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%I #5 Mar 30 2012 18:58:15
%S 1,1,1,1,4,1,1,4,9,1,1,4,15,16,1,1,4,15,44,25,1,1,4,15,56,105,36,1,1,
%T 4,15,56,185,216,49,1,1,4,15,56,209,524,399,64,1,1,4,15,56,209,732,
%U 1295,680,81,1,1,4,15,56,209,780,2303,2864,1089,100,1,1,4,15,56
%N Triangle of coefficients of polynomials u(n,x) jointly generated with A209572; see the Formula section.
%C Penultimate number in row n is (n-1)^2, for n>1.
%C Combinatorial limit of row n satisfies linear recurrence
%C r(n)=4*r(n-1)-r(n-2) with r(1)=1 and r(2)=4. For a
%C discussion and guide to related arrays, see A208510.
%F u(n,x)=x*u(n-1,x)+v(n-1,x),
%F v(n,x)=2x*u(n-1,x)+x*v(n-1,x)+1,
%F where u(1,x)=1, v(1,x)=1.
%e First five rows:
%e 1
%e 1...1
%e 1...4....1
%e 1...4....9....1
%e 1...4....15...16...1
%e First three polynomials v(n,x): 1, 1 + x, 1 + 4x + x^2.
%t u[1, x_] := 1; v[1, x_] := 1; z = 16;
%t u[n_, x_] := x*u[n - 1, x] + v[n - 1, x];
%t v[n_, x_] := 2 x*u[n - 1, x] + 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[%] (* A209571 *)
%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[%] (* A209572 *)
%Y Cf. A209572, A208510.
%K nonn,tabl
%O 1,5
%A _Clark Kimberling_, Mar 11 2012
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