## Tiling with Transcendental Inflation Multiplier

### Info

An one-dimensional substitution rule that uses an infinite number of proto tiles and yields a transcendental inflation multiplier.

The inflation factor is approximately $2.6113$.

The substitution rules are given by:

$T_{0}\rightarrow T_{0},T_{1}$

$T_{1}\rightarrow 2T_{0},T_{2}$

$T_{2}\rightarrow 2T_{0},T_{1},T_{3}$

$T_{3}\rightarrow T_{0},T_{2},T_{4}$

$T_{4}\rightarrow 2T_{0},T_{3},T_{5}$

$T_{5}\rightarrow T_{0},T_{4},T_{6}$

$T_{6}\rightarrow T_{0},T_{5},T_{7}$

$T_{k}\rightarrow (1+f\left(k\right))T_{0},T_{k-1},T_{k+1}$ with ‘$f\left(k\right)$ as the Thue-Morse sequence.

The corresponding substitution matrix can be written as:

$\begin{array}{cccccccccc} 1 & 2 & 2 & 1 & 2 & 1 & 1 & 2 & 2 & \cdots\ 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots\ 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & \cdots\ 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & \cdots\ 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & \cdots\ 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & \cdots\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & \cdots\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & \ddots\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & \ddots\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \ddots & \ddots \end{array}$

The lengths of the proto tiles are given by:

$length(T_{0})\approx1.0000$

$length(T_{1})\approx1.6113$

$length(T_{2})\approx2.2076$

$length(T_{3})\approx2.1534$

$length(T_{4})\approx2.4155$

$length(T_{5})\approx2.1543$

$length(T_{6})\approx2.2099$

### Patch

download vectorformat Tiling with Transcendental Inflation Multiplier

### References

[FGM2022]
Frettloeh, D. and Garber, A. and Manibo, N.
Substitution tilings with transcendental inflation factor
2022, 10.48550/ARXIV.2208.01327