A052931 -id:A052931 - cp's OEIS frontend

A214699 a(n) = 3*a(n-2) - a(n-3) with a(0)=0, a(1)=3, a(2)=0.

0, 3, 0, 9, -3, 27, -18, 84, -81, 270, -327, 891, -1251, 3000, -4644, 10251, -16932, 35397, -61047, 123123, -218538, 430416, -778737, 1509786, -2766627, 5308095, -9809667, 18690912, -34737096, 65882403, -122902200, 232384305, -434589003, 820055115, -1536151314

Offset: 0

Roman Witula, Jul 26 2012

All a(n) are divisible by 3.
The Ramanujan-type sequence number 1 for the argument 2*Pi/9 defined by the following identity:
3^(1/3)*a(n) = (c(1)/c(2))^(1/3)*c(1)^n + (c(2)/c(4))^(1/3)*c(2)^n + (c(4)/c(1))^(1/3)*c(4)^n = -( (c(1)/c(2))^(1/3)*c(2)^(n+1) + (c(2)/c(4))^(1/3)*c(4)^(n+1) + (c(4)/c(1))^(1/3)*c(1)^(n+1) ), where c(j) := 2*cos(2*Pi*j/9).
The definitions of other Ramanujan-type sequences, for the argument of 2*Pi/9 in one's, are given in the Crossrefs section.

			We have a(2) = a(1) + a(4) = a(4) + a(7) + a(8) = -a(3) + a(5) + a(6) = 0, which implies
(c(1)/c(2))^(1/3)*c(1)^2 + (c(2)/c(4))^(1/3)*c(2)^2 + (c(4)/c(1))^(1/3)*c(4)^2 = (c(1)/c(2))^(1/3)*(c(1) + c(1)^4) + (c(2)/c(4))^(1/3)*(c(2) + c(2)^4) + (c(4)/c(1))^(1/3)*(c(4) + c(4)^4) = (c(1)/c(2))^(1/3)*(c(1)^4 + c(1)^7 + c(1)^8) + (c(2)/c(4))^(1/3)*(c(2)^4 + c(2)^7 + c(2)^8) + (c(4)/c(1))^(1/3)*(c(4)^4 + c(4)^7 + c(4)^8) = 0.
Moreover we have 3000*3^(1/3) = (c(1)/c(2))^(1/3)*c(1)^13 + (c(2)/c(4))^(1/3)*c(2)^13 + (c(4)/c(1))^(1/3)*c(4)^13. - _Roman Witula_, Oct 06 2012

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

G. C. Greubel, Table of n, a(n) for n = 0..1000
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 (0,3,-1).

Cf. A006053, A052931, A214683, A214778, A214779.

Cf. A214951, A214954, A217053, A217052, A217069.

[n le 3 select 3*(1+(-1)^n)/2 else 3*Self(n-2) - Self(n-3): n in [1..40]]; // G. C. Greubel, Jan 08 2024

LinearRecurrence[{0,3,-1}, {0,3,0}, 30]
CoefficientList[Series[3*x/(1 - 3*x^2 + x^3),{x,0,34}],x] (* James C. McMahon, Jan 09 2024 *)

def a(n): # a=A214699
    if (n<3): return 3*(n%2)
    else: return 3*a(n-2) - a(n-3)
[a(n) for n in range(41)] # G. C. Greubel, Jan 08 2024

G.f.: 3*x/(1 - 3*x^2 + x^3).
From Roman Witula, Oct 06 2012: (Start)
a(n+1) = 3*(-1)^n*A052931(n), which from recurrence relations for a(n) and A052931 can easily be proved inductively.
a(n) = -A214779(n+1) - A214779(n). (End)

A187498 Let i be in {1,2,3,4} and let r >= 0 be an integer. Let p={p_1, p_2, p_3, p_4} = {-3,0,1,2}, n=3r+p_i, and define a(-3)=0. Then a(n)=a(3r+p_i) gives the quantity of H_(9,4,0) tiles in a subdivided H_(9,i,r) tile after linear scaling by the factor Q^r, where Q=sqrt(2*cos(Pi/9)).

