0,12

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.

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.

Triangle T begins:

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;

-1,-5,-11,-15,-15,-11,-5,-1;

1, 6, 16, 26, 30, 26, 16, 6, 1;

-1,-7,-22,-42,-56,-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; ...

The matrix square is A118401:

1;

0, 1;

2, 0, 1;

-2, 2, 0, 1;

4,-2, 2, 0, 1;

-6, 4,-2, 2, 0, 1;

8,-6, 4,-2, 2, 0, 1;

-10, 8,-6, 4,-2, 2, 0, 1;

12,-10, 8,-6, 4,-2, 2, 0, 1; ...

in which all columns are equal.

(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)} (PARI) {T(n, k)=if(n==1&k==0, 1, (-1)^n*(binomial(n, k)-2*binomial(n-2, k-1)))}

Cf. A118401 (matrix square), A084633 (row sums), A087698 (unsigned version); A118404 (matrix inverse).

Sequence in context: A144431 A053821 A076545 this_sequence A087698 A101677 A128084

Adjacent sequences: A118397 A118398 A118399 this_sequence A118401 A118402 A118403

sign,tabl

Paul D. Hanna (pauldhanna(AT)juno.com), Apr 27 2006