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Is there a second, possibly non-real rectangle shape, whose shape remains the same upon removing a square?

The rectangle in the golden ratio retains its shape when a square is truncated from it. In fact, if we allow $n$ truncations, the base-$2$ Lyndon words of length $n$ classify and enumerate the rectangles which are periodic when cutting off rectangles.

With two exceptions: There is only one real rectangle of period one, whereas there are two length-one Lyndon words in base $2$. Hence OEIS A059966 differs from the number of Lyndon words by firstly not allowing for the one "empty string" and secondly only having a single fixed point whereas there are two Lyndon words which are constant when rotated.

Is there a pair of rectangle shapes which theoretically correspond with these two missing cases, i.e. the empty string and the other fixed point, e.g., such as the ratios $1:0$ and $0:0$?

Motivation

I have an interest in the truncation of binary strings, and in particular in the topology of the 2-adic numbers you get when you repeatedly write base-2 Lyndon words as binary numbers. I'd like to gain more insight around the differences between the topologies that underpin OEIS A05996 and OEIS A001037 – for example can one glue the ends of the logistic map together to arrive at a system with one fixed point rather than two?

Wrzlprmft
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it's a hire car baby
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1 Answers1

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If you remove an $x$ by $x$ square from a $1$ by $x$ rectangle, you're left with a $x$ by $1-x$ rectangle, so we're led to the equation $1/x=x/(1-x)$, which is $x^2+x-1=0$, which has solutions $(-1\pm\sqrt5)/2$. So there is a "non-real" rectangle shape (whatever that means), $1$ by $(-1-\sqrt5)/2$.

Gerry Myerson
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  • Thanks. This was really helpful to think about. I'm inclined to think of your negative rectangle as a reflection. Given the truncation / scaling process is already modulo reflections and rotations I doubt this is the "other" fixed point of the process I'm looking for. I'm coming round to the idea that if you truncate the square you get the $0:0$ "rectangle" and that's the other fixed point because if you were to take zero away from either side, it would remain the same. I'll look closer at the bijection between these rectangles and Lyndon words and see if that favours either. – it's a hire car baby Jul 06 '23 at 11:26
  • Another perspective on this is that one could think of your two solutions as rectangles inscribed with either a left-handed or a right-handed golden spiral. But I need an answer which is unique to the golden rectangle, and again I think that property can be replicated for other "square-truncation-congruent" rectangle shapes. – it's a hire car baby Jul 07 '23 at 08:34