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A140070
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.
4
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, 1, 1, 1093, 543, 731, 235, 87, 13, 1, 1, 3280, 1636, 2736, 966, 496, 100, 16, 1, 1, 9841, 4916, 9844, 3702, 2454, 596, 148, 17, 1, 1, 29524, 14757, 34448, 13546, 11064, 3050, 1040, 165, 20, 1
OFFSET
1,5
COMMENTS
Row sums = A006012: (1, 2, 6, 20, 68, 232, 792, 2704,...).
Companion to triangle A140071.
FORMULA
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.
From Peter Bala, Jan 17 2014: (Start)
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 + ....
Recurrence equation: T(n,k) = 4*T(n-1,k) - 3*T(n-2,k) + T(n-2,k-2).
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.
Another recurrence equation: R(n,x) = (x + 2)*R(n-1,x) - R(n-1,-x) with R(0,x) = 1. Cf. A157751. (End)
EXAMPLE
First few rows of the triangle are:
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, 1;
1, 1093, 543, 731, 235, 87, 13, 1;
1, 3280, 1636, 2736, 966, 496, 100, 16, 1;
...
MAPLE
T:= proc(n, k) option remember;
`if`(k<0 or k>n, 0, `if`(k=0 or k=n, 1,
4*T(n-1, k) - 3*T(n-2, k) + T(n-2, k-2)))
end:
seq(seq(T(n, k), k=0..n), n=0..10); # Alois P. Heinz, Feb 18 2020
MATHEMATICA
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 *)
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Gary W. Adamson, May 04 2008
STATUS
approved