%I #8 Feb 17 2020 09:45:55
%S 1,1,1,1,2,2,1,3,5,4,1,4,9,13,4,1,5,14,28,18,12,1,6,20,50,49,56,8,1,7,
%T 27,80,105,161,56,32,1,8,35,119,195,366,210,200,16,1,9,44,168,329,721,
%U 581,732,160,80,1,10,54,228,518,1288,1337,2045,780,640,32,1,11,65,300,774,2142,2716,4824,2674,2884,432,192
%N Triangle, read by rows, of trinomial coefficients arranged so that there are n+1 terms in row n by setting T(n,k) equal to the coefficient of z^k in (1 + z + 2*z^2)^(n-[k/2]), for n>=k>=0, where [k/2] is the integer floor of k/2.
%C Row sums form A099515. In general if T(n,k) = coefficient of z^k in (a + b*z + c*z^2)^(n-[k/2]), then the resulting number triangle will have the o.g.f.: ((1-a*x-c*x^2*y^2) + b*x*y)/((1-a*x-c*x^2*y^2)^2 - x*(b*x*y)^2).
%F G.f.: (1-x+x*y-2*x^2*y^2)/((1-x)^2-4*x^2*y^2+3*x^3*y^2+4*x^4*y^4).
%e Rows begin:
%e [1],
%e [1,1],
%e [1,2,2],
%e [1,3,5,4],
%e [1,4,9,13,4],
%e [1,5,14,28,18,12],
%e [1,6,20,50,49,56,8],
%e [1,7,27,80,105,161,56,32],
%e [1,8,35,119,195,366,210,200,16],
%e [1,9,44,168,329,721,581,732,160,80],...
%e and can be derived from coefficients of (1+z+2*z^2)^n:
%e [1],
%e [1,1,2],
%e [1,2,5,4,4],
%e [1,3,9,13,18,12,8],
%e [1,4,14,28,49,56,56,32,16],
%e [1,5,20,50,105,161,210,200,160,80,32],...
%e by shifting each column k down by [k/2] rows.
%t With[{m = 11}, CoefficientList[CoefficientList[Series[(1-x+x*y-2*x^2*y^2)/ ((1-x)^2-4*x^2*y^2+3*x^3*y^2+4*x^4*y^4), {x, 0 , m}, {y, 0, m}], x], y]] // Flatten (* Georg Fischer, Feb 17 2020 *)
%o (PARI) T(n,k)=if(n<k || k<0,0,polcoeff((1+z+2*z^2+z*O(z^k))^(n-k\2),k,z))
%Y Cf. A099509, A099510, A099512, A099515.
%K nonn,tabl
%O 0,5
%A _Paul D. Hanna_, Oct 20 2004
%E a(50..51) corrected by _Georg Fischer_, Feb 17 2020