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Both papers works in higher dimensions and in hyperbolic spaces.

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By using a transformation, it is possible to give a counting formula for the hexagonal lattice, similar to the formula on MathWolf. Here are the details.

Suppose your lattice is given by $\Lambda=T(\mathbb{Z}^2)$, where $T$ is an invertible 2x2 matrix.

Let $E$ be an ellipse, and let $N_E(R)$ denote the number of points of $\mathbb{Z}^2$ that are in the dilate $R\cdot E$. That is, $$ N_E(R)=|R\cdot E\cap\mathbb{Z}^2|, $$ where the vertical bars denote cardinality.

Now if we let $D$ be the unit disk, you wish to count $$ |R\cdot D\cap \Lambda| = | R\cdot D\cap T(\mathbb{Z}^2)| = |R\cdot T^{-1}(D)\cap\mathbb{Z}^2| = N_E(R), $$ where $E=T^{-1}D$ is an ellipse.

If $$ T=\begin{pmatrix} a & b\\ c& d \\ \end{pmatrix} $$ then the boundary of $R\cdot E$ is given by the equation $$ (ax+by)^2+(cx+dy)^2=R^2. $$ In general, the explicit expression for $y$ is fairly messy. But for a hexagonal lattice, $a=1,b=0,c=1/2,d=\sqrt{3}/2$, so we have $$ x^2+\frac{(x+\sqrt 3 y)^2}{4} = R^2, $$ hence $$ y=\frac 1{\sqrt 3}[ -x\pm \sqrt{4R^2-x^2}]. $$

This explicit expression let's us write down a counting expression like the one on MathWorld.

Since $E$ is symmetric under $(x,y)\mapsto (-x,-y)$, it suffices to count the number of lattice points above the $x$-axis. The number of lattice points on the axes can be counted fairly easily. It is slightly tricky to count the number of lattice points within the ellipse. In the first quadrant, we may count as usual, but in the second quadrant, we must subtract lattice points that are below the ellipse: $$ N(R)=1 + 2\lfloor \sqrt{4/5}R\rfloor + 2\lfloor \sqrt{4/3}R\rfloor + 2\sum_{i=1}^{\lfloor \sqrt{4/5}R\rfloor} \left\lfloor \frac 1{\sqrt 3}[ - i+\sqrt{4R^2-i^2}]\right\rfloor +2\sum_{i=1}^{\lfloor 2R\rfloor}\left\lfloor \frac 1{\sqrt 3}[ i+\sqrt{4R^2-i^2}]\right\rfloor -2\sum_{i=\lceil\sqrt{4/5}R\rceil}^{\lfloor 2R\rfloor}\left\lfloor \frac 1{\sqrt 3}[ i-\sqrt{4R^2-i^2}]\right\rfloor $$

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