cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A208659 Triangle of coefficients of polynomials v(n,x) jointly generated with A185045; see the Formula section.

Original entry on oeis.org

1, 2, 2, 2, 6, 4, 2, 10, 16, 8, 2, 14, 36, 40, 16, 2, 18, 64, 112, 96, 32, 2, 22, 100, 240, 320, 224, 64, 2, 26, 144, 440, 800, 864, 512, 128, 2, 30, 196, 728, 1680, 2464, 2240, 1152, 256, 2, 34, 256, 1120, 3136, 5824, 7168, 5632, 2560, 512, 2, 38, 324
Offset: 1

Views

Author

Clark Kimberling, Mar 03 2012

Keywords

Comments

Alternating row sums: 1, 0, 0, 0, 0, 0, 0, 0, 0, ...
For a discussion and guide to related arrays, see A208510.
As triangle T(n,k) with 0 <= k <= n, it is (2, -1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 04 2012

Examples

			First five rows:
  1;
  2,  2;
  2,  6,  4;
  2, 10, 16,  8;
  2, 14, 36, 40, 16;
First five polynomials v(n,x):
  1
  2 +  2x = 2*(1+x)
  2 +  6x +  4x^2 = 2*(1+x)*(1+2x)
  2 + 10x + 16x^2 +  8x^3 = 2*(1+x)*(1+2x)^2
  2 + 14x + 36x^2 + 40x^3 + 16x^4 = 2*(1+x)*(1+2x)^3
		

Crossrefs

Programs

  • Mathematica
    u[1, x_] := 1; v[1, x_] := 1; z = 16;
    u[n_, x_] := u[n - 1, x] + 2 x*v[n - 1, x];
    v[n_, x_] := u[n - 1, x] + 2 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[%]    (* A185045 *)
    Table[Expand[v[n, x]], {n, 1, z}]
    cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
    TableForm[cv]
    Flatten[%]    (* A208659 *)
    (* Using the function RiordanSquare defined in A321620 we also have: *)
    A208659 = RiordanSquare[(1 + x)/(1 - x), 16] // Flatten (* Gerry Martens, Oct 16 2022 *)

Formula

u(n,x) = u(n-1,x) + 2x*v(n-1,x),
v(n,x) = u(n-1,x) + 2x*v(n-1,x) + 1,
where u(1,x)=1, v(1,x)=1.
As triangle T(n,k) with 0 <= k <= n: T(n,k) = A029653(n,k)*2^k. - Philippe Deléham, Mar 04 2012
Sum_{k=0..n} T(n,k)*x^k = 2*(1+x)*(1+2x)^(n-2) for n > 1. - Philippe Deléham, Mar 05 2012