A210748 Triangle of coefficients of polynomials v(n,x) jointly generated with A210747; see the Formula section.

1, 3, 2, 6, 10, 5, 11, 29, 32, 13, 19, 71, 118, 99, 34, 32, 156, 352, 437, 299, 89, 53, 322, 919, 1521, 1526, 887, 233, 87, 636, 2205, 4559, 6036, 5117, 2595, 610, 142, 1218, 4979, 12373, 20320, 22591, 16653, 7508, 1597, 231, 2279, 10751, 31233

Offset

1,2

Comments

Row n starts with -2+F(n+3) and ends with F(2n-1), where F=A000045 (Fibonacci numbers).

Row sums: A002450

Alternating row sums: 1,1,1,1,1,1,1,1,1,...(A000012)

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

Links

Table of n, a(n) for n=1..49.

Formula

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

v(n,x)=(x+1)*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....2
6....10...5
11...29...32....13
19...71...118...99...34
First three polynomials v(n,x): 1, 3 + 2x, 6 + 10x +5x^2

Mathematica

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := 2 x*u[n - 1, x] + (x + 1)*v[n - 1, x] + 1;
v[n_, x_] := (x + 1)*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[%]   (* A210747 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]   (* A210748 *)
Table[u[n, x] /. x -> 1, {n, 1, z}]  (* A002450 *)
Table[v[n, x] /. x -> 1, {n, 1, z}]  (* A002450 *)
Table[u[n, x] /. x -> -1, {n, 1, z}] (* A077925 *)
Table[v[n, x] /. x -> -1, {n, 1, z}] (* A000012 *)

Crossrefs

Cf. A210747, A208510.

Sequence in context: A365789 A072765 A210756 * A331889 A369247 A367544

Adjacent sequences: A210745 A210746 A210747 * A210749 A210750 A210751

Keyword

nonn,tabl

Author

Clark Kimberling, Mar 25 2012

Status

approved