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%I #6 Mar 12 2014 16:54:04
%S 1,0,-1,-1,0,1,0,-3,0,-1,-3,0,2,0,1,0,-5,0,-10,0,-1,-5,0,-5,0,9,0,1,0,
%T -7,0,-35,0,-21,0,-1,-7,0,-28,0,14,0,20,0,1,0,-9,0,-84,0,-126,0,-36,0,
%U -1,-9,0,-75,0,-42,0,90,0,35,0,1,0,-11,0,-165,0,-462,0,-330,0,-55,0,-1,-11,0,-154,0,-297,0,132,0,275,0,54,0,1,0,-13,0
%N Triangle read by rows: T(0,0)=1; for n>=1, 0<=k<=n, T(n,k) is the coefficient of x^k in the characteristic polynomial (-x)^n+... of the n X n matrix M(n)S(n), where M(n) is the n X n matrix with 0's on the diagonal and 1's elsewhere and S(n) is the n X n matrix whose (i,j) term is 0 for j=i, (-1)^(i+j) for i>j and (-1)^(i+j+1) for i<j.
%C Sum of terms in row 2n (n>=1) is 0. Sum of the absolute values of the terms in row 2n is C(2n,n) (A000984). All terms in row 2n-1 are nonpositive. Their sum is -4^(n-1). M(2n-1)S(2n-1)=-S(2n-1)
%e M(4)=[0,1,1,1/1,0,1,1/1,1,0,1/1,1,1,0], S(4)=[0,1,-1,1/-1,0,1,-1/1,-1,0,1/-1,1,-1,0], M(4)S(4)=[ -1,0,0,0/0,1,-2,2/-2,2,-1,0/0,0,0,1]; char. poly. of M(4)S(4) is x^4 + 2x^2 - 3, yielding row 4 of the triangle: -3,0,2,0,1.
%e Triangle starts:
%e 1;
%e 0,-1;
%e -1,0,1;
%e 0,-3,0,-1;
%e -3,0,2,0,1;
%e 0,-5,0,-10,0,-1
%p with(linalg): m:=proc(i,j) if i=j then 0 else 1 fi end: s:=proc(i,j) if i=j then 0 elif i>j then (-1)^(i+j) else (-1)^(i+j+1) fi end: for n from 1 to 14 do f[n]:=(-1)^n*sort(expand(charpoly(multiply(matrix(n,n,m),matrix(n,n,s)),x))) od: 1; for n from 1 to 14 do seq(coeff(f[n],x,j),j=0..n) od; # yields sequence in triangular form
%Y Cf. A000984.
%K sign,tabl
%O 0,8
%A _Roger L. Bagula_, Jan 01 2007
%E Edited by _N. J. A. Sloane_, Jan 07 2006
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