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

%I A100098 #21 Jan 16 2025 21:55:11
%S A100098 1,1,4,7,22,46,130,295,790,1870,4864,11782,30148,73984,187534,463687,
%T A100098 1168870,2902870,7293640,18161170,45541492,113576596,284470564,
%U A100098 710118262,1777323772,4439253196,11105933440,27749232700,69403169200
%N A100098 An inverse Chebyshev transform of (1-x)/(1-2x).
%C A100098 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)).
%C A100098 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
%C A100098 Unsigned version of A127361. - _Philippe Deléham_, Nov 25 2007
%H A100098 Vincenzo Librandi, <a href="/A100098/b100098.txt">Table of n, a(n) for n = 0..1000</a>
%F A100098 G.f.: sqrt(1-4x^2)*(sqrt(1-4x^2)-6x+3)/(2*(2-5x)*(1-4x^2));
%F A100098 a(n) = Sum_{k=0..floor(n/2)} binomial(n, k)*(2^(n-2k) + 0^(n-2k))/2.
%F A100098 From _Paul Barry_, Oct 01 2007: (Start)
%F A100098 G.f.: (1+2x+3*sqrt(1-4x^2))/(4-2x-20x^2);
%F A100098 a(n) = Sum_{k=0..floor((n+1)/2)} (C(n,k) - C(n,k-1))*A001045(n-2k+1). (End)
%F A100098 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
%F A100098 a(n) ~ 5^n / 2^(n+1). - _Vaclav Kotesovec_, Feb 08 2014
%t A100098 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 *)
%K A100098 easy,nonn
%O A100098 0,3
%A A100098 _Paul Barry_, Nov 04 2004