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%I #13 Jan 24 2020 03:25:13
%S 1,2,1,3,5,3,5,12,15,7,8,27,45,42,17,13,55,119,151,116,41,21,108,282,
%T 458,480,315,99,34,205,630,1228,1631,1467,845,239,55,381,1343,3054,
%U 4849,5502,4358,2244,577,89,696,2769,7173,13218,17895,17838,12666
%N Triangle of coefficients of polynomials u(n,x) jointly generated with A209140; see the Formula section.
%C Column 1: A000045 (Fibonacci numbers).
%C Alternating row sums: 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, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1, 2, -1, 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) = (x+1)*u(n-1,x) + 2x*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.: (1-2*y*x-y*x^2-y^2*x^2)/(1-x-x^2-2*y*x-y^2*x^2).
%F T(n,k) = T(n-1,k) + 2*T(n-1,k-1) + T(n-2,k) + T(n-2,k-2), T(0,0) = T(1,0) = T(2,1) = 1, T(2,0) = 2, 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 2, 1;
%e 3, 5, 3;
%e 5, 12, 15, 7;
%e 8, 27, 45, 42, 17;
%e First three polynomials u(n,x):
%e 1
%e 2 + x
%e 3 + 5x + 3x^2
%e From _Philippe Deléham_, Apr 11 2012: (Start)
%e (1, 1, -1, 0, 0, 0, ...) DELTA (0, 1, 2, -1, 0, 0, ...) begins:
%e 1;
%e 1, 0;
%e 2, 1, 0;
%e 3, 5, 3, 0;
%e 5, 12, 15, 7, 0;
%e 8, 27, 45, 42, 17, 0;
%e 13, 55, 119, 151, 116, 41, 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_] := (x + 1)*u[n - 1, x] + 2 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[%] (* A209139 *)
%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[%] (* A209140 *)
%Y Cf. A209140, A208510.
%K nonn,tabl
%O 1,2
%A _Clark Kimberling_, Mar 05 2012
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