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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.
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%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