Timeline for Do these rational sequences always reach an integer?

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15 events

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Dec 24, 2020 at 0:53	comment	added	Sebastien Palcoux		Thanks for pointing this out. I don't understand in what it could help. Could you be a bit more specific?
Dec 23, 2020 at 18:50	comment	added	katago		@Sebastien Palcoux, mathoverflow.net/questions/379323/… this is a problem related to this, and it seems the method focus on the calculation on newton polygon mentioned by David E Speyer can also make some progress in this problem, at least has some potential.
Dec 17, 2020 at 14:57	comment	added	katago		Sebastien Palcoux；Yes, and your proof is much more precise than mine for $II(2,1)$. And for general $II(p,k)$, I think this is a highly nontrivial problem, maybe the difficulty is already very close to Collestz conjecture.
Dec 17, 2020 at 14:51	comment	added	Sebastien Palcoux		So you only know a proof for $2$, but not yet for $2^k$ or $3$, right? Your framework is promising for a more general proof, but for $2$, there is the following shorter proof by contradiction: assume that $(u_n)$ never reach an integer, then for all $n$ we must have $u_n=k_n + \frac{1}{2}$, but $u_{n+1} = k_n + \frac{k_n}{2}$, so $k_n$ must be odd for all $n$. So $k_n=2h_n+1$, $u_n=2h_n+1 + \frac{1}{2}$, and $u_{n+1} = 3h_n+1+\frac{1}{2} $, which means that $2h_{n+1}=3h_n$, it follows that $h_n = (\frac{3}{2})^nh_0$, and so $2^n$ divides $h_0$ for all $n$, contradiction.
Dec 17, 2020 at 14:03	comment	added	katago		Yes, we can avoid it, it is just for convenience because then we can consider the problem in $\mathbb{N}^*$, but this is not necessary and is no help for making progress on the problem. And the Subsequent argument is also valid for $u_{0}=\sum_{k=-1}^{\infty} a_{k} 2^{k}$ for problem $I(2,1)$ and also valid for $u_{0}=\sum_{k=-t}^{\infty} a_{k} 2^{k}$ for the problem $I(2,t)$. Although in the latter case, only with such an argument is not enough to get a complete proof
Dec 17, 2020 at 14:01	comment	added	Sebastien Palcoux		I am confused with the rescaling $u_0 \to pu_0$. Is it possible to avoid it, and to just write $u_0=\sum_{k=-1}^{\infty}a_kp^k$ (and in general, $\sum_{k=-t}^{\infty}a_kp^k$)?
Dec 17, 2020 at 13:58	comment	added	katago		It should be $a_0(k)$, already corrected.
Dec 17, 2020 at 13:57	comment	added	katago		Sebastien Palcoux： yes I try to calculate for $a_k(n)$ where $a_k(n)$ is the k-th digit of p expansion in $u_n$, but the recurrence formula only true when there is no arithmetic carry in $a_0(n), a_1(n),...,a_k(n)$
Dec 17, 2020 at 13:54	history	edited	katago	CC BY-SA 4.0	added 278 characters in body
Dec 17, 2020 at 13:47	comment	added	katago		Sebastien Palcoux； yes but I do not claim for all $n\in \mathbb{N}^*$, $1\leq a_k(n)\leq p-1$, I just requre $a_k=a_k(0)$ is the k-th digit of $u_0$ in the base $p$ expension.
Dec 17, 2020 at 13:43	history	edited	katago	CC BY-SA 4.0	added 116 characters in body
Dec 17, 2020 at 13:36	history	edited	katago	CC BY-SA 4.0	added 13 characters in body
Dec 17, 2020 at 12:06	history	edited	katago	CC BY-SA 4.0	added 11 characters in body
Dec 17, 2020 at 11:55	history	edited	katago	CC BY-SA 4.0	added 181 characters in body
Dec 17, 2020 at 11:48	history	answered	katago	CC BY-SA 4.0