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.

A207628 Triangle of coefficients of polynomials v(n,x) jointly generated with A207627; see the Formula section.

Original entry on oeis.org

1, 1, 4, 1, 6, 8, 1, 8, 20, 16, 1, 10, 36, 56, 32, 1, 12, 56, 128, 144, 64, 1, 14, 80, 240, 400, 352, 128, 1, 16, 108, 400, 880, 1152, 832, 256, 1, 18, 140, 616, 1680, 2912, 3136, 1920, 512, 1, 20, 176, 896, 2912, 6272, 8960, 8192, 4352, 1024, 1, 22, 216
Offset: 1

Views

Author

Clark Kimberling, Feb 21 2012

Keywords

Comments

Column n is divisible by 2^(n-1); row n ends with 2^(n-1) for n > 2.
Also triangle T(n,k), k=0..n, read by rows, given by (1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (4, -2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Feb 25 2012
Also seems to be square array of unsigned coefficients of 3U-2T (with T and U the two sequences of Chebyshev polynomials). - Thomas Baruchel, Jun 03 2018

Examples

			First five rows:
  1;
  1,  4;
  1,  6,  8;
  1,  8, 20, 16;
  1, 10, 36, 56, 32;
		

Crossrefs

Cf. A207627.

Programs

  • Mathematica
    u[1, x_] := 1; v[1, x_] := 1; z = 16;
    u[n_, x_] := u[n - 1, x] + v[n - 1, x]
    v[n_, x_] := 2 x*u[n - 1, x] + 2 x*v[n - 1, x] + 1
    Table[Factor[u[n, x]], {n, 1, z}]
    Table[Factor[v[n, x]], {n, 1, z}]
    cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
    TableForm[cu]
    Flatten[%]    (* A207627 *)
    Table[Expand[v[n, x]], {n, 1, z}]
    cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
    TableForm[cv]
    Flatten[%]    (* A207628 *)

Formula

u(n,x) = u(n-1,x) + v(n-1,x),
v(n,x) = 2x*u(n-1,x) + 2x*v(n-1,x) + 1,
where u(1,x)=1, v(1,x)=1.
The polynomials v(n,x) seem to be v(n)=sum(k=0,n-1, (-1)^(k+n+1) * x^k * polcoeff( 3*polchebyshev(2*n-k-2,2)-2*polchebyshev(2*n-k-2,1), k)) by using the PARI syntax. - Thomas Baruchel, Jun 05 2018
As triangle T(n,k), k=0..n:
G.f.: (1+2*y*x)/(1-(1+2*y)*x). - Philippe Deléham, Feb 25 2012
T(n,k) = 2*T(n-1,k-1) + T(n-1,k) with T(0,0) = T(1,0) = 1, T(1,1) = 4. - Philippe Deléham, Feb 25 2012
As triangle T(n,k), k=0..n, it is given by T(n,k) = A029635(n,k)*2^k with T(0,0) = 1. - Philippe Deléham, Feb 25 2012