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

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

Offset: 0

Author

L. Edson Jeffery, Mar 15 2011

Keywords

• nonn
• easy

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 r-th "block" defined by A_r={a(3*r-1),a(3*r),a(3*r+1),a(3*r+2)} with a(-1)=1. Note that A_r-2*A_(r-1)-3*A_(r-2)+A_(r-3)+A_(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,1), where M=(m_(i,j))=(U_3)^r was defined above. Hence the block A_r corresponds component-wise 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 first-column 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.

Crossrefs

Cf. A187504, A187505, 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)}={0,0,0,0,0,1,1,1,1,2,3,4}, k=0,1,...,11.

G.f.: x^5(1+x+x^2-x^3+x^5-x^6)/(1-2*x^3-3*x^6+x^9+x^12).