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A187070
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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_(7,3,0) tiles in a subdivided H_(7,i,r) tile after linear scaling by the factor x^r, where x=sqrt((2*cos(Pi/7))^2-1).
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10
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0, 0, 1, 1, 1, 2, 3, 5, 6, 11, 14, 25, 31, 56, 70, 126, 157, 283, 353, 636, 793, 1429, 1782, 3211, 4004, 7215, 8997, 16212, 20216, 36428, 45425, 81853, 102069, 183922, 229347, 413269, 515338, 928607, 1157954, 2086561, 2601899
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OFFSET
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0,6
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COMMENTS
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Theory. (Start)
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))^2 - 1]; 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_2)^r, where
U_2=
(0 0 1)
(0 1 1)
(1 1 1).
2. The sequence. Let r>=0, and let C_r be the r-th "block" defined by C_r={a(2*r),a(2*r+1),a(2*r+2)}. Note that C_r-2*C_(r-1)-C_(r-2)+C_(r-3)={0,0,0}. Let n=2*r+i-1. Then a(n)=a(2*r+i-1)=m_(i,3), where M=(m_(i,j))=(U_2)^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)
Combining blocks A_r, B_r and C_r, from A187068, A187069 and this sequence, respectively, as matrix columns [A_r,B_r,C_r] generates the matrix (U_2)^r, and a negative index (-1)*r yields the corresponding inverse [A_(-r),B_(-r),C_(-r)]=(U_2)^(-r) of (U_2)^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 third-column entries m_(1,3) and m_(2,3) from successive matrices M=(U_2)^r.
This sequence is a trivial extension of A038196.
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LINKS
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FORMULA
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Recurrence: a(n) = 2*a(n-2) + a(n-4) - a(n-6).
G.f.: x^2*(1+x-x^2)/(1-2*x^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[(2cos(k*Pi/7))^2-1], X_k = (w_k)^3+(w_k)^2-w_k and Y_k = -(w_k)^3+(w_k)^2+w_k, k=1,2,3.
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EXAMPLE
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Suppose r=3. Then
C_r = C_3 = {a(2*r,a(2*r+1),a(2*r+2)} = {a(6),a(7),a(8)} = {3,5,6},
corresponding to the entries in the third column of
M = (U_2)^3 =
(1 2 3)
(2 4 5)
(3 5 6).
Choose i=2 and set n=2*r+i-1. Then a(n) = a(2*r+i-1) = a(6+2-1) = a(7) = 5, which equals the entry in row 2 and column 3 of M. Hence a subdivided H_(7,2,3) tile should contain a(7) = m_(2,3) = 5 H_(7,3,0) tiles.
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MATHEMATICA
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a[0] = a[1] = 0; a[2] = a[3] = a[4] = 1; a[_?Negative] = 0; a[n_] := a[n] = 2*a[n-2] + a[n-4] - a[n-6]; Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Jan 02 2013 *)
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PROG
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(PARI) x='x+O('x^50); concat([0, 0], Vec(x^2*(1+x-x^2)/(1-2*x^2-x^4+x^6))) \\ G. C. Greubel, Jul 05 2017
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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