%I #16 Aug 30 2024 21:41:23

%S 0,1,2,7,16,46,110,295,720,1870,4612,11782,29224,73984,184102,463687,

%T 1156000,2902870,7245020,18161170,45356736,113576596,283765132,

%U 710118262,1774619616,4439253196,11095532840,27749232700,69363052600

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

%C Image of x/(1-2x) under the transform g(x)->(1/sqrt(1-4x^2)g(xc(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 Hankel transform is A125905(n-1), the alternating sign version of A001353. - _Paul Barry_, Nov 25 2007

%H Vincenzo Librandi, <a href="/A100099/b100099.txt">Table of n, a(n) for n = 0..1000</a>

%F G.f.: sqrt(1-4x^2)(sqrt(1-4x^2)+4x-1)/(2(5x-2)(4x^2-1)).

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

%F a(n) = Sum_{k=0..n} C(n,floor(k/2))*A001045(n-k). - _Paul Barry_, Nov 25 2007

%F Conjecture: 2n*a(n) +(-13n+16)*a(n-1) +4(3n-8)*a(n-2) +4(13n-29)*a(n-3) +80(3-n)*a(n-4)=0. - _R. J. Mathar_, Dec 14 2011

%F a(n) ~ 5^n / 2^(n+1). - _Vaclav Kotesovec_, Feb 01 2014

%t CoefficientList[Series[Sqrt[1-4*x^2]*(Sqrt[1-4*x^2]+4*x-1)/(2*(5*x-2)*(4*x^2-1)), {x, 0, 20}], x] (* _Vaclav Kotesovec_, Feb 01 2014 *)

%K easy,nonn

%O 0,3

%A _Paul Barry_, Nov 04 2004