A187068 -id:A187068 - cp's OEIS frontend

A006054 a(n) = 2*a(n-1) + a(n-2) - a(n-3), with a(0) = a(1) = 0, a(2) = 1.

0, 0, 1, 2, 5, 11, 25, 56, 126, 283, 636, 1429, 3211, 7215, 16212, 36428, 81853, 183922, 413269, 928607, 2086561, 4688460, 10534874, 23671647, 53189708, 119516189, 268550439, 603427359, 1355888968, 3046654856, 6845771321, 15382308530, 34563733525

Offset: 0

N. J. A. Sloane

Let u(k), v(k), w(k) be defined by u(1)=1, v(1)=0, w(1)=0 and u(k+1)=u(k)+v(k)+w(k), v(k+1)=u(k)+v(k), w(k+1)=u(k); then {u(n)} = 1,1,3,6,14,31,... (A006356 with an extra initial 1), {v(n)} = 0,1,2,5,11,25,... (this sequence with its initial 0 deleted) and {w(n)} = {u(n)} prefixed by an extra 0 = A077998 with an extra initial 0. - Benoit Cloitre, Apr 05 2002. Also u(k)^2+v(k)^2+w(k)^2 = u(2k). - Gary W. Adamson, Dec 23 2003
Form the graph with matrix A=[1, 1, 1; 1, 0, 0; 1, 0, 1]. Then A006054 counts walks of length n between the vertex of degree 1 and the vertex of degree 3. - Paul Barry, Oct 02 2004
Form the digraph with matrix [1,1,0; 1,0,1; 1,1,1]. A006054(n) counts walks of length n between the vertices with loops. - Paul Barry, Oct 15 2004
Nonzero terms = INVERT transform of (1, 1, 2, 2, 3, 3, ...). Example: 56 = (1, 1, 2, 5, 11, 25) dot (3, 3, 2, 2, 1, 1) = (3 + 3 + 4 + 10 + 11 + 25). - Gary W. Adamson, Apr 20 2009
-a(n+1) appears in the formula for the nonpositive powers of rho:= 2*cos(Pi/7), the ratio of the smaller diagonal in the heptagon to the side length s=2*sin(Pi/7), when expressed in the basis <1,rho,sigma>, with sigma:=rho^2-1, the ratio of the larger heptagon diagonal to the side length, as follows. rho^(-n) = C(n)*1 + C(n-1)*rho - a(n+1)*sigma, n >= 0, with C(n)=A077998(n), C(-1):=0. See the Steinbach reference, and a comment under A052547.
If, with the above notations, the power basis of the field Q(rho) is taken one has for nonpositive powers of rho, rho^(-n) = a(n+2)*1 + A077998(n-1)*rho - a(n+1)*rho^2. For nonnegative powers see A006053. See also the Steinbach reference. - Wolfdieter Lang, May 06 2011
a(n) appears also in the nonnegative powers of sigma,(defined in the above comment, where also the basis is given). See a comment in A106803.
The sequence b(n):=(-1)^(n+1)*a(n) forms the negative part (i.e., with nonpositive indices) of the sequence (-1)^n*A006053(n+1). In this way we obtain what we shall call the Ramanujan-type sequence number 2a for the argument 2*Pi/7 (see the comment to Witula's formula in A006053). We have b(n) = -2*b(n-1) + b(n-2) + b(n-3) and b(n) * 49^(1/3) = (c(1)/c(4))^(1/3) * (c(1))^(-n) + (c(2)/c(1))^(1/3) * (c(2))^(-n) + (c(4)/c(2))^(1/3) * (c(4))^(-n) = (c(2)/c(1))^(1/3) * (c(1))^(-n+1) + (c(4)/c(2))^(1/3) * (c(2))^(-n+1) + (c(1)/c(4))^(1/3) * (c(4))^(-n+1), where c(j) := 2*cos(2*Pi*j/7) (for the proof, see the comments to A215112). - Roman Witula, Aug 06 2012
(1, 1, 2, 5, 11, 25, 56, ...) * (1, 0, 1, 0, 1, ...) = the variant of A006356: (1, 1, 3, 6, 14, 31, ...). - Gary W. Adamson, May 15 2013
The limit of a(n+1)/a(n) for n -> infinity is, for all generic sequences with this recurrence of signature (2,1,-1), sigma = rho^2-1, approximately 2.246979603, the length ratio (largest diagonal)/side in the regular heptagon (7-gon). For rho = 2*cos(Pi/7) and sigma see a comment above, and the P. Steinbach reference. Proof: a(n+1)/a(n) = 2 + 1/(a(n)/a(n-1)) - 1/((a(n)/a(n-1))*(a(n-1)/a(n-2))), leading in the limit to sigma^3 -2*sigma^2 - sigma + 1, which is solved by sigma = rho^2-1, due to C(7, rho) = 0 , with the minimal polynomial C(7, x) = x^3 - x^2 - 2*x + 1 of rho (see A187360). - Wolfdieter Lang, Nov 07 2013
Numbers of straight-chain aliphatic amino acids involving single, double or triple bonds (allowing adjacent double bonds) when cis/trans isomerism is neglected. - Stefan Schuster, Apr 19 2018
Let A(r,n) be the total number of ordered arrangements of an n+r tiling of r red squares and white tiles of total length n, where the individual tile lengths can range from 1 to n. A(r,0) corresponds to a tiling of r red squares only, and so A(r,0) = 1. Also, A(r,n)=0 for n<0. Let A_1(r,n) = Sum_{j=0..n} A(r,j). Then the expansion of 1/(1 - 2*x - x^2 + x^3) is A_1(0,n) + A_1(1,n-2) + A_1(n-4) + ... = a(n) without the initial two 0's. In general, the expansion of 1/(1 - 2*x -x^k + x^(k+1)) is equal to Sum_{j>=0} A_1(j, n-j*k). - Gregory L. Simay, May 25 2018
For n>1, a(n) is the number of ways to tile a strip of length n-1 with one color of squares and dominos, two colors of trominos and quadrominos, 3 colors of 5-minos and 6-minos, and so on. - Greg Dresden and Zhiyu Zhang, Jun 26 2025

