\phi^{2k-1} - \overline{\phi}^{2k-1}) \right\rfloor = \left\lfloor F_{2k+1} - \frac{1}{\sqrt{5}} \overline{\phi}^{2k-3} (1-\overline{\phi}^{2})\right\rfloor. $$ Since $-1 < \overline{\phi} < 0$, the second term in the expression above is positive and less than $1$. Thus, $n_{k} = F_{2k+1}$.

Hence $$ x_{k} = \phi^{2} F_{2k-1} - n_{k} = -\frac{1}{\sqrt{5}} \overline{\phi}^{2k-3}(1-\overline{\phi}^{2}) $$ which tends to $0$ exponentially as $k \to \infty$. In particular, $( x_{k} )_{k}$ is not dense in $[0,1)$.

$\endgroup$
2
  • $\begingroup$ Jeremy thank you for your answer. This was quite unexpected! Do you think that similar convergence (maybe to some other value on the interval) also happens for generic irrational $\lambda$ (or perhaps for Diophantine numbers). Can they be somehow roughly characterized? I also hope that my second claim is wrong, meaning that for generic rational number one would get periodic sequence. $\endgroup$ Commented Nov 11, 2020 at 8:52
  • $\begingroup$ I would guess that the sequence $(x_{k})_{k}$ fails to be dense for an measure zero subset of $\lambda$'s that one could choose, so the generic case is presumably what you found in your numerical experiments. I don't know about your claim 2, my intuition would be that it is true. $\endgroup$ Commented Nov 14, 2020 at 20:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .