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Conwayโ€™s musical sequences (2)

A Conway musical sequence is an infinite word in L and S, containing no two consecutive Sโ€™s nor three consecutive Lโ€™s, such that all its inflations remain musical sequences.

Weโ€™ve seen that such musical sequences encode an aperiodic tiling of the line in short (S) and long (L) intervals, and that such tilings are all finite locally isomorphic.

But, apart from the middle C-sequences (the one-dimensional cartwheel tilings) we gave no examples of such tilings (or musical sequences). Letโ€™s remedy this!

Take any real number c as long as it is not an integral combination of 1 and 1ฯ„ (with ฯ„ the golden ratio) and assign to any integer aโˆˆZ a tile
Pc(a)={SL iff โŒˆc+(a+1)1ฯ„โŒ‰โ€“โŒˆc+a1ฯ„โŒ‰={01
(instead of ceilings we might have taken floors, because of the restriction on c).

With a little bit of work we see that the deflated word determined by Pc is again of this type, more precisely def(Pc)=Pโˆ’(cโˆ’โŒŠcโŒ‹)1ฯ„. But then it also follows that inflated words are of this type, meaning that all Pc define a musical sequence.

Letโ€™s just check that these sequences satisfy the gluing restrictions. If there is no integer between c+a1ฯ„ and c+(a+1)1ฯ„, because 21ฯ„โ‰ˆ1.236 there must be an interval in the preceding and the following 1ฯ„-interval, showing that an S in the sequence has an L on its left and right, so there are no two consecutive Sโ€™s in the sequences.



Similarly, if two consecutive 1ฯ„-intervals have an integer in them, the next interval cannot contain an integer as 31ฯ„โ‰ˆ1.854<2.



Now we come to the essential point: these sequences can be obtained by the cut-and-project method.

Take the line L through the origin with slope 1ฯ„ and LโŠฅ the line perpendicular it.

Consider the unit square H and Hฮณโ†’=H+ฮณโ†’ its translation under a shift vector ฮณโ†’=(ฮณx,ฮณy) and let ฯ€ (or ฯ€โŠฅ) be the orthogonal projection of the plane onto L (or onto LโŠฅ). One quickly computes that
ฯ€(a,b)=(ฯ„2a+ฯ„b1+ฯ„2,ฯ„a+b1+ฯ„2)andฯ€โŠฅ(a,b)=(aโˆ’ฯ„b1+ฯ„2,ฯ„2bโˆ’ฯ„a1+ฯ„2)
In the picture, we take ฮณโ†’=(c,โˆ’ฯ„c).



The window W will be the strip, parallel with L with basis ฯ€โŠฅ(Hฮณโ†’).

We cut the standard lattice Z2, of all points with integer coordinates in the plane, by retricting to the window P=Z2โˆฉW.

Next, we project P onto the line L, and we get a set of endpoints of intervals which divide the line L into short intervals of length 11+ฯ„2 and long intervals of length ฯ„1+ฯ„2.

For (a,b)โˆˆW, the interval will be short if (a,b+1)โˆˆW and long if (a+1,b)โˆˆW.



Because these intervals differ by a factor ฯ„ in length, we get a tiling of the line by short intervals S and long intervals L. It is easy to see that they satisfy the gluing restrictions (remember, no two consecutive short intervals and no three consecutive long intervals): the horizontal width of the window W is 1+ฯ„โ‰ˆ2.618 (so there cannot be three consecutive long intervals in the projection) and the vertical width of the window W is 1+1ฯ„=ฯ„โ‰ˆ1.618 so there cannot be two consecutive short intervals in the projection.

Ready for the punchline?

The sequence obtained from projecting P is equal to the sequence P(1+ฯ„2)c. So, we get all musical sequences of this form from the cut-and-project method!

On LโŠฅ the two end-points of the window are
{ฯ€โŠฅ(c+1,โˆ’ฯ„c)=((1+ฯ„2)c+11+ฯ„2,โˆ’ฯ„(1+ฯ„2)c+11+ฯ„2)ฯ€โŠฅ(c,โˆ’ฯ„c+1)=((1+ฯ„2)cโˆ’ฯ„1+ฯ„2,โˆ’ฯ„(1+ฯ„2)cโˆ’ฯ„1+ฯ„2)
Therefore, a point (a,b)โˆˆZ2 lies in the window W if and only if
(1+ฯ„2)cโˆ’ฯ„<aโˆ’ฯ„b<(1+ฯ„2)c+1 or equivalently, if (1+ฯ„2)c+(bโˆ’1)ฯ„<a<(1+ฯ„2)c+bฯ„+1 Observe that โŒˆ(1+ฯ„2)c+bฯ„โŒ‰โˆ’โŒˆ(1+ฯ„2)c+(bโˆ’1)ฯ„โŒ‰=P(1+ฯ„2)c(bโˆ’1)+1โˆˆ{1,2} We separate the two cases: (1) : If โŒˆ(1+ฯ„2)c+(b+1)ฯ„โŒ‰โˆ’โŒˆ(1+ฯ„2)c+bฯ„โŒ‰=1, then there must be an integer a such that (1+ฯ„2)c+(b+1)ฯ„โˆ’1<a<(1+ฯ„2)b+1, and this forces โŒˆ(1+ฯ„2)c+(b+2)ฯ„โŒ‰โˆ’โŒˆ(1+ฯ„2)c+(b+1)ฯ„โŒ‰=2. With bi=(1+ฯ„2)c+(b+i)ฯ„ and di=bi+1 we have the situation



and from the inequalities above this implies that both (a+1,b+1) and (a+1,b+2) are in W, giving a short interval S in the projection.

(2) : If โŒˆ(1+ฯ„2)c+(b+1)ฯ„โŒ‰โ€“โŒˆ(1+ฯ„2)c+bฯ„โŒ‰=1, then there must be an integer a such that (1+ฯ„2)c+bฯ„<a<(1+ฯ„2)cv+(b+1)ฯ„โˆ’1, giving the situation



giving from the inequalities that both (a+1,b+1) and (a+2,b+1) are in W, giving a long interval L in the projection, finishing the proof.

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