cp's OEIS Frontend

A187496 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)=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(2*cos(Pi/9)).

%I A187496 #13 Sep 08 2022 08:45:56
%S A187496 1,0,0,0,1,0,2,0,1,0,3,1,5,1,4,1,9,5,14,6,14,7,28,20,42,27,48,34,90,
%T A187496 75,132,109,165,143,297,274,429,417,571,560,1000,988,1429,1548,1988,
%U A187496 2108,3417,3536,4846,5644,6953,7752,11799,12597,16645
%N A187496 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)=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(2*cos(Pi/9)).
%C A187496 (Start) See A187498 for supporting theory. Define the matrix
%C A187496 U_1=
%C A187496 (0 1 0 0)
%C A187496 (1 0 1 0)
%C A187496 (0 1 0 1)
%C A187496 (0 0 1 1).
%C A187496 Let r>=0, and let B_r be the r-th "block" defined by B_r={a(3*r-3),a(3*r),a(3*r+1),a(3*r+2)} with a(-3)=0. Note that B_r-B_(r-1)-3*B_(r-2)+2*B_(r-3)+B_(r-4)={0,0,0,0}, for r>=4, with initial conditions {B_k}={{0,1,0,0},{1,0,1,0},{0,2,0,1},{2,0,3,1}}, 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 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 A187496 Since a(3*r)=a(3*(r+1)-3) 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_1)^r.
%D A187496 L. E. Jeffery, Unit-primitive matrices and rhombus substitution tilings, (in preparation).
%H A187496 G. C. Greubel, <a href="/A187496/b187496.txt">Table of n, a(n) for n = 0..5000</a>
%H A187496 <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 A187496 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)}={1,0,0,0,1,0,2,0,1,0,3,1}, m=0,1,...,11.
%F A187496 G.f.: (1-x^3+x^4-x^6-x^7+x^8)/(1-x^3-3*x^6+2*x^9+x^12).
%t A187496 LinearRecurrence[{0,0,1,0,0,3,0,0,-2,0,0,-1}, {1,0,0,0,1,0,2,0,1,0,3,1}, 50] (* _G. C. Greubel_, Apr 20 2018 *)
%o A187496 (PARI) x='x+O('x^50); Vec((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 A187496 (Magma) I:=[1,0,0,0,1,0,2,0,1,0,3,1]; [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 A187496 Cf. A187495, A187497, A187498.
%K A187496 nonn,easy
%O A187496 0,7
%A A187496 _L. Edson Jeffery_, Mar 17 2011