A006053 -id:A006053 - cp's OEIS frontend

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)).

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

A215569 a(n) = 3a(n-1) + 46a(n-2) + a(n-3) with a(0)=0, a(1)=14, a(2)=49.

Original entry on oeis.org

0, 14, 49, 791, 4641, 50358, 365351, 3417162, 27107990, 238878773, 1967021021, 16916594611, 141471629572, 1204545261843, 10138247340452, 85965295695706, 725459810009753, 6140921279372187, 51879880394260905, 438847479843913070, 3709157858947113027

Offset: 0

Roman Witula, Aug 16 2012

The Ramanujan-type sequence number 7 for the argument 2Pi/7 (see also A214683, A215112, A006053, A006054, A215076, A215100, A120757, A215560 for the numbers: 1, 1a, 2, 2a, 3-6 respectively). The sequence a(n) is one of the three special sequences (the remaining two are A215560 and A215572) connected with the following recurrence relation:
(c(1)^4/c(2))^(n/3) + (c(2)^4/c(4))^(n/3) + (c(4)^4/c(1))^(n/3) = at(n) + bt(n)*7^(1/3) + ct(n)*49^(1/3), where c(j):=2*cos(2*Pi*j/7), and the sequences at(n), bt(n), and ct(n) satisfy the following system of recurrence equations: at(n)=7*bt(n-2)+at(n-3),
bt(n)=7*ct(n-2)+bt(n-3), ct(n)=at(n-2)+ct(n-3), with at(0)=3, at(1)=at(2)=bt(0)=bt(1)=bt(2)=ct(0)=ct(1)=0, ct(2)=2 - for details see the Witula's first paper (see also A215560). It follows that a(n)=bt(3*n+1), at(3*n+1)=ct(3*n+1)=0, which implies the first formula below.
We note that all numbers a(n) are divided by 7.

			We have  (c(1)^4/c(2))^(4/3) + (c(2)^4/c(4))^(4/3) + (c(4)^4/c(1))^(4/3) = (2/7)*(c(1)^4/c(2))^(7/3) + (c(2)^4/c(4))^(7/3) + (c(4)^4/c(1))^(7/3)).

R. Witula, E. Hetmaniok, D. Slota, Sums of the powers of any order roots taken from the roots of a given polynomial, Proceedings of the Fifteenth International Conference on Fibonacci Numbers and Their Applications, Eger, Hungary, 2012.

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, Full Description of Ramanujan Cubic Polynomials, Journal of Integer Sequences, Vol. 13 (2010), Article 10.5.7.
Roman Witula, Ramanujan Cubic Polynomials of the Second Kind, Journal of Integer Sequences, Vol. 13 (2010), Article 10.7.5.
Roman Witula, Ramanujan Type Trigonometric Formulae, Demonstratio Math. 45 (2012) 779-796.
Index entries for linear recurrences with constant coefficients, signature (3,46,1).

Cf. A214683, A215112, A006053, A006054, A215076, A215100, A120757, A215560.

LinearRecurrence[{3,46,1},{0,14,49},30] (* Harvey P. Dale, Jan 12 2015 *)

7^(1/3)*a(n) = (c(1)^4/c(2))^(n+1/3) + (c(2)^4/c(4))^(n+1/3) + (c(4)^4/c(1))^(n+1/3) = (c(1)*(c(1)/c(2))^(1/3))^(3*n+1) + (c(2)*(c(2)/c(4))^(1/3))^(3*n+1) + (c(4)*(c(4)/c(1))^(1/3))^(3*n+1).

G.f.: (14*x+7*x^2)/(1-3*x-46*x^2-x^3).

More terms from Harvey P. Dale, Jan 12 2015

A216054 Square array T, read by antidiagonals: T(n,k) = 0 if n-k >= 1 or if k-n >= 6, T(0,0) = T(0,1) = T(0,2) = T(0,3) = T(0,4) = T(0,5) = 1, T(n,k) = T(n-1,k) + T(n,k-1).

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 0, 0, 1, 3, 2, 0, 0, 1, 4, 5, 0, 0, 0, 0, 5, 9, 5, 0, 0, 0, 0, 5, 14, 14, 0, 0, 0, 0, 0, 0, 19, 28, 14, 0, 0, 0, 0, 0, 0, 19, 47, 42, 0, 0, 0, 0, 0, 0, 0, 0, 66, 89, 42, 0, 0, 0, 0, 0, 0, 0, 0, 66, 155, 131, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 221, 286, 131, 0, 0

Offset: 0

Philippe Deléham, Mar 16 2013

A hexagon arithmetic of E. Lucas.

