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^3-2*x) with x=2*cos(Pi/9).

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

Offset: 0

Author

L. Edson Jeffery, Mar 14 2011

Keywords

• nonn
• easy

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 r-th "block" defined by C_r={a(3*r-1),a(3*r),a(3*r+1),a(3*r+2)} with a(-1)=0. Note that C_r-2*C_(r-1)-3*C_(r-2)+C_(r-3)+C_(r-4)={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 component-wise 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 third-column entries m_(2,3), m_(3,3) and m_(4,3) of M=(U_3)^r.

Crossrefs

Cf. A187503, A187504, A187506.

Formula

Recurrence: a(n)=2*a(n-3)+3*a(n-6)-a(n-9)-a(n-12), 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^2-x^3+x^4-2*x^6+x^7-x^8)/(1-2*x^3-3*x^6+x^9+x^12).