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%I #13 May 13 2017 14:18:47
%S 1,3,2,6,9,4,10,25,24,8,15,55,85,60,16,21,105,231,258,144,32,28,182,
%T 532,833,728,336,64,36,294,1092,2241,2720,1952,768,128,45,450,2058,
%U 5301,8361,8280,5040,1728,256,55,660,3630,11385,22363,28610,23920
%N Triangle of coefficients of polynomials v(n,x) jointly generated with A210753; see the Formula section.
%C Column 1: triangular numbers, A000217
%C Coefficient of v(n,x): 2^(n-1)
%C Row sums: A035344
%C Alternating row sums: 1,1,1,1,1,1,1,1,1,...
%C For a discussion and guide to related arrays, see A208510.
%C Appears to be the reversed row polynomials of A165241 with the unit diagonal removed. If so, the o.g.f. is [1-(1+y)x]/[1-2(1+y)x+(1+y)x^2] - 1/(1-x) and the triangular matrix here may be formed by adding each column of the matrix of A056242, presented in the example section with the additional zeros, to its subsequent column with the first row ignored. - _Tom Copeland_, Jan 09 2017
%F u(n,x)=(x+1)*u(n-1,x)+x*v(n-1,x)+1,
%F v(n,x)=(x+1)*u(n-1,x)+(x+1)*v(n-1,x)+1,
%F where u(1,x)=1, v(1,x)=1.
%e First five rows:
%e 1
%e 3....2
%e 6....9....4
%e 10...25...24...8
%e 15...55...85...60...16
%e First three polynomials v(n,x): 1, 3 + 2x, 6 + 9x +4x^2
%t u[1, x_] := 1; v[1, x_] := 1; z = 16;
%t u[n_, x_] := (x + 1)*u[n - 1, x] + x*v[n - 1, x] + 1;
%t v[n_, x_] := (x + 1)*u[n - 1, x] + (x + 1)*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[%] (* A210753 *)
%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[%] (* A210754 *)
%t Table[u[n, x] /. x -> 1, {n, 1, z}] (* A007070 *)
%t Table[v[n, x] /. x -> 1, {n, 1, z}] (* A035344 *)
%Y Cf. A210753, A208510.
%Y Cf. A056242, A165241.
%K nonn,tabl
%O 1,2
%A _Clark Kimberling_, Mar 25 2012
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