A208332 - OEIS

%I #17 Jan 22 2020 05:15:16

%S 1,1,1,1,1,4,1,1,6,10,1,1,8,16,28,1,1,10,22,52,76,1,1,12,28,80,156,

%T 208,1,1,14,34,112,256,472,568,1,1,16,40,148,376,832,1408,1552,1,1,18,

%U 46,188,516,1296,2640,4176,4240,1,1,20,52,232,676,1872,4320

%N Triangle of coefficients of polynomials u(n,x) jointly generated with A208333; see the Formula section.

%C Subtriangle of the triangle given by (1, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1, 3, -2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - _Philippe Deléham_, Apr 09 2012

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

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

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

%F From _Philippe Deléham_, Apr 09 2012: (Start)

%F As DELTA-triangle T(n,k) with 0 <= k <= n:

%F G.f.: (1 - 2*y*x + y*x^2 - 2*y^2*x^2)/(1 - x - 2*y*x + 2*y*x^2 - 2*y^2*x^2).

%F T(n,k) = T(n-1,k) + 2*T(n-1,k-1) - 2*T(n-2,k-1) + 2*T(n-2,k-2), T(0,0) = T(1,0) = T(2,0) = T(2,1) = 1, 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   1, 1;

%e   1, 1, 4;

%e   1, 1, 6, 10;

%e   1, 1, 8, 16, 28;

%e First five polynomials u(n,x):

%e 1, 1 + x, 1 + x + 4x^2, 1 + x + 6x^2 + 10x^3, 1 + x + 8x^2 + 16x^3 + 28x^4.

%e From _Philippe Deléham_, Apr 09 2012: (Start)

%e (1, 0, -1, 1, 0, 0, 0, ...) DELTA (0, 1, 3, -2, 0, 0, 0, ...) begins:

%e   1;

%e   1, 0;

%e   1, 1, 0;

%e   1, 1, 4, 0;

%e   1, 1, 6, 10, 0;

%e   1, 1, 8, 16, 28, 0;

%e   1, 1, 10, 22, 52, 76, 0;

%e   1, 1, 12, 28, 80, 156, 208, 0;

%e   ... (End)

%t u[1, x_] := 1; v[1, x_] := 1; z = 13;

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

%t v[n_, x_] := 2 x*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[%]  (* A208332 *)

%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[%]  (* A208333 *)

%Y Cf. A208332.

%K nonn,tabl

%O 1,6

%A _Clark Kimberling_, Feb 26 2012