A124020 - OEIS

%I #4 Oct 12 2012 14:55:30

%S 1,1,-1,0,-3,1,-1,-4,5,-1,-1,-2,12,-7,1,0,2,17,-24,9,-1,1,4,13,-52,40,

%T -11,1,1,1,0,-76,115,-60,13,-1,0,-5,-10,-72,235,-214,84,-15,1,-1,-8,

%U -2,-34,352,-554,357,-112,17,-1,-1,-4,24,2,383,-1092,1113,-552,144,-19,1,0,4,46,-24,295,-1673,2688,-2008,807,-180,21,-1

%N New tetradiagonal form matrix as triangular sequence from solution of : X(n,m)=Steinbach(n,m)^(-1).tri-Antidiagonal_1(n,n).

%C Matrices {{1}}, {{1, -1}, {-2, 2}}, {{1, -1, 0}, {-2, 2, -1}, {1, -2, 2}}, {{1, -1, 0, 0}, {-2, 2, -1, 0}, {1, -2, 2, -1}, {0, 1, -2, 2}}, {{1, -1, 0, 0, 0}, {-2, 2, -1, 0, 0}, {1, -2, 2, -1, 0}, {0, 1, -2, 2, -1}, {0, 0, 1, -2, 2}} Large root sequence: a0 = Table[x /. NSolve[CharacteristicPolynomial[Xn[d], x] == 0, x][[d]], {d, 1, 20}] {1., 3., 3.91223, 4.37167, 4.62826, 4.78478, 4.88683, 4.95691, 5.00703, 5.0441, 5.07226, 5.09415, 5.1115, 5.12547, 5.1369, 5.14636, 5.15428, 5.16097, 5.16668, 5.17159} Determinant sequence is: Table[Det[Xn[d]], {d, 1, 20}] {1,0, -1, -1, 0, 1, 1, 0, -1, -1, 0, 1, 1, 0, -1, -1, 0, 1, 1, 0}

%F Steinbach(n,m)=If[n + m - 1 > d, 0, 1]; tri-Antidiagonal_1(n,m)=If[n + m - 1 == d, 1, If[n + m == d, -1, If[n + m - 2 == d, -1, 0]]]; X(n,m)=Steinbach(n,m)^(-1).tri-Antidiagonal2(n,n)

%e triangular sequence:

%e {1},

%e {1, -1},

%e {0, -3, 1},

%e {-1, -4,5, -1},

%e {-1, -2, 12, -7, 1},

%e {0, 2, 17, -24, 9, -1},

%e {1, 4,13, -52, 40, -11,1},

%e {1, 1, 0, -76, 115, -60, 13, -1},

%e {0, -5, -10, -72, 235, -214, 84, -15,1}

%t An[d_] := Table[If[n + m - 1 > d, 0, 1], {n, 1, d}, {m, 1, d}]; Bn[d_] := Table[If[n + m - 1 == d, 1, If[n + m == d, -1, If[n + m - 2 == d, -1, 0]]], {n, 1, d}, {m, 1, d}]; Xn[d_] := MatrixPower[An[d], -1].Bn[d]; a = Join[Xn[1], Table[CoefficientList[CharacteristicPolynomial[Xn[d], x], x], {d, 1, 20}]]; Flatten[%]

%K sign,tabl,uned

%O 1,5

%A _Roger L. Bagula_ and _Gary W. Adamson_, Oct 31 2006