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%I #13 Jan 24 2020 03:25:25
%S 1,1,3,1,8,5,1,15,23,11,1,24,66,66,21,1,35,150,240,165,43,1,48,295,
%T 678,747,404,85,1,63,525,1631,2547,2157,947,171,1,80,868,3500,7246,
%U 8560,5864,2182,341,1,99,1356,6888,18126,28018,26592,15318,4929,683
%N Triangle of coefficients of polynomials v(n,x) jointly generated with A209135; see the Formula section.
%C Alternating row sums: 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...
%C For a discussion and guide to related arrays, see A208510.
%C Subtriangle of the triangle given by (1, 0, 2/3, 1/3, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 3, -4/3, -2/3, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - _Philippe Deléham_, Apr 11 2012
%F u(n,x) = u(n-1,x) + (x+1)*v(n-1,x),
%F v(n,x) = 2x*u(n-1,x) + (x+1)*v(n-1,x),
%F where u(1,x)=1, v(1,x)=1.
%F From _Philippe Deléham_, Apr 11 2012: (Start)
%F As DELTA-triangle T(n,k) with 0 <= k <= n:
%F G.f.: (1-x-y*x+y*x^2-2*y^2*x^2)/(1-2*x-y*x+x^2-y*x^2-2*y^2*x^2).
%F T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k) + T(n-2,k-1) + 2*T(n-2,k-2), T(0,0) = T(1,0) = T(2,0) = 1, T(2,1) = 3, T(1,1) = T(2,2) = 0 and T(n,k) = 0 if k < 0 or if k > n. (End)
%e First five rows:
%e 1;
%e 1, 3;
%e 1, 8, 5;
%e 1, 15, 23, 11;
%e 1, 24, 66, 66, 21;
%e First three polynomials v(n,x):
%e 1
%e 1 + 3x
%e 1 + 8x + 5x^2.
%e From _Philippe Deléham_, Apr 11 2012: (Start)
%e (1, 0, 2/3, 1/3, 0, 0, 0, ...) DELTA (0, 3, -4/3, -2/3, 0, 0, 0, ...) begins:
%e 1;
%e 1, 0;
%e 1, 3, 0;
%e 1, 8, 5, 0;
%e 1, 15, 23, 11, 0;
%e 1, 24, 66, 66, 21, 0;
%e 1, 35, 150, 240, 165, 43, 0; (End)
%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_] := 2 x*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[%] (* A209135 *)
%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[%] (* A209136 *)
%Y Cf. A209135, A208510.
%K nonn,tabl
%O 1,3
%A _Clark Kimberling_, Mar 05 2012
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