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

1, 3, 1, 6, 6, 1, 11, 18, 10, 1, 19, 45, 41, 15, 1, 32, 100, 130, 80, 21, 1, 53, 208, 352, 310, 141, 28, 1, 87, 413, 866, 994, 652, 231, 36, 1, 142, 794, 1991, 2828, 2429, 1253, 358, 45, 1, 231, 1490, 4358, 7391, 7871, 5348, 2248, 531, 55, 1, 375, 2745

Offset

1,2

Comments

Column 1: -2+F(n+3), where F=000045 (Fibonacci numbers)

Row sums: A003462

Alternating row sums: 1,2,1,2,1,2,1,2,1,2,1,2,1,2,...

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
3....1
6....6....1
11...18...10...1
19...45...41...15...1
First three polynomials v(n,x): 1, 3 + x , 6 + 6x + x^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. A209999, A208510.

Sequence in context: A325005 A325013 A152685 * A116412 A089511 A246257

Adjacent sequences: A210284 A210285 A210286 * A210288 A210289 A210290

Keyword

nonn,tabl

Author

Clark Kimberling, Mar 23 2012

Status

approved