			Square array begins:
1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ... row n=0
0, 1, 2, 3, 4, 5, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, ... row n=1
0, 0, 2, 5, 9, 14, 19, 19, 0, 0, 0, 0, 0, 0, 0, ... row n=2
0, 0, 0, 5, 14, 28, 47, 66, 66, 0, 0, 0, 0, 0, 0, ... row n=3
0, 0, 0, 0, 14, 42, 89, 155, 221, 221, 0, 0, 0, 0, ... row n=4
0, 0, 0, 0, 0, 0, 42, 131, 286, 507, 728, 728, 0, 0, ... row n=5
0, 0, 0, 0, 0, 0, 131, 417, 924, 1652, 2380, 2380, 0, ... row n=6
...

E. Lucas, Théorie des nombres, A.Blanchard, Paris, 1958, Tome 1, p.89

Cf. A006053, A052547, A096976, A187066,

Cf. Similar sequences A216230, A216228, A216226, A216238

Clear[t]; t[0, k_ /; k <= 5] = 1; t[n_, k_] /; k < n || k > n+5 = 0; t[n_, k_] := t[n, k] = t[n-1, k] + t[n, k-1]; Table[t[n-k, k], {n, 0, 12}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Mar 18 2013 *)

T(n,n) = A080937(n).
T(n,n+1) = A080937(n+1).
T(n,n+2) = A094790(n+1).
T(n,n+3) = A094789(n+1).
T(n,n4) = T(n,n+5) = A005021(n).
Sum_{k, 0<=k<=n} T(n-k,k) = A028495(n).

A309896 Generalized Fibonacci numbers. Square array read by ascending antidiagonals. F(n,k) for n >= 0 and k >= 0.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 2, 1, 0, 1, 1, 3, 3, 1, 0, 1, 1, 4, 4, 5, 1, 0, 1, 1, 5, 5, 9, 8, 1, 0, 1, 1, 6, 6, 14, 14, 13, 1, 0, 1, 1, 7, 7, 20, 20, 28, 21, 1, 0, 1, 1, 8, 8, 27, 27, 48, 47, 34, 1, 0, 1, 1, 9, 9, 35, 35, 75, 75, 89, 55, 1, 0

Offset: 0

Peter Luschny, Aug 21 2019

			Array starts:
[0] 1, 0, 0,  0,  0,  0,   0,   0,    0,    0,    0,    0, ...
[1] 1, 1, 1,  1,  1,  1,   1,   1,    1,    1,    1,    1, ...
[2] 1, 1, 2,  3,  5,  8,  13,  21,   34,   55,   89,  144, ...
[3] 1, 1, 3,  4,  9, 14,  28,  47,   89,  155,  286,  507, ...
[4] 1, 1, 4,  5, 14, 20,  48,  75,  165,  274,  571,  988, ...
[5] 1, 1, 5,  6, 20, 27,  75, 110,  275,  429, 1001, 1637, ...
[6] 1, 1, 6,  7, 27, 35, 110, 154,  429,  637, 1638, 2548, ...
[7] 1, 1, 7,  8, 35, 44, 154, 208,  637,  910, 2548, 3808, ...
[8] 1, 1, 8,  9, 44, 54, 208, 273,  910, 1260, 3808, 5508, ...
[9] 1, 1, 9, 10, 54, 65, 273, 350, 1260, 1700, 5508, 7752, ...

Genki Shibukawa, New identities for some symmetric polynomials and their applications, arXiv:1907.00334 [math.CA], 2019.

Cf. A000007 (n=0), A000012 (n=1), A000045 (n=2), A006053 (n=3), A188021 (n=4), A231181 (n=5).

@cached_function
def F(n, k):
    if k <  0: return 0
    if k == 0: return 1
    a = sum((-1)^j*binomial(n-1-j,j  )*F(n,k-1-2*j) for j in (0..(n-1)/2))
    b = sum((-1)^j*binomial(n-1-j,j+1)*F(n,k-2-2*j) for j in (0..(n-2)/2))
    return a + b
print([F(n-k, k) for n in (0..11) for k in (0..n)])

F(n, k) = Sum_{j=0..(n-1)/2} (-1)^j*binomial(n-1-j,j)*F(n, k-1-2*j) + Sum_{j=0..(n-2)/2} (-1)^j*binomial(n-1-j, j+1)*F(n, k-2-2*j) for k > 0; F(n, 0) = 1 and F(n, k) = 0 if k < 0.

