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There is a circle with center $(0, 0)$ and radius $r$. Let $n$ be the number of grid points inside or on the circle that at least one of its neighboring (up, down, left, right) grid points is outside the circle.

With my computer, I got some $r-n$ pairs:

$$

\begin{array}{c|lcr}

r&n\\

\hline

1&4\\

2&8\\

3&16\\

4&20\\

5&28\\

10&56\\

10^2&564\\

10^3&5656\\

10^4&56568\\

10^5&565684\\

10^6&5656852\\

10^7&56568540\\

10^8&565685424\\

10^9&5656854248\\

10^{10}&56568542492\\

\end{array}

$$

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And I found that $$\lim_{r\to\infty}\frac nr\approx5.6568542\approx4\sqrt2$$

My question is: how to prove the following equation? $$\lim_{r\to\infty}\frac nr=4\sqrt2$$

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Let’s look at the 1st quadrant (and then multiply by 4). For each $x$ value from $1$ to $r\sqrt2/2$, there is exactly one $y$-value, and it’s greater than $r\sqrt2/2$. For each $y$-value from $1$ to $\sqrt2/2$, there’s exactly one $x$-value, and it’s greater than $r\sqrt2/2$. So that gives you $r\sqrt2$ points in the first quadrant, and proves your observation.

Here’s a slightly different perspective. The $L^\infty$ arc length of the (ordinary) circle of radius $r$ is exactly $(4\sqrt2)r$. So when we define $n$ using a discrete grid, we get the approximation $n=(4\sqrt2)r+O(1)$, and $$\lim_{r\to\infty}\frac{(4\sqrt2)r+O(1)}{r}=4\sqrt2.$$

(This is a suggestive argument, rather than a full proof. I don’t know the precise hypotheses we would need to work with more general curves.)

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