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 1, 1, 2, 1, 3, 3, 4, 4, 6, 5, 10, 10, 14, 15, 20, 20, 34, 35, 48, 55, 69, 75, 117, 124, 165, 199, 241, 274, 406, 440, 571, 714, 846, 988, 1417, 1560, 1988, 2548, 2977, 3536, 4965, 5525, 6953, 9061, 10490, 12597, 17443, 19551

Offset: 0

L. Edson Jeffery, Mar 16 2011

Theory: (Start)
1. Definitions. Let T_(9,j,0) denote the rhombus with sides of unit length (=1), interior angles given by the pair (j*Pi/9,(9-j)*Pi/9) and Area(T_(9,j,0))=sin(j*Pi/9), j in {1,2,3,4}. Associated with T_(9,j,0) are its angle coefficients (j, 9-j) in which one coefficient is even while the other is odd. A half-tile is created by cutting T_(9,j,0) along a line extending between its two corners with even angle coefficient; let H_(9,j,0) denote this half-tile. Similarly, a T_(9,j,r) tile is a linearly scaled version of T_(9,j,0) with sides of length Q^r and Area(T_(9,j,r))=Q^(2*r)*sin(j*Pi/9), r>=0 an integer, where Q is the positive, constant square root Q=sqrt(2*cos(Pi/9)); likewise let H_(9,j,r) denote the corresponding half-tile. Often H_(9,i,r) (i in {1,2,3,4}) can be subdivided into an integral number of each equivalence class H_(9,j,0). But regardless of whether or not H_(9,j,r) subdivides, in theory such a proposed subdivision for each j can be represented by the matrix M=(m_(i,j)), i,j=1,2,3,4, in which the entry m_(i,j) gives the quantity of H_(9,j,0) tiles that should be present in a subdivided H_(9,i,r) tile. The number Q^(2*r) (the square of the scaling factor) is an eigenvalue of M=(U_1)^r, where
U_1=
(0 1 0 0)
(1 0 1 0)
(0 1 0 1)
(0 0 1 1).
2. The sequence. Let r>=0, and let D_r be the r-th "block" defined by D_r={a(3*r-3),a(3*r),a(3*r+1),a(3*r+2)} with a(-3)=0. Note that D_r-D_(r-1)-3*D_(r-2)+2*D_(r-3)+D_(r-4)={0,0,0,0}, for r>=4, with initial conditions {D_k}={{0,0,0,1},{0,0,1,1},{0,1,1,2},{1,1,3,3}}, k=0,1,2,3. Let p={p_1,p_2,p_3,p_4}={-3,0,1,2} and n=3*r+p_i. Then a(n)=a(3*r+p_i)=m_(i,4), where M=(m_(i,j))=(U_1)^r was defined above. Hence the block D_r corresponds component-wise to the fourth column of M, and a(3*r+p_i)=m_(i,4) gives the quantity of H_(9,4,0) tiles that should appear in a subdivided H_(9,i,r) tile. (End)
Combining blocks A_r, B_r, C_r and D_r, from A187495, A187496, A187497 and this sequence, respectively, as matrix columns [A_r,B_r,C_r,D_r] generates the matrix (U_1)^r, and a negative index (-1)*r yields the corresponding inverse [A_(-r),B_(-r),C_(-r),D_(-r)]=(U_1)^(-r) of (U_1)^r.. Therefore the four sequences need not be causal.
Since U_1 is symmetric, so is M=(U_1)^r, so the block D_r also corresponds to the fourth row of M. Therefore, alternatively, for j=1,2,3,4, a(3r+p_j)=m_(4,j) gives the quantity of H_(9,j,0) tiles that should be present in a H_(9,4,r) tile.
Since a(3*r)=a(3*(r+1)-3) for all r, this sequence arises by concatenation of fourth-column entries m_(2,4), m_(3,4) and m_(4,4) (or fourth-row entries m_(4,2), m_(4,3) and m_(4,4)) from successive matrices M=(U_1)^r.

L. E. Jeffery, Unit-primitive matrices and rhombus substitution tilings, (in preparation).

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

Cf. A187495, A187496, A187497.

A052931 := proc(n) if n < 0 then 0; else coeftayl(1/(1-3*x^2-x^3),x=0,n) ; end if; end proc:
A052931a := proc(n) if n mod 3 = 0 then A052931(n/3) ; else 0 ; end if; end proc:
A057078 := proc(n) op(1+(n mod 3),[1,0,-1]) ; end proc:
A187498 := proc(n) -A057078(n) +A052931a(n) +2*A052931a(n-2) +A052931a(n-3) +3*A052931a(n-4) +2*A052931a(n-5) +A052931a(n-6) +3*A052931a(n-7) -A052931a(n-8) ; %/3 ; end proc:
seq(A187498(n),n=0..20) ; # R. J. Mathar, Mar 22 2011

CoefficientList[Series[-x^2*(1 + x)*(x^6 + 3*x^4 + 2*x^2 + 1)/((1 + x + x^2)*(x^9 + 3*x^6 - 1)), {x, 0, 1000}], x] (* G. C. Greubel, Sep 23 2017 *)

x='x+O('x^50); Vec(-x^2*(1+x)*(x^6+3*x^4+2*x^2+1)/((1+x+x^2)*(x^9+3*x^6-1))) \\ G. C. Greubel, Sep 23 2017

Recurrence: a(n) = a(n-3) +3*a(n-6) -2*a(n-9) -a(n-12), for n >= 12, with initial conditions {a(m)} = {0,0,1,0,1,1,1,1,2,1,3,3}, m=0,1,...,11.

