A008346 -id:A008346 - cp's OEIS frontend

A078024 Expansion of (1-x)/(1-2*x^2-x^3).

1, -1, 2, -1, 3, 0, 5, 3, 10, 11, 23, 32, 57, 87, 146, 231, 379, 608, 989, 1595, 2586, 4179, 6767, 10944, 17713, 28655, 46370, 75023, 121395, 196416, 317813, 514227, 832042, 1346267, 2178311, 3524576, 5702889, 9227463, 14930354, 24157815, 39088171, 63245984, 102334157

Offset: 0

N. J. A. Sloane, Nov 17 2002

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,2,1).

Cf. A000032, A000045, A008346, A112469.

List([0..50], n-> Fibonacci(n-2) + 2*(-1)^n); # G. C. Greubel, Aug 04 2019

[Fibonacci(n-2) +2*(-1)^n: n in [0..50]]; // G. C. Greubel, Aug 04 2019

LinearRecurrence[{0,2,1},{1,-1,2},50] (* Harvey P. Dale, Jan 14 2015 *)
Table[Fibonacci[n-2] +2*(-1)^n, {n,0,50}] (* G. C. Greubel, Aug 04 2019 *)

Vec((1-x)/(1-2*x^2-x^3)+O(x^50)) \\ Charles R Greathouse IV, Sep 26 2012

[fibonacci(n-2) + 2*(-1)^n for n in (0..50)] # G. C. Greubel, Aug 04 2019

a(n) = Fibonacci(n+2) - Lucas(n) + 2*(-1)^n.
a(n) = (-1)^n*A112469(n). - Philippe Deléham, Apr 19 2013
a(n) = A008346(n) - A008346(n-1), n>=1. - Philippe Deléham, Apr 19 2013
a(n) = Fibonacci(n-2) + 2*(-1)^n. - Philippe Deléham, Apr 19 2013

A104578 A Padovan convolution triangle.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 1, 2, 0, 1, 1, 2, 3, 0, 1, 2, 3, 3, 4, 0, 1, 2, 6, 6, 4, 5, 0, 1, 3, 7, 12, 10, 5, 6, 0, 1, 4, 12, 16, 20, 15, 6, 7, 0, 1, 5, 17, 30, 30, 30, 21, 7, 8, 0, 1, 7, 24, 45, 60, 50, 42, 28, 8, 9, 0, 1, 9, 36, 70, 95, 105, 77, 56, 36, 9, 10, 0, 1, 12, 50, 111, 160, 175, 168, 112, 72

Offset: 0

Paul Barry, Mar 16 2005

A Padovan convolution triangle. See A000931 for the Padovan sequence.
Row sums are tribonacci numbers A000073(n+2). Antidiagonal sums are A008346. The first columns are A000931(n+3), A228577.
From Wolfdieter Lang, Oct 30 2018: (Start)
The alternating row sums give A001057(n+1), for n >= 0.
The inverse of this Riordan triangle is given in A319203.
The row polynomials R(n, x) := Sum_{k=0..n} T(n, k)*x^k, with R(-1, x) = 0, appear in the Cayley-Hamilton formula for nonnegative powers of a 3 X 3 matrix with Det M = sigma(3;3) = x1*x2*x3 = +1, sigma(3; 2) := x1*x2 + x1*x*3 + x2*x^3  = -1 and Tr M = sigma(3; 1) = x1 + x2 = x, where x1, x2, and  x3, are the eigenvalues of M, and sigma the elementary symmetric functions, as M^n = R(n-2, x)*M^2  + (R(n-3, x) + R(n-4, x))*M + R(n-3, x)*1_3, for n >= 3, where M^0 = 1_3 is the 3 X 3 unit matrix.
For the Cayley-Hamilton formula for 3 X 3 matrices with Det M = +1, sigma(3,2) = +1 and Tr(M) = x see A321196.
(End)

