A132148 Triangular array T(n,k) = C(n,k)*Lucas(n-k), 0 <= k <= n.

2, 1, 2, 3, 2, 2, 4, 9, 3, 2, 7, 16, 18, 4, 2, 11, 35, 40, 30, 5, 2, 18, 66, 105, 80, 45, 6, 2, 29, 126, 231, 245, 140, 63, 7, 2, 47, 232, 504, 616, 490, 224, 84, 8, 2, 76, 423, 1044, 1512, 1386, 882, 336, 108, 9, 2, 123, 760, 2115, 3480, 3780, 2772, 1470, 480, 135, 10, 2

Offset: 0

Author

Peter Bala, Aug 17 2007

Keywords

• easy
• nonn
• tabl

Examples

			Triangle starts
   2;
   1,   2;
   3,   2,    2;
   4,   9,    3,   2;
   7,  16,   18,   4,   2;
  11,  35,   40,  30,   5,  2;
  18,  66,  105,  80,  45,  6,  2;
  ...
		

Crossrefs

Cf. A000032 (T(n,0)), A000045, A005248 (row sums), A061084 (alter. row sums), A094440.

Programs

Maple

with(combinat): lucas := n -> fibonacci(n-1) + fibonacci(n+1): T := (n,k) -> binomial(n,k)*lucas(n-k): for n from 0 to 10 do seq( T(n,k), k = 0..n) od;

Mathematica

Flatten[Table[Binomial[n,k]LucasL[n-k],{n,0,10},{k,0,n}]] (* Harvey P. Dale, Nov 06 2011 *)

Formula

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

The row polynomials L(n,x) = Sum_{k = 0..n} C(n,k)*Lucas(n-k)*x^k satisfy L(n,x)*F(n,x) = F(2*n,x), where F(n,x) = Sum_{k = 0..n} C(n,k)*Fibonacci(n-k)*x^k.

Other identities and formulas include:

L(n+1,x)^2 - L(n,x)*L(n+2,x) = -5*(x^2 + x - 1)^n;

L(n+1,x) - (x^2 + x - 1)*L(n-1,x) = 5*F(n,x) for n >= 1;

L(2*n,x) - 2*(x^2 + x - 1)^n = 5*F(n,x)^2;

L(n,2*x) = Sum_{ k = 0..n} C(n,k)*L(n-k,x)*x^k;

L(n,3*x) = Sum_{ k = 0..n} C(n,k)*L(n-k,2*x)*x^k etc.

Sum_{k = 0..n} C(n,k)*L(k,x)*F(n-k,x) = 2^n*F(n,x).

Row sums: L(n,1) = Lucas(2*n); Alternating row sums: L(n,-1) = (-1)^n*Lucas(n); L(n,1/phi) = (-1)^n*L(n,-phi) = sqrt(5)^n for n >= 1, where phi = (1 + sqrt(5))/2.

The polynomials L(n,-x) satisfy a Riemann hypothesis: the zeros of L(n,-x) lie on the vertical line Re x = 1/2 in the complex plane.

From Peter Bala, Jun 29 2016: (Start)

L(n,x) = (x + phi)^n + (x - 1/phi)^n, where phi = (1 + sqrt(5))/2. The zeros of L(n,x) are given by -1/2 - i*sqrt(5)/2*cot( (2*k - 1)*Pi/(2*n) ) for k = 1..n.

d/dx(L(n,x)) = n*L(n-1,x).

L(-n,x) = L(n,x)/(x^2 + x - 1)^n.

L(n,x - 1) = (-1)^n*L(n,-x).

L(n,x)^2 - L(2*n,x) = 2*(x^2 + x - 1)^n.

L(n,x)^3 - L(3*n,x) = 3*(x^2 + x - 1)^n*L(n,x).

L(n,x)^4 - L(4*n,x) = 4*(x^2 + x - 1)^n*L(n,x)^2 - 2*(x^2 + x - 1)^(2*n).

If n divides m and m/n is odd then L(n,x) divides L(m,x) in the polynomial ring Z[x].

L(n,x) = F(n+1,x) - (x^2 + x - 1)*F(n-1,x) = 2*F(n+1,x) - (2*x + 1)*F(n,x), where F(n,x) = Sum_{k = 0..n} binomial(n,k)*Fibonacci(n-k)*x^k denotes the n-th row polynomial of A094440 (taken with an offset of 0).

exp( Sum_{n >= 1} L(n,x)*z^n/n ) = Sum_{n >= 0} F(n+1,x)*z^n.

exp( Sum_{n >= 1} L(n,x)*L(2*n,x)*z^n/n ) = 1/( F(1,x)*F(2*x)*F(3,x) ) * Sum_{n >= 0} F(n+1,x)*F(n+2,x)* F(n+3,x)*z^n. (End)

From Peter Bala, Dec 31 2023: (Start)

For n >= 1, the n-th row polynomial L(n,x) is the numerator of 1/(n-1)! * (d/dx)^(n-1) (2*x + 1)/(1 - x - x^2).

Recurrence: for n >= 1, n*L(n+1,x) = n*(2*x + 1)*L(n,x) + (1 - x - x^2)* d/dx(L(n,x)) with L(1,x) = 2*x + 1. (End)