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 A187502 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_(9,4,0) 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

%S 0,0,1,1,1,1,2,2,3,5,6,7,12,15,18,30,39,45,75,99,114,189,252,288,477,

%T 639,729,1206,1620,1845,3051,4104,4671,7722,10395,11826,19548,26325,

%U 29943,49491,66663,75816,125307,168804,191970,317277

%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_(9,4,0) 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 Theory. (Start)

%C 1. Definitions. Let T_(9,j,0) denote the rhombus with sides of unit length (=1), interior angles given by the pair (j*Pi/9,(9-j)*Pi/9) and Area(T_(9,j,0))=sin(j*Pi/9), j in {1,2,3,4}. Associated with T_(9,j,0) are its angle coefficients (j, 9-j) in which one coefficient is even while the other is odd. A half-tile is created by cutting T_(9,j,0) along a line extending between its two corners with even angle coefficient; let H_(9,j,0) denote this half-tile. Similarly, a T_(9,j,r) tile is a linearly scaled version of T_(9,j,0) with sides of length Q^r and Area(T_(9,j,r))=Q^(2*r)*sin(j*Pi/9), r>=0 an integer, where Q is the positive, constant square root Q=sqrt(x^2-1) with x=2*cos(Pi/9); likewise let H_(9,j,r) denote the corresponding half-tile. Often H_(9,i,r) (i in {1,2,3,4}) can be subdivided into an integral number of each equivalence class H_(9,j,0). But regardless of whether or not H_(9,j,r) subdivides, in theory such a proposed subdivision for each j can be represented by the matrix M=(m_(i,j)), i,j=1,2,3,4, in which the entry m_(i,j) gives the quantity of H_(9,j,0) tiles that should be present in a subdivided H_(9,i,r) tile. The number Q^(2*r) (the square of the scaling factor) is an eigenvalue of M=(U_2)^r, where

%C U_2=

%C (0 0 1 0)

%C (0 1 0 1)

%C (1 0 1 1)

%C (0 1 1 1).

%C 2. The sequence. Let r>=0, and let D_r be the r-th "block" defined by D_r={a(3*r-2),a(3*r),a(3*r+1),a(3*r+2)} with a(-2)=0. Note that D_r-3*D_(r-1)+3*D_(r-3)={0,0,0,0}, for r>=4, with initial conditions {D_k}={{0,0,0,1},{0,1,1,1},{1,2,2,3},{2,5,6,7}}, k=0,1,2,3. Let p={p_1,p_2,p_3,p_4}={-2,0,1,2} and n=3*r+p_i. Then a(n)=a(3*r+p_i)=m_(i,4), where M=(m_(i,j))=(U_2)^r was defined above. Hence the block D_r corresponds component-wise to the fourth column of M, and a(3*r+p_i)=m_(i,4) gives the quantity of H_(9,4,0) tiles that should appear in a subdivided H_(9,i,r) tile. (End)

%C Combining blocks A_r, B_r, C_r and D_r, from A187499, A187500, A187501 and this sequence, respectively, as matrix columns [A_r,B_r,C_r,D_r] generates the matrix (U_2)^r, which is singular for all r>0, that is, the four sequences are strictly causal.

%C Since U_2 is symmetric, so is M=(U_2)^r, so the block D_r also corresponds to the fourth row of M. Therefore, alternatively, for j=1,2,3,4, a(3r+p_j)=m_(4,j) gives the quantity of H_(9,j,0) tiles that should be present in a H_(9,4,r) tile.

%C Since a(3*r+1)=a(3*(r+1)-2) for all r, this sequence arises by concatenation of fourth-column entries m_(2,4), m_(3,4) and m_(4,4) (or fourth-row entries m_(4,2), m_(4,3) and m_(4,4)) from successive matrices M=(U_2)^r.

%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)}={0,0,1,1,1,1,2,2,3,5,6,7}, m=0,1,...,11.

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

%t Join[{0,0,1},LinearRecurrence[{0,0,3,0,0,0,0,0,-3},{1,1,1,2,2,3,5,6,7},50]] (* _Harvey P. Dale_, Mar 29 2013 *)

%Y Cf. A187499, A187500, A187501.

%K nonn,easy

%O 0,7

%A _L. Edson Jeffery_, Mar 15 2011

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