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%I #8 Mar 30 2012 18:58:17
%S 1,2,3,4,9,8,7,24,33,21,12,54,109,111,55,20,114,297,435,355,144,33,
%T 228,736,1383,1606,1098,377,54,441,1697,3912,5813,5625,3316,987,88,
%U 831,3723,10158,18419,22779,18962,9837,2584,143,1536,7859,24798
%N Triangle of coefficients of polynomials u(n,x) jointly generated with A210748; see the Formula section.
%C Row n starts with -1+F(n+2) and ends with F(2n), where F=A000045 (Fibonacci numbers).
%C Row sums: A002450
%C Alternating row sums: A077925
%C For a discussion and guide to related arrays, see A208510.
%F u(n,x)=2x*u(n-1,x)+(x+1)*v(n-1,x)+1,
%F v(n,x)=(x+1)*u(n-1,x)+(x+1)*v(n-1,x)+1,
%F where u(1,x)=1, v(1,x)=1.
%e First five rows:
%e 1
%e 2....3
%e 4....9....8
%e 7....24...33....21
%e 12...54...109...111...55
%e First three polynomials u(n,x): 1, 2+ 3x, 4 + 9x + 8x^2.
%t u[1, x_] := 1; v[1, x_] := 1; z = 16;
%t u[n_, x_] := 2 x*u[n - 1, x] + (x + 1)*v[n - 1, x] + 1;
%t v[n_, x_] := (x + 1)*u[n - 1, x] + (x + 1)*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[%] (* A210747 *)
%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[%] (* A210748 *)
%t Table[u[n, x] /. x -> 1, {n, 1, z}] (* A002450 *)
%t Table[v[n, x] /. x -> 1, {n, 1, z}] (* A002450 *)
%t Table[u[n, x] /. x -> -1, {n, 1, z}] (* A077925 *)
%t Table[v[n, x] /. x -> -1, {n, 1, z}] (* A000012 *)
%Y Cf. A208510.
%K nonn,tabl
%O 1,2
%A _Clark Kimberling_, Mar 25 2012
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