

A187503


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} = {1,0,1,2}, n=3*r+p_i, and define a(1)=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^32*x) with x=2*cos(Pi/9).


4



0, 0, 0, 0, 0, 1, 1, 1, 1, 2, 3, 4, 7, 9, 10, 19, 26, 30, 56, 75, 85, 160, 216, 246, 462, 622, 707, 1329, 1791, 2037, 3828, 5157, 5864, 11021, 14849, 16886, 31735, 42756, 48620, 91376, 123111, 139997, 263108, 354484, 403104, 757588, 1020696, 1160693
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OFFSET

0,10


COMMENTS

See A187506 for supporting theory. Define the matrix
U_3 = (0 0 0 1)
(0 0 1 1)
(0 1 1 1)
(1 1 1 1). Let r>=0 and M=(m_(i,j))=(U_3)^r, i,j=1,2,3,4. Let A_r be the rth "block" defined by A_r={a(3*r1),a(3*r),a(3*r+1),a(3*r+2)} with a(1)=1. Note that A_r2*A_(r1)3*A_(r2)+A_(r3)+A_(r4)={0,0,0,0}, for r>=4. Let p={p_1,p_2,p_3,p_4}={1,0,1,2} and n=3*r+p_i. Then a(n)=a(3*r+p_i)=m_(i,1), where M=(m_(i,j))=(U_3)^r was defined above. Hence the block A_r corresponds componentwise to the first column of M, and 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.
Since a(3*r+2)=a(3*(r+1)1) for all r, this sequence arises by concatenation of firstcolumn entries m_(2,1), m_(3,1) and m_(4,1) of M=(U_3)^r.
This sequence is a nontrivial extension of both A038197 and A187506.


LINKS

Table of n, a(n) for n=0..47.


FORMULA

Recurrence: a(n)=2*a(n3)+3*a(n6)a(n9)a(n12), for n>=12, with initial conditions {a(k)}={0,0,0,0,0,1,1,1,1,2,3,4}, k=0,1,...,11.
G.f.: x^5(1+x+x^2x^3+x^5x^6)/(12*x^33*x^6+x^9+x^12).


CROSSREFS

Cf. A187504, A187505, A187506.
Sequence in context: A277066 A138576 A259483 * A038197 A187506 A348443
Adjacent sequences: A187500 A187501 A187502 * A187504 A187505 A187506


KEYWORD

nonn,easy


AUTHOR

L. Edson Jeffery, Mar 15 2011


STATUS

approved



