Timeline for Consecutive rising sequence of largest prime factors
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Sep 23, 2017 at 4:42 | comment | added | Gerry Myerson | See also mathoverflow.net/questions/281780/k-factorazy-tuples/… | |
| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
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| Feb 11, 2017 at 5:37 | comment | added | so-called friend Don | As partial progress towards question 2, it might be noted that in the same paper Gerry mentions, Erdos and Pomerance show that the inequality $f(n) < f(n+1)$ holds on a set of positive lower density, and the same for $f(n) > f(n+1)$. | |
| Feb 9, 2017 at 3:28 | answer | added | Noam D. Elkies | timeline score: 5 | |
| Feb 8, 2017 at 23:11 | answer | added | Gerry Myerson | timeline score: 1 | |
| Feb 8, 2017 at 3:01 | comment | added | Gerry Myerson | Erdos and Pomerance, On the largest prime factors of $n$ and $n+1$, Aequationes Mathematicae 17 (1978) 311-321 give the same construction of infinitely many triples with $f(n)<f(n+1)<f(n+2)$. "On the other hand, we cannot find infinitely many $n$ for which [$f(n)>f(n+1)>f(n+2)$], but perhaps we overlook a simple proof." A proof is given by Balog, On triplets with descending largest prime factors, Studia Sci Math Hungar 38 (2001) 45-50 (but I wouldn't call it a simple proof). | |
| Feb 8, 2017 at 0:18 | comment | added | Andy | @KevinBuzzard Oh I absolutely agree it's not convincing, I just felt bad not to provide any progress because I had none :P | |
| Feb 7, 2017 at 22:54 | comment | added | Kevin Buzzard | @user104593 I'm not so confident about the CRT argument for the general case. If you're using CRT to solve congruences modulo 10 primes then in general the size of the smallest solution will be of the order of magnitude of the product of these primes, so you have lost control of the size of the other prime factors. It seems from OEIS that no run is known with $k=15$ (and the smallest example, assuming one exists, starts with a number bigger than $10^{13}$). One might instead wonder whether existence of long runs follows from standard conjectures on primes. | |
| Feb 7, 2017 at 22:28 | comment | added | Kevin Buzzard | Or possibly oeis.org/A100384 , the difference being whether or not you count e.g. the smallest run of 8 as being the same as the smallest run of 9 (the smallest run of 9 occurs before the smallest run of 8!) | |
| Feb 7, 2017 at 22:25 | comment | added | Kevin Buzzard | oeis.org/A079749 is the sequence this question is asking about. | |
| Feb 7, 2017 at 22:23 | comment | added | Gerry Myerson | Also related are oeis.org/A070087 ($P(n) > P(n+1)$ where $P(n)$ is the largest prime factor of $n$.) and oeis.org/A070089 ($P(n) < P(n+1)$ where $P(n)$ is the largest prime factor of $n$.). | |
| Feb 7, 2017 at 22:19 | comment | added | Gerry Myerson | The function $f(n)$ is tabulated at oeis.org/A006530 with many references and links. | |
| Feb 7, 2017 at 22:16 | history | edited | Gerry Myerson | CC BY-SA 3.0 |
link
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| S Feb 7, 2017 at 22:15 | history | suggested | Andy | CC BY-SA 3.0 |
added a short small progress
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| Feb 7, 2017 at 21:57 | review | Suggested edits | |||
| S Feb 7, 2017 at 22:15 | |||||
| Feb 7, 2017 at 21:52 | review | First posts | |||
| Feb 7, 2017 at 21:53 | |||||
| Feb 7, 2017 at 21:52 | history | asked | user104593 | CC BY-SA 3.0 |