G.f.: -x^2*(1+x)*(x^6+3*x^4+2*x^2+1) / ( (1+x+x^2)*(x^9+3*x^6-1) ).

A099097 Riordan array (1, 3+x).

Original entry on oeis.org

1, 0, 3, 0, 1, 9, 0, 0, 6, 27, 0, 0, 1, 27, 81, 0, 0, 0, 9, 108, 243, 0, 0, 0, 1, 54, 405, 729, 0, 0, 0, 0, 12, 270, 1458, 2187, 0, 0, 0, 0, 1, 90, 1215, 5103, 6561, 0, 0, 0, 0, 0, 15, 540, 5103, 17496, 19683, 0, 0, 0, 0, 0, 1, 135, 2835, 20412, 59049, 59049, 0, 0, 0, 0, 0, 0, 18, 945, 13608, 78732, 196830, 177147

Offset: 0

Paul Barry, Sep 25 2004

Row sums are A006190(n+1). Diagonal sums are A052931. The Riordan array (1, s+tx) defines T(n,k) = binomial(k,n-k)*s^k*(t/s)^(n-k). The row sums satisfy a(n) = s*a(n-1) + t*a(n-2) and the diagonal sums satisfy a(n) = s*a(n-2) + t*a(n-3).

Triangle T(n,k), 0 <= k <= n, read by rows given by [0, 1/3, -1/3, 0, 0, 0, 0, 0, ...] DELTA [3, 0, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 10 2008

			Triangle begins:
  1;
  0, 3;
  0, 1, 9;
  0, 0, 6, 27;
  0, 0, 1, 27,  81;
  0, 0, 0,  9, 108, 243;
  ...

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A027465.

Diagonals are of the form 3^n*binomial(n+m, m): A000244 (m=0), A027471 (m=1), A027472 (m=2), A036216 (m=3), A036217 (m=4), A036219 (m=5), A036220 (m=6), A036221 (m=7), A036222 (m=8), A036223 (m=9), A172362 (m=10).

Table[3^(2*k-n)*Binomial[k, n-k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, May 19 2021 *)

flatten([[3^(2*k-n)*binomial(k, n-k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 19 2021

Triangle: T(n, k) = binomial(k, n-k)*3^k*(1/3)^(n-k).
G.f. of column k: (3*x + x^2)^k.
G.f.: 1/(1 - 3*y*x - y*x^2). - Philippe Deléham, Nov 21 2011
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A006190(n+1), A135030(n+1), A181353(n+1) for x = 0,1,2,3 respectively. - Philippe Deléham, Nov 21 2011

A188048 Expansion of (1 - x^2)/(1 - 3*x^2 - x^3).

Original entry on oeis.org

1, 0, 2, 1, 6, 5, 19, 21, 62, 82, 207, 308, 703, 1131, 2417, 4096, 8382, 14705, 29242, 52497, 102431, 186733, 359790, 662630, 1266103, 2347680, 4460939, 8309143, 15730497, 29388368, 55500634, 103895601, 195890270, 367187437, 691566411, 1297452581

Offset: 0

L. Edson Jeffery, Mar 19 2011

Sequence is related to rhombus substitution tilings.

Robert Israel, Table of n, a(n) for n = 0..3286
L. Edson Jeffery, Unit-primitive matrix
Index entries for linear recurrences with constant coefficients, signature (0,3,1).

Cf. A052931.

I:=[1,0,2,1]; [n le 4 select I[n] else Self(n-1)+3*Self(n-2)-2*Self(n-3)-Self(n-4): n in [1..40]]; // Vincenzo Librandi, Jun 22 2015

F:= gfun:-rectoproc({a(n)=3*a(n-2)+a(n-3),a(0)=1,a(1)=0,a(2)=2},a(n),remember):
map(F, [$0..100]); # Robert Israel, Jun 21 2015

