A187499 - OEIS

%I #13 Feb 03 2019 16:47:04

%S 0,0,0,0,1,0,0,1,1,1,3,2,3,6,6,9,15,15,24,36,39,63,90,99,162,225,252,

%T 414,567,639,1053,1431,1620,2673,3618,4104,6777,9153,10395,17172,

%U 23166,26325,43497,58644,66663,110160,148473,168804,278964

%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)=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(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 A_r be the r-th "block" defined by A_r={a(3*r-2),a(3*r),a(3*r+1),a(3*r+2)} with a(-2)=1. Note that A_r-3*A_(r-1)+3*A_(r-3)={0,0,0,0}, for r>=4, with initial conditions {A_k}={{1,0,0,0},{0,0,1,0},{1,0,1,1},{1,1,3,2}}, 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 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 Since a(3*r+1)=a(3*(r+1)-2) 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_2)^r.

%C This sequence is a nontrivial extension of A187501.

%D L. E. Jeffery, Unit-primitive matrices and rhombus substitution tilings, (in preparation).

%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (0, 0, 3, 0, 0, 0, 0, 0, -3).

%F Recurrence: a(n)=3*a(n-3)-3*a(n-9), for n>=12, with initial conditions {a(m)}={0,0,0,0,1,0,0,1,1,1,3,2}, m=0,1,...,11.

%F G.f.: x^4(1-2*x^3+x^4+x^5-x^7)/(1-3*x^3+3*x^9).

%t Join[{0,0,0},LinearRecurrence[{0,0,3,0,0,0,0,0,-3},{0,1,0,0,1,1,1,3,2},50]] (* _Harvey P. Dale_, Jun 22 2013 *)

%Y Cf. A187500, A187501, A187502.

%K nonn,easy

%O 0,11

%A _L. Edson Jeffery_, Mar 16 2011