			From _Wolfdieter Lang_, Oct 30 2018: (Start)
The triangle T begins:
    n\k   0  1  2  3  4  5  6  7  8  9 10 ...
    --------------------------------------
    0:    1
    1:    0  1
    2:    1  0  1
    3:    1  2  0  1
    4:    1  2  3  0  1
    5:    2  3  3  4  0  1
    6:    2  6  6  4  5  0  1
    7:    3  7 12 10  5  6  0  1
    8:    4 12 16 20 15  6  7  0  1
    9:    5 17 30 30 30 21  7  8  0  1
   10:    7 24 45 60 50 42 28  8  9  0  1
   ...
Cayley-Hamilton formula for the tribonacci Q-matrix TQ(x) =[[x,1,1], [1,0,0], [0,1,0]] with Det(TQ) = +1, sigma(3, 2) = -1, and Tr(TQ) = x. For n = 3: TQ(x)^3 = R(1, x)*TQ(x)^2  + (R(0 x) + R(-1, x))*TQ(x) + R(0, x)*1_3 = x*TQ(x)^2 + TQ(x) + 1_3. For x = 1 see also A058265 (powers of the tribonacci constant).
Recurrence: T(6, 2) = T(5, 1) + T(4, 2) + T(3, 2) = 3 + 3 + 0 = 6.
Z- and A- recurrence with A319202 = {1, 0, 1, 1, -1, -3, 0, ...}:
  T(5, 0) = 0*1 + 1*2 + 1*3 + (-1)*0 + (-3)*1 = 2; T(5,2) = 1*2 + 0*3 + 1*0 + 1*1 = 3.
Boas-Buck type recurrence with b = {0, 2, 3, ...}: T(5, 2) = ((1+2)/(5-2)) * (3*1 + 2*0 + 0*3) = 1*3 = 3.
(End)

Tomislav Došlic and Luka Podrug, Tilings of a Honeycomb Strip and Higher Order Fibonacci Numbers, arXiv:2203.11761 [math.CO], 2022.
Milan Janjić, Words and Linear Recurrences, J. Int. Seq. 21 (2018), #18.1.4.

Cf. A000073, A000931, A001057, A001608, A008346, A058265, A228577, A319202, A319203 (inverse), A321196.

T[n_, k_] /; 0 <= k <= n := T[n, k] = T[n-1, k-1] + T[n-2, k] + T[n-3, k]; T[0, 0] = 1; T[, ] = 0; Table[T[n, k], {n, 0, 12}, {k, 0, n}] (* Jean-François Alcover, Jun 11 2019 *)

# uses[riordan_array from A256893]
riordan_array( 1/(1 - x^2 - x^3), x/(1 - x^2 - x^3), 8) # Peter Luschny, Nov 09 2018

Riordan array (1/(1 - x^2 - x^3), x/(1 - x^2 - x^3)).
T(n,k) = T(n-1,k-1) + T(n-2,k) + T(n-3,k), T(0,0)=1, T(n,k)=0 if k > n or if k < n. - Philippe Deléham, Jan 08 2014
From Wolfdieter Lang, Oct 30 2018: (Start)
The Riordan property  T = (G(x), x*G(x)) with G(x)= 1/(1-x^2-x^3) implies the following.
G.f. of row polynomials R(n, x) is G(x,z) = 1/(1- x*z - z^2 - z^3).
G.f. of column sequence k:  x^k/(1 - x^2 - x^3)^(k+1), k >= 0.
Boas-Buck recurrence (see the Aug 10 2017 remark in A046521, also for the reference):
T(n, k) = ((k+1)/(n-k))*Sum_{j=k..n-1} b(n-1-j)*T(j, k), for n >= 1, k = 0,1, ..., n-1, and input T(n,n) = 1, for n >= 0. Here b(n) = [x^n]*(d/dx)log(G(x)) = A001608(n+1), for n >= 0.
Recurrences from the A- and Z- sequences (see the W. Lang link under A006232 with references), which are A(n) = A319202(n) and Z(n) = A(n+1).
  T(0, 0) = 1, T(n, k) = 0 for n < k, and
  T(n, 0) = Sum_{j=0..n-1} Z(j)*T(n-1, j), for n >= 1, and
  T(n, k) = Sum_{j=0..n-k} A(j)*T(n-1, k-1+j), for n >= m >= 1.
(End)

A181716 a(n) = a(n-1) + a(n-2) + (-1)^n, with a(0)=0 and a(1)=1.

