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$\begingroup$ To me, the fact of being a convergent is a restatement of, rather than an explanation for, being a close approximation. And it doesn't say anything about why 12288 is so smooth. $\endgroup$
– David Eppstein
Commented Nov 19, 2011 at 21:27
2

$\begingroup$ Well of course, the two statements are equivalent- maybe I should have assumed that you already knew that. When you say that $12288$ is smooth, well it is also not very large. $\endgroup$
– Alan Haynes
Commented Nov 19, 2011 at 21:49
2

$\begingroup$ After all, some denominator of some rational approximation of some irrational number like $e^3$ or $\pi^\pi$ should be smooth. Wasn't $12288$ so smooth, the subject of this question would have been different. $\endgroup$
– Pietro Majer
Commented Nov 20, 2011 at 9:19

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