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.

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

Author

Roger L. Bagula, Mar 23 2008

Keywords

• tabl
• sign

Examples

			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;
		

Links

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

Crossrefs

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

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

Programs

Magma

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