Timeline for Reference request for $(1,2^n-1,2^n)$ example related to abc-conjecture
Current License: CC BY-SA 3.0
9 events
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| Sep 20, 2013 at 16:46 | comment | added | Greg Martin | Regarding "it cannot get any lower than that" - Stewart-Tijdeman have a construction that gives much better bounds on the radical, better than $c/(\log c)^A$ for any $A$. It's referenced in van der Horst's thesis, I believe. | |
| Sep 20, 2013 at 16:45 | comment | added | Greg Martin | The bound $c/\log c$ is indeed equally good in van der Horst's work as in the example I'm looking for. But I'm publishing a variant of the example I'm looking for, so I'd like to cite that specific example. (I'll include these other examples as background, though, so it's helpful to be pointed to them.) | |
| Sep 20, 2013 at 7:13 | comment | added | Carlo Beenakker | hmm, I thought the key improvement here, relative to Granville/Tucker, was the reduction of the bound from $c/\sqrt{\log c}$ down to $c/\log c$. It cannot get any lower than that, can it? | |
| Sep 20, 2013 at 5:06 | comment | added | Greg Martin | I'm glad for this pointer as well! But it's still not quite the example I'm looking for (I admit it's getting harder to distinguish from what I wrote in the question). This example is a generalization of the one from your first answer (Granville/Tucker), which is the $n=2$ case. But it still deals with only one prime dividing $2^k-1$ to a high power, rather than lots of squares of different primes dividing $2^k-1$. | |
| Sep 19, 2013 at 20:12 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
small edit
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| Sep 19, 2013 at 20:04 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
triple
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| Sep 19, 2013 at 15:50 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
expanded
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| Sep 19, 2013 at 15:29 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
smiley
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| Sep 19, 2013 at 15:07 | history | answered | Carlo Beenakker | CC BY-SA 3.0 |