

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^21) with x=2*cos(Pi/9).


3



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, 639, 729, 1206, 1620, 1845, 3051, 4104, 4671, 7722, 10395, 11826, 19548, 26325, 29943, 49491, 66663, 75816, 125307, 168804, 191970, 317277
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OFFSET

0,7


COMMENTS

Theory. (Start)
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,(9j)*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, 9j) in which one coefficient is even while the other is odd. A halftile 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 halftile. 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^21) with x=2*cos(Pi/9); likewise let H_(9,j,r) denote the corresponding halftile. 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
U_2=
(0 0 1 0)
(0 1 0 1)
(1 0 1 1)
(0 1 1 1).
2. The sequence. Let r>=0, and let D_r be the rth "block" defined by D_r={a(3*r2),a(3*r),a(3*r+1),a(3*r+2)} with a(2)=0. Note that D_r3*D_(r1)+3*D_(r3)={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 componentwise 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)
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.
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.
Since a(3*r+1)=a(3*(r+1)2) for all r, this sequence arises by concatenation of fourthcolumn entries m_(2,4), m_(3,4) and m_(4,4) (or fourthrow entries m_(4,2), m_(4,3) and m_(4,4)) from successive matrices M=(U_2)^r.


REFERENCES

L. E. Jeffery, Unitprimitive matrices and rhombus substitution tilings, (in preparation).


LINKS



FORMULA

Recurrence: a(n)=3*a(n3)3*a(n9), for n>=12, with initial conditions {a(m)}={0,0,1,1,1,1,2,2,3,5,6,7}, m=0,1,...,11.
G.f.: x^2*(1+x+x^22*x^3x^4x^5x^7+x^9)/(13*x^3+3*x^9).


MATHEMATICA

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 *)


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



