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%I #11 Feb 03 2019 16:59:45
%S 1,0,0,0,1,1,2,2,2,4,6,7,13,17,19,36,49,56,105,141,160,301,406,462,
%T 868,1169,1329,2498,3366,3828,7194,9692,11021,20713,27907,31735,59642,
%U 80355,91376,171731,231373,263108,494481,666212,757588
%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} = {-1,0,1,2}, n=3*r+p_i, and define a(-1)=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^3-2*x) with x=2*cos(Pi/9).
%C See A187506 for supporting theory. Define the matrix
%C U_3 = (0 0 0 1)
%C (0 0 1 1)
%C (0 1 1 1)
%C (1 1 1 1). Let r>=0 and M=(m_(i,j))=(U_3)^r, i,j=1,2,3,4. Let B_r be the r-th "block" defined by B_r={a(3*r-1),a(3*r),a(3*r+1),a(3*r+2)} with a(-1)=0. Note that B_r-2*B_(r-1)-3*B_(r-2)+B_(r-3)+B_(r-4)={0,0,0,0}, for r>=4. Let p={p_1,p_2,p_3,p_4}={-1,0,1,2} and n=3*r+p_i. Then a(n)=a(3*r+p_i)=m_(i,2), where M=(m_(i,j))=(U_3)^r was defined above. Hence the block B_r corresponds component-wise to the second column of M, and 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.
%C Since a(3*r+2)=a(3*(r+1)-1) for all r, this sequence arises by concatenation of second-column entries m_(2,2), m_(3,2) and m_(4,2) of M=(U_3)^r.
%F Recurrence: a(n)=2*a(n-3)+3*a(n-6)-a(n-9)-a(n-12), for n>=12, with initial conditions {a(k)}={1,0,0,0,1,1,2,2,2,4,6,7}, k=0,1,...,11.
%F G.f.: (1-2*x^3+x^4+x^5-x^6+x^9-x^10)/(1-2*x^3-3*x^6+x^9+x^12).
%Y Cf. A187503, A187505, A187506.
%K nonn,easy
%O 0,7
%A _L. Edson Jeffery_, Mar 15 2011
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