%I #18 Jun 08 2019 12:27:56
%S 1,1,2,4,10,26,68,176,454,1174,3052,7976,20932,55108,145448,384704,
%T 1019462,2706214,7194956,19155896,51065260,136284236,364097912,
%U 973654240,2605983772,6980545276,18712478072,50196568144,134739960904,361892443592,972537193168
%N a(n) = Sum_{k=0..n} binomial(n,2k)*A002426(k).
%C Hankel transform is (-1)^binomial(n,2)*(-2)^A128054(n) (see A128055).
%H G. C. Greubel, <a href="/A162533/b162533.txt">Table of n, a(n) for n = 0..1000</a>
%F G.f.: (1-x)/((1-x)^2-x^2-2x^4/((1-x)^2-x^2-x^4/((1-x)^2-x^2-x^4/(1-... (continued fraction).
%F G.f.: (1-x)/sqrt(1-4*x+4*x^2-4*x^4) = (1-x)/sqrt((1-2*x)^2-4*x^4) = (1-x)/sqrt((1-x-2*x^2)*(1-x+2*x^2)). - _Paul Barry_, Oct 13 2009
%F Conjecture: n*a(n) + (4-5*n)*a(n-1) + 2*(4*n-7)*a(n-2) + 4*(3-n)*a(n-3) + 4*(2-n)*a(n-4) + 4*(n-4)*a(n-5) = 0. - _R. J. Mathar_, Nov 16 2011
%F a(n) ~ 3^(1/4) * (1 + sqrt(3))^(n + 1/2) / (2^(3/2) * sqrt(Pi*n)). - _Vaclav Kotesovec_, Jun 08 2019
%t b[n_] := If[n < 0, 0, 3^n Hypergeometric2F1[1/2, -n, 1, 4/3]]; Table[Sum[Binomial[n, 2*k]*b[k], {k, 0, n}], {n, 0, 50}] (* or *) CoefficientList[Series[(1-x)/sqrt(1-4*x+4*x^2-4*x^4), {x, 0, 50}], x] (* _G. C. Greubel_, Feb 27 2017 *)
%o (PARI) x='x+O('x^50); Vec((1-x)/sqrt(1-4*x+4*x^2-4*x^4)) \\ _G. C. Greubel_, Feb 27 2017
%Y Cf. A027826.
%K easy,nonn
%O 0,3
%A _Paul Barry_, Jul 05 2009