A209571 Triangle of coefficients of polynomials u(n,x) jointly generated with A209572; see the Formula section.

1, 1, 1, 1, 4, 1, 1, 4, 9, 1, 1, 4, 15, 16, 1, 1, 4, 15, 44, 25, 1, 1, 4, 15, 56, 105, 36, 1, 1, 4, 15, 56, 185, 216, 49, 1, 1, 4, 15, 56, 209, 524, 399, 64, 1, 1, 4, 15, 56, 209, 732, 1295, 680, 81, 1, 1, 4, 15, 56, 209, 780, 2303, 2864, 1089, 100, 1, 1, 4, 15, 56

Offset: 1

Author

Clark Kimberling, Mar 11 2012

Keywords

• nonn
• tabl

Comments

Penultimate number in row n is (n-1)^2, for n>1.

Combinatorial limit of row n satisfies linear recurrence

r(n)=4*r(n-1)-r(n-2) with r(1)=1 and r(2)=4. For a

discussion and guide to related arrays, see A208510.

Examples

			First five rows:
1
1...1
1...4....1
1...4....9....1
1...4....15...16...1
First three polynomials v(n,x): 1, 1 + x, 1 + 4x + x^2.
		

Crossrefs

Cf. A209572, A208510.

Programs

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_] := 2 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[%]   (* A209571 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]   (* A209572 *)

Formula

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

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

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