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%I #12 Sep 18 2013 10:17:54
%S 1,2,4,12,38,116,360,1144,3670,11836,38392,125160,409628,1345000,
%T 4428752,14618608,48356838,160260332,532009688,1768729736,5888250996,
%U 19626282328,65489004464,218743627408,731311554044,2447018357208,8194289368240,27459924376592
%N G.f.: Sum_{n>=0} x^n * Sum_{k=0..n} binomial(n,k)^2 * x^k*(2-x)^(n-k).
%C Radius of convergence of g.f. is r = 1-sqrt(2)/2 = 0.2928932...
%C More generally, given
%C A(x) = Sum_{n>=1} x^n * Sum_{k=0..n} binomial(n,k)^2 * x^k*(t-x)^(n-k),
%C then A(x) = 1 / sqrt( (1 - t*x + 2*x^2)^2 - 4*x^2 )
%C and the radius of convergence r satisfies: (1-r)^2 = r*(t-r) for t>0.
%H Vincenzo Librandi, <a href="/A217616/b217616.txt">Table of n, a(n) for n = 0..1000</a>
%F G.f.: A(x) = 1 / sqrt( (1 - 2*x + 2*x^2)^2 - 4*x^2 ).
%F G.f.: A(x) = 1 / sqrt( (1-4*x+2*x^2)*(1+2*x^2) ).
%F G.f. satisfies: A(x) = [1 + 2*x^2*Sum_{n>=0} A000108(n)*(-x*A(x))^(2*n)] / (1-2*x+2*x^2) where A000108(n) = binomial(2*n,n)/(n+1) forms the Catalan
%F numbers.
%F Recurrence: n*a(n) = 2*(2*n-1)*a(n-1) - 4*(n-1)*a(n-2) + 4*(2*n-3)*a(n-3) - 4*(n-2)*a(n-4). - _Vaclav Kotesovec_, Sep 17 2013
%F a(n) ~ sqrt(8+6*sqrt(2)) * (2+sqrt(2))^n/(4*sqrt(Pi*n)). - _Vaclav Kotesovec_, Sep 17 2013
%e G.f.: A(x) = 1 + 2*x + 4*x^2 + 12*x^3 + 38*x^4 + 116*x^5 + 360*x^6 +...
%e where the g.f. equals the series:
%e A(x) = 1 +
%e x*((2-x) + x) +
%e x^2*((2-x)^2 + 2^2*x*(2-x) + x^2) +
%e x^3*((2-x)^3 + 3^2*x*(2-x)^2 + 3^2*x^2*(2-x) + x^3) +
%e x^4*((2-x)^4 + 4^2*x*(2-x)^3 + 6^2*x^2*(2-x)^2 + 4^2*x^3*(2-x) + x^4) +
%e x^5*((2-x)^5 + 5^2*x*(2-x)^4 + 10^2*x^2*(2-x)^3 + 10^2*x^3*(2-x)^2 + 5^2*x^4*(2-x) + x^5) +...
%t CoefficientList[Series[1/Sqrt[(1-2*x+2*x^2)^2-4*x^2], {x, 0, 20}], x] (* _Vaclav Kotesovec_, Sep 17 2013 *)
%o (PARI) {a(n)=polcoeff(sum(m=0, n+1, x^m*sum(k=0, m, binomial(m, k)^2*x^k*(2-x)^(m-k) + x*O(x^n))), n)}
%o for(n=0,40,print1(a(n),", "))
%Y Cf. A217615, A217617, A217461, A216434.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Oct 09 2012
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