

A187499


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)=1. Then a(n)=a(3*r+p_i) gives the quantity of H_(9,1,0) 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, 0, 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
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OFFSET

0,11


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 A_r be the rth "block" defined by A_r={a(3*r2),a(3*r),a(3*r+1),a(3*r+2)} with a(2)=1. Note that A_r3*A_(r1)+3*A_(r3)={0,0,0,0}, for r>=4, with initial conditions {A_k}={{1,0,0,0},{0,0,1,0},{1,0,1,1},{1,1,3,2}}, 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 A_r corresponds componentwise to the first column of M, and a(n)=a(3*r+p_i)=m_(i,1) gives the quantity of H_(9,1,0) 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 firstcolumn entries m_(2,1), m_(3,1) and m_(4,1) from successive matrices M=(U_2)^r.
This sequence is a nontrivial extension of A187501.


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)}={0,0,0,0,1,0,0,1,1,1,3,2}, m=0,1,...,11.
G.f.: x^4(12*x^3+x^4+x^5x^7)/(13*x^3+3*x^9).


MATHEMATICA

Join[{0, 0, 0}, LinearRecurrence[{0, 0, 3, 0, 0, 0, 0, 0, 3}, {0, 1, 0, 0, 1, 1, 1, 3, 2}, 50]] (* Harvey P. Dale, Jun 22 2013 *)


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



