Consider covering the plane by means of the classical Penrose tiles (i.e. the "fat" and "thin" rhombi) in a spiraling fashion, adding step by step a new tile around a given one, as introduced in this post.
At each step, we need one of the two tiles (and only one), as illustrated in the following picture:
- The starting tile is the pink one.
- The segments connecting the centers of the tiles represent the consecutive steps of the spiral walk around the starting tile.
- The numbers inside the tiles (1, fat rhombus; $\color{red} 0$, thin rhombus), highlight the alternation of the two kinds of tiles as a function of the walk step.
Displaying these numbers in a sequence, we find:
$s(n)=1$,$1$,$1$,$1$,$\color{red}0$,$1$,$1$,$\color{red}0$,$1$,$\color{red}0$,$\color{red}0$,$1$,$\color{red}0$,$\color{red}0$,$1$,$1\ldots$
My question is:
Is there a closed formula for $s(n)$?
Thanks for your suggestions!
There are two other sequences relevant to this tiling. In both cases it can be treated like a $1$-dimensional (Elementary) cellular automaton. Firstly, take all units associated by edge with a particular unit and take the result modulo $2$.
Secondly to draw a line radially though it from any point and at any given angle and take the bit value of the units intersected.
My guess is that you could yield a generative formula for determining the next state.
Also, there are typically “rules” regarding the orientation of the tiles (using markings on them to exclude certain orientations and permutations), these add a non-binary aspect to the continuation of the sequence. It might be a good idea to make an assessment of those before simply collecting the bit value of each cell.