A187066 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)=0. Then, a(n)=a(2*r+p_i) gives the quantity of H_(7,2,0) tiles in a subdivided H_(7,i,r) tile after linear scaling by the factor x^r, where x=sqrt(2*cos(Pi/7)).

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

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 B_r be the r-th "block" defined by B_r={a(2*r-2),a(2*r),a(2*r+1)} with a(-2)=0. Note that B_r-B_(r-1)-2*B_(r-2)+B_(r-3)={0,0,0}, with B_0={a(-2),a(0),a(1)}={0,1,0}. 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,2), where M=(m_(i,j))=(U_1)^r was defined above. Hence the block B_r corresponds component-wise to the second column of M, and a(n)=m_(i,2) gives the quantity of H_(7,2,0) tiles that should appear in a subdivided H_(7,i,r) tile. (End)

Combining blocks A_r, B_r and C_r, from A187065, this sequence 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 second-column entries m_(2,2) and m_(3,2) from successive matrices M=(U_1)^r.

Examples

			Suppose r=3. Then
B_r = B_3 = {a(2*r-2),a(2*r),a(2*r+1)}={a(4),a(6),a(7)} = {2,1,3},
corresponding to the entries in the third 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) = 1, which equals the entry in row 2 and column 2 of M. Hence a subdivided H_(7,2,3) tile should contain a(6) = m_(2,2) = 1 H_(7,2,0) 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. A187065, A187067, A187068, A187069, A187070.

Programs

Mathematica

LinearRecurrence[{0,1,0,2,0,-1},{1,0,0,1,2,1},50] (* Harvey P. Dale, Aug 16 2012 *)
CoefficientList[Series[(1 - x^2 + x^3)/(1 - x^2 - 2*x^4 + x^6), {x, 0, 50}], x] (* G. C. Greubel, Oct 20 2017 *)

PARI

my(x='x+O('x^50)); Vec((1-x^2+x^3)/(1-x^2-2*x^4+x^6)) \\ G. C. Greubel, Oct 20 2017

Formula

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

a(2*n) = A052547(n), a(2n+1) = A006053(n+1).

G.f.: (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)^5-(w_k)^3+(w_k)^2 and Y_k = -(w_k)^5+(w_k)^3+(w_k)^2, k=1,2,3.