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%I #40 Jan 08 2023 14:05:07
%S 1,1,3,7,21,65,215,735,2585,9281,33883,125383,469229,1772801,6752623,
%T 25902975,99978865,388001025,1513077235,5926139207,23301146501,
%U 91942524481,363957103303,1444966207967,5752187960841,22955311342145
%N Expansion of (1/(1-x^2))*c(x/(1-x^2)), where c(x) is the g.f. of A000108.
%C Binomial transform is A059279.
%C Hankel transform is A134751. - _Paul Barry_, Oct 07 2008
%C The radius of convergence r of the g.f. A(x) satisfies: r = (1-r^2)/4 = lim_{n->inf} a(n)/a(n+1) = sqrt(5) - 2 = 0.2360679... with A(r) = 1/(2*r) = (sqrt(5) + 2)/2 = 2.1180339... - _Paul D. Hanna_, Sep 06 2011
%H Paul Barry, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL12/Barry3/barry93.html">Continued fractions and transformations of integer sequences</a>, JIS 12 (2009) 09.7.6.
%F G.f.: (1 - sqrt((1 - 4*x - x^2)/(1 - x^2)))/(2*x).
%F a(n) = Sum_{k=0..floor(n/2)} C(n-k, k) * C(n-2*k)}.
%F G.f. satisfies: A(x) = 1/(1-x^2) + x*A(x)^2. - _Paul D. Hanna_, Sep 06 2011
%F Conjecture: (n+1)*a(n) + 2*(1-2*n)*a(n-1) + 2*(1-n)*a(n-2) + 2*(2*n-3)*a(n-3) + (n-3)*a(n-4) = 0. - _R. J. Mathar_, Nov 15 2011
%F G.f.: (1-1/G(0))/(2*x), where G(k) = 1 + 4*x*(4*k+1)/( (1-x^2)*(4*k+2) - x*(1-x^2)*(4*k+2)*(4*k+3)/(x*(4*k+3) + (1-x^2)*(k+1)/G(k+1))); (continued fraction). - _Sergei N. Gladkovskii_, Jun 26 2013
%F a(n) ~ 5^(1/4)*(2+sqrt(5))^(n+1)/(4*sqrt(2*Pi)*n^(3/2)). - _Vaclav Kotesovec_, Sep 16 2013
%F G.f.: 1/G(x), where G(x) = 1 - x^2 - (x - x^3)/ G(x) (continued fraction). - _Nikolaos Pantelidis_, Jan 08 2023
%t a[0] = a[1] = 1; a[2] = 3; a[3] = 7; a[n_] := a[n] = (-((n-3)*a[n-4]) - 2*(2*n-3)*a[n-3] + 2*(n-1)*a[n-2] + 2*(2*n-1)*a[n-1])/(n+1); Table[a[n], {n, 0, 25}] (* _Jean-François Alcover_, Sep 09 2017, using "FindSequenceFunction" *)
%o (PARI) {a(n)=polcoeff((1-sqrt(1-4*x/(1-x^2 +O(x^(n+2)))))/(2*x), n)} /* _Paul D. Hanna_ */
%Y Partial sums of A128750.
%K easy,nonn
%O 0,3
%A _Paul Barry_, Apr 23 2005
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