A187497 Let i be in {1,2,3,4} and r>=0 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 number 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(2*cos(Pi/9)).

0, 1, 0, 1, 0, 1, 0, 2, 1, 3, 1, 3, 1, 6, 4, 9, 5, 10, 6, 19, 15, 28, 21, 34, 27, 62, 55, 90, 82, 117, 109, 207, 199, 297, 308, 406, 417, 703, 714, 1000, 1131, 1417, 1548, 2417, 2548, 3417, 4096, 4965, 5644, 8382, 9061, 11799, 14705

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 C_r be the r-th "block" defined by C_r = {a(3*r-3), a(3*r), a(3*r+1), a(3*r+2)} with a(-3)=0. Note that C_r - C_(r-1) - 3*C_(r-2) + 2*C_(r-3) + C_(r-4) = {0,0,0,0}, for r>=4, with initial conditions {C_k}={{0,0,1,0},{0,1,0,1},{1,0,2,1},{0,3,1,3}}, 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 C_r corresponds component-wise to the third column of M, and a(n)=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)=a(3*(r+1)-3) for all r, this sequence arises by concatenation of third-column entries m_(2,3), m_(3,3) and m_(4,3) 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..1000
• Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,3,0,0,1).

Crossrefs

Cf. A187495, A187496, A187498.

Programs

Mathematica

CoefficientList[Series[x*(1+x^2-x^3+x^4-x^5-x^6+x^9-x^10)/(1-x^3-3*x^6+ 2*x^9+ x^12),{x,0,70}],x] (* or *) LinearRecurrence[{0,0,0,0,0,3,0,0,1},{0,1,0,1,0,1,0,2,1},70] (* Harvey P. Dale, Feb 10 2013 *)

PARI

x='x+O('x^30); concat([0], Vec(x*(1 +x^2 -x^3 +x^4 -x^5 -x^6 +x^9 -x^10)/(1 -x^3 -3*x^6 +2*x^9 +x^12))) \\ G. C. Greubel, Nov 28 2017

Formula

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

a(0)=0, a(1)=1, a(2)=0, a(3)=1, a(4)=0, a(5)=1, a(6)=0, a(7)=2, a(8)=1, a(n) = 3*a(n-6) + a(n-9). - Harvey P. Dale, Feb 10 2013