A124034 Tri-antidiagonal matrices of central ones with upper negative one to give a triangular sequence: first element is negative one. k=1;m(n,m,d)=If[n + m - 1 == d && n > 1, k, If[n + m ==d, -1, If[n + m - 2 == d, -1, If[n == 1 && m == d, -k, 0]]]].

-1, -1, -1, 2, 2, 1, 1, 1, 1, -1, 1, 0, -1, 2, 1, 2, 0, 0, 4, 1, -1, -1, 0, -1, -6, -4, 2, 1, 1, 1, -1, -7, -3, 7, 1, -1, -2, -2, -1, 8, 6, -12, -7, 2, 1, -1, -1, -2, 10, 3, -23, -6, 10, 1, -1, -1, 0, 2, -12, -7, 34, 22, -18, -10, 2, 1, -2, 0, 0, -16, -4, 52, 16, -48, -9, 13, 1, -1, 1, 0, 2, 20, 13, -70, -46, 78, 47, -24, -13, 2, 1, -1, -1, 2, 22, 9

Offset

1,4

Comments

Matrices: 1 X 1 {{-1}}, 2 X 2 {{-1, -1}, {1, -1}}, 3 X 3 {{0, -1, -1}, {-1, 1, -1}, {1, -1, 0}}, 4 X 4 {{0, 0, -1, -1}, {0, -1, 1, -1}, {-1, 1, -1, 0}, {1, -1, 0, 0}}, 5 X 5 {{0,0, 0, -1, -1}, {0, 0, -1, 1, -1}, {0, -1, 1, -1, 0}, {-1, 1, -1, 0, 0}, {1, -1, 0, 0, 0}}

Links

Table of n, a(n) for n=1..96.

Formula

k=1; m(n,m,d)=If[n + m - 1 == d && n > 1, k, If[n + m ==d, -1, If[n + m - 2 == d, -1, If[n == 1 && m == d, -k, 0]]]]

Example

Triangular sequence:
{-1},
{-1, -1},
{2, 2, 1},
{1, 1, 1, -1},
{1, 0, -1, 2, 1},
{2, 0, 0, 4,1, -1},
{-1, 0, -1, -6, -4, 2, 1},
{1, 1, -1, -7, -3, 7, 1, -1},
{-2, -2, -1, 8, 6, -12, -7, 2, 1},
{-1, -1, -2, 10, 3, -23, -6, 10, 1, -1},
{-1, 0, 2, -12, -7, 34, 22, -18, -10, 2, 1}

Mathematica

k = 1; An[d_] := Table[If[n + m - 1 == d && n > 1, k, If[n + m == d, -1, If[n + m - 2 == d, -1, If[n == 1 &&m == d, -k, 0]]]], {n, 1, d}, {m, 1, d}]; Join[An[1], Table[CoefficientList[CharacteristicPolynomial[An[d], x], x], {d, 1, 20}]]; Flatten[%]

Crossrefs

Sequence in context: A237422 A102552 A131341 * A332029 A211312 A085978

Adjacent sequences: A124031 A124032 A124033 * A124035 A124036 A124037

Keyword

uned,sign,tabl

Author

Roger L. Bagula, Nov 02 2006

Status

approved