

A187500


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_(9,2,0) 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



1, 0, 0, 1, 0, 1, 2, 1, 2, 4, 3, 5, 9, 9, 12, 21, 24, 30, 51, 63, 75, 126, 162, 189, 315, 414, 477, 792, 1053, 1206, 1998, 2673, 3051, 5049, 6777, 7722, 12771, 17172, 19548, 32319, 43497, 49491, 81810, 110160, 125307, 207117, 278964
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OFFSET

0,7


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 B_r be the rth "block" defined by B_r={a(3*r2),a(3*r),a(3*r+1),a(3*r+2)} with a(2)=0. Note that B_r3*B_(r1)+3*B_(r3)={0,0,0,0}, for r>=4, with initial conditions {B_k}={{0,1,0,0},{0,1,0,1},{0,2,1,2},{1,4,3,5}}, 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 B_r corresponds componentwise 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+1)=a(3*(r+1)2) for all r, this sequence arises by concatenation of secondcolumn entries m_(2,2), m_(3,2) and m_(4,2) from successive matrices M=(U_2)^r.


REFERENCES

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


LINKS

Table of n, a(n) for n=0..46.


FORMULA

Recurrence: a(n)=3*a(n3)3*a(n9), for n>=12, with initial conditions {a(m)}={1,0,0,1,0,1,2,1,2,4,3,5}, m=0,1,...,11.
G.f.: (12*x^3+x^5x^6+x^7x^8+x^9x^11)/(13*x^3+3*x^9).


CROSSREFS

Cf. A187499, A187501, A187502.
Sequence in context: A316997 A323465 A124904 * A323460 A129144 A295313
Adjacent sequences: A187497 A187498 A187499 * A187501 A187502 A187503


KEYWORD

nonn,easy


AUTHOR

L. Edson Jeffery, Mar 16 2011


STATUS

approved



