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%I #10 Aug 09 2015 01:15:19
%S 0,0,-1,-1,-1,1,-1,1,2,-1,0,3,0,-3,1,1,2,-5,-2,4,-1,1,-2,-7,6,5,-5,1,
%T 0,-5,0,15,-5,-9,6,-1,-1,-3,12,9,-25,1,14,-7,1,-1,3,15,-18,-29,35,7,
%U -20,8,-1,0,7,0,-42,14,63,-42,-20,27,-9,1
%N Sequence of tridiagonal matrices with one center zero terminal that give a triangular sequence from the characteristic polynomials based on the 3 X 3 matrix type: {{1, -1, 0}, {-1, 1, -1}, {0, -1, 0}}.
%C Matrices: {{0}}, {{1, -1}, {-1, 0}}, {{1, -1, 0}, {-1, 1, -1}, {0, -1, 0}}, {{1, -1, 0, 0}, {-1, 1, -1, 0}, {0, -1, 1, -1}, {0, 0, -1, 0}}, {{1, -1, 0, 0, 0}, {-1, 1, -1, 0, 0}, {0, -1, 1, -1, 0}, {0, 0, -1, 1, -1}, {0, 0, 0, -1, 0}}, {{1, -1, 0, 0, 0, 0}, {-1, 1, -1, 0, 0, 0}, {0, -1, 1, -1, 0, 0}, {0, 0, -1, 1, -1, 0}, { 0, 0, 0, -1, 1, -1}, {0, 0, 0, 0, -1, 0}} Determinants ( not all Sl(3,Z) and invertable): Table[Det[M[d]], {d, 1, 10}] {0, -1, -1, 0, 1, 1, 0, -1, -1, 0}
%F t(n,m,d)=If[ n == m && n < d && m < d, 1, If[n == m - 1 || n == m + 1, -1, If[n == m == d, 0, 0]]]
%e Triangle begins:
%e {0},
%e {0, -1},
%e {-1, -1, 1},
%e {-1, 1, 2, -1},
%e {0, 3, 0, -3, 1},
%e {1, 2, -5, -2, 4, -1},
%e {1, -2, -7, 6, 5, -5, 1},
%e {0, -5, 0, 15, -5, -9, 6, -1},
%e {-1, -3, 12, 9, -25, 1, 14, -7, 1},
%e {-1, 3, 15, -18, -29, 35,7, -20, 8, -1},
%e {0, 7, 0, -42, 14, 63, -42, -20, 27, -9, 1}
%e Some of the polynomials are Steinbach.
%t T[n_, m_, d_] := If[ n == m && n < d && m < d, 1, If[n == m - 1 || n == m + 1, -1, If[n == m == d, 0, 0]]] M[d_] := Table[T[n, m, d], {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] MatrixForm[a]
%K uned,tabl,sign
%O 1,9
%A _Gary W. Adamson_ and _Roger L. Bagula_, Oct 30 2006
%E Looking at the triangle suggests that the very first term should be 1, not 0. - _N. J. A. Sloane_, Nov 01 2006
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