All of the polygons in this tessellation are regular. There are only three regular tessellations, and they use, respectively, equilateral triangles, squares, and regular hexagons to tile a plane. There is also a set of eight semi-regular (or Archimedean) tessellations, which you may see here. Archimedean tessellations include more than one type of polygon, but they are vertex-transitive, meaning that each vertex has the same set of polygons surrounding it.

This is a tessellation of regular polygons, but it lacks vertex-transitivity, so it cannot be called a semi-regular (or Archimedean) tessellation. In other words, in this tessellation, there is more than one type of vertex.

There are many such tessellations with an indefinitely repeating pattern. Has this particular one been seen before? I do not know the answer to this question — but if you do, please let me know, in a comment.

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Nice tessellation! This is one of the 61 3-uniform tilings (edge-to-edge tilings of regular polygons having three classes of vertices), first enumerated by D. P. Chavey in 1984. See https://oeis.org/A068599 for the OEIS sequence giving the number of n-uniform tilings up through 6. It is also one of the 140 5-isohedral tilings (which have 5 classes of faces), which are enumerated in https://oeis.org/A268184. This particular drawing uses two colors for two of the face classes, so is one of 15,294 5-isohedral edge-to-edge colorings of regular polygons (https://oeis.org/A269630). If only tilings in which each face class (with respect to the color-preserving symmetry) uses a different color are considered, then there are only 484 such 5-isohedral colorings (https://oeis.org/A268591).

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I did not know any of that — much thanks!

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Thanks! I slightly misspoke in the description of the third linked sequence. I meant to say that your coloring uses *one color* for two of the classes, not two colors.

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