

A187505


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)=0. Then a(n)=a(3*r+p_i) gives the quantity of H_(9,3,0) 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, 1, 0, 1, 1, 1, 2, 3, 3, 6, 8, 9, 17, 23, 26, 49, 66, 75, 141, 190, 216, 406, 547, 622, 1169, 1575, 1791, 3366, 4535, 5157, 9692, 13058, 14849, 27907, 37599, 42756, 80355, 108262, 123111, 231373, 311728, 354484, 666212, 897585, 1020696
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OFFSET

0,7


COMMENTS

(Start) 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).
2. Let r>=0 and M=(m_(i,j))=(U_3)^r, i,j=1,2,3,4. Let C_r be the rth "block" defined by C_r={a(3*r1),a(3*r),a(3*r+1),a(3*r+2)} with a(1)=0. Note that C_r2*C_(r1)3*C_(r2)+C_(r3)+C_(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,3), where M=(m_(i,j))=(U_3)^r was defined above. Hence the block C_r corresponds componentwise to the third column of M, and a(3*r+p_i)=m_(i,3) gives the quantity of H_(9,3,0) tiles that should appear in a subdivided H_(9,i,r) tile. (End)
Since a(3*r+2)=a(3*(r+1)1) for all r, this sequence arises by concatenation of thirdcolumn entries m_(2,3), m_(3,3) and m_(4,3) of M=(U_3)^r.


LINKS



FORMULA

Recurrence: a(n)=2*a(n3)+3*a(n6)a(n9)a(n12), for n>=12, with initial conditions {a(k)}={0,1,0,1,1,1,2,3,3,6,8,9}, k=0,1,...,11.
G.f.: x*(1+x^2x^3+x^42*x^6+x^7x^8)/(12*x^33*x^6+x^9+x^12).


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



