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%I #25 Jul 23 2020 20:35:11
%S 1,1,1,0,-5,-24,-90,-312,-1053,-3536,-11934,-40664,-140114,-488240,
%T -1719380,-6113200,-21921245,-79200160,-288045110,-1053728920,
%U -3874721030,-14313562480,-53093391980,-197669347600,-738398308850,-2766700765024
%N Expansion of (1+sqrt(1-4x))/(2-4x).
%C Hankel transform is A056594 with g.f. 1/(1+x^2).
%C Row sums of the Riordan array (sqrt(1-4x)/(1-2x),xc(x)^2), c(x) the g.f. of A000108.
%C The inverse Catalan transform yields A146559. - _R. J. Mathar_, Mar 20 2009
%H Matthew House, <a href="/A158499/b158499.txt">Table of n, a(n) for n = 0..1669</a>
%H Paul Barry, <a href="https://arxiv.org/abs/2004.04577">On a Central Transform of Integer Sequences</a>, arXiv:2004.04577 [math.CO], 2020.
%F a(n) = Sum_{k=0..n} binomial(2k,k)*A158495(n-k).
%F Conjecture: n*a(n) +6*(1-n)*a(n-1) +4*(2*n-3)*a(n-2)=0. - _R. J. Mathar_, Nov 14 2011
%F This conjecture has been proven. - _Matthew House_, Nov 08 2015
%t CoefficientList[ Series[(1 + Sqrt[1 - 4x])/(2 - 4x), {x, 0, 26}], x] (* _Robert G. Wilson v_, Nov 08 2015 *)
%o (PARI) x='x+O('x^33); Vec(((1-4*x)+sqrt(1-4*x))/(2*(1-2*x)*sqrt(1-4*x))) \\ _Altug Alkan_, Nov 08 2015
%K easy,sign
%O 0,5
%A _Paul Barry_, Mar 20 2009
%E Name edited by _Matthew House_, Nov 08 2015
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