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%I #20 Oct 23 2022 22:54:50
%S 1,2,2,2,6,4,2,10,16,8,2,14,36,40,16,2,18,64,112,96,32,2,22,100,240,
%T 320,224,64,2,26,144,440,800,864,512,128,2,30,196,728,1680,2464,2240,
%U 1152,256,2,34,256,1120,3136,5824,7168,5632,2560,512,2,38,324
%N Triangle of coefficients of polynomials v(n,x) jointly generated with A185045; see the Formula section.
%C Alternating row sums: 1, 0, 0, 0, 0, 0, 0, 0, 0, ...
%C For a discussion and guide to related arrays, see A208510.
%C As triangle T(n,k) with 0 <= k <= n, it is (2, -1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - _Philippe Deléham_, Mar 04 2012
%F u(n,x) = u(n-1,x) + 2x*v(n-1,x),
%F v(n,x) = u(n-1,x) + 2x*v(n-1,x) + 1,
%F where u(1,x)=1, v(1,x)=1.
%F As triangle T(n,k) with 0 <= k <= n: T(n,k) = A029653(n,k)*2^k. - _Philippe Deléham_, Mar 04 2012
%F Sum_{k=0..n} T(n,k)*x^k = 2*(1+x)*(1+2x)^(n-2) for n > 1. - _Philippe Deléham_, Mar 05 2012
%e First five rows:
%e 1;
%e 2, 2;
%e 2, 6, 4;
%e 2, 10, 16, 8;
%e 2, 14, 36, 40, 16;
%e First five polynomials v(n,x):
%e 1
%e 2 + 2x = 2*(1+x)
%e 2 + 6x + 4x^2 = 2*(1+x)*(1+2x)
%e 2 + 10x + 16x^2 + 8x^3 = 2*(1+x)*(1+2x)^2
%e 2 + 14x + 36x^2 + 40x^3 + 16x^4 = 2*(1+x)*(1+2x)^3
%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_] := u[n - 1, x] + 2 x*v[n - 1, x] + 1;
%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[%] (* A185045 *)
%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[%] (* A208659 *)
%t (* Using the function RiordanSquare defined in A321620 we also have: *)
%t A208659 = RiordanSquare[(1 + x)/(1 - x), 16] // Flatten (* _Gerry Martens_, Oct 16 2022 *)
%Y Cf. A185045, A208510.
%K nonn,tabl
%O 1,2
%A _Clark Kimberling_, Mar 03 2012
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