160 N. Rivier and J.-F. Sadoc(1) On an axis of 12-fold symmetry. It is given by the lattice vectorsni≈
p
√
3 iin the two, orthogonal triangular lattices. Thus,√
3 ≈n/p=Cj/Aj,
where Cj = BjAj andBj/Aj are successive convergents of 1 +√
3,
obtained by truncation of its continued fraction.^10 TheAjare given by
recursion,Aj=qjAj− 1 +Aj− 2 , withA 0 =1,A− 1 = 0, and similarly for
BjandCj, withB 0 =q 0 =2,B− 1 = 1. On a single triangular lattice,
the two vectors are orthogonal, with nearly the same length. They form
a square with a slight orthorhombic distortion. In the overlap, one recog-
nizes the “square” with approximant of√
3 ≈ 2 /1, the unit cell made of
orthogonal mirror planes with√
3 ≈ 5 /3. In the gap structure, there is
the unit cell with√
3 ≈ 7 /4.
(2) Other coincidence points do not lie on a symmetry axis. They are rep-
resented on a single triangular lattice by the two lattice vectorsb 1 =
(1/2)pi+(
√
3 /2)qjandb 2 =(1/2)mi−(√
3 /2)nj, withpandq, resp.
mandn, integers of same parity. The two vectors have the same length
|b 1 |=|b 2 |, and are nearly orthogonal. They form a rhombus that is almost
a square. Its diagonalsb 1 +b 2 andb 2 −b 1 lie on me axes of symmetry of
the triangular lattice, thusp−m=q+n, (8.1)3(n−q)=p+m. (8.2)
Moreover,b 1 ·b 2 =(3q^2 −m^2 )/ 2. (8.3)Note thatm>n, sincem=nwould implyq= 0. (The two vectors form
also an equilateral triangle, so thatα+β=π/6, whereαandβare the angles
between theb’s and the symmetry axes. For example, tanβ=m/(n√
3), so
thatm<n.)
This yields two approximants for√
3,p/nandm/q, withq<m<n<p.
The solution of (8.1) and (8.2) is,^11(^10) 1+√3=[2, 1 ]=[q 0 ,q 1 ,q 2 ,q 3 ,...], whereqj=2forjeven,qj=1forjodd (see
footnote 2).
(^11) Proof (by induction): (p−m)−(n+q)=qj+2Bj+1− 2 Aj+2=qj+2[qj+1Bj−
2 Aj+1]+[qj+2Bj− 1 − 2 Aj], withqj+2=qjfor
√
- The two [...] = 0 by induction,
thusqj+2Bj+1=2Aj+2. Similarly for (8.2): 3(n−q)−(p+m)=3qj+2Aj+1−
2 Cj+2+qj+2Cj+1=2qj+2Aj+1− 2 Cj+2+qj+Bj+1=2qj+2Aj+1−2(Cj+2−Aj+2),
using the result in the proof of (8.1). Then,qj+2Aj+1−(Cj+2−Aj+2)=
q√j+2[qj+1Aj−(Cj+1−Aj+1)] + [qj+2Aj−(Cj+1−Aj+1)], withqj+2=qjfor - Once again, the two [...] vanish by induction. Note that we have only used
the fact thatqj+2=qj, i.e. a continuous fraction expansion of period 2.