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%I #16 Feb 18 2020 10:23:33
%S 1,1,1,1,4,1,1,13,5,1,1,40,18,8,1,1,121,58,42,9,1,1,364,179,184,51,12,
%T 1,1,1093,543,731,235,87,13,1,1,3280,1636,2736,966,496,100,16,1,1,
%U 9841,4916,9844,3702,2454,596,148,17,1,1,29524,14757,34448,13546,11064,3050,1040,165,20,1
%N Triangle read by rows, iterates of matrix X * [1,0,0,0,...], where X = an infinite lower bidiagonal matrix with [1,3,1,3,1,3,...] in the main diagonal and [1,1,1,...] in the subdiagonal.
%C Row sums = A006012: (1, 2, 6, 20, 68, 232, 792, 2704,...).
%C Companion to triangle A140071.
%F Triangle read by rows, iterates of matrix X * [1,0,0,0,...], where X = an infinite lower bidiagonal matrix with [1,3,1,3,1,3,...] in the main diagonal and [1,1,1,...] in the subdiagonal; with the rest zeros.
%F From _Peter Bala_, Jan 17 2014: (Start)
%F O.g.f.: (1 + (x - 3)*z)/(1 - 4*z - (x^2 - 3)*z^2) = 1 + (x + 1)*z + (x^2 + 4*x + 1)*z^2 + ....
%F Recurrence equation: T(n,k) = 4*T(n-1,k) - 3*T(n-2,k) + T(n-2,k-2).
%F Recurrence equation for row polynomials: R(n,x) = 4*R(n-1,x) + (x^2 - 3)*R(n-2,x) with R(0,x) = 1 and R(1,x) = 1 + x.
%F Another recurrence equation: R(n,x) = (x + 2)*R(n-1,x) - R(n-1,-x) with R(0,x) = 1. Cf. A157751. (End)
%e First few rows of the triangle are:
%e 1;
%e 1, 1;
%e 1, 4, 1;
%e 1, 13, 5, 1;
%e 1, 40, 18, 8, 1;
%e 1, 121, 58, 42, 9, 1;
%e 1, 364, 179, 184, 51, 12, 1;
%e 1, 1093, 543, 731, 235, 87, 13, 1;
%e 1, 3280, 1636, 2736, 966, 496, 100, 16, 1;
%e ...
%p T:= proc(n, k) option remember;
%p `if`(k<0 or k>n, 0, `if`(k=0 or k=n, 1,
%p 4*T(n-1, k) - 3*T(n-2, k) + T(n-2, k-2)))
%p end:
%p seq(seq(T(n, k), k=0..n), n=0..10); # _Alois P. Heinz_, Feb 18 2020
%t With[{m = 10}, CoefficientList[CoefficientList[Series[(1 + (y - 3)*x)/(1 - 4*x - (y^2 - 3)*x^2), {x, 0, m}, {y, 0, m}], x], y]] // Flatten (* _Georg Fischer_, Feb 18 2020 *)
%Y Cf. A006012, A140071, A157751.
%K nonn,tabl
%O 1,5
%A _Gary W. Adamson_, May 04 2008
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