A208333 Triangle of coefficients of polynomials v(n,x) jointly generated with A208332; see the Formula section.

1, 0, 4, 0, 2, 10, 0, 2, 6, 28, 0, 2, 6, 24, 76, 0, 2, 6, 28, 80, 208, 0, 2, 6, 32, 100, 264, 568, 0, 2, 6, 36, 120, 360, 840, 1552, 0, 2, 6, 40, 140, 464, 1232, 2624, 4240, 0, 2, 6, 44, 160, 576, 1680, 4128, 8064, 11584, 0, 2, 6, 48, 180, 696, 2184, 5952

Offset: 1

Author

Clark Kimberling, Feb 26 2012

Keywords

• nonn
• tabl

Comments

As triangle T(n,k) with 0 <= k <= n, it is (0, 1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (4, -3/2, -1/2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Feb 28 2012

Examples

			First five rows:
  1;
  0,  4;
  0,  2, 10;
  0,  2,  6, 28;
  0,  2,  6, 24, 76;
First five polynomials u(n,x):
  1
      4x
      2x + 10x^2
      2x +  6x^2 + 28x^3
      2x +  6x^2 + 24x^3 + 76x^4.
		

Crossrefs

Cf. A003946, A026150, A208332.

Programs

Mathematica

u[1, x_] := 1; v[1, x_] := 1; z = 13;
u[n_, x_] := u[n - 1, x] + x*v[n - 1, x];
v[n_, x_] := 2 x*u[n - 1, x] + 2 x*v[n - 1, x];
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[%]  (* A208332 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]  (* A208333 *)

Formula

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

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

where u(1,x)=1, v(1,x)=1.

From Philippe Deléham, Feb 28 2012: (Start)

As triangle with 0 <= k <= n:

T(n,k) = T(n-1,k) + 2*T(n-1,k-1) - 2*T(n-2,k-1) + 2*T(n-2,k-2) with T(0,0) = 1, T(1,0) = 0, T(1,1) = 4 and T(n,k) = 0 if k < 0 or if k > n.

G.f.: (1-x+2*y*x)/(1-x-2*y*x+2*y*x^2-2*y^2*x^2).

T(n,n) = A026150(n+1).

Sum_{k=0..n} T(n,k) = A003946(n). (End)