%I #7 Feb 03 2019 16:46:30
%S 1,0,0,1,0,1,2,1,2,4,3,5,9,9,12,21,24,30,51,63,75,126,162,189,315,414,
%T 477,792,1053,1206,1998,2673,3051,5049,6777,7722,12771,17172,19548,
%U 32319,43497,49491,81810,110160,125307,207117,278964
%N Let i be in {1,2,3,4} and let r >= 0 be an integer. Let p = {p_1, p_2, p_3, p_4} = {2,0,1,2}, n=3*r+p_i, and define a(2)=0. Then a(n)=a(3*r+p_i) gives the quantity of H_(9,2,0) tiles in a subdivided H_(9,i,r) tile after linear scaling by the factor Q^r, where Q=sqrt(x^21) with x=2*cos(Pi/9).
%C (Start) See A187502 for supporting theory. Define the matrix
%C U_2=
%C (0 0 1 0)
%C (0 1 0 1)
%C (1 0 1 1)
%C (0 1 1 1).
%C Let r>=0, and let B_r be the rth "block" defined by B_r={a(3*r2),a(3*r),a(3*r+1),a(3*r+2)} with a(2)=0. Note that B_r3*B_(r1)+3*B_(r3)={0,0,0,0}, for r>=4, with initial conditions {B_k}={{0,1,0,0},{0,1,0,1},{0,2,1,2},{1,4,3,5}}, k=0,1,2,3. Let p={p_1,p_2,p_3,p_4}={2,0,1,2}, n=3*r+p_i and M=(U_2)^r. Then B_r corresponds componentwise to the second column of M, and a(n)=a(3*r+p_i)=m_(i,2) gives the quantity of H_(9,2,0) tiles that should appear in a subdivided H_(9,i,r) tile. (End)
%C Since a(3*r+1)=a(3*(r+1)2) for all r, this sequence arises by concatenation of secondcolumn entries m_(2,2), m_(3,2) and m_(4,2) from successive matrices M=(U_2)^r.
%D L. E. Jeffery, Unitprimitive matrices and rhombus substitution tilings, (in preparation).
%F Recurrence: a(n)=3*a(n3)3*a(n9), for n>=12, with initial conditions {a(m)}={1,0,0,1,0,1,2,1,2,4,3,5}, m=0,1,...,11.
%F G.f.: (12*x^3+x^5x^6+x^7x^8+x^9x^11)/(13*x^3+3*x^9).
%Y Cf. A187499, A187501, A187502.
%K nonn,easy
%O 0,7
%A _L. Edson Jeffery_, Mar 16 2011
