%I #4 Mar 30 2012 18:36:43
%S 1,3,6,17,45,116,305,799,2090,5473,14329,37512,98209,257115,673134,
%T 1762289,4613733,12078908,31622993,82790071,216747218,567451585,
%U 1485607537,3889371024,10182505537,26658145587,69791931222,182717648081
%N Row sums of triangle A099510, so that a(n) = Sum_{k=0..n} coefficient of z^k in (1 + 2*z + z^2)^(n-[k/2]), where [k/2] is the integer floor of k/2.
%F G.f.: (1+x-x^2)/(1-2*x-x^2-2*x^3+x^4). a(n) = Sum_{k=0..n} binomial(2*n-2*[k/2], k).
%o (PARI) a(n)=sum(k=0,n,polcoeff((1+2*x+x^2+x*O(x^k))^(n-k\2),k))
%Y Cf. A099510.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Oct 21 2004