			G.f. = x^2 + 2*x^3 + 5*x^4 + 11*x^5 + 25*x^6 + 56*x^7 + 126*x^8 + 283*x^9 + ... - _Michael Somos_, Jun 25 2018

Jay Kappraff, Beyond Measure, A Guided Tour Through Nature, Myth and Number, World Scientific, 2002.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..150
C. P. de Andrade, J. P. de Oliveira Santos, E. V. P. da Silva and K. C. P. Silva, Polynomial Generalizations and Combinatorial Interpretations for Sequences Including the Fibonacci and Pell Numbers, Open Journal of Discrete Mathematics, 2013, 3, 25-32 doi:10.4236/ojdm.2013.31006. - From _N. J. A. Sloane_, Feb 20 2013
Maximilian Fichtner, K. Voigt, and S. Schuster, The tip and hidden part of the iceberg: Proteinogenic and non-proteinogenic aliphatic amino acids, Biochimica et Biophysica Acta (BBA)-General, 2016, Volume 1861, Issue 1, Part A, January 2017, Pages 3258-3269.
Brian Hopkins and Hua Wang, Restricted Color n-color Compositions, arXiv:2003.05291 [math.CO], 2020.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 434
S. Morier-Genoud, V. Ovsienko, and S. Tabachnikov, Introducing supersymmetric frieze patterns and linear difference operators, Math. Z. 281 (2015) 1061.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
R. Sachdeva and A. K. Agarwal, Combinatorics of certain restricted n-color composition functions, Discrete Mathematics, 340, (2017), 361-372.
P. Steinbach, Golden fields: a case for the heptagon, Math. Mag. 70 (1997), no. 1, 22-31.
Alexey Ustinov, Supercontinuants, arXiv:1503.04497 [math.NT], 2015.
Kai Wang, Fibonacci Numbers And Trigonometric Functions Outline, (2019).
Roman Witula, Ramanujan Type Trigonometric Formulas: The General Form for the Argument 2*Pi/7, Journal of Integer Sequences, Vol. 12 (2009), Article 09.8.5.
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 (2,1,-1).

Cf. A005578, A006053, A006356, A007583, A080937, A094790, A214683, A214699, A214779, A215112, A306334.

