A209999 Triangle of coefficients of polynomials u(n,x) jointly generated with A210287; see the Formula section.

1, 2, 2, 4, 6, 3, 7, 16, 13, 4, 12, 36, 44, 24, 5, 20, 76, 122, 100, 40, 6, 33, 152, 306, 332, 201, 62, 7, 54, 294, 712, 968, 783, 370, 91, 8, 88, 554, 1573, 2572, 2614, 1666, 637, 128, 9, 143, 1024, 3339, 6392, 7829, 6296, 3277, 1040, 174, 10, 232, 1864

Offset

1,2

Comments

Column 1: -1+F(n+2), where F=000045 (Fibonacci numbers)

Row sums: A003462

Alternating row sums: 1,0,1,0,1,0,1,0,1,0,1,0,...

For a discussion and guide to related arrays, see A208510.

Links

Table of n, a(n) for n=1..57.

Formula

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

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

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

Example

First five rows:
1
2....2
4....6....3
7....16...13...4
12...36...44...24...5
First three polynomials u(n,x): 1, 2 + 2x, 4 + 6x + 3x^2.

Mathematica

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := x*u[n - 1, x] + (x + 1)*v[n - 1, x] + 1;
v[n_, x_] := u[n - 1, x] + (x + 1)*v[n - 1, x] + 1;
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[%]    (* A209999 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]    (* A210287 *)

Crossrefs

Cf. A210287, A208510.

Sequence in context: A368259 A305125 A260095 * A127718 A115068 A051495

Adjacent sequences: A209996 A209997 A209998 * A210000 A210001 A210002

Keyword

nonn,tabl

Author

Clark Kimberling, Mar 23 2012

Status

approved