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A187068 Let i be in {1,2,3}, let r >= 0 be an integer and n=2*r+i-1. Then a(n)=a(2*r+i-1) 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))^2-1).

%I A187068 #77 Feb 12 2025 05:46:19
%S A187068 1,0,0,0,1,1,1,2,3,5,6,11,14,25,31,56,70,126,157,283,353,636,793,1429,
%T A187068 1782,3211,4004,7215,8997,16212,20216,36428,45425,81853,102069,183922,
%U A187068 229347,413269,515338,928607,1157954,2086561,2601899
%N A187068 Let i be in {1,2,3}, let r >= 0 be an integer and n=2*r+i-1. Then a(n)=a(2*r+i-1) 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))^2-1).
%C A187068 (Start) See A187070 for supporting theory. Define the matrix
%C A187068 U_2=
%C A187068 (0 0 1)
%C A187068 (0 1 1)
%C A187068 (1 1 1).
%C A187068 Let r>=0, and let A_r be the r-th "block" defined by A_r={a(2*r),a(2*r+1),a(2*r+2)}. Note that A_r-2*A_(r-1)-A_(r-2)+A_(r-3)={0,0,0}. Let n=2*r+i-1 and M=(m_(i,j))=(U_2)^r. Then A_r corresponds component-wise to the first column of M, and a(n)=a(2*r+i-1)=m_(i,1) gives the quantity of H_(7,1,0) tiles that should appear in a subdivided H_(7,i,r) tile.  (End)
%C A187068 Since a(2*r+2)=a(2*(r+1)) for all r, this sequence arises by concatenation of first-column entries m_(1,1) and m_(2,1) from successive matrices M=(U_2)^r.
%C A187068 This sequence is a nontrivial extension of both A038196 and A187070.
%H A187068 G. C. Greubel, <a href="/A187068/b187068.txt">Table of n, a(n) for n = 0..1000</a>
%H A187068 L. E. Jeffery, <a href="/wiki/User:L._Edson_Jeffery/Unit-Primitive_Matrices">Unit-primitive matrices</a>
%H A187068 Roman Witula, Damian Slota and Adam 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 A187068 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,2,0,1,0,-1).
%F A187068 {a(n+2)} = A187070.
%F A187068 a(n) = 2*a(n-2) + a(n-4) - a(n-6).
%F A187068 G.f.: (1-2*x^2+x^5)/(1-2*x^2-x^4+x^6).
%F A187068 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[(2cos(k*Pi/7))^2-1], X_k = (w_k)^5-2*(w_k)^3+1 and Y_k = -(w_k)^5+2*(w_k)^3+1, k=1,2,3.
%e A187068 (Start) Suppose r=3. Then
%e A187068 A_r = A_3 = {a(2*r),a(2*r+1),a(2*r+2)} = {a(6),a(7),a(8)} = {1,2,3},
%e A187068 corresponding to the entries in the first column of
%e A187068 M = m_(i,j) = (U_2)^3 =
%e A187068 (1 2 3)
%e A187068 (2 4 5)
%e A187068 (3 5 6).
%e A187068 Suppose i=2. Setting n=2*r+i-1, then a(n) = a(2*r+i-1) = a(6+2-1) = a(7) = m_(2,1) = 2. Hence a subdivided H_(7,2,3) tile should contain a(7) = m_(2,1) = 2 H_(7,1,0) tiles. (End)
%t A187068 a[0] = 1; a[1] = a[2] = a[3] = 0; a[4] = a[5] = 1; a[_?Negative] = 0; a[n_] := a[n] = 2*a[n-2] + a[n-4] - a[n-6]; Table[a[n], {n, 0, 42}] (* _Jean-François Alcover_, Jan 02 2013 *)
%t A187068 CoefficientList[Series[(1 - 2*x^2 + x^5)/(1 - 2*x^2 - x^4 + x^6), {x, 0, 50}], x] (* _G. C. Greubel_, Jul 06 2017 *)
%o A187068 (PARI) x='x+O('x^50); Vec((1-2*x^2+x^5)/(1-2*x^2-x^4+x^6)) \\ _G. C. Greubel_, Jul 06 2017
%Y A187068 Cf. A038196, A187069, A187070.
%K A187068 nonn,easy
%O A187068 0,8
%A A187068 _L. Edson Jeffery_, Mar 06 2011