A209563 Triangle of coefficients of polynomials u(n,x) jointly generated with A209564; see the Formula section.

1, 1, 1, 1, 3, 1, 1, 3, 6, 1, 1, 3, 8, 10, 1, 1, 3, 8, 19, 15, 1, 1, 3, 8, 21, 40, 21, 1, 1, 3, 8, 21, 53, 76, 28, 1, 1, 3, 8, 21, 55, 125, 133, 36, 1, 1, 3, 8, 21, 55, 142, 273, 218, 45, 1, 1, 3, 8, 21, 55, 144, 354, 554, 339, 55, 1, 1, 3, 8, 21, 55, 144, 375, 839, 1053

Offset

1,5

Comments

A209563: first k terms of row n are F(2),...,F(2k), where F = A000045 (Fibonacci numbers) and k=floor ((n+1)/2).

A209564: first k terms of row n are F(1), ..., F(2k-1), where k=floor ((n+2)/2).

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

Links

Table of n, a(n) for n=1..75.

Formula

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

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

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

Example

First five rows:
1
1...1
1...3...1
1...3...6...1
1...3...8...10...1
First three polynomials v(n,x): 1, 1 + x, 1 + 3x + x^2.

Mathematica

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := x*u[n - 1, x] + v[n - 1, x];
v[n_, x_] := x*u[n - 1, x] + x*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[%]   (* A209563 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]   (* A209564 *)

Crossrefs

Cf. A209564, A208510.

Sequence in context: A094644 A113046 A245541 * A308624 A133825 A365968

Adjacent sequences: A209560 A209561 A209562 * A209564 A209565 A209566

Keyword

nonn,tabl

Author

Clark Kimberling, Mar 10 2012

Status

approved