A122374 - OEIS

%I #16 Apr 30 2018 11:41:38

%S 1,1,-1,-1,-1,1,-1,1,2,-1,1,-1,-4,-1,1,1,-3,-3,4,2,-1,-1,3,7,-2,-7,-1,

%T 1,-1,5,4,-11,-5,7,2,-1,1,-5,-10,9,18,-3,-10,-1,1,1,-7,-5,22,9,-24,-7,

%U 10,2,-1,-1,7,13,-20,-34,18,34,-4,-13,-1,1,-1,9,6,-37,-14,58,16,-42,-9,13,2,-1,1,-9,-16,35,55,-50,-80,30,55,-5,-16,-1,1

%N Triangle in which row n gives the coefficients of det(A-xI), where A is the n X n matrix with 1's on antidiagonal and last row and column, 0's elsewhere.

%e Triangle starts:

%e { 1},

%e { 1,  -1},

%e {-1,  -1,   1},

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

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

%e { 1,  -3,  -3,   4,   2,  -1},

%e {-1,   3,   7,  -2,  -7,  -1,   1},

%e {-1,   5,   4, -11,  -5,   7,   2,  -1},

%e { 1,  -5, -10,   9,  18,  -3, -10,  -1,   1},

%e { 1,  -7,  -5,  22,   9, -24,  -7,  10,   2,  -1}

%e ...

%e From _M. F. Hasler_, Apr 30 2018: (Start)

%e For n = 0, the determinant of the 0 X 0 matrix is 1 by convention, which yields row 0 = [ 1 ].

%e For n = 1, we have det [1 - x] = 1 - x, which yields row 1 = [1, -1].

%e For n = 2, we have det [-x, 1; 1, 1 - x] = x(x - 1) - 1 = x^2 - x - 1; in order of increasing powers this yields row 2 = [-1, -1, +1]. (End)

%o (PARI) A122374_row(n)=(-1)^n*Vecrev(charpoly(matrix(n,n,i,j,i==n||j==n||i+j==n+1),x)) \\ Yields the n-th row. - _M. F. Hasler_, Apr 26 2018

%K sign,easy,less,tabl

%O 0,9

%A _Roger L. Bagula_ and _Gary W. Adamson_, Oct 19 2006

%E Edited by _M. F. Hasler_, Apr 26 2018