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Assume the lattice is generated by vectors $(1,0)$ and $(1/2,\sqrt{3}/2$.

Then the number of lattice points in the column with $x$ coordinate $k/2$ is $1+2 * Floor[ ((4r^2-k^2)/3)^{1/2}]$ if $k$ is even and $2*Floor[ ((4r^2-k^2)/3)^{1/2}+1/2]$ if $k$ is odd.

So we are going to write two sums, one for $k=2i$ and $i=1$ to $Floor[r]$ and one for $k=2i+1$ and $i=0$ to $Floor[r-1/2]$, of these rows.

Then we're going to add the column at $0$. This gives the exact counting formula

$1+2*Floor[r]+2*Floor[2*r/3^{1/2}]+4* 1+2*Floor[r]+2*Floor[r/ \sqrt{3}]+4* \sum_{i=1}^{Floor[r]} Floor[ ((4r^2-4i^2)/3)^{1/2}]$\sqrt{(4r^2-4i^2)/3}]+4*\sum_{i=0}^{Floor[r-1/2]} Floor[ ((4r^2-(2i+1)^2)/3)^{1/2}+1/2]$\sqrt{(4r^2-(2i+1)^2)/3}+1/2]$

I didn't double-check these calculations so there might be some mistakes, but it's clear that some version of this formula is correct.

1

Assume the lattice is generated by vectors $(1,0)$ and $(1/2,\sqrt{3}/2$.

Then the number of lattice points in the column with $x$ coordinate $k/2$ is $1+2 * Floor[ ((4r^2-k^2)/3)^{1/2}]$ if $k$ is even and $2*Floor[ ((4r^2-k^2)/3)^{1/2}+1/2]$ if $k$ is odd.

So we are going to write two sums, one for $k=2i$ and $i=1$ to $Floor[r]$ and one for $k=2i+1$ and $i=0$ to $Floor[r-1/2]$, of these rows.

Then we're going to add the column at $0$. This gives the exact counting formula

$1+2*Floor[r]+2*Floor[2*r/3^{1/2}]+4* \sum_{i=1}^{Floor[r]} Floor[ ((4r^2-4i^2)/3)^{1/2}]$

$+4*\sum_{i=0}^{Floor[r-1/2]} Floor[ ((4r^2-(2i+1)^2)/3)^{1/2}+1/2]$

I didn't double-check these calculations so there might be some mistakes, but it's clear that some version of this formula is correct.