Original entry on oeis.org

0, 1, 2, 2, 5, 6, 12, 17, 30, 46, 77, 122, 200, 321, 522, 842, 1365, 2206, 3572, 5777, 9350, 15126, 24477, 39602, 64080, 103681, 167762, 271442, 439205, 710646, 1149852, 1860497, 3010350, 4870846, 7881197, 12752042, 20633240, 33385281, 54018522, 87403802

Offset: 0

Robert G. Wilson v, Nov 07 2010

Aside from the first term, duplicate of A098600.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Aleksandar Petojević, Marjana Gorjanac Ranitović, Dragan Rastovac, and Milinko Mandić, The Golden Ratio, Factorials, and the Lambert W Function, Journal of Integer Sequences, Vol. 27 (2024), Article 24.5.7. See p. 4.
Index entries for linear recurrences with constant coefficients, signature (0,2,1).

Cf. A000032, A000045, A008346, A119282.

First differences of A014217.

I:=[0, 1, 2]; [n le 3 select I[n] else 2*Self(n-2)+Self(n-3): n in [1..40]]; // Vincenzo Librandi, Jan 09 2012

[Lucas(n-1)+(-1)^n: n in [0..40]]; // G. C. Greubel, Mar 25 2024

a[0]= 0; a[1]= 1; a[n_]:= a[n]= a[n-1] +a[n-2] +(-1)^n; Array[a,38,0]
LinearRecurrence[{0,2,1},{0,1,2},40] (* Vincenzo Librandi, Jan 09 2012 *)

[lucas_number2(n-1,1,-1)+(-1)^n for n in range(41)] # G. C. Greubel, Mar 25 2024

a(n) = a(n-1) + a(n-2) + (-1)^n.
a(n) = 2*a(n-2) + a(n-3).
a(n) - A000045(n) = A008346(n-2).
G.f.: x*(1+2*x)/(1-2*x^2-x^3). - Colin Barker, Jan 09 2012
a(n) = A000032(n-1) + (-1)^n. - G. C. Greubel, Mar 25 2024
E.g.f.: exp(x/2)*(sqrt(5)*sinh(sqrt(5)*x/2) - cosh(sqrt(5)*x/2)) + exp(-x). - Stefano Spezia, Jun 18 2024

A215037 a(n) = Sum_{k=0..n} fibonomial(k+3,3), n >= 0.

Original entry on oeis.org

1, 4, 19, 79, 339, 1431, 6072, 25707, 108922, 461362, 1954426, 8278978, 35070483, 148560678, 629313573, 2665814361, 11292572005, 47836100785, 202636977730, 858384007525, 3636173014596, 15403076054964, 65248477252164

Offset: 0

Wolfdieter Lang, Aug 09 2012

This sum is obtained from the m=2 member of the m-family of sums s(m;n) := Sum_{k=0..n} F(k+m)*F(k+1)*F(k), n>=0, given by
  (F(n+m)*F(n+2)*F(n+1) - (-1)^n*F(m)*A008346(n))/2 with A008346(n) = (F(n) + (-1)^n), where F = A000045.
The formula for s(m;n), m>=0, n>=0, follows by induction on m, using the sums for m=0 and m=1. s(0,n) = F(n+1)*(F(n+1)^2 - (-1)^n)/2 = F(n+2)*F(n+1)*F(n)/2 (see A001655(n-1)), and s(1,n) = (F(n+2)*F(n+1)^2 - (-1)^n*A008346(n))/2 (see A215038). For the formulas for s(0,n) and s(1,n) see also the link on partial summation, eqs. (6) and (7). There Sum_{k=0..n} fibonomial(k+2,k) is obtained more directly in eq. (5) with the help of the partial summation formula.

			a(3) = 2*1*1/2 + 3*2*1/2 + 5*3*2/2 + 8*5*3/2 = 1 + 3 + 15 + 60 = 79.

