OFFSET
1,5
COMMENTS
Matrices: {{1}}, {{1, -1}, {-1, 1}}, {{1, -1, 0}, {-1, 1, -1}, {0, -1, 2}}, {{1, -1, 0, 0}, {-1, 1, -1, 0}, {0, -1, 2, -1}, {0, 0, -1, 3}}, {{1, -1, 0, 0, 0}, {-1, 1, -1, 0, 0}, {0, -1, 2, -1, 0}, {0, 0, -1, 3, -1}, {0, 0, 0, -1, 5}}, {{1, -1, 0, 0, 0, 0}, {-1, 1, -1, 0, 0, 0}, {0, -1, 2, -1, 0, 0}, {0, 0, -1, 3, -1, 0}, {0, 0, 0, -1, 5, -1}, {0, 0, 0, 0, -1, 8}} The Dombrowski paper defines a recursive polynomial form from the tridiagonal matrices: p[1,x]=1,p[2,x]=(x-b[1])/a[1] p[n,x]=((x-b[n-1])*p[n-1,x]-a[n-2]*p[n-2,x])/a[n-1] As long as b[n-1]/a[n-1] and a[n-2]/a[n-1] behave well ( rationally or like Integers) this definition is a good recursive polynomial on a tridiagonal matrix. Here I use: a[n]=-1 and b[n]=Fibonacci[n]
REFERENCES
Joanne Dombrowski, Tridiagonal matrix representations of cyclic selfadjoint operators, Pacific J. Math. 114, no. 2 (1984), 325-334
FORMULA
M(n,m)=If[ n == m, Fibonacci[n], If[n == m - 1 || n == m + 1, -1, 0]]
EXAMPLE
Triangular sequence:
{1},
{1, -1},
{0, -2, 1},
{-1, -3, 4, -1},
{-3, -6, 14, -7, 1},
{-14, -24, 72, -48, 12, -1},
{-109, -172, 586, -449, 143, -20, 1},
{-1403, -2103, 7718, -6375,2296, -402, 33, -1},
{-29354, -42588, 163595, -141144, 54448, -10718, 1094, -54, 1}
MATHEMATICA
T[n_, m_] := If[ n == m, Fibonacci[n], If[n == m - 1 || n == m + 1, -1, 0]]; M[d_] := Table[T[n, m], {n, 1, d}, {m, 1, d}]; Table[M[d], {d, 1, 10}] Table[Det[M[d]], {d, 1, 10}] Table[Det[M[d] - x*IdentityMatrix[d]], {d, 1, 10}] a = Join[M[1], Table[CoefficientList[Det[M[d] - x*IdentityMatrix[d]], x], {d, 1, 10}]]; Flatten[a]
CROSSREFS
KEYWORD
uned,sign
AUTHOR
Roger L. Bagula and Gary W. Adamson, Oct 30 2006
STATUS
approved
