A010049 -id:A010049 - cp's OEIS frontend

A134410 Second-order Lucas numbers; a(n) = (2n+3)Lucas(n) - nLucas(n-1).

6, 3, 19, 27, 61, 108, 204, 367, 661, 1173, 2069, 3622, 6306, 10923, 18839, 32367, 55421, 94608, 161064, 273527, 463481, 783753, 1322869, 2229002, 3749886, 6299283, 10567579, 17705667, 29630461, 49532148, 82715844, 137997247, 230015581, 383064573, 637434389

Offset: 0

Peter Bala, Oct 24 2007

This sequence is defined by analogy with the sequence of second-order Fibonacci numbers A010049.

Inverse binomial transform is (-1)^n * a(n). - Michael Somos, Jun 02 2014

			G.f. = 6 + 3*x + 19*x^2 + 27*x^3 + 61*x^4 + 108*x^5 + 204*x^6 + 367*x^7 + ...

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

Cf. A000032, A010049, A132148.

a[ n_] := 3 (n + 1) LucasL[n] - n LucasL[n + 1]; (* Michael Somos, Jun 02 2014 *)
LinearRecurrence[{2,1,-2,-1},{6,3,19,27},40] (* Harvey P. Dale, Jun 26 2017 *)

{a(n) = (6 + 5*n) * fibonacci(n+1) - (3 + 5*n) * fibonacci(n)}; /* Michael Somos, Jun 02 2014 */

Vec((2-x)*(3-3*x+2*x^2)/(1-x-x^2)^2 + O(x^40)) \\ Colin Barker, Jun 02 2016

Defining equation: a(n) = (2n+3)*Lucas(n) - n*Lucas(n-1).
Recurrence: a(0) = 6, a(1) = 3, a(n+2) = a(n+1) + a(n) + 5*Lucas(n).
O.g.f.: (2-x)*(3-3x+2x^2)/(1-x-x^2)^2.
Set A(n) = (a(n-1) + a(n+1))/5, B(n) = a(n+1) - a(n-1). Then A(n+2) = A(n+1) + A(n) + 5*Fibonacci(n) and B(n+2) = B(n+1) + B(n) + 5*Lucas(n). The polynomials L_2(n,-x) = sum {k = 0..n} C(n,k)*a(n-k)*(-x)^k appear to satisfy a Riemann hypothesis; their zeros appear to lie on the vertical line Re x = 1/2 in the complex plane. Compare with the polynomials L(n,-x) defined in A132148.
0 = a(n)*(-a(n) - 4*a(n+1) + a(n+2)) + a(n+1)*(-3*a(n+1) + 6*a(n+2) - a(n+3)) + a(n+2)*(+3*a(n+2) - 4*a(n+3)) + a(n+3)*(+a(n+3)) for all n in Z. - Michael Somos, Jun 02 2014
a(n) = 2^(-1-n)*(6*((1-sqrt(5))^n+(1+sqrt(5))^n)+(-(-5+sqrt(5))*(1+sqrt(5))^n+(1-sqrt(5))^n*(5+sqrt(5)))*n). - Colin Barker, Jun 02 2016

A136531 Coefficients of polynomials B(x,n) = ((1+a+b)x - c)B(x,n-1) - abB(x,n-2) where B(x,0) = 1, B(x,1) = x, a=-b, b=1, c=1.

Original entry on oeis.org

1, 0, 1, 1, -1, 1, -1, 3, -2, 1, 2, -5, 6, -3, 1, -3, 10, -13, 10, -4, 1, 5, -18, 29, -26, 15, -5, 1, -8, 33, -60, 65, -45, 21, -6, 1, 13, -59, 122, -151, 125, -71, 28, -7, 1, -21, 105, -241, 338, -321, 217, -105, 36, -8, 1, 34, -185, 468, -730, 784, -609, 350, -148, 45, -9, 1

Offset: 0

Roger L. Bagula, Mar 23 2008

			Triangle begins
        k=0  k=1  k=2  k=3  k=4  k=5  k=6
  n=0:   1;
  n=1:   0,   1;
  n=2:   1,  -1,   1;
  n=3:  -1,   3,  -2,   1;
  n=4:   2,  -5,   6,  -3,   1;
  n=5:  -3,  10, -13,  10,  -4,   1;
  n=6:   5, -18,  29, -26,  15,  -5,   1;

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

Cf. A000027, A000217, A008728, A010049, A039834.

