Timeline for Finding the min of a sequence related with factorials
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Dec 7, 2013 at 14:34 | vote | accept | mathlove | ||
| Dec 4, 2013 at 19:12 | answer | added | Gerhard Paseman | timeline score: 1 | |
| Dec 4, 2013 at 6:39 | answer | added | mathlove | timeline score: 0 | |
| Dec 3, 2013 at 20:54 | comment | added | The Masked Avenger | Consider prime powers greater than and closer to N/2 instead. You will see that min(N) is bounded above by these numbers, so is closer to N/2. | |
| Dec 3, 2013 at 17:53 | comment | added | Thanh Vu | Since I cannot comment, so I have to write it here. Let $p$ be the largest prime at most $N$ ($N/2 < p \le N$). So for any $n < p$, there is no $m$ so that $N!$ divided $n!^m$. When $p\le n <N$, then $a_n \ge 2$, so $na_n > N$, when $n = N$, $a_n = 1$. Should it imply that $\min(N) = N$? | |
| Dec 3, 2013 at 17:14 | comment | added | mathlove | Thank you for your information. I think that your third comment must be true, but I can't prove it. Could you please prove it? | |
| Dec 3, 2013 at 16:43 | comment | added | The Masked Avenger | You should look up distribution of smooth numbers. The answer here is likely next nonsmoothnumber after n/2. | |
| Dec 3, 2013 at 16:42 | comment | added | The Masked Avenger | More precisely, let c be smallest such that c + n/2 is an integer with prime factor larger than square root of n. Then c + n/2 is an upper bound of min(n). | |
| Dec 3, 2013 at 16:38 | comment | added | mathlove | @TheMaskedAvenger : Thank you for pointing it out. I edited. I agree with your conjecture, but I'm interested in the 'near that'. For $N=2008$, $1005\cdot a_{1005}=1005$ is the min because $1005$ has a big prime number $67$. | |
| Dec 3, 2013 at 16:33 | history | edited | mathlove | CC BY-SA 3.0 |
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| Dec 3, 2013 at 16:31 | comment | added | The Masked Avenger | Unless I am missing something, it should be the smallest prime power larger than n/2, or near that. | |
| Dec 3, 2013 at 16:25 | comment | added | The Masked Avenger | Why is min(14) 16? I thought 11 is an upper bound. | |
| Dec 3, 2013 at 16:08 | history | asked | mathlove | CC BY-SA 3.0 |