CoefficientList[Series[(1-x^2)/(1-3x^2-x^3),{x,0,40}],x]  (* Harvey P. Dale, Mar 31 2011 *)
LinearRecurrence[{0,3,1}, {1,0,2}, 50] (* Roman Witula, Aug 20 2012 *)

abs(polsym(1-3*x+x^3,66)/3) /* Joerg Arndt, Aug 19 2012 */

G.f.: (1 - x^2)/(1 - 3*x^2 - x^3).
a(n) = 3*a(n-2)+a(n-3), for n>=3, with a(0)=1, a(1)=0, a(2)=2.
a(n) = a(n-1)+3*a(n-2)-2*a(n-3)-a(n-4), for n>=4, with {a(k)}={1,0,2,1}, k=0,1,2,3.
a(n) = A187497(3*n+1).
a(n) = m_(3,3), where (m_(i,j)) = (U_1)^n, i,j=1,2,3,4 and U_1 is the tridiagonal unit-primitive matrix [0, 1, 0, 0; 1, 0, 1, 0; 0, 1, 0, 1; 0, 0, 1, 1].
3*(-1)^n*a(n) = A215664(n). - Roman Witula, Aug 20 2012
a(2n) = A094831(n); a(2n+1) = A094834(n). - John Blythe Dobson, Jun 20 2015
a(n) = A052931(n)-A052931(n-2). - R. J. Mathar, Nov 03 2020
a(n) = (2^n/3)*(cos^n(Pi/9) + cos^n(5*Pi/9) + cos^n(7*Pi/9)). - Greg Dresden, Sep 24 2022

A099916 Expansion of (1+x^2)^2/(1-x^3+x^6).

Original entry on oeis.org

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

Offset: 0

Paul Barry, Oct 30 2004

The denominator is the 18th cyclotomic polynomial. The g.f. is a Chebyshev transform of that of A052931, by the Chebyshev mapping g(x)->(1/(1+x^2))g(x/(1+x^2)). The reciprocal of the 18th cyclotomic polynomial A014027 is given by sum{k=0..n, A099916(n-k)(k/2+1)(-1)^(k/2)(1+(-1)^k)/2}.

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

a(n)=sum{k=0..floor(n/2), C(n-k, k)(-1)^k*sum{j=0..n-2k, C(j, n-2k-2j)3^(3j-n+2k)}}; a(n)=sum{k=0..n, A014027(n-k)C(2, k/2)(1+(-1)^k)/2}.

A117724 Triangle T(n,k) = coefficient [x^n] of x^2/(1-(k+1)*x^2-x^3) for row n, and columns k = 0..n, read by rows.

Original entry on oeis.org

0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 2, 3, 4, 5, 1, 1, 1, 1, 1, 1, 1, 4, 9, 16, 25, 36, 49, 2, 4, 6, 8, 10, 12, 14, 16, 2, 9, 28, 65, 126, 217, 344, 513, 730, 3, 12, 27, 48, 75, 108, 147, 192, 243, 300, 4, 22, 90, 268, 640, 1314, 2422, 4120, 6588, 10030, 14674

Offset: 0

Roger L. Bagula, Apr 13 2006

			The table starts:
  0;
  0,  0;
  1,  1,  1;
  0,  0,  0,  0;
  1,  2,  3,  4,   5;
  1,  1,  1,  1,   1,   1;
  1,  4,  9, 16,  25,  36,  49;
  2,  4,  6,  8,  10,  12,  14,  16;
  2,  9, 28, 65, 126, 217, 344, 513, 730;
  3, 12, 27, 48,  75, 108, 147, 192, 243, 300;

Nathaniel Johnston, Rows n = 0..50, flattened

Cf. A000931, A008346, A052931, A117716, A119282.

m:=12;
R:=PowerSeriesRing(Integers(), m+2);
A117724:= func< n, k | Coefficient(R!( x^2/(1-(k+1)*x^2-x^3) ), n) >;
[A117724(n, k): k in [0..n], n in [0..m]]; // G. C. Greubel, Jul 23 2023

t:=taylor(x^2/(1-(k+1)*x^2-x^3), x, 15):
seq(seq(coeff(t,x,n), k=0..n),n=0..12); # Nathaniel Johnston, Apr 27 2011

T[n_, k_]:= T[n, k]= Coefficient[Series[x^2/(1-(k+1)*x^2-x^3), {x,0,n+ 2}], x, n];
Table[T[n,k], {n,0,12}, {k,0,n}]//Flatten

def A117724(n, k):
    P. = PowerSeriesRing(QQ)
    return P( x^2/(1-(k+1)*x^2-x^3) ).list()[n]
flatten([[A117724(n, k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Jul 23 2023

T(n,k) = coefficient [x^n] ( x^2/(1-(k+1)*x^2-x^3) ).
T(n, 0) = A000931(n+1).
T(n, 1) = A008346(n-2) = (-1)^(n-1)*A119282(n-1).
T(n, 2) = A052931(n-2).

