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Dec 28, 2014 at 3:20 comment added Alexey Ustinov There is a result on exceptional set for sum of three such numbers, see "Exceptional set of a representation with fractional powers" E. P. Balanzario, M. Z. Garaev, R. Zuazua, link.springer.com/article/10.1007%2Fs10474-006-0516-8
Dec 27, 2014 at 5:34 comment added Brendan Murphy Sergei Konyagin proved that every large integer is the sum of two elements of the form $[n^\alpha]$ for any $\alpha<3/2$ (see cds.cern.ch/record/720991/files/sis-2004-086.ps?version=1). Earlier Deshouillers proved the same for $\alpha<4/3$. I believe the conjecture is that the same is true for $\alpha<2$, although $\alpha=3/2$ might be a natural breaking point as well (e.g. I can reprove some of Deshouillers results in a different way, but the method couldn't get similar results past 3/2's).
Dec 27, 2014 at 4:04 comment added The Masked Avenger Altwrnatively, consider $k(\alpha,\beta)$, which represents the order of building $S(\beta)$ from $S(\alpha)$. That might lead to weak but easy upper bounds.
Dec 27, 2014 at 3:04 comment added Qiaochu Yuan If $\alpha = \frac{p}{q}$ is rational then certainly $S(\alpha)$ contains $S(p)$, so if $S(p)$ is known to be a basis of order $k$ then so is $S(\alpha)$. So you get bounds coming from bounds on Waring's problem in this case.
Dec 27, 2014 at 2:05 history edited David S. Newman CC BY-SA 3.0
I changed the word smallest to largest. I hope I got it right this time.
Dec 27, 2014 at 1:59 comment added Włodzimierz Holsztyński Are the truncated $\frac 32$ powers a basis of order $4?\ $ Also, I would be even more interested if the set consisting of squares, cubes and $7$ is a basis of order 3? One may ask about similar sums of standard sets too.
Dec 27, 2014 at 1:49 history edited Arturo Magidin CC BY-SA 3.0
LaTeX, notation.
Dec 27, 2014 at 1:01 history asked David S. Newman CC BY-SA 3.0