|
| |
|
|
A187067
|
|
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)).
|
|
5
|
|
|
|
0, 1, 1, 1, 1, 2, 3, 3, 4, 6, 9, 10, 14, 19, 28, 33, 47, 61, 89, 108, 155, 197, 286, 352, 507, 638, 924, 1145, 1652, 2069, 2993, 3721, 5373, 6714, 9707, 12087, 17460, 21794, 31501, 39254, 56714, 70755, 102256
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,6
|
|
|
COMMENTS
|
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)); 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
U_1=
(0 1 0)
(1 0 1)
(0 1 1).
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)
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.
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.
|
|
|
LINKS
|
|
|
|
FORMULA
|
Recurrence: a(n) = a(n-2) + 2*a(n-4) - a(n-6).
G.f.: x(1 + x - x^4)/(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)^4 + (w_k)^3 - 1 and Y_k = (w_k)^4 - (w_k)^3 - 1, k=1,2,3.
|
|
|
EXAMPLE
|
Suppose r=3. Then
C_r = C_3 = {a(2*r-2, a(2*r), a(2*r+1)} = {a(4), a(6), a(7)} = {1,3,3},
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) = 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.
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,easy
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