Sign in definition corrected, offset set to -1 by Assoc. Eds. of the OEIS, Jun 15 2010

Edited by G. C. Greubel, Jul 23 2023

A099917 Expansion of (1+x^2)^2/(1+x^3+x^6).

Original entry on oeis.org

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

Offset: 0

Paul Barry, Oct 30 2004

The denominator is the 9th cyclotomic polynomial. The g.f. is a Chebyshev transform of that of (-1)^n*A052931(n) by the Chebyshev mapping g(x)->(1/(1+x^2))g(x/(1+x^2)). The reciprocal of the 9th cyclotomic polynomial A014018 is given by sum{k=0..n, A099917(n-k)(k/2+1)(-1)^(k/2)(1+(-1)^k)/2}.

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

Cf. A099916.

a(n)=sum{k=0..floor(n/2), C(n-k, k)(-1)^k*sum{j=0..n-2k, C(j, n-2k-2j)3^k(-1/3)^(n-2k)}}; a(n)=sum{k=0..n, A014018(n-k)C(2, k/2)(1+(-1)^k)/2}.

A188022 Expansion of x(1+x) / (1-3x^2-x^3).

Original entry on oeis.org

0, 1, 1, 3, 4, 10, 15, 34, 55, 117, 199, 406, 714, 1417, 2548, 4965, 9061, 17443, 32148, 61390, 113887, 216318, 403051, 762841, 1425471, 2691574, 5039254, 9500193, 17809336, 33539833, 62928201, 118428835, 222324436, 418214706, 785402143, 1476968554

Offset: 0

L. Edson Jeffery, Mar 18 2011

Define the 4 X 4 tridiagonal unit-primitive matrix (see [Jeffery]) M=A_{9,1}=[0,1,0,0; 1,0,1,0; 0,1,0,1; 0,0,1,1]; then a(n)=[M^n](3,4)=[M^n](4,3).

Michael De Vlieger, Table of n, a(n) for n = 0..3650
Paul Barry, Centered polygon numbers, heptagons and nonagons, and the Robbins numbers, arXiv:2104.01644 [math.CO], 2021.
L. E. Jeffery, Unit-primitive matrices
Kai Wang, Fibonacci Numbers And Trigonometric Functions Outline, (2019).
Index entries for linear recurrences with constant coefficients, signature (0,3,1).

Cf. A094832 (bisection), A094833 (bisection).

Cf. A052931, A187498.

LinearRecurrence[{0, 3, 1}, {0, 1, 1}, 36] (* or *)
CoefficientList[Series[x (1 + x)/(1 - 3 x^2 - x^3), {x, 0, 35}], x] (* Michael De Vlieger, Mar 10 2020 *)

a(n) = 3*a(n-2)+a(n-3).
a(n) = A187498(3*n+1).
a(n) = A052931(n-2)+A052931(n-1). - R. J. Mathar, Mar 22 2011

cp's OEIS Frontend

A214699 a(n) = 3*a(n-2) - a(n-3) with a(0)=0, a(1)=3, a(2)=0.

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A187498 Let i be in {1,2,3,4} and let r >= 0 be an integer. Let p={p_1, p_2, p_3, p_4} = {-3,0,1,2}, n=3*r+p_i, and define a(-3)=0. Then a(n)=a(3*r+p_i) gives the quantity of H_(9,4,0) tiles in a subdivided H_(9,i,r) tile after linear scaling by the factor Q^r, where Q=sqrt(2*cos(Pi/9)).

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A099097 Riordan array (1, 3+x).

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A188048 Expansion of (1 - x^2)/(1 - 3*x^2 - x^3).

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A099916 Expansion of (1+x^2)^2/(1-x^3+x^6).

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A117724 Triangle T(n,k) = coefficient [x^n] of x^2/(1-(k+1)*x^2-x^3) for row n, and columns k = 0..n, read by rows.

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A099917 Expansion of (1+x^2)^2/(1+x^3+x^6).

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A187498 Let i be in {1,2,3,4} and let r >= 0 be an integer. Let p={p_1, p_2, p_3, p_4} = {-3,0,1,2}, n=3r+p_i, and define a(-3)=0. Then a(n)=a(3r+p_i) gives the quantity of H_(9,4,0) tiles in a subdivided H_(9,i,r) tile after linear scaling by the factor Q^r, where Q=sqrt(2*cos(Pi/9)).

A188022 Expansion of x(1+x) / (1-3x^2-x^3).