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%I #5 Jun 13 2017 22:13:17
%S 1,1,1,1,2,1,1,3,3,2,1,4,6,7,1,1,5,10,16,6,3,1,6,15,30,19,16,1,1,7,21,
%T 50,45,51,10,4,1,8,28,77,90,126,45,30,1,1,9,36,112,161,266,141,126,15,
%U 5,1,10,45,156,266,504,357,393,90,50,1,1,11,55,210,414,882,784,1016,357
%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 + z^2)^(n-[k/2]), for n>=k>=0, where [k/2] is the integer floor of k/2.
%C Row sums form absolute values of A078039. 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-x^2*y^2)/((1-x)^2-2*x^2*y^2+x^3*y^2+x^4*y^4).
%e Rows begin:
%e [1],
%e [1,1],
%e [1,2,1],
%e [1,3,3,2],
%e [1,4,6,7,1],
%e [1,5,10,16,6,3],
%e [1,6,15,30,19,16,1],
%e [1,7,21,50,45,51,10,4],
%e [1,8,28,77,90,126,45,30,1],
%e [1,9,36,112,161,266,141,126,15,5],...
%e and can be derived from coefficients of (1+z+z^2)^n:
%e [1],
%e [1,1,1],
%e [1,2,3,2,1],
%e [1,3,6,7,6,3,1],
%e [1,4,10,16,19,16,10,4,1],
%e [1,5,15,30,45,51,45,30,15,5,1],...
%e by shifting each column k down by [k/2] rows.
%o (PARI) T(n,k)=if(n<k || k<0,0,polcoeff((1+z+z^2+z*O(z^k))^(n-k\2),k,z))
%Y Cf. A027907, A078039, A099510.
%K nonn,tabl
%O 0,5
%A _Paul D. Hanna_, Oct 20 2004
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