A187065 Let i be in {1,2,3} and let r >= 0 be an integer. Let p = {p_1, p_2, p_3} = {-2,0,1}, n=2*r+p_i, and define a(-2)=1. Then, a(n)=a(2*r+p_i) gives the quantity of H_(7,1,0) tiles in a subdivided H_(7,i,r) tile after linear scaling by the factor x^r, where x=sqrt(2*cos(Pi/7)).

0, 0, 1, 0, 0, 1, 2, 1, 1, 3, 5, 4, 5, 9, 14, 14, 19, 28, 42, 47, 66, 89, 131, 155, 221, 286, 417, 507, 728, 924, 1341, 1652, 2380, 2993, 4334, 5373, 7753, 9707, 14041, 17460, 25213, 31501, 45542, 56714, 81927, 102256, 147798, 184183, 266110, 331981, 479779

Offset: 0

Author

L. Edson Jeffery, Mar 09 2011

Keywords

• nonn
• easy

Comments

(Start) See A187067 for supporting theory. Define the matrix

U_1= (0 1 0)

(1 0 1)

(0 1 1).

Let r>=0 and M=(m_(i,j))=(U_1)^r, i,j=1,2,3. Let A_r be the r-th "block" defined by A_r={a(2*r-2),a(2*r),a(2*r+1)} with a(-2)=1. Note that A_r-A_(r-1)-2*A_(r-2)+A_(r-3)={0,0,0}, with A_0={a(-2),a(0),a(1)}={1,0,0}. Let p={p_1,p_2,p_3}=(-2,0,1) and n=2*r+p_i. Then a(n)=a(2*r+p_i)=m_(i,1), where M=(m_(i,j))=(U_1)^r was defined above. Hence the block A_r corresponds component-wise to the first column of M, and a(n)=m_(i,1) gives the quantity of H_(7,1,0) tiles that should appear in a subdivided H_(7,i,r) tile. (End)

Combining blocks A_r, B_r and C_r, from this sequence, A187066 and A187067, respectively, as matrix columns [A_r,B_r,C_r] generates the matrix (U_1)^r, and a negative index (-1)*r yields the corresponding inverse [A_(-r),B_(-r),C_(-r)]=(U_1)^(-r) of (U_1)^r. Therefore, the three sequences need not be causal.

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

a(n+2)=A187067(n), a(2*n)=A096976(n+1), a(2*n+1)=A006053(n).

Examples

			Suppose r=3. Then
A_r = A_3 = {a(2*r-2),a(2*r),a(2*r+1)} = {a(4),a(6),a(7)} = {0,2,1},
corresponding to the entries in the first column of
  M = (U_2)^3 = (0 2 1)
                (2 1 3)
                (1 3 3).
Choose i=2 and set n=2*r+p_i. Then a(n) = a(2*r+p_i) = a(6+0) = a(6) = 2, which equals the entry in row 2 and column 1 of M. Hence a subdivided H_(7,2,3) tile should contain a(6) = m_(2,1) = 2 H_(7,1,0) tiles.
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• L. Edson Jeffery, Unit-primitive matrices
• Roman Witula, Damian Slota and Adam Warzynski, Quasi-Fibonacci Numbers of the Seventh Order, J. Integer Seq., 9 (2006), Article 06.4.3.
• Index entries for linear recurrences with constant coefficients, signature (0,1,0,2,0,-1).

Crossrefs

Cf. A187066, A187067, A187068, A187069, A187070.

Programs

Magma

I:=[0,0,1,0,0,1]; [n le 6 select I[n] else Self(n-2)+2*Self(n-4)-Self(n-6): n in [1..60]]; // Vincenzo Librandi, Sep 18 2015

Mathematica

LinearRecurrence[{0,1,0,2,0,-1},{0,0,1,0,0,1},50] (* Harvey P. Dale, Aug 15 2012 *)
CoefficientList[Series[x^2 (1 - x^2 + x^3)/(1 - x^2 - 2 x^4 + x^6), {x, 0, 50}], x] (* Vincenzo Librandi, Sep 18 2015 *)

PARI

my(x='x+O('x^50)); concat([0,0], Vec(x^2*(1-x^2+x^3)/(1-x^2-2*x^4 +x^6))) \\ G. C. Greubel, Jan 29 2018

Formula

Recurrence: a(n) = a(n-2) + 2*a(n-4) - a(n-6).

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

Closed-form: a(n) = -(1/14)*((X_1 + Y_1*(-1)^(n-1))*((w_2)^2 - (w_3)^2)*(w_1)^(n-1) + (X_2 + Y_2*(-1)^(n-1))*((w_3)^2 - (w_1)^2)*(w_2)^(n-1) + (X_3 + Y_3*(-1)^(n-1))*((w_1)^2 - (w_2)^2)*(w_3)^(n-1)), where w_k = sqrt(2*(-1)^(k-1)*cos(k*Pi/7)), X_k = (w_k)^3 - w_k + 1 and Y_k = -(w_k)^3 + w_k + 1, k=1,2,3.

Extensions

More terms from Vincenzo Librandi, Sep 18 2015