A187069 -id:A187069 - cp's OEIS frontend

A187070 Let i be in {1,2,3}, let r >= 0 be an integer and n=2r+i-1. Then a(n)=a(2r+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).

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

Offset: 0

L. Edson Jeffery, Mar 05 2011

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.

			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.

G. C. Greubel, Table of n, a(n) for n = 0..1000
L. Edson Jeffery, Unit-primitive matrices
Roman Witula, D. Slota and A. 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 (0,2,0,1,0,-1).

Cf. A038196, A187068, A187069.

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 *)

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

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).
a(2*n)=A106803(n); a(2*n+1)=A006054(n+1); a(2*n+2)=A077998(n).
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.

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

Original entry on oeis.org

1, 0, 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

Offset: 0

L. Edson Jeffery, Mar 06 2011

(Start) See A187070 for supporting theory. Define the matrix
U_2=
(0 0 1)
(0 1 1)
(1 1 1).
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)
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.
This sequence is a nontrivial extension of both A038196 and A187070.

			(Start) Suppose r=3. Then
A_r = A_3 = {a(2*r),a(2*r+1),a(2*r+2)} = {a(6),a(7),a(8)} = {1,2,3},
corresponding to the entries in the first column of
M = m_(i,j) = (U_2)^3 =
(1 2 3)
(2 4 5)
(3 5 6).
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)

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 (0,2,0,1,0,-1).

Cf. A038196, A187069, A187070.

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 *)
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 *)

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

{a(n+2)} = A187070.
a(n) = 2*a(n-2) + a(n-4) - a(n-6).
G.f.: (1-2*x^2+x^5)/(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)^5-2*(w_k)^3+1 and Y_k = -(w_k)^5+2*(w_k)^3+1, k=1,2,3.

A187066 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=2r+p_i, and define a(-2)=0. Then, a(n)=a(2r+p_i) gives the quantity of H_(7,2,0) tiles in a subdivided H_(7,i,r) tile after linear scaling by the factor x^r, where x=sqrt(2*cos(Pi/7)).

Original entry on oeis.org

1, 0, 0, 1, 2, 1, 1, 3, 5, 4, 5, 9, 14, 14, 19, 28, 42, 47, 66, 89, 131, 155, 221, 286, 417, 507, 728, 924, 1341, 1652, 2380, 2993, 4334, 5373, 7753, 9707, 14041, 17460, 25213, 31501, 45542, 56714, 81927, 102256, 147798, 184183

Offset: 0

L. Edson Jeffery, Mar 09 2011

(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 B_r be the r-th "block" defined by B_r={a(2*r-2),a(2*r),a(2*r+1)} with a(-2)=0. Note that B_r-B_(r-1)-2*B_(r-2)+B_(r-3)={0,0,0}, with B_0={a(-2),a(0),a(1)}={0,1,0}. 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,2), where M=(m_(i,j))=(U_1)^r was defined above. Hence the block B_r corresponds component-wise to the second column of M, and a(n)=m_(i,2) gives the quantity of H_(7,2,0) tiles that should appear in a subdivided H_(7,i,r) tile. (End)
Combining blocks A_r, B_r and C_r, from A187065, this sequence 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 second-column entries m_(2,2) and m_(3,2) from successive matrices M=(U_1)^r.

			Suppose r=3. Then
B_r = B_3 = {a(2*r-2),a(2*r),a(2*r+1)}={a(4),a(6),a(7)} = {2,1,3},
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) = 1, which equals the entry in row 2 and column 2 of M. Hence a subdivided H_(7,2,3) tile should contain a(6) = m_(2,2) = 1 H_(7,2,0) tiles.

G. C. Greubel, Table of n, a(n) for n = 0..1000
L. Edson 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).

Cf. A187065, A187067, A187068, A187069, A187070.

LinearRecurrence[{0,1,0,2,0,-1},{1,0,0,1,2,1},50] (* Harvey P. Dale, Aug 16 2012 *)
CoefficientList[Series[(1 - x^2 + x^3)/(1 - x^2 - 2*x^4 + x^6), {x, 0, 50}], x] (* G. C. Greubel, Oct 20 2017 *)

my(x='x+O('x^50)); Vec((1-x^2+x^3)/(1-x^2-2*x^4+x^6)) \\ G. C. Greubel, Oct 20 2017

Recurrence: a(n) = a(n-2)+2*a(n-4)-a(n-6).
a(2*n) = A052547(n), a(2n+1) = A006053(n+1).
G.f.: (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)^5-(w_k)^3+(w_k)^2 and Y_k = -(w_k)^5+(w_k)^3+(w_k)^2, k=1,2,3.

A187065 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=2r+p_i, and define a(-2)=1. Then, a(n)=a(2r+p_i) 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)).

