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Triangle of coefficients of polynomials u(n,x) jointly generated with A210859; see the Formula section.
3

%I #7 Mar 30 2012 18:58:17

%S 1,1,1,1,3,2,1,6,8,3,1,10,24,20,5,1,15,59,82,45,8,1,21,125,271,251,98,

%T 13,1,28,237,765,1073,701,204,21,1,36,413,1898,3860,3802,1842,414,34,

%U 1,45,674,4235,12115,17011,12325,4589,820,55,1,55,1044,8662

%N Triangle of coefficients of polynomials u(n,x) jointly generated with A210859; see the Formula section.

%C Row n starts with 1 and ends with F(n), where F=A000045 (Fibonacci numbers).

%C Alternating row sums: 1,0,0,0,0,0,0,0,...

%C For a discussion and guide to related arrays, see A208510.

%F u(n,x)=u(n-1,x)+x*v(n-1,x),

%F v(n,x)=(x+n)*u(n-1,x)+x*v(n-1,x),

%F where u(1,x)=1, v(1,x)=1.

%e First five rows:

%e 1

%e 1...1

%e 1...3....2

%e 1...6....8....3

%e 1...10...24...20...5

%e First three polynomials u(n,x): 1, 1 + x, 1 + 3x + 2x^2.

%t u[1, x_] := 1; v[1, x_] := 1; z = 14;

%t u[n_, x_] := u[n - 1, x] + x*v[n - 1, x];

%t v[n_, x_] := (x + n)*u[n - 1, x] + 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[%] (* A210858 *)

%t cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];

%t TableForm[cv]

%t Flatten[%] (* A210859 *)

%Y Cf. A210806, A208510.

%K nonn,tabl

%O 1,5

%A _Clark Kimberling_, Mar 28 2012