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%I #10 Sep 08 2013 19:59:31
%S 1,0,2,0,1,4,0,1,3,8,0,1,3,9,16,0,1,3,11,23,32,0,1,3,13,31,57,64,0,1,
%T 3,15,39,87,135,128,0,1,3,17,47,121,227,313,256,0,1,3,19,55,159,339,
%U 579,711,512,0,1,3,21,63,201,471,933,1431,1593,1024,0,1,3,23,71
%N Triangle of coefficients of polynomials v(n,x) jointly generated with A208755; see the Formula section.
%C For a discussion and guide to related arrays, see A208510.
%C As triangle T(n,k) with 0<=k<=n, it is (0, 1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (2, 0, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - _Philippe Deléham_, Mar 02 2012
%F u(n,x)=u(n-1,x)+2x*v(n-1,x),
%F v(n,x)=x*u(n-1,x)+x*v(n-1,x),
%F where u(1,x)=1, v(1,x)=1.
%F As triangle with 0<=k<=n : G.f.: (1-x+y*x)/(1-(1+y)*x-(2*y^2-y)*x^2). - _Philippe Deléham_, Mar 02 2012
%F T(n,k) = T(n-1,k) + T(n-1,k-1) - T(n-2,k-1) + 2*T(n-2,k-2). - _Philippe Deléham_, Mar 02 2012
%e First five rows:
%e 1
%e 0...2
%e 0...1...4
%e 0...1...3...8
%e 0...1...3...9...16
%e First five polynomials v(n,x):
%e 1
%e 2x
%e x + 4x^2
%e x + 3x^2 + 8x^3
%e x + 3x^2 + 9x^3 + 16^4
%t u[1, x_] := 1; v[1, x_] := 1; z = 16;
%t u[n_, x_] := u[n - 1, x] + 2 x*v[n - 1, x];
%t v[n_, x_] := x*u[n - 1, x] + x*v[n - 1, x];
%t Table[Expand[u[n, x]], {n, 1, z/2}]
%t Table[Expand[v[n, x]], {n, 1, z/2}]
%t cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
%t TableForm[cu]
%t Flatten[%] (* A208755 *)
%t Table[Expand[v[n, x]], {n, 1, z}]
%t cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
%t TableForm[cv]
%t Flatten[%] (* A208756 *)
%Y Cf. A208755, A208510.
%K nonn,tabl
%O 1,3
%A _Clark Kimberling_, Mar 01 2012
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