Original entry on oeis.org

0, 0, 1, 0, 0, 1, 2, 1, 1, 3, 5, 4, 5, 9, 14, 14, 19, 28, 42, 47, 66, 89, 131, 155, 221, 286, 417, 507, 728, 924, 1341, 1652, 2380, 2993, 4334, 5373, 7753, 9707, 14041, 17460, 25213, 31501, 45542, 56714, 81927, 102256, 147798, 184183, 266110, 331981, 479779

Offset: 0

L. Edson Jeffery, Mar 09 2011

(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.
a(n+2)=A187067(n), a(2*n)=A096976(n+1), a(2*n+1)=A006053(n).

			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.

G. C. Greubel, Table of n, a(n) for n = 0..1000
L. Edson 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).

Cf. A187066, A187067, A187068, A187069, A187070.

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

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 *)

my(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

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.

More terms from Vincenzo Librandi, Sep 18 2015

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 = 2r + p_i and define a(-2)=0. Then, a(n) = a(2r + 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)).

Original entry on oeis.org

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

Offset: 0

L. Edson Jeffery, Mar 09 2011

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.
a(2*n) = A006053(n+1), a(2*n+1) = A028495(n).

			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.

Matthew House, Table of n, a(n) for n = 0..7784
L. Edson Jeffery, Unit-primitive matrices
Roman Witula, D. Slota and A. 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).

Cf. A187065, A187066, A187068, A187069, A187070.

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.
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

A188106 Triangle T(n,k) with the coefficient [x^k] of 1/(1-2*x-x^2+x^3)^(n-k+1) in row n, column k.

Original entry on oeis.org

1, 1, 2, 1, 4, 5, 1, 6, 14, 11, 1, 8, 27, 42, 25, 1, 10, 44, 101, 119, 56, 1, 12, 65, 196, 342, 322, 126, 1, 14, 90, 335, 770, 1080, 847, 283, 1, 16, 119, 526, 1495, 2772, 3248, 2180, 636, 1, 18, 152, 777, 2625, 6032, 9366, 9414, 5521, 1429, 1, 20, 189, 1096, 4284, 11718, 22590, 30148, 26517, 13804, 3211

Offset: 0

L. Edson Jeffery, Mar 20 2011

Modified versions of the generating function for D(0)={1,2,5,11,...}=A006054(m+2), m=0,1,2,..., are related to rhombus substitution tilings (see A187068, A187069 and A187070). The columns of the triangle have generating functions 1/(1-x), 2*x/(1-x)^2, x^2*(5-x)/(1-x)^3, x^3*(11-2*x-x^2)/(1-x)^4, x^4*(25-6*x-3*x^2)/(1-x)^5, ..., for which the sum of the signed coefficients in the n-th numerator equals 2^n. The diagonals {1,2,5,...}, {1,4,14,...}, ..., are generated by successive series expansion of F(n+1,x), n=0,1,..., where F(n,x)=1/(1-2*x-x^2+x^3)^n. For example, the second diagonal is {T{1,0},T{2,1},...}={1,4,14,...}=A189426, for which successive partial sums give A189427 (excluding the zero terms). Moreover, the diagonals correspond to successive convolutions of A006054 (= the first diagonal) with itself.

			1;
1, 2;
1, 4, 5;
1, 6, 14, 11;
1, 8, 27, 42, 25;
1, 10, 44, 101, 119, 56;
1, 12, 65, 196, 342, 322, 126;
1, 14, 90, 335, 770, 1080, 847, 283;
1, 16, 119, 526, 1495 ...

Cf. A006054, A033505, A189426, A189427.

A188106 := proc(n,k) 1/(1-2*x-x^2+x^3)^(n-k+1) ; coeftayl(%,x=0,k) ; end proc:
seq(seq(A188106(n,k),k=0..n),n=0..10) ; # R. J. Mathar, Mar 22 2011

Sum_{k=0..n} T(n,k) = A033505(n).
T(n,0) = 1.
T(n,2) = A014106(n-1).
T(n,3) = (n-2)*(4*n^2+2*n-9)/3.
T(n,4) = (n-2)*(n-3)*(2*n+7)*(2*n-3)/6.

a(43) and following corrected by Georg Fischer, Oct 14 2023

cp's OEIS Frontend

A187070 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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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).

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A187066 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_(7,2,0) tiles in a subdivided H_(7,i,r) tile after linear scaling by the factor x^r, where x=sqrt(2*cos(Pi/7)).

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A187065 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)=1. Then, a(n)=a(2*r+p_i) 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)).

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Magma

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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)).

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A188106 Triangle T(n,k) with the coefficient [x^k] of 1/(1-2*x-x^2+x^3)^(n-k+1) in row n, column k.

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Extensions

A187070 Let i be in {1,2,3}, let r >= 0 be an integer and n=2r+i-1. Then a(n)=a(2r+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).

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

A187066 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=2r+p_i, and define a(-2)=0. Then, a(n)=a(2r+p_i) gives the quantity of H_(7,2,0) tiles in a subdivided H_(7,i,r) tile after linear scaling by the factor x^r, where x=sqrt(2*cos(Pi/7)).

A187065 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=2r+p_i, and define a(-2)=1. Then, a(n)=a(2r+p_i) 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)).

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 = 2r + p_i and define a(-2)=0. Then, a(n) = a(2r + 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)).