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A138020 G.f. satisfies A(x) = sqrt( (1 + 2*x*A(x)) / (1 - 2*x*A(x)) ).

%I A138020 #24 Aug 10 2025 02:49:37
%S A138020 1,2,6,24,110,544,2828,15232,84246,475648,2730068,15882240,93438540,
%T A138020 554967040,3323125528,20039827456,121597985254,741871845376,
%U A138020 4548193111428,28004975116288,173113004348580,1073893324357632,6683288759506856,41715337804120064
%N A138020 G.f. satisfies A(x) = sqrt( (1 + 2*x*A(x)) / (1 - 2*x*A(x)) ).
%H A138020 Vaclav Kotesovec, <a href="/A138020/b138020.txt">Table of n, a(n) for n = 0..1000</a>
%F A138020 a(n) ~ 2^(n - 1/2) * phi^((5*n + 3)/2) / (sqrt(Pi) * 5^(1/4) * n^(3/2)), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - _Vaclav Kotesovec_, Oct 04 2020
%F A138020 From _Alexander Burstein_, Nov 26 2021: (Start)
%F A138020 G.f.: A(x) = 1 + 2*x*A(x)*(1 + A(x)^2)/(1 + A(x)).
%F A138020 G.f.: A(-x*A(x)^2) = 1/A(x). (End)
%F A138020 D-finite with recurrence +n*(n+1)*(5*n-11) *a(n) +4*(-55*n^3 +231*n^2 -263*n +51)*a(n-2) -16*(n-3)*(n-4)*(5*n-1)*a(n-4)=0. - _R. J. Mathar_, Mar 25 2024
%F A138020 From _Seiichi Manyama_, Dec 22 2024: (Start)
%F A138020 a(n) = (2^n/(n+1)) * Sum_{k=0..n} binomial(n/2+k-1/2,k) * binomial(n/2+1/2,n-k).
%F A138020 a(n) = 2^n * Sum_{k=0..n} binomial(n,k) * binomial(n/2+k+1/2,n)/(n+2*k+1). (End)
%p A138020 A138020 := proc(n)
%p A138020     option remember ;
%p A138020     if n < 5 then
%p A138020         op(n+1,[1,2,6,24,110]) ;
%p A138020     else
%p A138020         4*(-55*n^3 +231*n^2 -263*n +51)*procname(n-2) -16*(n-3)*(n-4)*(5*n-1)*procname(n-4) ;
%p A138020         -%/n/(n+1)/(5*n-11)
%p A138020     end if;
%p A138020 end proc:
%p A138020 seq(A138020(n),n=0..30) ; # _R. J. Mathar_, Sep 27 2024
%t A138020 CoefficientList[y/.AsymptoticSolve[y^2-1-2x(y+y^3) ==0,y->1,{x,0,23}][[1]],x] (* _Alexander Burstein_, Nov 26 2021 *)
%o A138020 (PARI) a(n)=polcoeff((1/x)*serreverse(x*sqrt((1-2*x)/(1+2*x+x^2*O(x^n)))),n)
%Y A138020 Cf. A078531, A151374.
%K A138020 nonn
%O A138020 0,2
%A A138020 _Paul D. Hanna_, Feb 28 2008