Wolfdieter Lang, Partial summation formula applied to sums over cubes of Fibonacci numbers.
Index entries for linear recurrences with constant coefficients, signature (4, 3, -9, 2, 1).

Cf. A000045, A008346, A001655, A215038, A363753.

LinearRecurrence[{4, 3, -9, 2, 1}, {1, 4, 19, 79, 339}, 23] (* Hans J. H. Tuenter, Jun 26 2023 *)

Let F(n) be the Fibonacci number A000045(n).
a(n) = Sum_{k=0..n} F(k+3)*F(k+2)*F(k+1)/2.
a(n) = (F(n+3)^2*F(n+2) + (-1)^n*A008346(n+1))/4, n>=0, with A008346(n) = F(n) + (-1)^n. See a comment above.
G.f.: 1/((1+x-x^2)*(1-4*x-x^2)*(1-x)) (from the g.f. of the Fibonomials A001655).
From Hans J. H. Tuenter, Jun 26 2023: (Start)
a(n) = (F(n+3)^2*F(n+2) + (-1)^n*F(n+1)-1)/4.
a(n) = (F(n+3)^3 + F(n+2)^3 + (-1)^n*F(n+1) - 2)/8.
a(n) = (F(3*n+8) + 4*(-1)^n*F(n+1) - 5)/20.
a(n) = 4*a(n-1) + 3*a(n-2) - 9*a(n-3) + 2*a(n-4) + a(n-5).
a(-n) = A363753(n-3).
(End)

A094687 Convolution of Fibonacci and Jacobsthal numbers.

Original entry on oeis.org

0, 0, 1, 2, 6, 13, 30, 64, 137, 286, 594, 1221, 2498, 5084, 10313, 20858, 42094, 84797, 170582, 342760, 688105, 1380390, 2767546, 5546037, 11109786, 22248228, 44542825, 89160674, 178442742, 357081901, 714481614, 1429477456, 2859786953

Offset: 0

Paul Barry, May 19 2004

Also convolution of A008346(n-1) and A000079(n).

Also difference of Fibonacci and Jacobsthal numbers shifted left: a(n) = A000045(n+1) - A001045(n+1). - David Callan, Jul 22 2008

			a(2) =   0 + 2*0  +  1 =   1
a(3) =   1 + 2*0  +  1 =   2
a(4) =   2 + 2*1  +  2 =   6
a(5) =   6 + 2*2  +  3 =  13
a(6) =  13 + 2*6  +  5 =  30
a(7) =  30 + 2*13 +  8 =  64
a(8) =  64 + 2*30 + 13 = 137
a(9) = 137 + 2*64 + 21 = 286
... - _Philippe Deléham_, Mar 06 2013

G. C. Greubel, Table of n, a(n) for n = 0..1000
Tamás Szakács, Linear recursive sequences and factorials, Ph. D. Thesis, Univ. Debrecen (Hungary, 2024). See p. 28.
Index entries for linear recurrences with constant coefficients, signature (2,2,-3,-2).

