

A187495


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)=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(2*cos(Pi/9)).


3



0, 0, 0, 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
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OFFSET

0,10


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 A_r be the rth "block" defined by A_r={a(3*r3),a(3*r),a(3*r+1),a(3*r+2)} with a(3)=1. Note that A_rA_(r1)3*A_(r2)+2*A_(r3)+A_(r4)={0,0,0,0}, for r>=4, with initial conditions {A_k}={{1,0,0,0},{0,1,0,0},{1,0,1,0},{0,2,0,1}}, 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 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)=a(3*(r+1)3) 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_1)^r.
This sequence is a nontrivial extension of A187496.


REFERENCES

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


LINKS

Table of n, a(n) for n=0..56.


FORMULA

Recurrence: a(n)=a(n3)+3*a(n6)2*a(n9)a(n12), for n>=12, with initial conditions {a(m)}={0,0,0,1,0,0,0,1,0,2,0,1}, m=0,1,...,11.
G.f.: x^3*(1x^3+x^4x^6x^7+x^8)/(1x^33*x^6+2*x^9+x^12).


CROSSREFS

Cf. A187496, A187497, A187498.
Sequence in context: A125924 A283929 A082513 * A187496 A193056 A244417
Adjacent sequences: A187492 A187493 A187494 * A187496 A187497 A187498


KEYWORD

nonn,easy


AUTHOR

L. Edson Jeffery, Mar 17 2011


STATUS

approved



