A100098 An inverse Chebyshev transform of (1-x)/(1-2x).

1, 1, 4, 7, 22, 46, 130, 295, 790, 1870, 4864, 11782, 30148, 73984, 187534, 463687, 1168870, 2902870, 7293640, 18161170, 45541492, 113576596, 284470564, 710118262, 1777323772, 4439253196, 11105933440, 27749232700, 69403169200

Offset: 0

Author

Paul Barry, Nov 04 2004

Keywords

• easy
• nonn

Comments

Image of (1-x)/(1-2*x) under the transform g(x)->(1/sqrt(1-4*x^2))*g(x*c(x^2)), where c(x) is the g.f. of the Catalan numbers A000108. This is the inverse of the Chebyshev transform which takes A(x) to ((1-x^2)/(1+x^2))*A(x/(1+x^2)).

Transform of the Jacobsthal numbers A001045(n+1) under the Riordan array (c(x^2),xc(x^2)). Hankel transform is 3^n. - Paul Barry, Oct 01 2007

Unsigned version of A127361. - Philippe Deléham, Nov 25 2007

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Programs

Mathematica

CoefficientList[Series[Sqrt[1-4*x^2]*(Sqrt[1-4*x^2]-6*x+3)/(2*(2-5*x)*(1-4*x^2)), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 08 2014 *)

Formula

G.f.: sqrt(1-4x^2)*(sqrt(1-4x^2)-6x+3)/(2*(2-5x)*(1-4x^2));

a(n) = Sum_{k=0..floor(n/2)} binomial(n, k)*(2^(n-2k) + 0^(n-2k))/2.

From Paul Barry, Oct 01 2007: (Start)

G.f.: (1+2x+3*sqrt(1-4x^2))/(4-2x-20x^2);

a(n) = Sum_{k=0..floor((n+1)/2)} (C(n,k) - C(n,k-1))*A001045(n-2k+1). (End)

Conjecture: 2*n*a(n) + (-5*n+4)*a(n-1) + 2*(-4*n+3)*a(n-2) + 20*(n-2)*a(n-3) = 0. - R. J. Mathar, Nov 22 2012

a(n) ~ 5^n / 2^(n+1). - Vaclav Kotesovec, Feb 08 2014