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 A187500 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^2-1) with x=2*cos(Pi/9). 3

%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^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 B_r be the r-th "block" defined by B_r={a(3*r-2),a(3*r),a(3*r+1),a(3*r+2)} with a(-2)=0. Note that B_r-3*B_(r-1)+3*B_(r-3)={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 component-wise 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 second-column entries m_(2,2), m_(3,2) and m_(4,2) 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)}={1,0,0,1,0,1,2,1,2,4,3,5}, m=0,1,...,11.

%F G.f.: (1-2*x^3+x^5-x^6+x^7-x^8+x^9-x^11)/(1-3*x^3+3*x^9).

%Y Cf. A187499, A187501, A187502.

%K nonn,easy

%O 0,7

%A _L. Edson Jeffery_, Mar 16 2011

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