Timeline for Consecutive integers of the form $2^a 3^b 5^c$
Current License: CC BY-SA 4.0
16 events
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| Apr 16 at 18:52 | comment | added | W. Edwin Clark | Just to note that such numbers are 5-smooth or regular numbers. See smooth numbers and regular numbers in Wikipedia A formula for the number of such numbers less than n is given in the second link. | |
| Apr 12 at 9:36 | comment | added | Jakub Konieczny | On a related note: I have had some luck in using small values of $\{ b \log_2 3 + c \log_2 5\}$ (with $b,c \in \mathbb{Z}$) to show that, given $k$, at least one of $N_{k-1},N_{k+1}$ needs to be reasonably close to $N_k$. Unfortunately, for all I know, it could happen that $N_{k-1}$ is close to $N_k$ but $N_{k+1}$ is far. | |
| Apr 12 at 9:31 | comment | added | Jakub Konieczny | @mathworker21 Yes, this would definitely suffice! The two statements are not exactly equivalent, since the answer to my question would still be positive if, say, there was exactly one "exceptional" interval of length $X/\log \log X$, while all other exceptional intervals are short. I'm no expert in diophantine approximation, but I have not been able to find estimates on distribution of $\{ b \log_2 3 + c \log_2 5\}$ to make this approach work. | |
| Apr 11 at 22:50 | comment | added | mathworker21 | You just want every interval $[Y,Y+\epsilon\frac{X}{\log X}) \subseteq [X,2X)$ to contain an integer of the form $2^a 3^b 5^c$ (for any fixed $\epsilon > 0$, for large$_{\epsilon}$ enough $X$). Probably you can do $\frac{100X}{\log^2 X}$ in place of $\epsilon \frac{X}{\log X}$. In any event, it suffices to show for any $M \in \mathbb{N}$ and any interval $I \subseteq [0,1)$ of length $|I| = \epsilon\frac{1}{M}$, that there are $b,c \le M$ with $$\bigl\{b\log_2(3)+c\log_2(5)\bigr\} \in I,$$ where $\{\cdot\}$ denotes fractional part. It's quite possible this is known. | |
| Apr 11 at 1:04 | history | edited | Daniel Asimov | CC BY-SA 4.0 |
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| Apr 10 at 23:19 | comment | added | Max Lonysa Muller | @JakubKonieczny Interesting, thank you. | |
| Apr 10 at 23:05 | comment | added | Jakub Konieczny | Since automatic sequences are a bit of an esoteric subject, and since the origin does not seem to shed much light on the problem, I did not mention them in the post. (More discussion on asymptotically automatic sequences can be found in arxiv.org/abs/2305.09885 and arxiv.org/abs/2209.09588). | |
| Apr 10 at 23:04 | comment | added | Jakub Konieczny | @MaxMuller: My motivation comes from a generalisation of the notion of an automatic sequence, which is tentatively dubbed "asymptotically automatic". It turns out that the asymptotic variant of Cobham's theorem is significantly more complicated than the original. If the answer to my question is positive, then one can construct an interesting (at least to me!) example of a sequence that is asymptotically automatic in bases 2, 3 and 5, which is impossible for automatic sequences. | |
| Apr 10 at 22:50 | comment | added | Max Lonysa Muller | @SamHopkins I'm not suggesting he should add more motivation to the question body -- I'm just curious. | |
| Apr 10 at 22:21 | comment | added | Sam Hopkins | @MaxMuller: it's possible there's other context but the "rationale" paragraph at the end seems like good enough explanation/motivation for this problem to me. | |
| Apr 10 at 22:05 | comment | added | Max Lonysa Muller | @JakubKonieczny I wonder whether this problem arises in a broader context, and if so, where and how? | |
| S Apr 10 at 21:55 | history | suggested | CommunityBot | CC BY-SA 4.0 |
some small copyediting corrections
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| Apr 10 at 21:38 | review | Suggested edits | |||
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| Apr 10 at 20:31 | comment | added | GH from MO | You are right, my bad. Tricky logarithms. I will delete my comment. | |
| Apr 10 at 20:26 | comment | added | Jakub Konieczny | @GHfromMO I might be going crazy, but it seems to me that the estimates in the question are correct. If I want $n = 2^a 3^b 5^c$ to be between $X$ and $2X$ then I can choose $a$ and $b$ in $O(\log X)$ ways each, and then I have $O(1)$ ways to choose $c$ (in fact, I have either $0$ or $1$ way). I think you would have been right if we were looking at $[1,X)$ instead of $[X,2X)$, maybe that's the source of the confusion? | |
| Apr 10 at 19:46 | history | asked | Jakub Konieczny | CC BY-SA 4.0 |