Timeline for How $a+b$ can grow when $a!b! \mid n!$
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 29, 2013 at 0:24 | comment | added | S. Carnahan♦ | The first example of virtue 10 is (416, 187, 239). Before that, (385, 155, 239) is the only other example of virtue 9. | |
| Jul 28, 2013 at 20:39 | history | edited | Aaron Meyerowitz | CC BY-SA 3.0 |
added 134 characters in body
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| Jul 28, 2013 at 20:37 | comment | added | The Masked Avenger | To show the 5n/2 bound, replace (c!) in a decomposition by (c-4)! 2!2!3! or better, and similarly for 4! and 5!. This gives that a decomposition with optimal sum has only 2's and 3's, from which 5n/2 as an upper bound on the sum follows quickly. | |
| Jul 28, 2013 at 20:08 | comment | added | The Masked Avenger | Actually it doesn't,as evidenced by 12! being a multiple of 12^5 giving a sum of 25; it looks like the upper bound will be more like 5n/2. | |
| Jul 28, 2013 at 19:05 | comment | added | The Masked Avenger | Indeed it does Will. | |
| Jul 28, 2013 at 19:03 | comment | added | The Masked Avenger | I need more care. Not only do I want an upper bound on a+b+...+c when (a!b!...c!) divides n!, I need that the summands are all greater than 1. | |
| Jul 28, 2013 at 19:01 | comment | added | Will Sawin | $2n$ is what falls out of GH's argument in the general case. | |
| Jul 28, 2013 at 18:55 | comment | added | The Masked Avenger | Further, 3!5!7! = 10!; I suspect a bound for the general problem is arbitraily close to 2n. (Actually I can get 2n-2 as a lower bound on the upper bound.) | |
| Jul 28, 2013 at 15:25 | comment | added | Noam D. Elkies | There's also the old observation that $6!7!=10!$, though this has "virtue" only $3$. | |
| Jul 28, 2013 at 10:27 | history | answered | Aaron Meyerowitz | CC BY-SA 3.0 |