%I #2 Mar 30 2012 17:34:35
%S -1,0,-1,1,-1,1,2,-1,2,-1,3,-1,3,-3,1,4,-1,4,-6,4,-1,5,-1,5,-10,10,-5,
%T 1,6,-1,6,-15,20,-15,6,-1,7,-1,7,-21,35,-35,21,-7,1,8,-1,8,-28,56,-70,
%U 56,-28,8,-1,9,-1,9,-36,84,-126,126,-84,36,-9,1
%N Coefficients of characteristic polynomials for a two diagonal Matrix type that has determinant equal to trace:M(n)=Table[If[ k == m && m < n, 1, If[k == m + 1, 1, If[k == 1 && m == n, (-1)^(n + 1)*(n - 1), 0]]], {k, n}, {m, n}]
%C This set of matrices was constructed so that the determinant would be equal to the trace.
%C This system is interesting because the result contains a signed Pascal's triangle.
%C The initial term is adjusted to {-1} to fit the first column.
%C Row sums are:{-1, -1, 1, 2, 3, 4, 5, 6, 7, 8, 9,...}.
%e {-1},
%e {0, -1},
%e {1, -1, 1},
%e {2, -1, 2, -1},
%e {3, -1, 3, -3, 1},
%e {4, -1, 4, -6, 4, -1},
%e {5, -1, 5, -10, 10, -5, 1},
%e {6, -1, 6, -15, 20, -15, 6, -1},
%e {7, -1, 7, -21, 35, -35, 21, -7, 1},
%e {8, -1, 8, -28, 56, -70, 56, -28, 8, -1},
%e {9, -1, 9, -36, 84, -126, 126, -84, 36, -9, 1}
%t Clear[M, n, m, k]
%t M[n_] := Table[If[ k == m && m < n, 1, If[k == m + 1, 1, If[k == 1 && m == n, (-1)^(n + 1)*(n - 1), 0]]], {k, n}, {m, n}]
%t TableForm[Table[M[n], {n, 1, 10}]]
%t Table[Det[M[n]], {n, 1, 10}]
%t Table[Sum[M[n][[k, k]], {k, 1, n}], {n, 1, 10}]
%t a = Join[{{-1}}, Table[CoefficientList[CharacteristicPolynomial[M[n], x], x], {n, 1, 10}]]
%t Flatten[a]
%K uned,sign
%O 0,7
%A _Roger L. Bagula_, Dec 02 2009