A187067 -id:A187067 - cp's OEIS frontend

A006053 a(n) = a(n-1) + 2*a(n-2) - a(n-3), with a(0) = a(1) = 0, a(2) = 1.

0, 0, 1, 1, 3, 4, 9, 14, 28, 47, 89, 155, 286, 507, 924, 1652, 2993, 5373, 9707, 17460, 31501, 56714, 102256, 184183, 331981, 598091, 1077870, 1942071, 3499720, 6305992, 11363361, 20475625, 36896355, 66484244, 119801329, 215873462, 388991876, 700937471

Offset: 0

N. J. A. Sloane

a(n+1) = S(n) for n>=1, where S(n) is the number of 01-words of length n, having first letter 1, in which all runlengths of 1's are odd. Example: S(4) counts 1000, 1001, 1010, 1110. See A077865. - Clark Kimberling, Jun 26 2004
For n>=1, number of compositions of n into floor(j/2) kinds of j's (see g.f.). - Joerg Arndt, Jul 06 2011
Counts walks of length n between the first and second nodes of P_3, to which a loop has been added at the end. Let A be the adjacency matrix of the graph P_3 with a loop added at the end. A is a 'reverse Jordan matrix' [0,0,1; 0,1,1; 1,1,0]. a(n) is obtained by taking the (1,2) element of A^n. - Paul Barry, Jul 16 2004
Interleaves A094790 and A094789. - Paul Barry, Oct 30 2004
a(n) appears in the formula for the nonnegative 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)*sigma, n>=0, with C(n) = A052547(n-2). See the Steinbach reference, and a comment under A052547. - Wolfdieter Lang, Nov 25 2010
If with the above notations the power basis <1,rho,rho^2> of Q(rho) is used, nonnegative powers of rho are given by rho^n  = -a(n-1)*1 + A052547(n-1)*rho + a(n)*rho^2. For negative powers see A006054. - Wolfdieter Lang, May 06 2011
-a(n-1) also appears in the formula for the nonpositive powers of sigma (see the above comment for the definition, and the Steinbach basis <1,rho,sigma>) as follows: sigma^(-n) = A(n)*1 -a(n+1)*rho -A(n-1)*sigma, with A(n) = A052547(n), A(-1):=0. - Wolfdieter Lang, Nov 25 2010

			G.f. = x^2 + x^3 + 3*x^4 + 4*x^5 + 9*x^6 + 14*x^7 + 28*x^8 + 47*x^9 + ...
Regarding the description "number of compositions of n into floor(j/2) kinds of j's," the a(6)=9 compositions of 6 are (2a, 2a, 2a), (3a, 3a), (2a, 4a), (2a, 4b), (4a, 2a), (4b, 2a), (6a), (6b), (6c). - _Bridget Tenner_, Feb 25 2022

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
R. Witula, E. Hetmaniok and D. Slota, Sums of the powers of any order roots taken from the roots of a given polynomial, Proceedings of the 15th International Conference on Fibonacci Numbers and Their Applications (2012).

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Robin Chapman and Nicholas C. Singer, Eigenvalues of a bidiagonal matrix, Amer. Math. Monthly, 111 (2004), p. 441.
Nachum Dershowitz, Between Broadway and the Hudson: A Bijection of Corridor Paths, arXiv:2006.06516 [math.CO], 2020.
Tomislav Došlić, Mate Puljiz, Stjepan Šebek, and Josip Žubrinić, On a variant of Flory model, arXiv:2210.12411 [math.CO], 2022.
Jia Huang, A coin flip game and generalizations of Fibonacci numbers, arXiv:2501.07463 [math.CO], 2025. See p. 9.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 433
Mohammed L. Nadji, Moussa Ahmia, Daniel F. Checa, and José L. Ramírez, Arndt Compositions with Restricted Parts, Palindromes, and Colored Variants, J. Int. Seq. (2025) Vol. 28, Issue 3, Article 25.3.6. See p. 11.
László Németh and Dragan Stevanović, Graph solution of system of recurrence equations, Research Gate, 2023. See Table 2 at p. 6.
László Németh and László Szalay, Sequences Involving Square Zig-Zag Shapes, J. Int. Seq., Vol. 24 (2021), Article 21.5.2.
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.
Genki Shibukawa, New identities for some symmetric polynomials and their applications, arXiv:1907.00334 [math.CA], 2019.
P. Steinbach, Golden fields: a case for the heptagon, Math. Mag. 70 (1997), no. 1, 22-31.
Kai Wang, Fibonacci Numbers And Trigonometric Functions Outline, (2019).
Roman Witula, Ramanujan Type Trigonometric Formulas: The General Form for the Argument 2Pi/7, J. Integer Seq., 12 (2009), Article 09.8.5.
Index entries for linear recurrences with constant coefficients, signature (1,2,-1).

Cf. A006054, A052547, A077865, A094789, A094790, A096975, A096976.