A078806 Triangular array T given by T(n,k)= number of 01-words of length n having exactly k 1's, every runlength of 1's odd and initial letter 1.

Original entry on oeis.org

1, 1, 0, 1, 1, 1, 1, 2, 1, 0, 1, 3, 2, 2, 1, 1, 4, 4, 4, 1, 0, 1, 5, 7, 7, 4, 3, 1, 1, 6, 11, 12, 10, 6, 1, 0, 1, 7, 16, 20, 20, 13, 7, 4, 1, 1, 8, 22, 32, 36, 28, 19, 8, 1, 0, 1, 9, 29, 49, 61, 56, 42, 22, 11, 5, 1, 1, 10, 37, 72, 99, 104, 86, 56, 31, 10, 1, 0, 1, 11, 46, 102, 155, 182

Offset: 1

Clark Kimberling, Dec 07 2002

Row sums: A006053.

			T(5,2) counts the words 10100, 10010, 10001. Top of triangle T:
1 = T(1,1)
1 0 = T(2,1) T(2,2)
1 1 1 = T(3,1) T(3,2) T(3,3)
1 2 1 0
1 3 2 2 1

Clark Kimberling, Binary words with restricted repetitions and associated compositions of integers, in Applications of Fibonacci Numbers, vol.10, Proceedings of the Eleventh International Conference on Fibonacci Numbers and Their Applications, William Webb, editor, Congressus Numerantium, Winnipeg, Manitoba 194 (2009) 141-151.

Cf. A078804, A078805.

A099860 A Chebyshev transform related to the knot 7_1.

Original entry on oeis.org

1, 1, 2, 2, 1, 1, 0, -1, -1, -2, -2, -1, -1, 0, 1, 1, 2, 2, 1, 1, 0, -1, -1, -2, -2, -1, -1, 0, 1, 1, 2, 2, 1, 1, 0, -1, -1, -2, -2, -1, -1, 0, 1, 1, 2, 2, 1, 1, 0, -1, -1, -2, -2, -1, -1, 0, 1, 1, 2, 2, 1, 1, 0, -1, -1, -2, -2, -1, -1, 0, 1, 1, 2, 2, 1, 1, 0, -1, -1, -2, -2

Offset: 0

Paul Barry, Oct 28 2004

The g.f. is the transform of the g.f. of A006053(n+1) under the Chebyshev mapping G(x)-> (1/(1+x^2))G(x/(1+x^2)). The denominator of the g.f. is a parameterization of the Alexander polynomial of 7_1. It is also the 14th cyclotomic polynomial.

Index entries for linear recurrences with constant coefficients, signature (1,-1,1,-1,1,-1).

Cf. A099859.

LinearRecurrence[{1,-1,1,-1,1,-1},{1,1,2,2,1,1},100] (* Harvey P. Dale, May 21 2019 *)

G.f.: (1+x^2)^2/(1-x+x^2-x^3+x^4-x^5+x^6); a(n)=sum{k=0..floor(n/2), binomial(n-k, k)(-1)^k*A006053(n-2k+1)}.

A122514 Expansion of x/(1 - 2*x^2 - x^3 + x^4).

Original entry on oeis.org

0, 1, 0, 2, 1, 3, 4, 5, 10, 11, 21, 27, 43, 64, 92, 144, 205, 316, 462, 693, 1035, 1532, 2301, 3406, 5099, 7581, 11303, 16855, 25088, 37432, 55728, 83097, 123800, 184490, 274969, 409683, 610628, 909845, 1355970, 2020635, 3011157, 4487395, 6686979

Offset: 0

Roger L. Bagula, Sep 16 2006

a(n) is the number of compositions of n+2 such that: i) the first part is odd, ii) the last part is even, and iii) no two consecutive parts have the same parity. - Geoffrey Critzer, Mar 04 2012

			a(7) = 5 because there are 5 such compositions of the integer 9: 1+8, 7+2, 3+6, 5+4, 1+2+1+2+1+2. - _Geoffrey Critzer_, Mar 04 2012

Index entries for linear recurrences with constant coefficients, signature (0,2,1,-1).

Cf. A006054, A006053.

nn = 44; a = x/(1 - x^2); b = x^2/(1 - x^2); Drop[ CoefficientList[Series[1/(1 - a b), {x, 0, nn}], x], 2] (* Geoffrey Critzer, Mar 04 2012 *)
CoefficientList[Series[x/(1-2x^2-x^3+x^4),{x,0,50}],x] (* Harvey P. Dale, Jul 17 2019 *)

A143787 Number of compositions of n into floor((3*j)/2) kinds of j's for all j>=1.

