A092879 Triangle of coefficients of the product of two consecutive Fibonacci polynomials.

1, 1, 1, 1, 3, 2, 1, 5, 7, 2, 1, 7, 16, 13, 3, 1, 9, 29, 40, 22, 3, 1, 11, 46, 91, 86, 34, 4, 1, 13, 67, 174, 239, 166, 50, 4, 1, 15, 92, 297, 541, 553, 296, 70, 5, 1, 17, 121, 468, 1068, 1461, 1163, 496, 95, 5, 1, 19, 154, 695, 1912, 3300, 3544, 2269, 791, 125, 6, 1, 21, 191

Offset: 0

Author

Michael Somos, Mar 10 2004

Keywords

• nonn
• tabl

Comments

The Fibonacci polynomials are defined by F(0,x) = 1, F(1,x) = 1 and F(n, x) = F(n-1, x) + x*F(n-2, x).

This is also the reflected triangle of coefficients of the polynomials defined by the recursion: c0=-1; p(x, n) = (2 + c0 - x)*p(x, n - 1) + (-1 - c0*(2 - x))*p(x, n - 2) + c0*p(x, n - 3). - Roger L. Bagula, Apr 09 2008

Examples

			Triangle begins;
  1;
  1,1;
  1,3,2;
  1,5,7,2;
  1,7,16,13,3;
  1,9,29,40,22,3;
  ...
F(3,x) = 1 + 2*x and F(4,x) = 1 + 3*x + x^2 so F(3,x)*F(4,x)=(1 + 3*x + x^2)*(1 + 2*x) = 1 + 5*x + 7*x^2 + 2*x^3 leads to T(3,k) = [1,5,7,2].
		

Crossrefs

Row sums are A001654(n+1).

Cf. A109954, A136674.

Programs

Maple

T:=proc(n,k): add((-1)^(i+k)*binomial(i+2*n-2*k+1,i), i=0..k) end: seq(seq(T(n,k), k=0..n), n=0..10); # Johannes W. Meijer, Jul 20 2011
T:=proc(n,k): coeff(F(n, x)*F(n+1, x), x, k) end: F:=proc(n, x) option remember: if n=0 then 1 elif n=1 then 1 else procname(n-1, x) + x*procname(n-2, x) fi: end: seq(seq(T(n,k), k=0..n), n=0..10); # Johannes W. Meijer, Jul 20 2011

Mathematica

c0 = -1; p[x, -1] = 0; p[x, 0] = 1; p[x, 1] = 2 - x + c0; p[x_, n_] :=p[x, n] = (2 + c0 -x)*p[x, n - 1] + (-1 - c0 (2 - x))*p[x, n - 2] + c0*p[x, n - 3]; Table[ExpandAll[p[x, n]], {n, 0, 10}]; a = Table[Reverse[CoefficientList[p[x, n], x]], {n, 0, 10}]; Flatten[a] (* Roger L. Bagula, Apr 09 2008 *)

PARI

T(n,k)=local(m);if(k<0 || k>n,0,n++; m=contfracpnqn(matrix(2,n,i,j,x)); polcoeff(m[1,1]*m[2,1]/x^n,n-k))

Formula

From Johannes W. Meijer, Jul 20 2011: (Start)

T(n, k) = Sum_{i=0..k} (-1)^(i+k)*binomial(i+2*n-2*k+1, i).

T(n, k) = A035317(2*n-k, k) = A158909(n, n-k.) (End)

T(n,k) = T(n-1,k) + T(n-1,k-1) + T(n-2,k-1) + T(n-2,k-2) - T(n-3,k-3), T(0,0) = 1, T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Nov 12 2013

Extensions

Edited and information added by Johannes W. Meijer, Jul 20 2011