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a(n) = (n+1)*Sum_{k=0..(n-1)/2}((binomial(2*n-3*k-2,n-k-1))/(n-k)).
1

%I #24 Jun 04 2017 18:22:50

%S 0,2,3,10,30,101,350,1250,4548,16782,62579,235273,890331,3387204,

%T 12943353,49643762,191010623,736946570,2850013623,11044973890,

%U 42882986660,166770990377,649526893537,2533096497017,9890766366030,38662031939117

%N a(n) = (n+1)*Sum_{k=0..(n-1)/2}((binomial(2*n-3*k-2,n-k-1))/(n-k)).

%H G. C. Greubel, <a href="/A270363/b270363.txt">Table of n, a(n) for n = 0..1000</a>

%F G.f.: (1-sqrt(1-4*x))/(sqrt(1-4*x)*x-x+2)*((6*x+sqrt(1-4*x)-1)/(4*x+sqrt(1-4*x)-1)).

%F Conjecture: n*(7*n^2-17*n-2) *a(n) +(-35*n^3+99*n^2+20*n-120) *a(n-1) +2* (2*n-5) *(7*n^2-3*n-12)*a(n-2) +n*(7*n^2-17*n-2) *a(n-3) +2*-(2*n-5) *(7*n^2-3*n-12) *a(n-4)=0. - _R. J. Mathar_, Mar 22 2016

%F a(n) ~ 2^(2*n+1) / (7*sqrt(Pi*n)). - _Vaclav Kotesovec_, Mar 22 2016

%t CoefficientList[Series[(1 - Sqrt[1 - 4*x])/(Sqrt[1 - 4*x]*x - x + 2)* ((6*x + Sqrt[1 - 4*x] - 1)/(4*x + Sqrt[1 - 4*x] - 1)), {x,0,50}], x] (* _G. C. Greubel_, Jun 04 2017 *)

%o (Maxima)

%o taylor((1-sqrt(1-4*x))/(sqrt(1-4*x)*x-x+2)*((6*x+sqrt(1-4*x)-1)/(4*x+sqrt(1-4*x)-1)),x,0,15);

%o a(n):=(n+1)*sum((binomial(2*n-3*k-2,n-k-1))/(n-k),k,0,(n-1)/2);

%o (PARI) x='x+O('x^100); concat(0, Vec((1-sqrt(1-4*x))/(sqrt(1-4*x)*x-x+2)*((6*x+sqrt(1-4*x)-1)/(4*x+sqrt(1-4*x)-1)))) \\ _Altug Alkan_, Mar 25 2016

%Y Cf. A000108, A033184.

%K nonn

%O 0,2

%A _Vladimir Kruchinin_, Mar 22 2016