A187504 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_(9,2,0) 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).

1, 0, 0, 0, 1, 1, 2, 2, 2, 4, 6, 7, 13, 17, 19, 36, 49, 56, 105, 141, 160, 301, 406, 462, 868, 1169, 1329, 2498, 3366, 3828, 7194, 9692, 11021, 20713, 27907, 31735, 59642, 80355, 91376, 171731, 231373, 263108, 494481, 666212, 757588

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 B_r be the r-th "block" defined by B_r={a(3*r-1),a(3*r),a(3*r+1),a(3*r+2)} with a(-1)=0. Note that B_r-2*B_(r-1)-3*B_(r-2)+B_(r-3)+B_(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,2), where M=(m_(i,j))=(U_3)^r was defined above. Hence the block B_r corresponds component-wise to the second column of M, and a(3*r+p_i)=m_(i,2) gives the quantity of H_(9,2,0) 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 second-column entries m_(2,2), m_(3,2) and m_(4,2) of M=(U_3)^r.

Crossrefs

Cf. A187503, 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)}={1,0,0,0,1,1,2,2,2,4,6,7}, k=0,1,...,11.

G.f.: (1-2*x^3+x^4+x^5-x^6+x^9-x^10)/(1-2*x^3-3*x^6+x^9+x^12).