Timeline for Are the terms of a linear recurrence integral?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 11, 2016 at 9:19 | vote | accept | CommunityBot | ||
| Nov 14, 2016 at 13:00 | answer | added | Sidney Raffer | timeline score: 13 | |
| Sep 16, 2016 at 18:55 | comment | added | Luc Guyot | @Fry and Todd Leason: Thanks for addressing this case! I certainly made a very poor choice of parameters, as my point is the opposite of Fry's: Isn't the case of recurrences of order $2$ just challenging enough? This is Vesselin Dimitrov's point in his first comment to the other post mathoverflow.net/questions/214839/… : $u_n = (5^n - 2)/3^n$ used to be a challenge. By the way, what is the essential difference between these two posts? If the recurrences of order $2$ can be entirely ruled out here, this should be mentioned in the question. | |
| Sep 16, 2016 at 16:49 | comment | added | user40023 | @ToddLeason Of course, in the case proposed by LucGuyot just looking modulo 4 is enough. Anyway, I pointed out that for Lucas sequences the problem seems almost always decidible. | |
| Sep 16, 2016 at 15:00 | comment | added | Todd Leason | @LucGuyot: $1+3^n$ is 2-divisible at most by 4. For, the binomial formula of $(1+2)^n$ gives $1+3^n=2(1+n^2+4(...))$ and $1+n^2\equiv 2(4)$ for $n$ odd. | |
| Sep 16, 2016 at 14:44 | comment | added | user40023 | @LucGuyot $3^n + 1 = u_{2n} / u_n$ where $u_n := (3^n - 1) /(3 - 1)$ is a Lucas sequence. Formulas for the $p$-adic valuation of Lucas sequence are given in C. Sanna, The p-adic valuation of Lucas sequences, The Fibonacci Quarterly 54, 118–124 (a preprint here: researchgate.net/publication/…), in particular $v_2(u_n) = v_2(n) + 1$ if $2 \nmid n$ and $v_2(u_n) = 0$ is $2 \nmid n$. Therefore, $v_2(3^n + 1) \leq v_2(u_{2n}) = v_2(n) + 2 \ll \log n$ so that $(1+3^n) / 2^n$ is not an integer for all large $n$. | |
| Sep 16, 2016 at 14:00 | comment | added | Luc Guyot | I am already stuck with a linear recurrence of order $2$ such as $u_n = \lambda_1^n + \lambda_2^n$ with $\lambda_i \in \mathbb{Q}$ for which the Skolem problem is solvable. Indeed, I have no idea whether $(1 + 3^n)/2^n$ is an integer for infinitely many $n$ or not. In other words, what do we know about the limit points of $3^n$ in the ring $\mathbb{Z}_2$ of the $2$-adic integers? | |
| Sep 16, 2016 at 12:43 | comment | added | Emil Jeřábek | This sounds similar to the Skolem problem or the positivity problem. I would thus guess the problem is open. | |
| Sep 16, 2016 at 10:57 | comment | added | Sylvain JULIEN | Perhaps one should study the polynomial $\sum_{i}a_{i}X^{i}$. | |
| Sep 16, 2016 at 10:41 | history | asked | user40023 | CC BY-SA 3.0 |