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A set of integers is called a basis of order h if every nonnegative integer can be written as the sum of h not necessarily distinct elements of the set.(Nathanson, Additive Number Theory, page 7)

Thus, for example, the squares are a basis of order 4. (LaGrange's Theorem)

Consider the set of numbers $S=\{ [n^{3/2}] \mid n\text{ an integer}\}$, where the square brackets indicate the familiar floor function. This begins $\{0,1,2,5,8,11,14,18,22,\ldots\}$. $S$ is certainly not a basis of order 2; there is for example no way to write 17 as a sum of two numbers from $S$. Is $S$ a basis of order 3?

More generally, for $\alpha$ greater than 1 , let $S(\alpha)=\{ [n^{\alpha}] \mid n\text{ an integer}\}$. For a given integer $k$ is there an $\alpha$ such that $S(\alpha)$ is a basis of order $k$? If so, what is the largest such $\alpha$?

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  • $\begingroup$ 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. $\endgroup$
    – Włodzimierz Holsztyński
    Commented Dec 27, 2014 at 1:59
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    $\begingroup$ 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. $\endgroup$
    – Qiaochu Yuan
    Commented Dec 27, 2014 at 3:04
  • $\begingroup$ 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. $\endgroup$
    – The Masked Avenger
    Commented Dec 27, 2014 at 4:04
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    $\begingroup$ 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). $\endgroup$
    – Brendan Murphy
    Commented Dec 27, 2014 at 5:34
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    $\begingroup$ 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 $\endgroup$
    – Alexey Ustinov
    Commented Dec 28, 2014 at 3:20

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