%I #21 Oct 14 2016 08:07:18

%S 0,1,2,8,20,64,172,512,1416,4096,11468,32768,92248,262144,739832,

%T 2097152,5925520,16777216,47429900,134217728,379536440,1073741824,

%U 3036661032,8589934592,24294699120,68719476736,194363001272

%N An inverse Chebyshev transform of the Pell numbers.

%C Image of x/(1-2x-x^2) 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)).

%H Vincenzo Librandi, <a href="/A100097/b100097.txt">Table of n, a(n) for n = 0..1000</a>

%F G.f.: x*sqrt(1-4*x^2)*(sqrt(1-4*x^2)+2*x)/((1-4*x^2)*(1-8*x^2)).

%F a(n) = sum{k=0..floor(n/2), binomial(n, k)*A000129(n-2k)}.

%F Conjecture: (-n+2)*a(n) +(-n+3)*a(n-1) +4*(3*n-7)*a(n-2) +4*(3*n-10)*a(n-3) +32*(-n+3)*a(n-4) +32*(-n+4)*a(n-5)=0. - _R. J. Mathar_, Nov 24 2012

%F Recurrence: (n-2)*a(n) = 4*(3*n-7)*a(n-2) - 32*(n-3)*a(n-4). - _Vaclav Kotesovec_, Feb 12 2014

%F a(n) ~ 2^((3*n-3)/2). - _Vaclav Kotesovec_, Feb 12 2014

%F a(2*n) = 8^n/(2*sqrt(2)) - 2^n * (2*n-1)!! * hypergeom([1, n+1/2], [n+1], 1/2)/(4*n!), a(2*n+1) = 8^n. - _Vladimir Reshetnikov_, Oct 13 2016

%t CoefficientList[Series[x*Sqrt[1-4*x^2]*(Sqrt[1-4*x^2]+2*x)/((1-4*x^2)*(1-8*x^2)), {x, 0, 20}], x] (* _Vaclav Kotesovec_, Feb 12 2014 *)

%Y Cf. A100095, A100096.

%K nonn,easy

%O 0,3

%A _Paul Barry_, Nov 03 2004