Cf. A000045, A000079, A001045, A008346.

a:=[0,0,1,2];; for n in [5..40] do a[n]:=2*a[n-1]+2*a[n-2] - 3*a[n-3]-2*a[n-4]; od; a; # G. C. Greubel, Mar 06 2019

I:=[0,0,1,2]; [n le 4 select I[n] else 2*Self(n-1) + 2*Self(n-2) -3*Self(n-3) -2*Self(n-4): n in [1..40]]; // G. C. Greubel, Mar 06 2019

with(combstruct):
TSU := [T, { T = Sequence(S, card > 1), S = Sequence(U, card > 0), U = Sequence(Z, card > 1)}, unlabeled]:
seq(count(TSU, size = j+2), j=0..32); # Peter Luschny, Jan 04 2020

LinearRecurrence[{2,2,-3,-2}, {0,0,1,2}, 40] (* G. C. Greubel, Mar 06 2019 *)

my(x='x+O('x^40)); concat([0,0], Vec(x^2/((1-x-x^2)*(1-x-2*x^2)))) \\ G. C. Greubel, Mar 06 2019

(x^2/((1-x-x^2)*(1-x-2*x^2))).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Mar 06 2019

G.f.: x^2/((1-x-x^2)*(1-x-2*x^2)).
a(n) = 2*a(n-1) + 2*a(n-2) - 3*a(n-3) - 2*a(n-4).
a(n) = Sum_{k=0..n} A000045(k)*A001045(n-k).
a(n+1) = a(n) + 2*a(n-1) + A000045(n). - Philippe Deléham, Mar 06 2013
a(n) = J(n+1) - F(n+1) = Sum_{k=0..n} F(k)*J(n-k), where J=A001045, F=A000045. - Yuchun Ji, Mar 05 2019
E.g.f.: exp(x/2)*(5*(3*cosh(3*x/2) - 3*cosh(sqrt(5)*x/2) + sinh(3*x/2)) - 3*sqrt(5)*sinh(sqrt(5)*x/2))/15. - Stefano Spezia, Aug 28 2025

A099925 a(n) = Lucas(n) + (-1)^n.

Original entry on oeis.org

3, 0, 4, 3, 8, 10, 19, 28, 48, 75, 124, 198, 323, 520, 844, 1363, 2208, 3570, 5779, 9348, 15128, 24475, 39604, 64078, 103683, 167760, 271444, 439203, 710648, 1149850, 1860499, 3010348, 4870848, 7881195, 12752044, 20633238, 33385283, 54018520, 87403804

Offset: 0

Ralf Stephan, Nov 02 2004

Let phi = 1/2*(1 + sqrt(5)) denote the golden ratio and put c = sum {n = 1..inf} 1/2^floor(n*(phi + 2)). The bicimal expansion of the constant c begins 0.001000100100010001001.... The binary digits are the generalized Fibonacci word A221150.

The sequence 2^a(n) for n >= 1 gives the partial quotients, apart from the first, in the simple continued fraction expansion of the constant 1/2*c = 0.06692 72114 83804 90296 ... = 1/(14 + 1/(2^0 + 1/(2^4 + 1/(2^3 + 1/(2^8 + 1/(2^10 + 1/(2^19 + ...))))))). Cf. A008346. - Peter Bala, Nov 06 2013

Colin Barker, Table of n, a(n) for n = 0..1000
Yüksel Soykan, Generalized Pell-Padovan Numbers, Asian Journal of Advanced Research and Reports (2020) Vol. 11, No. 2, 8-28, Article No. 57839.
Index entries for linear recurrences with constant coefficients, signature (0,2,1).

Cf. A000032, A008346, A098600, A068397.

