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A187495 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} = {-3,0,1,2}, n=3*r+p_i, and define a(-3)=1. Then a(n)=a(3*r+p_i) gives the quantity of H_(9,1,0) tiles in a subdivided H_(9,i,r) tile after linear scaling by the factor Q^r, where Q=sqrt(2*cos(Pi/9)).

%I A187495 #16 Sep 08 2022 08:45:56
%S A187495 0,0,0,1,0,0,0,1,0,2,0,1,0,3,1,5,1,4,1,9,5,14,6,14,7,28,20,42,27,48,
%T A187495 34,90,75,132,109,165,143,297,274,429,417,571,560,1000,988,1429,1548,
%U A187495 1988,2108,3417,3536,4846,5644,6953,7752,11799,12597
%N A187495 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} = {-3,0,1,2}, n=3*r+p_i, and define a(-3)=1. Then a(n)=a(3*r+p_i) gives the quantity of H_(9,1,0) tiles in a subdivided H_(9,i,r) tile after linear scaling by the factor Q^r, where Q=sqrt(2*cos(Pi/9)).
%C A187495 (Start) See A187498 for supporting theory. Define the matrix
%C A187495 U_1=
%C A187495 (0 1 0 0)
%C A187495 (1 0 1 0)
%C A187495 (0 1 0 1)
%C A187495 (0 0 1 1).
%C A187495 Let r>=0, and let A_r be the r-th "block" defined by A_r={a(3*r-3),a(3*r),a(3*r+1),a(3*r+2)} with a(-3)=1. Note that A_r-A_(r-1)-3*A_(r-2)+2*A_(r-3)+A_(r-4)={0,0,0,0}, for r>=4, with initial conditions {A_k}={{1,0,0,0},{0,1,0,0},{1,0,1,0},{0,2,0,1}}, k=0,1,2,3. Let p={p_1,p_2,p_3,p_4}={-3,0,1,2}, n=3*r+p_i and M=(m_(i,j))=(U_1)^r, i,j=1,2,3,4. Then A_r corresponds component-wise to the first column of M, and a(n)=a(3*r+p_i)=m_(i,1) gives the quantity of H_(9,1,0) tiles that should appear in a subdivided H_(9,i,r) tile. (End)
%C A187495 Since a(3*r)=a(3*(r+1)-3) for all r, this sequence arises by concatenation of first-column entries m_(2,1), m_(3,1) and m_(4,1) from successive matrices M=(U_1)^r.
%C A187495 This sequence is a nontrivial extension of A187496.
%D A187495 L. E. Jeffery, Unit-primitive matrices and rhombus substitution tilings, (in preparation).
%H A187495 G. C. Greubel, <a href="/A187495/b187495.txt">Table of n, a(n) for n = 0..5000</a>
%H A187495 <a href="/index/Rec#order_12">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,1,0,0,3,0,0,-2,0,0,-1).
%F A187495 Recurrence: a(n) = a(n-3) +3*a(n-6) -2*a(n-9) -a(n-12), for n>=12, with initial conditions {a(m)}={0,0,0,1,0,0,0,1,0,2,0,1}, m=0,1,...,11.
%F A187495 G.f.: x^3*(1-x^3+x^4-x^6-x^7+x^8)/(1-x^3-3*x^6+2*x^9+x^12).
%t A187495 LinearRecurrence[{0,0,1,0,0,3,0,0,-2,0,0,-1}, {0,0,0,1,0,0,0,1,0,2,0,1}, 50] (* _G. C. Greubel_, Apr 20 2018 *)
%o A187495 (PARI) x='x+O('x^50); concat([0,0,0], Vec(x^3*(1-x^3+x^4-x^6-x^7+x^8)/(1-x^3-3*x^6+2*x^9+x^12))) \\ _G. C. Greubel_, Apr 20 2018
%o A187495 (Magma) I:=[0,0,0,1,0,0,0,1,0,2,0,1]; [n le 12 select I[n] else Self(n-3) + 3*Self(n-6) - 2*Self(n-9) - Self(n-12): n in [1..50]]; // _G. C. Greubel_, Apr 20 2018
%Y A187495 Cf. A187496, A187497, A187498.
%K A187495 nonn,easy
%O A187495 0,10
%A A187495 _L. Edson Jeffery_, Mar 17 2011