The asymptotic of the number of integers that are sums of three nonnegative cubes

Question

Let $c(n) $ be the number of distinct integers between $0 $ and $n $ of the form $ a^3 + b^3 + c^3$, meaning the sum of $3$ nonnegative cubes.

$C(n) = O( n \space \ln(n)^x ) $

Find and prove the optimal real value of $x$.

Answer 1

Maybe this could help:

The amount of cubes lower than or equal to $n$ is $\lfloor\sqrt[3]{n}\rfloor$. If we just have $A^3+B^3$ instead of $A^3+B^3+C^3$ we get: $$C(n)=\dfrac{1}{2}\sum_{i=1}^{\lfloor\sqrt[3]{n-1}\rfloor}\lfloor\sqrt[3]{n-i^3}\rfloor$$ And for $A^3+B^3+C^3$ (which is the result you want): $$C(n)=\dfrac{1}{6}\sum_{i=1}^{\lfloor\sqrt[3]{n-2}\rfloor}\sum_{j=1}^{\lfloor\sqrt[3]{n-i^3-1}\rfloor}\lfloor\sqrt[3]{n-i^3-j^3}\rfloor$$ This formula basically just checks whether there exists a $C$ for every combination of $A$ and $B$.