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%I #26 Feb 03 2019 16:52:05
%S 0,1,1,1,1,2,3,3,4,6,9,10,14,19,28,33,47,61,89,108,155,197,286,352,
%T 507,638,924,1145,1652,2069,2993,3721,5373,6714,9707,12087,17460,
%U 21794,31501,39254,56714,70755,102256
%N 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_(7,3,0) tiles in a subdivided H_(7,i,r) tile after linear scaling by the factor x^r, where x = sqrt(2*cos(Pi/7)).
%C Theory. (Start)
%C 1. Definitions. Let T_(7,j,0) denote the rhombus with sides of unit length (=1), interior angles given by the pair (j*Pi/7,(7-j)*Pi/7) and Area(T_(7,j,0)) = sin(j*Pi/7), j in {1,2,3}. Associated with T_(7,j,0) are its angle coefficients (j, 7-j) in which one coefficient is even while the other is odd. A half-tile is created by cutting T_(7,j,0) along a line extending between its two corners with even angle coefficient; let H_(7,j,0) denote this half-tile. Similarly, a T_(7,j,r) tile is a linearly scaled version of T_(7,j,0) with sides of length x^r and Area(T_(7,j,r)) = x^(2*r)*sin(j*Pi/7), r>=0 an integer, where x is the positive, constant square root x = sqrt(2*cos(j*Pi/7)); likewise let H_(7,j,r) denote the corresponding half-tile. Often H_(7,i,r) (i in {1,2,3}) can be subdivided into an integral number of each equivalence class H_(7,j,0). But regardless of whether or not H_(7,j,r) subdivides, in theory such a proposed subdivision for each j can be represented by the matrix M=(m_(i,j)), i,j=1,2,3, in which the entry m_(i,j) gives the quantity of H_(7,j,0) tiles that should be present in a subdivided H_(7,i,r) tile. The number x^(2*r) (the square of the scaling factor) is an eigenvalue of M=(U_1)^r, where
%C U_1=
%C (0 1 0)
%C (1 0 1)
%C (0 1 1).
%C 2. The sequence. Let r>=0, and let C_r be the r-th "block" defined by C_r = {a(2*r-2), a(2*r), a(2*r+1)}. Note that C_r - C_(r-1) - 2*C_(r-2) + C_(r-3) = {0,0,0}, with C_0 = {a(-2),a(0),a(1)} = {0,0,1}. Let n = 2*r + p_i. Then a(n) = a(2*r + p_i) = m_(i,3), where M = (m_(i,j)) = (U_1)^r was defined above. Hence the block C_r corresponds component-wise to the third column of M, and a(n) = m_(i,3) gives the quantity of H_(7,3,0) tiles that should appear in a subdivided H_(7,i,r) tile. (End)
%C Combining blocks A_r, B_r and C_r, from A187065, A187066 and this sequence, 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.
%C Since a(2*r-2) = a(2*(r-1)) for all r, this sequence arises by concatenation of third-column entries m_(2,3) and m_(3,3) from successive matrices M = (U_1)^r.
%C a(2*n) = A006053(n+1), a(2*n+1) = A028495(n).
%H Matthew House, <a href="/A187067/b187067.txt">Table of n, a(n) for n = 0..7784</a>
%H L. E. Jeffery, <a href="/wiki/User:L._Edson_Jeffery/Unit-Primitive_Matrices">Unit-primitive matrices</a>
%H R. Witula, D. Slota and A. Warzynski, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL9/Slota/slota57.html">Quasi-Fibonacci Numbers of the Seventh Order</a>, J. Integer Seq., 9 (2006), Article 06.4.3.
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,1,0,2,0,-1).
%F Recurrence: a(n) = a(n-2) + 2*a(n-4) - a(n-6).
%F G.f.: x(1 + x - x^4)/(1 - x^2 - 2*x^4 + x^6).
%F 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)^4 + (w_k)^3 - 1 and Y_k = (w_k)^4 - (w_k)^3 - 1, k=1,2,3.
%F For n>1, a(2n) = a(2n-1) + a(2n-4), a(2n+1) = a(2n-1) + a(2n-2). - _Franklin T. Adams-Watters_, Jan 06 2014
%e Suppose r=3. Then
%e C_r = C_3 = {a(2*r-2, a(2*r), a(2*r+1)} = {a(4), a(6), a(7)} = {1,3,3},
%e corresponding to the entries in the third column of
%e M = (U_2)^3 =
%e (0 2 1)
%e (2 1 3)
%e (1 3 3).
%e Choose i=2 and set n = 2*r + p_i. Then a(n) = a(2*r + p_i) = a(6+0) = a(6) = 3, which equals the entry in row 2 and column 3 of M. Hence a subdivided H_(7,2,3) tile should contain a(6) = m_(2,3) = 3 H_(7,3,0) tiles.
%Y Cf. A187065, A187066, A187068, A187069, A187070.
%K nonn,easy
%O 0,6
%A _L. Edson Jeffery_, Mar 09 2011
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