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%I #8 Jul 31 2015 13:25:03
%S 1,4,8,27,89,257,784,2421,7336,22324,68147,207549,632177,1926608,
%T 5870089,17884476,54493120,166034731,505883825,1541369745,4696373312,
%U 14309268413,43598614528,132839740908,404746601923,1233213978037
%N Row sums of triangle A099512, so that a(n) = Sum_{k=0..n} coefficient of z^k in (1 + 3*z + z^2)^(n-[k/2]), where [k/2] is the integer floor of k/2.
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2, 1, 7, -1).
%F G.f.: (1+2*x-x^2)/(1-2*x-x^2-7*x^3+x^4).
%F a(0)=1, a(1)=4, a(2)=8, a(3)=27, a(n)=2*a(n-1)+a(n-2)+7*a(n-3)- a(n-4) [From Harvey P. Dale, Jul 12 2011]
%t LinearRecurrence[{2,1,7,-1},{1,4,8,27},30] (* or *) CoefficientList[ Series[ (1+2x-x^2)/(1-2x-x^2-7x^3+x^4),{x,0,30}],x] (* _Harvey P. Dale_, Jul 12 2011 *)
%o (PARI) a(n)=sum(k=0,n,polcoeff((1+3*x+x^2+x*O(x^k))^(n-k\2),k))
%Y Cf. A099512.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Oct 21 2004
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