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a(n) = (1/2)*s(n+3), where s = A025244.
0

%I #15 Nov 11 2022 13:17:30

%S 1,2,5,11,26,65,163,416,1081,2837,7516,20089,54077,146478,398997,

%T 1092215,3003014,8289569,22964919,63828252,177931665,497367721,

%U 1393768952,3914793457,11019379609,31079140922,87818240869,248571086403,704722488690

%N a(n) = (1/2)*s(n+3), where s = A025244.

%H Paul Barry, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL14/Barry4/barry142.html">On a Generalization of the Narayana Triangle</a>, J. Int. Seq. 14 (2011) # 11.4.5.

%F G.f.: (1-x-x^2-4*x^3-sqrt(1-2*x-x^2-6*x^3+x^4))/(4*x^3). - _Michael Somos_, Jun 08 2000

%F Conjecture: (n+3)*a(n) +(-2*n-3)*a(n-1) -n*a(n-2) +3*(-2*n+3)*a(n-3) +(n-3)*a(n-4)=0. - _R. J. Mathar_, Feb 25 2015

%o (PARI) a(n)=polcoeff((-sqrt(1-2*x-x^2-6*x^3+x^4+x^4*O(x^n)))/4,n+3)

%K nonn

%O 1,2

%A _Clark Kimberling_