OFFSET
0,7
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.
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..1000
L. E. 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).
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.
EXAMPLE
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.
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 *)
PROG
(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
(PARI) 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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
L. Edson Jeffery, Mar 09 2011
EXTENSIONS
More terms from Vincenzo Librandi, Sep 18 2015
STATUS
approved