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In latex: when is $ \left\lceil \frac{2}{2^{1/n}-1}\right\rceil $ different from $ \left\lfloor \frac{2n}{\log 2} \right\rfloor $?
If n belongs to this sequence and m=ceiling(2/(2^{1/n}-1)), then 0 < m/(2n) - 1/ln(2) < ln(2)/3 * 1/(2n)^2 implying that m/(2n) is a convergent of 1/ln(2) (note that m and 2n are not necessary coprime). - Max Alekseyev (maxal(AT)cs.ucsd.edu), Jun 06 2007
Comment from David Applegate, Jun 07, 2007: (Start) "Some background to Max Alekseyev's comments: The key point is that the Laurent series for 2/(2^(1/n)-1) about n=infinity is 2/ln(2)*n-1+(1/6)*ln(2)/n+O(1/n^3).
"Also, since 2/log(2) is irrational, 2n/log(2) is never integral, so floor(2n/log(2)) = ceil(2n/log(2)-1).
"So the question becomes: when is 2n/log(2)-1 so close to an integer that 2/(2^(1/n)-1) is on the other side of the integer? That is why the continued fraction expansion of 2/log(2) is relevant." (End)
Comment from David Applegate, Jun 08 2007, edited Jun 11 2007: The appropriate generalization of ceil(2/(2^(1/n)-1)) =? floor(2n/log(2)) is floor(a/(b^(1/n)-1)+a/2) = ceil(an/log(b)). When a=2, the a/2 can be hidden in floor() + 1 = ceil().
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