Timeline for Conjecture: $ x\leqslant f(x)\leqslant x+x^{\log_{113}13} \{x|1\leqslant x\leqslant+\infty,x\in \mathrm{positive~integer}\}$?
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Oct 25, 2018 at 6:58 | comment | added | scibee | I truly apprecuate your helps | |
| Oct 24, 2018 at 1:18 | answer | added | Stanley Yao Xiao | timeline score: 4 | |
| Oct 23, 2018 at 21:41 | comment | added | Steven Landsburg | @GerryMyerson : Agreed all around --- and my apologies for mistakenly referring to Bullet's comment as "your" comment. | |
| Oct 23, 2018 at 21:40 | history | edited | GH from MO |
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| Oct 23, 2018 at 21:40 | answer | added | GH from MO | timeline score: 3 | |
| Oct 23, 2018 at 21:02 | comment | added | Gerry Myerson | @Steven, the link in Bullet's comment actually says there is a prime between $x$ and $x+x^{21/40}$ for all large $x$, so the first half of Bullet's comment is misleading, and scibee is correct to object. But the second part of Bullet's comment is correct, and is what I had in mind when I posted my comment. | |
| Oct 23, 2018 at 16:09 | comment | added | Steven Landsburg | It would be very good to include an explanation of why you expect this might be true. | |
| Oct 23, 2018 at 15:56 | comment | added | Steven Landsburg | @GerryMyerson : Not that it actually matters, but I read your comment as saying that there is a prime between $x$ and $x^{21/40}$ for all $x$ and that this implies the conjecture for large $x$. So scibee's response is in fact on target and (at least given my reading, which I think is the natural one), "114 is not a large $x$" is not relevant. (Of course my reading is not the same as what you actually meant.) | |
| Oct 23, 2018 at 11:02 | comment | added | Gerry Myerson | 114 is not a large $x$, scibee. | |
| Oct 23, 2018 at 8:12 | comment | added | scibee | There is not a prime between 114 and 114+114^(21/40) | |
| Oct 23, 2018 at 7:57 | comment | added | LeechLattice | Since there is a prime between $x$ and $x+x^{(21/40)}$, your conjecture holds for large $x$. | |
| S Oct 23, 2018 at 7:53 | history | suggested | Glorfindel | CC BY-SA 4.0 |
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| Oct 23, 2018 at 7:28 | review | Suggested edits | |||
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| Oct 23, 2018 at 7:12 | comment | added | Greg Martin | You could add the context that the upper bound in the conjecture is attained for $x=114$. | |
| Oct 23, 2018 at 7:10 | comment | added | Greg Martin | I assume you'll be editing the post to fix the poor formatting...? | |
| Oct 23, 2018 at 7:10 | review | First posts | |||
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| Oct 23, 2018 at 7:07 | history | asked | scibee | CC BY-SA 4.0 |