Help calculating $\sum^{R}_{k=1} \bigl\lfloor{\sqrt { R^2-k^2}}\bigr\rfloor$

Question

I'm trying to calculate, or at least approximate, $$\sum^{R}_{k=1} \left\lfloor{\sqrt { R^2-k^2}}\right\rfloor,$$ where $R$ is a natural number.

I have tried factoring this as $$\sum_{k=1}^R \left\lfloor \sqrt {(R+k)(R-k)} \right\rfloor,$$ but then I don't know where to go from there.

Answer 1

Since $\sqrt{R^2-k^2}$ is a decreasing function of $k$, we have $$\frac{\pi}{4}R^2=\int_0^R\sqrt{R^2-k^2}dk > \sum_{k=1}^R \sqrt{R^2-k^2} > \int_1^{R}\sqrt{R^2-k^2}dk>\frac{\pi}{4}R^2-R$$

Answer 2

This is known as Gauss's circle problem, and is essentially the same as counting the number of points with integer coordinates inside a circle.

There doesn't appear to be any known way of calculating this sum faster than by summing the terms one by one.