Row sums of A144159 and A180264.

a006054 n = a006053_list !! n
a006054_list = 0 : 0 : 1 : zipWith (+) (map (2 *) $ drop 2 a006054_list)
   (zipWith (-) (tail a006054_list) a006054_list)
-- Reinhard Zumkeller, Oct 14 2011

A006054:=z**2/(1-2*z-z**2+z**3); # Simon Plouffe in his 1992 dissertation

LinearRecurrence[{2, 1, -1}, {0, 0, 1}, 60] (* Vladimir Joseph Stephan Orlovsky, Feb 10 2012 *)

a(n):=if n<2 then 0 else if n=2 then 1 else b(n-2);
b(n):=sum(sum(binomial(j,n-3*k+2*j)*(-1)^(j-k)*binomial(k,j)*2^(-n+3*k-j),j,0,k),k,1,n); /* Vladimir Kruchinin, May 05 2011 */

x='x+O('x^66);
concat([0, 0], Vec(x^2/(1-2*x-x^2+x^3))) \\ Joerg Arndt, May 05 2011

G.f.: x^2/(1-2*x-x^2+x^3).
Sum_{k=0..n+2} a(k) = A077850(n). - Philippe Deléham, Sep 07 2006
Let M = the 3 X 3 matrix [1,1,0; 1,2,1; 0,1,2], then M^n*[1,0,0] = [A080937(n-1), A094790(n), A006054(n-1)]. E.g., M^3*[1,0,0] = [5,9,5] = [A080937(2), A094790(3), A006054(2)]. - Gary W. Adamson, Feb 15 2006
a(n) = round(k*A006356(n-1)), for n>1, where k = 0.3568958678... = 1/(1+2*cos(Pi/7)). - Gary W. Adamson, Jun 06 2008
a(n+1) = A187070(2n+1) = A187068(2n+3). - L. Edson Jeffery, Mar 10 2011
a(n+3) = Sum_{k=1..n} Sum_{j=0..k} binomial(j,n-3*k+2*j)*(-1)^(j-k)*binomial(k,j)*2^(-n+3*k-j); a(0)=0, a(1)=0, a(2)=1. - Vladimir Kruchinin, May 05 2011
7*a(n) = (c(2)-c(4))*(1+c(1))^n + (c(4)-c(1))*(1+c(2))^n + (c(1)-c(2))*(1+c(4))^n, where c(j):=2*cos(2*Pi*j/7) - for the proof see Witula et al. papers. - Roman Witula, Aug 07 2012
a(n) = -A006053(1-n) for all n in Z. - Michael Somos, Jun 25 2018

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).

Original entry on oeis.org

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.

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.

A187069 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_(7,2,0) tiles in a subdivided H_(7,i,r) tile after linear scaling by the factor x^r, where x=sqrt((2*cos(Pi/7))^2-1).

Original entry on oeis.org

0, 1, 0, 1, 1, 2, 2, 4, 5, 9, 11, 20, 25, 45, 56, 101, 126, 227, 283, 510, 636, 1146, 1429, 2575, 3211, 5786, 7215, 13001, 16212, 29213, 36428, 65641, 81853, 147494, 183922, 331416, 413269, 744685, 928607, 1673292, 2086561, 3759853, 4688460, 8448313, 10534874

Offset: 0

L. Edson Jeffery, Mar 06 2011

See A187070 for supporting theory. Define the matrix
U_2 = (0 0 1)
      (0 1 1)
      (1 1 1).
Let r>=0, and let B_r be the r-th "block" defined by B_r={a(2*r),a(2*r+1),a(2*r+2)}. Note that B_r-2*B_(r-1)-B_(r-2)+B_(r-3)={0,0,0}. Let n=2*r+i-1 and M=(m_(i,j))=(U_2)^r. Then B_r corresponds component-wise to the second column of M, and a(n)=a(2*r+i-1)=m_(i,2) gives the quantity of H_(7,2,0) tiles that should appear in a subdivided H_(7,i,r) tile.
Since a(2*r+2)=a(2*(r+1)) for all r, this sequence arises by concatenation of second-column entries m_(1,2) and m_(2,2) from successive matrices M=(U_2)^r.

			Suppose r=3.
Then B_r = B_3 = {a(2*r),a(2*r+1),a(2*r+2)} = {a(6),a(7),a(8)} = {2,4,5}, corresponding to the entries in the second column of
  M = (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_(2,2) = 4. Hence a subdivided H_(7,2,3) tile should contain a(7) = m_(2,2) = 4 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 (0,2,0,1,0,-1).

Cf. A038196, A187068, A187070, A006054.