CoefficientList[Series[(3 - 2 x^2)/((1 + x) (1 - x - x^2)), {x, 0, 38}], x] (* Michael De Vlieger, Sep 16 2020 *)

Vec((3-2*x^2)/((1+x)*(1-x-x^2)) + O(x^40)) \\ Colin Barker, Jun 03 2016

G.f.: (3-2*x^2)/((1+x)*(1-x-x^2)).
a(0) = 3, a(1) = 0, a(2) = 4 and a(n) = 2*a(n-2) + a(n-3) for n >= 3. - Peter Bala, Nov 06 2013
a(n) = A068397(n) - 1 for n>2.
a(n) = ((-1)^n+(1/2*(1-sqrt(5)))^n+(1/2*(1+sqrt(5)))^n). - Colin Barker, Jun 03 2016

A164925 Array, binomial(j-i,j), read by rising antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 0, 1, 1, -1, 0, 1, 1, -2, 0, 0, 1, 1, -3, 1, 0, 0, 1, 1, -4, 3, 0, 0, 0, 1, 1, -5, 6, -1, 0, 0, 0, 1, 1, -6, 10, -4, 0, 0, 0, 0, 1, 1, -7, 15, -10, 1, 0, 0, 0, 0, 1, 1, -8, 21, -20, 5, 0, 0, 0, 0, 0, 1, 1, -9, 28, -35, 15, -1, 0, 0, 0, 0, 0, 1, 1, -10, 36, -56, 35, -6, 0, 0, 0, 0, 0, 0, 1

Offset: 0

Mark Dols, Aug 31 2009

Inverse of A052509, or A004070???

			Array, A(n, k), begins as:
  1,  1,  1,   1,  1,   1,  1,  1,  1, ...
  1,  0,  0,   0,  0,   0,  0,  0,  0, ...
  1, -1,  0,   0,  0,   0,  0,  0,  0, ...
  1, -2,  1,   0,  0,   0,  0,  0,  0, ...
  1, -3,  3,  -1,  0,   0,  0,  0,  0, ...
  1, -4,  6,  -4,  1,   0,  0,  0,  0, ...
  1, -5, 10, -10,  5,  -1,  0,  0,  0, ...
  1, -6, 15, -20, 15,  -6,  1,  0,  0, ...
  1, -7, 21, -35, 35, -21,  7, -1,  0, ...
Antidiagonal triangle, T(n, k), begins as:
  1;
  1,  1;
  1,  0,  1;
  1, -1,  0,  1;
  1, -2,  0,  0,  1;
  1, -3,  1,  0,  0,  1;
  1, -4,  3,  0,  0,  0,  1;
  1, -5,  6, -1,  0,  0,  0,  1;
  1, -6, 10, -4,  0,  0,  0,  0,  1;

G. C. Greubel, Antidiagonals n = 0..50, flattened

Cf. A004070, A008346, A010892, A052509, A052992.

Cf. A109466, A130595, A164899, A164915, A164965.

A164925:= func< n,k | k eq 0 or k eq n select 1 else Binomial(2*k-n,k) >;
[A164925(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Feb 10 2023

T[n_, k_]:= If[k==0 || k==n, 1, Binomial[2*k-n, k]];
Table[T[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 10 2023 *)

{A(i, j) = if( i<0, 0, if(i==0 || j==0, 1, binomial(j-i, j)))}; /* Michael Somos, Jan 25 2012 */

def A164925(n,k): return 1 if (k==0 or k==n) else binomial(2*k-n, k)
flatten([[A164925(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Feb 10 2023

Sum_{k=0..n} T(n, k) = A164965(n). - Mark Dols, Sep 02 2009
From G. C. Greubel, Feb 10 2023: (Start)
A(n, k) = binomial(k-n, k), with A(0, k) = A(n, 0) = 1 (array).
T(n, k) = binomial(2*k-n, k), with T(n, 0) = T(n, n) = 1 (antidiagonal triangle).
Sum_{k=0..n} (-1)^k*T(n, k) = A008346(n).
Sum_{k=0..n} (-2)^k*T(n, k) = (-1)^n*A052992(n). (End)

Edited by Michael Somos, Jan 26 2012

Offset changed by G. C. Greubel, Feb 10 2023

A210797 Triangle of coefficients of polynomials u(n,x) jointly generated with A210798; see the Formula section.

