A210219 Triangle of coefficients of polynomials u(n,x) jointly generated with A210220; see the Formula section.

1, 2, 1, 3, 4, 1, 4, 9, 7, 1, 5, 16, 22, 11, 1, 6, 25, 50, 46, 16, 1, 7, 36, 95, 130, 86, 22, 1, 8, 49, 161, 295, 296, 148, 29, 1, 9, 64, 252, 581, 791, 610, 239, 37, 1, 10, 81, 372, 1036, 1792, 1897, 1163, 367, 46, 1, 11, 100, 525, 1716, 3612, 4900, 4166, 2083, 541, 56, 1

Offset: 1

Author

Clark Kimberling, Mar 19 2012

Keywords

• nonn
• tabl

Comments

First two terms in row n: n,(n-1)^2; last term: 1.

Period of alternating row sums: (1,1,0).

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

Apparently this is A071920 without the marginal zeros, read by downwards antidiagonals, or T(n,k) = A071922(n,k). - R. J. Mathar, May 17 2014

Examples

			First five rows:
  1
  2...1
  3...4....1
  4...9....7....1
  5...16...22...11...1
First three polynomials u(n,x): 1, 2 + x, 3 + 4x + x^2.
		

Links

• Andrew Howroyd, Table of n, a(n) for n = 1..1275 (rows 1..50)

Crossrefs

Cf. A210220, A208510, A001906 (row sums).

Programs

Mathematica

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := x*u[n - 1, x] + v[n - 1, x] + 1;
v[n_, x_] := 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[%]     (* A210219 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]     (* A210220 *)
(* alternative program *)
T[n_,k_] := Sum[Binomial[k, 2*j]*Binomial[n-j, k], {j, 0, Floor[k/2]}]; Flatten[Table[T[n, k],{n, 1, 11}, {k, 1, n}]] (* Detlef Meya, Dec 05 2023 *)

PARI

T(n,k) = sum(j=0, k\2, binomial(k,2*j)*binomial(n-j,k)) \\ Andrew Howroyd, Jan 01 2024

Formula

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

T(n,k) = Sum_{j=0..floor(k/2)} binomial(k,2*j)*binomial(n-j,k). - Detlef Meya, Dec 05 2023