Timeline for Consecutive integers each of which has a large prime factor
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 25, 2019 at 20:13 | comment | added | Will Jagy | @GerhardPaseman thanks. I decided to finish arranging the apartment, so I did not notice this until now. Good work. | |
| Feb 25, 2019 at 17:59 | comment | added | Gerhard Paseman | I see now in a comment to 255269 that I ran such a program and provided partial results (so replace (650 660) by (364 374)) with an interval of 40 rough numbers somewhere below 27 million. Gerhard "Is K Big Enough Now?" Paseman, 2019.02.25. | |
| Feb 25, 2019 at 17:46 | comment | added | Gerhard Paseman | (112 120) (650 660) (1105 1116) are intervals of rough numbers, gotten from visually scanning the text file for OEIS sequence 63539. I may write a program later today. My program for Happy New Prime Year (50691) can be tweaked to check for this sequence though, if you feel like tweaking AWK. Gerhard "Hoping You Feel Better Soon" Paseman, 2019.02.25. | |
| Feb 25, 2019 at 17:31 | comment | added | Will Jagy | @GerhardPaseman thank you. If you can stand it, please go a bit higher, increase $k$ in a loop until it slows down. I'm busy today, ill and I need to arrange the apartment so the handyman can check my stove, water heater, and wall hanging room heater for possible CO emissions, in any case service the wall heater, which is old | |
| Feb 25, 2019 at 17:16 | comment | added | Gerhard Paseman | Consider the following dynamic on positive integers: n goes to n/P(n) and 1 goes to 1. Every n reaches 1 in about log log n steps (with some exceptions). Call n quite rough if at every step P(n_i) is larger than the square root of n_i. So primes, products of two distinct primes, and 42 are quite rough. What is the density/distribution of quite rough numbers? I suspect a question post about this hasn't appeared on MathOverflow. Can you show there is no sequence of 4 consecutive numbers which are all quite rough? Gerhard "Feel Free To Ask It" Paseman, 2019.02.25. | |
| Feb 25, 2019 at 16:41 | comment | added | Gerhard Paseman | @Will, for k=3 my computer says n=4. For k=5 my computer says n=18. Gerhard "Check OEIS For Bigger Numbers" Paseman, 2019.02.25. | |
| Feb 25, 2019 at 16:38 | comment | added | Gerhard Paseman | You might check out mathoverflow.net/questions/255269 (of which this question is almost a duplicate) and similar questions. My belief is that not much is known, and that your upper bound on n will be exponential in k, and that proving this is hard. Gerhard "But Go Check For Yourself" Paseman, 2019.02.25. | |
| Feb 25, 2019 at 16:37 | comment | added | Will Jagy | fine for $k=2$ with Chinese Remainder Theorem. For $k \geq3,$ need to get a better search for nearly consecutive $p_j$ such that $ n \equiv -1 \pmod p_1, $ $ n \equiv -2 \pmod p_2, $ $ n \equiv -3 \pmod p_3.$ but $n$ is especially small. I'd say it is already worth trying $k=3$ by computer, see how difficult it might be. | |
| Feb 25, 2019 at 16:00 | comment | added | Chris Wuthrich | Ooops, I'd better delete that comment :) | |
| Feb 25, 2019 at 15:29 | history | asked | Penchez | CC BY-SA 4.0 |