Cf. A120757, A187065, A187066, A187067, A214683, A215076, A215100.

a006053 n = a006053_list !! n
a006053_list = 0 : 0 : 1 : zipWith (+) (drop 2 a006053_list)
   (zipWith (-) (map (2 *) $ tail a006053_list) a006053_list)
-- Reinhard Zumkeller, Oct 14 2011

[ n eq 1 select 0 else n eq 2 select 0 else n eq 3 select 1 else Self(n-1) +2*Self(n-2) -Self(n-3): n in [1..40] ]; // Vincenzo Librandi, Aug 19 2011

a[0]:=0: a[1]:=0: a[2]:=1: for n from 3 to 40 do a[n]:=a[n-1]+2*a[n-2]-a[n-3] od:seq(a[n], n=0..40); # Emeric Deutsch
A006053:=z**2/(1-z-2*z**2+z**3); # conjectured by Simon Plouffe in his 1992 dissertation

LinearRecurrence[{1,2,-1}, {0,0,1}, 50]  (* Vladimir Joseph Stephan Orlovsky, May 25 2011 *)

{a(n) = if( n<0, n = -1-n; polcoeff( -1 / (1 - 2*x - x^2 + x^3) + x * O(x^n), n), polcoeff( x^2 / (1 - x - 2*x^2 + x^3) + x * O(x^n), n))}; /* Michael Somos, Nov 30 2014 */

@CachedFunction
def a(n): # a = A006053
    if (n<3): return (n//2)
    else: return a(n-1) + 2*a(n-2) - a(n-3)
[a(n) for n in range(41)] # G. C. Greubel, Feb 12 2023

G.f.: x^2/(1 - x - 2*x^2 + x^3). - Emeric Deutsch, Dec 14 2004
a(n) = c^(n-2) - a(n-1)*(c-1) + (1/c)*a(n-2) for n > 3 where c = 2*cos(Pi/7). Example: a(7) = 14 = c^5 - 9*(c-1) + 4/c = 18.997607... - 7.21743962... + 2.219832528... - Gary W. Adamson, Jan 24 2010
G.f.: -1 + 1/(1 - Sum_{j>=1} floor(j/2)*x^j). - Joerg Arndt, Jul 06 2011
a(n+2) = A094790(n/2+1)*(1+(-1)^n)/2 + A094789((n+1)/2)*(1-(-1)^n)/2. - Paul Barry, Oct 30 2004
First differences of A028495. - Floor van Lamoen, Nov 02 2005
a(n) = A187065(2*n+1); a(n+1) = A187066(2*n+1) = A187067(2*n). - L. Edson Jeffery, Mar 16 2011
a(n) = 2^n*(c(1)^(n-1)*(c(1)+c(2)) + c(3)^(n-1)*(c(3)+c(6)) + c(5)^(n-1)*(c(5)+c(4)) )/7, with c(j):=cos(Pi*j/7). - Herbert Kociemba, Dec 18 2011
a(n+1)*(-1)^n*49^(1/3) = (c(1)/c(4))^(1/3)*(2*c(1))^n + (c(2)/c(1))^(1/3)*(2*c(2))^n + (c(4)/c(2))^(1/3)*(2c(4))^n = (c(2)/c(1))^(1/3)*(2*c(1))^(n+1) + (c(4)/c(2))^(1/3)*(c(2))^(n+1) + (c(1)/c(4))^(1/3)*(2*c(4))^(n+1), where c(j) := cos(2Pi*j/7); for the proof, see Witula et al.'s papers. - Roman Witula, Jul 21 2012
The previous formula connects the sequence a(n) with A214683, A215076, A215100, A120757. We may call a(n) the Ramanujan-type sequence number 2 for the argument 2*Pi/7. - Roman Witula, Aug 02 2012
a(n) = -A006054(1-n) for all n in Z. - Michael Somos, Nov 30 2014
G.f.: x^2 / (1 - x / (1 - 2*x / (1 + 5*x / (2 - x / (5 - 2*x))))). - Michael Somos, Jan 20 2017
a(n) ~ r*c^n, where r=0.241717... is one of the roots of 49*x^3-7*x+1, and c=2*cos(Pi/7) (as in Gary W. Adamson's formula). - Daniel Checa, Nov 04 2022
a(2n-1) = 2*a(n+1)*a(n) - a(n)^2 - a(n-1)^2. - Richard Peterson, May 25 2023

More terms from Emeric Deutsch, Dec 14 2004

Typo in definition fixed by Reinhard Zumkeller, Oct 14 2011

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

cp's OEIS Frontend

A006053 a(n) = a(n-1) + 2*a(n-2) - a(n-3), with a(0) = a(1) = 0, a(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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A188107 Triangle T(n,k) with the coefficient [x^k] of 1/(1 - x - 2*x^2 + x^3)^(n-k+1) in row n, column k.

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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=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)).