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Triangle of coefficients of polynomials u(n,x) jointly generated with A164975; see the Formula section.
3

%I #16 Jan 24 2020 03:26:16

%S 1,2,1,3,4,2,5,9,9,4,8,20,25,20,8,13,40,65,65,44,16,21,78,150,190,162,

%T 96,32,34,147,331,490,521,392,208,64,55,272,697,1192,1473,1368,928,

%U 448,128,89,495,1425,2745,3888,4185,3480,2160,960,256,144,890

%N Triangle of coefficients of polynomials u(n,x) jointly generated with A164975; see the Formula section.

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

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

%C Subtriangle of the triangle given by (1, 1, -1, 0, 0, 0, 0, 0, 0, 0, ....) DELTA (0, 1, 1, 0, 0, 0, 0, 0, 0, 0, ....) where DELTA is the operator defined in A084938. - _Philippe Deléham_, Mar 21 2012

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

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

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

%F From _Philippe Deléham_, Mar 21 2012: (Start)

%F As DELTA-triangle with 0 <= k <= n:

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

%F T(n,k) = T(n-1,k) + 2*T(n-1,k-1) + T(n-2,k) - T(n-2,k-1), T(0,0) = T(1,0) = T(2,1) = 1, T(1,1) = T(2,2) = 0, T(2,0) = 2, T(n,k) = 0 f k < 0 or if k > n. (End)

%e First five rows:

%e 1;

%e 2, 1;

%e 3, 4, 2;

%e 5, 9, 9, 4;

%e 8, 20, 25, 20, 8;

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

%e 1

%e 2 + x

%e 3 + 4x + 2x^2

%e From _Philippe Deléham_, Mar 21 2012: (Start)

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

%e 1;

%e 1, 0;

%e 2, 1, 0;

%e 3, 4, 2, 0;

%e 5, 9, 9, 4, 0;

%e 8, 20, 25, 20, 8, 0; (End)

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

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

%t v[n_, 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[%] (* A209125 *)

%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[%] (* A164975 *)

%Y Cf. A164975, A208510.

%K nonn,tabl

%O 1,2

%A _Clark Kimberling_, Mar 05 2012