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Jul 25, 2019 at 20:22 comment added Christian Gaetz For $m$ even you're certainly not going to get $\Theta(n)$ for small $k$. Since $m$-th powers are positive, there are at most $n^{1/m}$ possible summands, and so there are $O(n^{k/m})$ possible sums of $k$ $m$-th powers which are at most $n$. Thus your bound in the second bullet point can be improved a lot.
Jul 25, 2019 at 18:40 review First posts
Jul 25, 2019 at 19:37
Jul 25, 2019 at 18:37 history asked aras CC BY-SA 4.0