Cf. A055243, A078008, A136526, A151575, A212804.

C := ComplexField(); // T = A136531
T:= func< n,k | k eq n select 1 else Round(i^(k-n-1)*(i*Evaluate(GegenbauerPolynomial(n-k, k+1), 1/(2*i)) - Evaluate(GegenbauerPolynomial(n-k-1, k+1), 1/(2*i)))) >;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Sep 26 2022

Mathematica

(* First program *) a = -b; c = 1; b = 1; B[x_, n_]:= B[x, n]= If[n<2, x^n, ((1+a+b)*x -c)*B[x, n-1] -a*b*B[x, n-2]]; Table[CoefficientList[B[x,n], x], {n,0,10}]//Flatten (* Second program *) B[x_, n_]:= (-1)^n*(Fibonacci[n+1, 1-x] - Fibonacci[n, 1-x]); Table[CoefficientList[B[x, n], x], {n,0,16}]//Flatten (* G. C. Greubel, Sep 22 2022 *)

SageMath

def T(n,k): # T = A136531 if k==n: return 1 else: return i^(k-n-1)*(i*gegenbauer(n-k, k+1, 1/(2*i)) - gegenbauer(n-k-1, k+1, 1/(2*i))) flatten([[T(n,k) for k in range(n+1)] for n in range(12)]) # G. C. Greubel, Sep 26 2022

Formula

G.f.: (1+y) / (1 + (1-x)*y - y^2). - Kevin Ryde, Sep 21 2022

From G. C. Greubel, Sep 22 2022: (Start)

T(n, k) = coefficients of i^n*(ChebyshevU(n, (x-1)/(2*i)) - i*ChebyshevU(n-1, (x-1)/(2*i))).

T(n, k) = coefficients of (-1)^n*( Fibonacci(n+1, 1-x) - Fibonacci(n, 1-x) ).

T(n, k) = i^(k-n-1)*(i*GegenbauerC(n-k, k+1, 1/(2*i)) - GegenbauerC(n-k-1, k+1, 1/(2*i))).

T(n, 0) = Fibonacci(1-n) = (-1)^n*A212804(n) = A039834(n-1).

T(n, 1) = (-1)^(n-1)*A010049(n), n >= 1.

T(n, 2) = (-1)^n*A055243(n-2), n >= 2.

T(n, n) = 1.

T(n, n-1) = -(n-1).

T(n, n-2) = A000217(n-1), n >= 2.

T(n, n-3) = -A008728(n-3), n >= 3.

Sum_{k=0..n-2} T(n, k) = A000027(n-1), n >= 2.

Sum_{k=0..n} T(n, k) = 1.

Sum_{k=0..floor(n/2)} T(n-k, k) = A151575(n) = (-1)^n*A078008(n). (End)

Extensions

Offset corrected by Kevin Ryde, Sep 21 2022

A135830 A000012(signed) * A049310 * A000012.

Original entry on oeis.org

1, 0, 1, 2, 0, 1, 1, 3, 0, 1, 4, 1, 4, 0, 1, 4, 7, 1, 5, 0, 1, 9, 5, 11, 1, 6, 0, 1, 12, 16, 6, 16, 1, 7, 0, 1, 22, 17, 27, 7, 22, 1, 8, 0, 1, 33, 38, 23, 43, 8, 29, 1, 9, 0, 1

Offset: 1

Gary W. Adamson, Nov 30 2007

Row sums A010049: (1, 1, 3, 5, 10, 18, 33, ...).

A135831 = A000012 * A000012(signed) * A049310.

			First few rows of the triangle:
   1;
   0,  1;
   2,  0,  1;
   1,  3,  0,  1;
   4,  1,  4,  0,  1;
   4,  7,  1,  5,  0,  1;
   9,  5, 11,  1,  6,  0,  1;
  12, 16,  6, 16,  1,  7,  0,  1;
  ...

