A208761 - OEIS

%I #20 Jan 24 2020 03:27:59

%S 1,1,2,1,6,4,1,12,18,8,1,20,52,50,16,1,30,120,186,126,32,1,42,240,534,

%T 576,306,64,1,56,434,1302,1986,1654,718,128,1,72,728,2828,5712,6632,

%U 4484,1650,256,1,90,1152,5628,14436,21912,20508,11682,3726,512

%N Triangle of coefficients of polynomials u(n,x) jointly generated with A208762; see the Formula section.

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

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

%C Subtriangle of the triangle given by [1,0,1,0,0,0,0,...] DELTA [0,2,0,-1,0,0,0,0,...] where DELTA is the operator defined in A084938. - _Philippe Deléham_, Mar 04 2012

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

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

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

%F Recurrence: T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k) + T(n-2,k-1) + 2*T(n-2,k-2). - _Philippe Deléham_, Mar 04 2012

%F G.f.: (-1-x*y+x)*x*y/(-1+x*y+2*x+2*x^2*y^2+x^2*y-x^2). - _R. J. Mathar_, Aug 12 2015

%e First five rows:

%e   1;

%e   1,  2;

%e   1,  6,  4;

%e   1, 12, 18,  8;

%e   1, 20, 52, 50, 16;

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

%e   1

%e   1 +  2x

%e   1 +  6x +  4x^2

%e   1 + 12x + 18x^2 +  8x^3

%e   1 + 20x + 52x^2 + 50x^3 + 16x^4

%e From _Philippe Deléham_, Mar 04 2012: (Start)

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

%e   1;

%e   1,   0;

%e   1,   2,   0;

%e   1,   6,   4,   0;

%e   1,  12,  18,   8,   0;

%e   1,  20,  52,  50,  16,   0;

%e   1,  30, 120, 186, 126,  32,  0; (End)

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

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

%t v[n_, x_] := (x + 1)*u[n - 1, x] + (x + 1) 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[%]    (* A208761 *)

%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[%]    (* A208762 *)

%Y Cf. A208762, A208510.

%K nonn,tabl

%O 1,3

%A _Clark Kimberling_, Mar 03 2012