%I #6 Jun 13 2015 17:41:48
%S 1,1,-1,1,0,1,-1,-1,-1,-1,1,2,2,2,1,-1,-3,-4,-4,-3,-1,1,4,7,8,7,4,1,
%T -1,-5,-11,-15,-15,-11,-5,-1,1,6,16,26,30,26,16,6,1,-1,-7,-22,-42,-56,
%U -56,-42,-22,-7,-1,1,8,29,64,98,112,98,64,29,8,1,-1,-9,-37,-93,-162,-210,-210,-162,-93,-37,-9,-1
%N Triangle T, read by rows, where all columns of T are different and yet all columns of the matrix square T^2 (A118401) are equal; a signed version of triangle A087698.
%C Matrix inverse equals A118404. Row sums equal A084633. Signed version of: A087698 = maximum number of Boolean inputs at Hamming distance 2 for symmetric Boolean functions. This is an example of the fact that special matrices (cf. A118401) can have more than 2 signed matrix square-roots if the main diagonal is allowed to be signed.
%F G.f.: A(x,y) = (1+2*x+2*x^2)/(1+x+x*y). G.f. of column k = (-1)^k*(1+2*x+2*x^2)/(1+x)^(k+1) for k>=0. T(n,k) = (-1)^n*[C(n,k) - 2*C(n-2,k-1)] for n>=k>=0 except that T(1,0)=1.
%e Triangle T begins:
%e 1;
%e 1,-1;
%e 1, 0, 1;
%e -1,-1,-1,-1;
%e 1, 2, 2, 2, 1;
%e -1,-3,-4,-4,-3,-1;
%e 1, 4, 7, 8, 7, 4, 1;
%e -1,-5,-11,-15,-15,-11,-5,-1;
%e 1, 6, 16, 26, 30, 26, 16, 6, 1;
%e -1,-7,-22,-42,-56,-56,-42,-22,-7,-1;
%e 1, 8, 29, 64, 98, 112, 98, 64, 29, 8, 1;
%e -1,-9,-37,-93,-162,-210,-210,-162,-93,-37,-9,-1; ...
%e The matrix square is A118401:
%e 1;
%e 0, 1;
%e 2, 0, 1;
%e -2, 2, 0, 1;
%e 4,-2, 2, 0, 1;
%e -6, 4,-2, 2, 0, 1;
%e 8,-6, 4,-2, 2, 0, 1;
%e -10, 8,-6, 4,-2, 2, 0, 1;
%e 12,-10, 8,-6, 4,-2, 2, 0, 1; ...
%e in which all columns are equal.
%o (PARI) T(n,k)=polcoeff(polcoeff((1+2*x+2*x^2)/(1+x+x*y+x*O(x^n)),n,x)+y*O(y^k),k,y)
%o (PARI) T(n,k)=if(n==1&k==0,1,(-1)^n*(binomial(n,k)-2*binomial(n-2,k-1)))
%Y Cf. A118401 (matrix square), A084633 (row sums), A087698 (unsigned version); A118404 (matrix inverse).
%K sign,tabl
%O 0,12
%A _Paul D. Hanna_, Apr 27 2006