Cf. A049310, A010049, A135831.

A000012(signed) * A049310 * A000012, as infinite lower triangular matrices, where A000012(signed) = (1; -1,1; 1,-1,1; ...).

A239366 Triangular array read by rows: T(n,k) is the number of palindromic compositions of n having exactly k 1's, n>=0, 0<=k<=n.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 2, 0, 1, 0, 1, 1, 1, 1, 0, 0, 1, 3, 0, 3, 0, 1, 0, 1, 2, 1, 1, 2, 1, 0, 0, 1, 5, 0, 5, 0, 4, 0, 1, 0, 1, 3, 2, 3, 2, 1, 3, 1, 0, 0, 1, 8, 0, 10, 0, 7, 0, 5, 0, 1, 0, 1, 5, 3, 5, 5, 4, 3, 1, 4, 1, 0, 0, 1, 13, 0, 18, 0, 16, 0, 9, 0, 6, 0, 1, 0, 1, 8, 5, 10, 8, 7, 9, 5, 4, 1, 5, 1, 0, 0, 1

Offset: 0

Geoffrey Critzer, Mar 20 2014

Row sums = 2^floor(n/2).

T(n,0) = A053602(n-1) for n>0, T(n,1) = A079977(n-5) for n>4, T(2n+1,3) = A006367(n-1) for n>0, both bisections of column k=2 contain A010049. - Alois P. Heinz, Mar 21 2014

			1,
0, 1,
1, 0, 1,
1, 0, 0, 1,
2, 0, 1, 0, 1,
1, 1, 1, 0, 0, 1,
3, 0, 3, 0, 1, 0, 1,
2, 1, 1, 2, 1, 0, 0, 1,
5, 0, 5, 0, 4, 0, 1, 0, 1,
3, 2, 3, 2, 1, 3, 1, 0, 0, 1
There are eight palindromic compositions of 6: T(6,0)=3 because we have: 6, 3+3, 2+2+2.  T(6,2)=3 because we have: 1+4+1, 2+1+1+2, 1+2+2+1.  T(6,4)=1 because we have: 1+1+2+1+1. T(6,6)=1 because we have: 1+1+1+1+1+1.

Alois P. Heinz, Rows n = 0..140, flattened

b:= proc(n) option remember;  `if`(n=0, 1, expand(
      add(b(n-j)*`if`(j=1, x^2, 1), j=1..n)))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..n))
    (add(b(i)*`if`(n-2*i=1, x, 1), i=0..n/2)):
seq(T(n), n=0..30);  # Alois P. Heinz, Mar 21 2014

nn=15;Table[Take[CoefficientList[Series[((1+x)*(1-x+x^2+x*y-x^2*y))/(1-x^2-x^4-x^2*y^2+x^4*y^2),{x,0,nn}],{x,y}][[n]],n],{n,1,nn}]//Grid

G.f.: G(x,y) = ((1 + x)*(1 - x + x^2 + x*y - x^2*y))/(1 - x^2 - x^4 - x^2*y^2 + x^4*y^2). Satisfies G(x,y) = 1/(1 - x) - x + y*x + (x^2/(1 - x^2) - x^2 +y^2*x^2)*G(x,y).

cp's OEIS Frontend

A134410 Second-order Lucas numbers; a(n) = (2n+3)*Lucas(n) - n*Lucas(n-1).

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A136531 Coefficients of polynomials B(x,n) = ((1+a+b)*x - c)*B(x,n-1) - a*b*B(x,n-2) where B(x,0) = 1, B(x,1) = x, a=-b, b=1, c=1.

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A135830 A000012(signed) * A049310 * A000012.

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A239366 Triangular array read by rows: T(n,k) is the number of palindromic compositions of n having exactly k 1's, n>=0, 0<=k<=n.

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Formula

A134410 Second-order Lucas numbers; a(n) = (2n+3)Lucas(n) - nLucas(n-1).

A136531 Coefficients of polynomials B(x,n) = ((1+a+b)x - c)B(x,n-1) - abB(x,n-2) where B(x,0) = 1, B(x,1) = x, a=-b, b=1, c=1.