|
%I #22 Jun 30 2020 18:50:54
%S 1,1,2,5,12,29,72,182,466,1207,3158,8334,22158,59299,159614,431838,
%T 1173710,3203244,8774780,24118522,66497316,183858411,509670494,
%U 1416231616,3944027402,11006186760,30772507128,86191006746,241815195292,679488418879,1912123070998
%N G.f.: exp( Sum_{n>=1} x^n/n * Sum_{k=0..n} binomial(n,k)^2 * x^k/(1-x)^k ).
%C The radius of convergence of g.f. A(x) is r = 0.339332122592393190... where 1-4*r+4*r^2-4*r^3+4*r^4 = 0, with A(r) = (1-2*r)/(2*r^3) = 4.112009743749...
%H Paul Barry, <a href="https://arxiv.org/abs/1807.05794">Riordan Pseudo-Involutions, Continued Fractions and Somos 4 Sequences</a>, arXiv:1807.05794 [math.CO], 2018.
%F G.f.: (1-2*x - sqrt(1-4*x+4*x^2-4*x^3+4*x^4))/(2*x^3).
%F Conjecture: (n+3)*a(n) +2*(-2*n-3)*a(n-1) +4*n*a(n-2) +2*(-2*n+3)*a(n-3) +4*(n-3)*a(n-4)=0. - _R. J. Mathar_, May 17 2019
%F G.f. A(x) satisfies: A(x) = 1 + x * (1 + x^2*A(x)^2) / (1 - 2*x). - _Ilya Gutkovskiy_, Jun 30 2020
%e G.f.: A(x) = 1 + x + 2*x^2 + 5*x^3 + 12*x^4 + 29*x^5 + 72*x^6 + 182*x^7 +...
%t (1 - 2x - Sqrt[1 - 4x + 4x^2 - 4x^3 + 4x^4])/(2x^3) + O[x]^31 // CoefficientList[#, x]& (* _Jean-François Alcover_, Oct 27 2018 *)
%o (PARI) {a(n)=polcoeff(exp(sum(m=1,n+1,x^m/m*sum(k=0,m,binomial(m,k)^2*x^k/(1-x+x*O(x^n))^k))),n)}
%o (PARI) {a(n)=polcoeff((1-2*x - sqrt(1-4*x+4*x^2-4*x^3+4*x^4 +x^4*O(x^n)))/(2*x^3),n)}
%o for(n=0,40,print1(a(n),", "))
%Y Cf. A025273, A217282.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Sep 30 2012
|
