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^2-1) with x=2*cos(Pi/9).

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

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 r-th "block" defined by B_r={a(3*r-2),a(3*r),a(3*r+1),a(3*r+2)} with a(-2)=0. Note that B_r-3*B_(r-1)+3*B_(r-3)={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 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+1)=a(3*(r+1)-2) 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_2)^r.

References

L. E. Jeffery, Unit-primitive matrices and rhombus substitution tilings, (in preparation).

Links

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

Formula

Recurrence: a(n)=3*a(n-3)-3*a(n-9), 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.: (1-2*x^3+x^5-x^6+x^7-x^8+x^9-x^11)/(1-3*x^3+3*x^9).

Crossrefs

Cf. A187499, A187501, A187502.

Sequence in context: A323465 A364780 A124904 * A323460 A129144 A360594

Adjacent sequences: A187497 A187498 A187499 * A187501 A187502 A187503

Keyword

nonn,easy

Author

L. Edson Jeffery, Mar 16 2011

Status

approved