A187496 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} = {-3,0,1,2}, n=3*r+p_i, and define a(-3)=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(2*cos(Pi/9)).

1, 0, 0, 0, 1, 0, 2, 0, 1, 0, 3, 1, 5, 1, 4, 1, 9, 5, 14, 6, 14, 7, 28, 20, 42, 27, 48, 34, 90, 75, 132, 109, 165, 143, 297, 274, 429, 417, 571, 560, 1000, 988, 1429, 1548, 1988, 2108, 3417, 3536, 4846, 5644, 6953, 7752, 11799, 12597, 16645

Offset: 0

Author

L. Edson Jeffery, Mar 17 2011

Keywords

• nonn
• easy

Comments

(Start) See A187498 for supporting theory. Define the matrix

U_1=

(0 1 0 0)

(1 0 1 0)

(0 1 0 1)

(0 0 1 1).

Let r>=0, and let B_r be the r-th "block" defined by B_r={a(3*r-3),a(3*r),a(3*r+1),a(3*r+2)} with a(-3)=0. Note that B_r-B_(r-1)-3*B_(r-2)+2*B_(r-3)+B_(r-4)={0,0,0,0}, for r>=4, with initial conditions {B_k}={{0,1,0,0},{1,0,1,0},{0,2,0,1},{2,0,3,1}}, k=0,1,2,3. Let p={p_1,p_2,p_3,p_4}={-3,0,1,2}, n=3*r+p_i and M=(m_(i,j))=(U_1)^r, i,j=1,2,3,4. Then B_r corresponds component-wise to the second column of M, and a(n)=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. (End)

Since a(3*r)=a(3*(r+1)-3) for all r, this sequence arises by concatenation of second-column entries m_(2,2), m_(3,2) and m_(4,2) from successive matrices M=(U_1)^r.

References

• L. E. Jeffery, Unit-primitive matrices and rhombus substitution tilings, (in preparation).

Links

• G. C. Greubel, Table of n, a(n) for n = 0..5000
• Index entries for linear recurrences with constant coefficients, signature (0,0,1,0,0,3,0,0,-2,0,0,-1).

Crossrefs

Cf. A187495, A187497, A187498.

Programs

Magma

I:=[1,0,0,0,1,0,2,0,1,0,3,1]; [n le 12 select I[n] else Self(n-3) + 3*Self(n-6) - 2*Self(n-9) - Self(n-12): n in [1..50]]; // G. C. Greubel, Apr 20 2018

Mathematica

LinearRecurrence[{0,0,1,0,0,3,0,0,-2,0,0,-1}, {1,0,0,0,1,0,2,0,1,0,3,1}, 50] (* G. C. Greubel, Apr 20 2018 *)

PARI

x='x+O('x^50); Vec((1-x^3+x^4-x^6-x^7+x^8)/(1-x^3-3*x^6 +2*x^9 +x^12)) \\ G. C. Greubel, Apr 20 2018

Formula

Recurrence: a(n) = a(n-3) +3*a(n-6) -2*a(n-9) -a(n-12), for n>=12, with initial conditions {a(m)}={1,0,0,0,1,0,2,0,1,0,3,1}, m=0,1,...,11.

G.f.: (1-x^3+x^4-x^6-x^7+x^8)/(1-x^3-3*x^6+2*x^9+x^12).