Original entry on oeis.org

1, 1, 4, 11, 33, 95, 278, 808, 2355, 6856, 19969, 58151, 169353, 493190, 1436288, 4182793, 12181260, 35474611, 103310209, 300862991, 876181998, 2551642760, 7430968523, 21640683328, 63022629465, 183536340391, 534499885849, 1556586163406, 4533135643968, 13201529892305, 38445880553108, 111963215139163, 326062542045345

Offset: 0

Joerg Arndt, Jul 06 2011

nonn

The g.f. for compositions of k_1 kinds of 1's, k_2 kinds of 2's, ..., k_j kinds of j's, ... is 1/(1-sum(j>=1, k_j * x^j )).

Paolo Xausa, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,3,-1).

Cf. A121907 (floor((3*j-1)/2)), A055841 (3*j-1), A052156 (2*j-1), A006053 (floor(j/2)), A176848 (floor(j/3)).

LinearRecurrence[{2,3,-1},{1,1,4,11},50] (* Paolo Xausa, Nov 14 2023 *)

a(n) = +2*a(n-1) +3*a(n-2) -1*a(n-3).
G.f.: ((1-x)^2*(1+x))/(1-2*x-3*x^2+x^3).
G.f.: 1/(1-sum(j>=1, floor((3*j)/2)*x^j )).

A176848 Number of compositions of n into floor(j/3) kinds of j's for all j>=1.

Original entry on oeis.org

1, 0, 0, 1, 1, 1, 3, 4, 5, 10, 15, 21, 36, 56, 83, 134, 210, 320, 505, 791, 1221, 1911, 2988, 4639, 7240, 11305, 17595, 27436, 42806, 66691, 103968, 162144, 252720, 393965, 614285, 957581, 1492791, 2327396, 3628273, 5656274, 8818275, 13747425, 21431700, 33411976, 52088551, 81204526, 126596778, 197361904, 307682405

Offset: 0

Joerg Arndt, Jul 06 2011

nonn

The g.f. for compositions of k_1 kinds of 1's, k_2 kinds of 2's, ..., k_j kinds of j's, ... is 1/(1-sum(j>=1, k_j * x^j )).

Jarib R. Acosta, Yadira Caicedo, Juan P. Poveda, José L. Ramírez, Mark Shattuck, Some New Restricted n-Color Composition Functions, J. Int. Seq., Vol. 22 (2019), Article 19.6.4.
Index entries for linear recurrences with constant coefficients, signature (1, 0, 2, -1).

Cf. A121907 (floor((3*j-1)/2)), A055841 (3*j-1), A052156 (2*j-1), A006053 (floor(j/2)), A143787 (floor((3*j)/2)).

N=66; x='x+O('x^N) /* that many terms */
gf= 1/(1-sum(j=1,N, floor(j/3)*x^j ))
Vec(gf) /* show terms */

G.f.: 1/(1-sum(j>=1, floor(j/3)*x^j )).
Conjectural g.f.: (x-1)^2*(x^2+x+1) / (x^4-2*x^3-x+1). - Colin Barker, May 15 2013
G.f.: 1 + x^3*Q(0)/2 , where Q(k) = 1 + 1/(1 - x*(4*k+1 + 2*x^2 - x^3)/( x*(4*k+3 + 2*x^2 - x^3 ) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 11 2013

cp's OEIS Frontend

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)).

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A215569 a(n) = 3*a(n-1) + 46*a(n-2) + a(n-3) with a(0)=0, a(1)=14, a(2)=49.

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A216054 Square array T, read by antidiagonals: T(n,k) = 0 if n-k >= 1 or if k-n >= 6, T(0,0) = T(0,1) = T(0,2) = T(0,3) = T(0,4) = T(0,5) = 1, T(n,k) = T(n-1,k) + T(n,k-1).

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A309896 Generalized Fibonacci numbers. Square array read by ascending antidiagonals. F(n,k) for n >= 0 and k >= 0.

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A078806 Triangular array T given by T(n,k)= number of 01-words of length n having exactly k 1's, every runlength of 1's odd and initial letter 1.

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A099860 A Chebyshev transform related to the knot 7_1.

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A122514 Expansion of x/(1 - 2*x^2 - x^3 + x^4).

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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=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)).

A215569 a(n) = 3a(n-1) + 46a(n-2) + a(n-3) with a(0)=0, a(1)=14, a(2)=49.