# A note on the number of vertices of the Archimedean tiling

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## Abstract

There are 11 Archimedean tilings in \(\mathbb {R}^{2}\). Let *E*(*n*) denote the ellipse of short half axis length *n*\((n\in \mathbb {Z}^{+})\) centered at an arbitrary vertex of the Archimedean tiling by regular polygons of edge length 1, and let \(\mathcal {N}(E(n))\) denote the number of vertices of the Archimedean tiling that lie inside or on the boundary of *E*(*n*). In this paper, we present an algorithm to calculate the number \(\mathcal {N}(E(n))\), and get a unified formula \(\displaystyle \lim _{n\rightarrow \infty }\frac{\mathcal {N}(E(n))}{n^{2}}=m\cdot \frac{\pi }{S}\), where *S* is the area of the central polygon, and *m* is the ratio of long half axis length and short half axis length of the ellipse. Let \(\mathcal {C}\) be a cube-tiling by cubes of edge length 1 in \(\mathbb {R}^{3}\), and the vertex of cube-tiling is called a *C*-point. Let *S*(*n*) denote the sphere of radius \(n(n\in \mathbb {Z}^{+})\) centered at an arbitrary *C*-point, and let \(\mathcal {N}_{C}(S(n))\) denote the number of *C*-points that lie inside or on the surface of *S*(*n*). In this paper, we present an algorithm to calculate the number \(\mathcal {N}_{C}(S(n))\) and get a formula \(\displaystyle \lim _{n\rightarrow \infty }\frac{\mathcal {N}_{C}(S(n))}{n^{3}}=\frac{4\pi }{3V}\), where *V* is the volume of the cube.

## Keywords

Discrete geometry Cube-tiling Archimedean tiling Central polygon## Mathematics Subject Classification

52C20 52A10## References

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