## 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 $2.7899$.

The substitution rules are given by:

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

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

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

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

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

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

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

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

The corresponding substitution matrix can be written as:

$1 3 2 1 2 1 1 2 2 …$ $1 0 1 0 0 0 0 0 0 …$ $0 1 0 1 0 0 0 0 0 …$ $0 0 1 0 1 0 0 0 0 …$ $0 0 0 1 0 1 0 0 0 …$ $0 0 0 0 1 0 1 0 0 …$ $0 0 0 0 0 1 0 1 0 …$ $0 0 0 0 0 0 1 0 1 …$ $0 0 0 0 0 0 0 1 0 …$ $…$

The lengths of the proto tiles are given by:

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

$length(T_{1})\approx1.7899$

$length(T_{2})\approx1.9937$

$length(T_{3})\approx1.7724$

$length(T_{4})\approx1.9510$

$length(T_{5})\approx1.6708$

$length(T_{6})\approx1.7104$