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

my(x='x+O('x^50)); concat([0], Vec(x*(1-x^2+x^3-x^4)/(1-2*x^2-x^4+x^6))) \\ G. C. Greubel, Oct 20 2017

Recurrence: a(n) = 2*a(n-2) + a(n-4) - a(n-6).
G.f.: x*(1-x^2+x^3-x^4)/(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)^4-(w_k)^2+w_k-1 and Y_k = (w_k)^4+(w_k)^2-w_k-1, k=1,2,3.
a(2*n) = A006054(n), a(2*n+3) = A052534(n).

A106803 Expansion of x(1-x)/(1-2x-x^2+x^3).

Original entry on oeis.org

0, 1, 1, 3, 6, 14, 31, 70, 157, 353, 793, 1782, 4004, 8997, 20216, 45425, 102069, 229347, 515338, 1157954, 2601899, 5846414, 13136773, 29518061, 66326481, 149034250, 334876920, 752461609, 1690765888, 3799116465, 8536537209

Offset: 0

Roger L. Bagula, May 17 2005

Essentially a duplicate of A077998: a(n) = A077998(n-1). - Joerg Arndt, Aug 14 2015
a(n) appears in the formula for the nonnegative powers of sigma, the ratio of the smaller diagonal in the heptagon to the side length s=2*sin(Pi/7), when expressed in the basis <1,rho,sigma>, with rho = 2*cos(Pi/7), the ratio of the smaller heptagon diagonal to the side length, as follows. sigma^n = a(n-1)*1 + B(n)*rho + a(n)*sigma, n>=0, with B(n)=A006054(n). Put a(-1):= 1. See the Steinbach reference, and a comment under A052547.
a(n-1) is the top left entry of the n-th power of the 3X3 matrix [0, 1, 0; 1, 1, 1; 0, 1, 1] or of the 3X3 matrix [0, 0, 1; 0, 1, 1; 1, 1, 1]. - R. J. Mathar, Feb 03 2014

Michael De Vlieger, Table of n, a(n) for n = 0..2845
P. Steinbach, Golden fields: a case for the heptagon, Math. Mag. 70 (1997), p. 22-31.
Kai Wang, Fibonacci Numbers And Trigonometric Functions Outline, (2019).
Index entries for linear recurrences with constant coefficients, signature (2,1,-1).

Cf. A006356, A077998, A187068, A187070, A199853.

m = {{0, 0, 1}, {1, 2, 0}, {1, 1, 0}}; v[0] = {0, 1, 1}; v[n_] := m.v[n - 1]; Table[v[n][[1]], {n, 0, 30}] (* Edited and corrected by L. Edson Jeffery, Oct 18 2017 *)
RecurrenceTable[{a[1]== 0, a[2]== 1, a[3]== 1, a[n]== 2*a[n-1]  + a[n-2] - a[n-3]}, a, {n,30}] (* G. C. Greubel, Aug 14 2015 *)

concat(0,Vec((1-x)/(x^3-2*x-x^2+1)+O(x^99))) \\ Charles R Greathouse IV, Sep 25 2012

a(n) = A077998(n-1). - R. J. Mathar, Aug 07 2008
a(n) = A187070(2*n), a(n) = A187068(2*n+2). - L. Edson Jeffery, Mar 10 2011
a(n+1) = - A199853(n+1). - G. C. Greubel, Aug 14 2015
a(n) = 2*a(n-1) + a(n-2) - a(n-3), a(0)=0, a(1)=a(2)=1. - G. C. Greubel, Aug 14 2015
a(n) = A006356(n-2) for n > 1. - Georg Fischer, Oct 21 2018

Edited by N. J. A. Sloane, Aug 08 2008

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

A006054 a(n) = 2*a(n-1) + a(n-2) - a(n-3), with a(0) = a(1) = 0, a(2) = 1.

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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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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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A187069 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_(7,2,0) tiles in a subdivided H_(7,i,r) tile after linear scaling by the factor x^r, where x=sqrt((2*cos(Pi/7))^2-1).

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A106803 Expansion of x*(1-x)/(1-2*x-x^2+x^3).

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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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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)).

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).

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)).

A187069 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_(7,2,0) tiles in a subdivided H_(7,i,r) tile after linear scaling by the factor x^r, where x=sqrt((2*cos(Pi/7))^2-1).

A106803 Expansion of x(1-x)/(1-2x-x^2+x^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)).

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)).