

A187496


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)).


3



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
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,7


COMMENTS

(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 rth "block" defined by B_r={a(3*r3),a(3*r),a(3*r+1),a(3*r+2)} with a(3)=0. Note that B_rB_(r1)3*B_(r2)+2*B_(r3)+B_(r4)={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 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)=a(3*(r+1)3) 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_1)^r.


REFERENCES

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


LINKS

Table of n, a(n) for n=0..54.


FORMULA

Recurrence: a(n)=a(n3)+3*a(n6)2*a(n9)a(n12), 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.: (1x^3+x^4x^6x^7+x^8)/(1x^33*x^6+2*x^9+x^12).


CROSSREFS

Cf. A187495, A187497, A187498.
Sequence in context: A125924 A082513 A187495 * A193056 A086780 A158612
Adjacent sequences: A187493 A187494 A187495 * A187497 A187498 A187499


KEYWORD

nonn,easy


AUTHOR

L. Edson Jeffery, Mar 17 2011


STATUS

approved



