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A187506
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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} = {-1,0,1,2}, n=3*r+p_i, and define a(-1)=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^3-2*x) with x=2*cos(Pi/9).
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4
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0, 0, 1, 1, 1, 1, 2, 3, 4, 7, 9, 10, 19, 26, 30, 56, 75, 85, 160, 216, 246, 462, 622, 707, 1329, 1791, 2037, 3828, 5157, 5864, 11021, 14849, 16886, 31735, 42756, 48620, 91376, 123111, 139997, 263108, 354484, 403104, 757588, 1020696, 1160693, 2181389
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OFFSET
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0,7
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COMMENTS
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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,(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^3-2*x) 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_3)^r, where
U_3=
(0 0 0 1)
(0 0 1 1)
(0 1 1 1)
(1 1 1 1).
2. The sequence. Let r>=0, and let D_r be the r-th "block" defined by D_r={a(3*r-1),a(3*r),a(3*r+1),a(3*r+2)} with a(-1)=0. Note that D_r-2*D_(r-1)-3*D_(r-2)+D_(r-3)+D_(r-4)={0,0,0,0}, for r>=4. Let p={p_1,p_2,p_3,p_4}={-1,0,1,2} and n=3*r+p_i. Then a(n)=a(3*r+p_i)=m_(i,4), where M=(m_(i,j))=(U_3)^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)
Combining blocks A_r, B_r, C_r and D_r, from A187503, A187504, A187505 and this sequence, respectively, as matrix columns [A_r,B_r,C_r,D_r] generates the matrix (U_3)^r, and a negative index (-1)*r yields the corresponding inverse [A_(-r),B_(-r),C_(-r),D(-r)]=(U_3)^(-r) of (U_3)^r. Therefore, the four sequences need not be causal.
Since U_3 is symmetric, so is M=(U_3)^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+2)=a(3*(r+1)-1) for all r>0, 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.
This sequence is a nontrivial extension of A038197.
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REFERENCES
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L. E. Jeffery, Unit-primitive matrices and rhombus substitution tilings, (in preparation).
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LINKS
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Index entries for linear recurrences with constant coefficients, signature (1,-1,3,-3,3,0,0,0,-1,1,-1).
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FORMULA
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Recurrence: a(n)=2*a(n-3)+3*a(n-6)-a(n-9)-a(n-12), with initial conditions {a(m)}={0,0,1,1,1,2,3,4,7,10,19,26}, m=0,1,...,11.
G.f.: x^2*(1+x+x^2-x^3+x^5-x^6)/(1-2*x^3-3*x^6+x^9+x^12).
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MATHEMATICA
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LinearRecurrence[{1, -1, 3, -3, 3, 0, 0, 0, -1, 1, -1}, {0, 0, 1, 1, 1, 1, 2, 3, 4, 7, 9}, 50] (* Harvey P. Dale, May 19 2015 *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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