Original entry on oeis.org

1, 1, 1, 1, 3, 2, 1, 4, 5, 3, 1, 6, 10, 10, 5, 1, 7, 16, 22, 18, 8, 1, 9, 24, 42, 47, 33, 13, 1, 10, 33, 69, 98, 95, 59, 21, 1, 12, 44, 108, 182, 220, 188, 105, 34, 1, 13, 56, 156, 308, 444, 472, 363, 185, 55, 1, 15, 70, 220, 490, 818, 1034, 985, 690, 324, 89, 1

Offset: 1

Clark Kimberling, Mar 26 2012

Row n starts with 1 and ends with F(n), where F=A000045 (Fibonacci numbers).
Column 2: A032766
Column 3: A001859
Row sums: A099232
Alternating row sums: A008346
For a discussion and guide to related arrays, see A208510.

			First five rows:
1
1...1
1...3...2
1...4...5....3
1...6...10...10...5
First three polynomials u(n,x): 1, 1 + x, 1 + 3x + 2x^2.

Cf. A210798, A208510.

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + (x + j)*v[n - 1, x] + c;
d[x_] := h + x; e[x_] := p + x;
v[n_, x_] := d[x]*u[n - 1, x] + e[x]*v[n - 1, x] + f;
j = 0; c = 0; h = 2; p = -1; f = 1;
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]    (* A210797 *)
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]    (* A210798 *)
Table[u[n, x] /. x -> 1, {n, 1, z}]   (* A099232 *)
Table[v[n, x] /. x -> 1, {n, 1, z}]   (* A006130 *)
Table[u[n, x] /. x -> -1, {n, 1, z}]  (* A008346 *)
Table[v[n, x] /. x -> -1, {n, 1, z}]  (* A039834 *)

u(n,x)=u(n-1,x)+x*v(n-1,x),
v(n,x)=(x+2)*u(n-1,x)+(x-1)*v(n-1,x)+1,
where u(1,x)=1, v(1,x)=1.

A210798 Triangle of coefficients of polynomials v(n,x) jointly generated with A210797; see the Formula section.

Original entry on oeis.org

1, 2, 2, 1, 3, 3, 2, 5, 7, 5, 1, 6, 12, 13, 8, 2, 8, 20, 29, 25, 13, 1, 9, 27, 51, 62, 46, 21, 2, 11, 39, 84, 125, 129, 84, 34, 1, 12, 48, 126, 224, 284, 258, 151, 55, 2, 14, 64, 182, 374, 562, 622, 505, 269, 89, 1, 15, 75, 250, 580, 1008, 1328, 1315, 969, 475

Offset: 1

Clark Kimberling, Mar 26 2012

Row n starts with A109613(n) and ends with F(n+1), where F=A000045 (Fibonacci numbers).
Column 2: A114113
For a discussion and guide to related arrays, see A208510.

			First five rows:
1
2...2
1...3...3
2...5...7....5
1...6...12...13...8
First three polynomials v(n,x): 1, 2 + 2x, 1 + 3x + 3x^2

Cf. A210797, A208510.

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + (x + j)*v[n - 1, x] + c;
d[x_] := h + x; e[x_] := p + x;
v[n_, x_] := d[x]*u[n - 1, x] + e[x]*v[n - 1, x] + f;
j = 0; c = 0; h = 2; p = -1; f = 1;
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]    (* A210797 *)
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]    (* A210798 *)
Table[u[n, x] /. x -> 1, {n, 1, z}]   (* A099232 *)
Table[v[n, x] /. x -> 1, {n, 1, z}]   (* A006130 *)
Table[u[n, x] /. x -> -1, {n, 1, z}]  (* A008346 *)
Table[v[n, x] /. x -> -1, {n, 1, z}]  (* A039834 *)

u(n,x)=u(n-1,x)+x*v(n-1,x),
v(n,x)=(x+2)*u(n-1,x)+(x-1)*v(n-1,x)+1,
where u(1,x)=1, v(1,x)=1.

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

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