%I #13 Aug 12 2015 04:08:16
%S 1,2,1,4,5,2,8,16,12,3,16,44,49,25,5,32,112,166,127,50,8,64,272,504,
%T 513,301,96,13,128,640,1424,1808,1408,670,180,21,256,1472,3824,5816,
%U 5641,3562,1427,331,34,512,3328,9888,17520,20330,15981,8494,2939
%N Triangle of coefficients of polynomials u(n,x) jointly generated with A209142; see the Formula section.
%C Each row begins with a power of 2 and ends with a Fibonacci number. Alternating row sums: all 1's. For a discussion and guide to related arrays, see A208510.
%C Subtriangle of the triangle T(n,k) given by (1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - _Philippe Deléham_, Mar 07 2012
%F u(n,x)=u(n-1,x)+(x+1)*v(n-1,x),
%F v(n,x)=(x+1)*u(n-1,x)+(x+1)*v(n-1,x),
%F where u(1,x)=1, v(1,x)=1.
%F T(n,k) = 2*T(n-1,k) + T(n-1,k-1) + T(n-2,k-1) + T(n-2,k-2), T(0,0) = 1, T(1,0) = 1, T(1,1) = 0, T(n,k) = 0 if k>n or if k<0. - _Philippe Deléham_, Mar 07 2012
%F G.f.: -x*y/(-1+x*y+x^2*y^2+2*x+x^2*y). - _R. J. Mathar_, Aug 12 2015
%e First five rows:
%e 1
%e 2....1
%e 4....5....2
%e 8....16...12...3
%e 16...44...49...25...5
%e First three polynomials u(n,x): 1, 2 + x, 4 + 5x + 2x^2
%e Triangle (1, 1, 0, 0, 0...) DELTA (0, 1, 1, -1, 0, 0, 0, ...) begins :
%e 1
%e 1, 0
%e 2, 1, 0
%e 4, 5, 2, 0
%e 8, 16, 12, 3, 0
%e 16, 44, 49, 25, 5, 0
%e 32, 112, 166, 127, 50, 8, 0
%t u[1, x_] := 1; v[1, x_] := 1; z = 16;
%t u[n_, x_] := u[n - 1, x] + (x + 1)*v[n - 1, x];
%t v[n_, x_] := (x + 1)*u[n - 1, x] + (x + 1)*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[%] (* A209141 *)
%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[%] (* A209142 *)
%Y Cf. A209142, A208510.
%K nonn,tabl
%O 1,2
%A _Clark Kimberling_, Mar 06 2012