%I #8 Sep 03 2021 13:58:15

%S 1,2,3,2,8,9,2,10,30,27,2,10,46,108,81,2,10,50,198,378,243,2,10,50,

%T 242,810,1296,729,2,10,50,250,1122,3186,4374,2187,2,10,50,250,1234,

%U 4986,12150,14580,6561,2,10,50,250,1250,5946,21330,45198,48114,19683

%N Triangle of coefficients of polynomials v(n,x) jointly generated with A209996; see the Formula section.

%C Row n starts 2, 2*5, 2*5^2,... ; ends with 3^(n-1).

%C Conjecture: penultimate term in row n is A199923(n).

%C Alternating row sums: A077925

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

%F u(n,x)=x*u(n-1,x)+2x*v(n-1,x)+1,

%F v(n,x)=(x+1)*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 2...3

%e 2...8....9

%e 2...10...30...27

%e 2...10...46...108...81

%e First three polynomials v(n,x): 1, 2 + 3x , 2 + 8x + 9x^2.

%t u[1, x_] := 1; v[1, x_] := 1; z = 16;

%t u[n_, x_] := x*u[n - 1, x] + 2 x*v[n - 1, x] + 1;

%t v[n_, x_] := (x + 1)*u[n - 1, x] + 2 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[%] (* A209996 *)

%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[%] (* A209998 *)

%Y Cf. A209996, A208510.

%K nonn,tabl

%O 1,2

%A _Clark Kimberling_, Mar 23 2012

%E a(55) corrected by _Georg Fischer_, Sep 03 2021