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A099510 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 + 2*z + z^2)^(n-[k/2]), for n>=k>=0, where [k/2] is the integer floor of k/2.

%I A099510 #6 Jun 13 2017 22:13:15
%S A099510 1,1,2,1,4,1,1,6,6,4,1,8,15,20,1,1,10,28,56,15,6,1,12,45,120,70,56,1,
%T A099510 1,14,66,220,210,252,28,8,1,16,91,364,495,792,210,120,1,1,18,120,560,
%U A099510 1001,2002,924,792,45,10,1,20,153,816,1820,4368,3003,3432,495,220,1,1,22,190
%N A099510 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 + 2*z + z^2)^(n-[k/2]), for n>=k>=0, where [k/2] is the integer floor of k/2.
%C A099510 Row sums form A099511. 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 A099510 G.f.: (1-x+2*x*y-x^2*y^2)/((1-x)^2-2*x^2*y^2-2*x^3*y^2+x^4*y^4). T(n, k) = binomial(2*n-2*(k\2), k).
%e A099510 Rows begin:
%e A099510 [1],
%e A099510 [1,2],
%e A099510 [1,4,1],
%e A099510 [1,6,6,4],
%e A099510 [1,8,15,20,1],
%e A099510 [1,10,28,56,15,6],
%e A099510 [1,12,45,120,70,56,1],
%e A099510 [1,14,66,220,210,252,28,8],
%e A099510 [1,16,91,364,495,792,210,120,1],
%e A099510 [1,18,120,560,1001,2002,924,792,45,10],...
%e A099510 and can be derived from coefficients of (1+2*z+z^2)^n:
%e A099510 [1],
%e A099510 [1,2,1],
%e A099510 [1,4,6,4,1],
%e A099510 [1,6,15,20,15,6,1],
%e A099510 [1,8,28,56,70,56,28,8,1],
%e A099510 [1,10,45,120,210,252,210,120,45,10,1],...
%e A099510 by shifting each column k down by [k/2] rows.
%o A099510 (PARI) T(n,k)=if(n<k || k<0,0,polcoeff((1+2*z+z^2+z*O(z^k))^(n-k\2),k,z))
%Y A099510 Cf. A034870, A099509, A099511, A099512.
%K A099510 nonn,tabl
%O A099510 0,3
%A A099510 _Paul D. Hanna_, Oct 20 2004