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%I #8 Feb 03 2019 16:45:57
%S 0,1,0,0,1,1,1,3,2,3,6,6,9,15,15,24,36,39,63,90,99,162,225,252,414,
%T 567,639,1053,1431,1620,2673,3618,4104,6777,9153,10395,17172,23166,
%U 26325,43497,58644,66663,110160,148473,168804,278964,375921
%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_(9,3,0) tiles in a subdivided H_(9,i,r) tile after linear scaling by the factor Q^r, where Q=sqrt(x^2-1) 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 C_r be the r-th "block" defined by C_r={a(3*r-2),a(3*r),a(3*r+1),a(3*r+2)} with a(-2)=0. Note that C_r-3*C_(r-1)+3*C_(r-3)={0,0,0,0}, for r>=4, with initial conditions {C_k}={{0,0,1,0},{1,0,1,1},{1,1,3,2},{3,3,6,6}}, 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 C_r corresponds component-wise to the third column of M, and a(n)=a(3*r+p_i)=m_(i,3) gives the quantity of H_(9,3,0) 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 third-column entries m_(2,3), m_(3,3) and m_(4,3) from successive matrices M=(U_2)^r.
%D L. E. Jeffery, Unit-primitive matrices and rhombus substitution tilings, (in preparation).
%F Recurrence: a(n)=3*a(n-3)-3*a(n-9), for n>=12, with initial conditions {a(m)}={0,1,0,0,1,1,1,3,2,3,6,6}, m=0,1,...,11.
%F G.f.: x*(1-2*x^3+x^4+x^5-x^7)/(1-3*x^3+3*x^9).
%Y Cf. A187499, A187500, A187502.
%K nonn,easy
%O 0,8
%A _L. Edson Jeffery_, Mar 16 2011
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