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%I #23 Jun 23 2018 02:33:08
%S 1,1,4,1,6,8,1,8,20,16,1,10,36,56,32,1,12,56,128,144,64,1,14,80,240,
%T 400,352,128,1,16,108,400,880,1152,832,256,1,18,140,616,1680,2912,
%U 3136,1920,512,1,20,176,896,2912,6272,8960,8192,4352,1024,1,22,216
%N Triangle of coefficients of polynomials v(n,x) jointly generated with A207627; see the Formula section.
%C Column n is divisible by 2^(n-1); row n ends with 2^(n-1) for n > 2.
%C Also triangle T(n,k), k=0..n, read by rows, given by (1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (4, -2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - _Philippe Deléham_, Feb 25 2012
%C Also seems to be square array of unsigned coefficients of 3U-2T (with T and U the two sequences of Chebyshev polynomials). - _Thomas Baruchel_, Jun 03 2018
%F u(n,x) = u(n-1,x) + v(n-1,x),
%F v(n,x) = 2x*u(n-1,x) + 2x*v(n-1,x) + 1,
%F where u(1,x)=1, v(1,x)=1.
%F The polynomials v(n,x) seem to be v(n)=sum(k=0,n-1, (-1)^(k+n+1) * x^k * polcoeff( 3*polchebyshev(2*n-k-2,2)-2*polchebyshev(2*n-k-2,1), k)) by using the PARI syntax. - _Thomas Baruchel_, Jun 05 2018
%F As triangle T(n,k), k=0..n:
%F G.f.: (1+2*y*x)/(1-(1+2*y)*x). - _Philippe Deléham_, Feb 25 2012
%F T(n,k) = 2*T(n-1,k-1) + T(n-1,k) with T(0,0) = T(1,0) = 1, T(1,1) = 4. - _Philippe Deléham_, Feb 25 2012
%F As triangle T(n,k), k=0..n, it is given by T(n,k) = A029635(n,k)*2^k with T(0,0) = 1. - _Philippe Deléham_, Feb 25 2012
%e First five rows:
%e 1;
%e 1, 4;
%e 1, 6, 8;
%e 1, 8, 20, 16;
%e 1, 10, 36, 56, 32;
%t u[1, x_] := 1; v[1, x_] := 1; z = 16;
%t u[n_, x_] := u[n - 1, x] + v[n - 1, x]
%t v[n_, x_] := 2 x*u[n - 1, x] + 2 x*v[n - 1, x] + 1
%t Table[Factor[u[n, x]], {n, 1, z}]
%t Table[Factor[v[n, x]], {n, 1, z}]
%t cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
%t TableForm[cu]
%t Flatten[%] (* A207627 *)
%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[%] (* A207628 *)
%Y Cf. A207627.
%K nonn,tabl
%O 1,3
%A _Clark Kimberling_, Feb 21 2012
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