A210561 Triangle of coefficients of polynomials u(n,x) jointly generated with A210562; see the Formula section.

1, 1, 2, 1, 3, 4, 1, 3, 8, 8, 1, 3, 9, 20, 16, 1, 3, 9, 26, 48, 32, 1, 3, 9, 27, 72, 112, 64, 1, 3, 9, 27, 80, 192, 256, 128, 1, 3, 9, 27, 81, 232, 496, 576, 256, 1, 3, 9, 27, 81, 242, 656, 1248, 1280, 512, 1, 3, 9, 27, 81, 243, 716, 1808, 3072, 2816, 1024, 1, 3, 9

Offset: 1

Author

Clark Kimberling, Mar 22 2012

Keywords

• nonn
• tabl
• easy

Comments

Last term in row n: 2^(n-1)

Limiting row: 3^(k-1)

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

Examples

			First five rows:
1
1...2
1...3...4
1...3...8...8
1...3...9...20...16
First three polynomials u(n,x): 1, 1 + 2x, 1 + 3x + 4x^2.
		

Links

• P. Bala, A note on the diagonals of a proper Riordan Array

Crossrefs

Cf. A210562, A208510, A004070, A048739.

Programs

Mathematica

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := x*u[n - 1, x] + x*v[n - 1, x] + 1;
v[n_, x_] := (x + 1)*u[n - 1, 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[%]  (* A210559 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]  (* A210560 *)

Formula

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

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

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

From Peter Bala, Mar 06 2017: (Start)

T(n,k) = 2*T(n-1,k-1) + T(n-2,k-1).

E.g.f. for n-th subdiagonal: exp(2*x)*(1 + x + x^2/2! + x^3/3! + ... + x^n/n!). Cf. A004070.

Riordan array (1/(1 - x), x*(2 + x)).

Row sums A048739.

(End)