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A187496
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Let i in {1,2,3,4} and r>=0 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)).
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3
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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, 75, 132, 109, 165, 143, 297, 274, 429, 417, 571, 560, 1000, 988, 1429, 1548, 1988, 2108, 3417, 3536, 4846, 5644, 6953, 7752, 11799, 12597, 16645
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OFFSET
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0,7
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COMMENTS
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(Start) See A187498 for supporting theory. Define the matrix
U_1=
(0 1 0 0)
(1 0 1 0)
(0 1 0 1)
(0 0 1 1).
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)
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.
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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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Table of n, a(n) for n=0..54.
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FORMULA
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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.
G.f.: (1-x^3+x^4-x^6-x^7+x^8)/(1-x^3-3*x^6+2*x^9+x^12).
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CROSSREFS
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Cf. A187495, A187497, A187498.
Sequence in context: A125924 A082513 A187495 * A193056 A086780 A158612
Adjacent sequences: A187493 A187494 A187495 * A187497 A187498 A187499
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KEYWORD
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nonn,easy
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AUTHOR
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L. Edson Jeffery, Mar 17 2011
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STATUS
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approved
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