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%I #5 Jun 13 2017 22:13:11
%S 1,2,3,4,12,1,8,36,13,6,16,96,66,63,1,32,240,248,360,33,9,64,576,800,
%T 1560,321,180,1,128,1344,2352,5760,1970,1683,62,12,256,3072,6496,
%U 19152,9420,10836,985,390,1,512,6912,17152,59136,38472,55692,8989,5418,100,15
%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 (2 + 3*z + z^2)^(n-[k/2]), for n>=k>=0, where [k/2] is the integer floor of k/2.
%C Row sums form A099528. 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*(2-3*y)-x^2*y^2)/(1-4*x+x^2*(4-2*y^2)-5*x^3*y^2+x^4*y^4).
%e Rows begin:
%e [1],
%e [2,3],
%e [4,12,1],
%e [8,36,13,6],
%e [16,96,66,63,1],
%e [32,240,248,360,33,9],
%e [64,576,800,1560,321,180,1],
%e [128,1344,2352,5760,1970,1683,62,12],
%e [256,3072,6496,19152,9420,10836,985,390,1],
%e [512,6912,17152,59136,38472,55692,8989,5418,100,15],...
%e and can be derived from the coefficients of (2+3*z+z^2)^n:
%e [1],
%e [2,3,1],
%e [4,12,13,6,1],
%e [8,36,66,63,33,9,1],
%e [16,96,248,360,321,180,62,12,1],
%e [32,240,800,1560,1970,1683,985,390,100,15,1],...
%e by shifting each column k down by [k/2] rows.
%o (PARI) T(n,k)=if(n<k || k<0,0,polcoeff((2+3*z+z^2+z*O(z^k))^(n-k\2),k,z))
%Y Cf. A099509, A099510, A099528.
%K nonn,tabl
%O 0,2
%A _Paul D. Hanna_, Oct 20 2004
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