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A187501 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_(9,3,0) 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
0, 1, 0, 0, 1, 1, 1, 3, 2, 3, 6, 6, 9, 15, 15, 24, 36, 39, 63, 90, 99, 162, 225, 252, 414, 567, 639, 1053, 1431, 1620, 2673, 3618, 4104, 6777, 9153, 10395, 17172, 23166, 26325, 43497, 58644, 66663, 110160, 148473, 168804, 278964, 375921 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,8
COMMENTS
(Start) See A187502 for supporting theory. Define the matrix
U_2=
(0 0 1 0)
(0 1 0 1)
(1 0 1 1)
(0 1 1 1).
Let r>=0, and let C_r be the r-th "block" defined by C_r={a(3*r-2),a(3*r),a(3*r+1),a(3*r+2)} with a(-2)=0. Note that C_r-3*C_(r-1)+3*C_(r-3)={0,0,0,0}, for r>=4, with initial conditions {C_k}={{0,0,1,0},{1,0,1,1},{1,1,3,2},{3,3,6,6}}, 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 C_r corresponds component-wise to the third column of M, and a(n)=a(3*r+p_i)=m_(i,3) gives the quantity of H_(9,3,0) tiles that should appear in a subdivided H_(9,i,r) tile. (End)
Since a(3*r+1)=a(3*(r+1)-2) for all r, this sequence arises by concatenation of third-column entries m_(2,3), m_(3,3) and m_(4,3) from successive matrices M=(U_2)^r.
REFERENCES
L. E. Jeffery, Unit-primitive matrices and rhombus substitution tilings, (in preparation).
LINKS
FORMULA
Recurrence: a(n)=3*a(n-3)-3*a(n-9), for n>=12, with initial conditions {a(m)}={0,1,0,0,1,1,1,3,2,3,6,6}, m=0,1,...,11.
G.f.: x*(1-2*x^3+x^4+x^5-x^7)/(1-3*x^3+3*x^9).
CROSSREFS
Sequence in context: A014679 A208454 A187499 * A274462 A050062 A058533
KEYWORD
nonn,easy
AUTHOR
L. Edson Jeffery, Mar 16 2011
STATUS
approved

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Last modified February 26 02:39 EST 2024. Contains 370335 sequences. (Running on oeis4.)