0,12

Signed version of A108561. Row sums equal A084247. The n-th unsigned row sum = A001045(n) + 1 (Jacobsthal numbers). Central terms of even-indexed rows are a signed version of A072547. Sums of squared terms in rows yields A112556, which equals the first differences of the unsigned central terms.

Equals row reversal of triangle A112468 up to sign, where A112468 is the Riordan array (1/(1-x),x/(1+x)). - Paul D. Hanna (pauldhanna(AT)juno.com), Jan 20 2006

G.f.: 1/(1-x*y) + x/((1-x*y)*(1+x+x*y)); the m-th matrix power T^m has the g.f.: 1/(1-x*y) + m*x/((1-x*y)*(1+x+x*y)). Recurrence: T(n, k) = [T^-1](n-1, k) + [T^-1](n-1, k-1), where T^-1 is the matrix inverse of T.

Triangle T begins:

1;

1,1;

-1,0,1;

1,1,1,1;

-1,-2,-2,0,1;

1,3,4,2,1,1;

-1,-4,-7,-6,-3,0,1;

1,5,11,13,9,3,1,1;

-1,-6,-16,-24,-22,-12,-4,0,1;

1,7,22,40,46,34,16,4,1,1;

-1,-8,-29,-62,-86,-80,-50,-20,-5,0,1; ...

Matrix log, LOG(T) = T - I, begins:

0;

1,0;

-1,0,0;

1,1,1,0;

-1,-2,-2,0,0;

1,3,4,2,1,0;

-1,-4,-7,-6,-3,0,0; ...

Matrix inverse, T^-1 = 2*I - T, begins:

1;

-1,1;

1,0,1;

-1,-1,-1,1;

1,2,2,0,1;

-1,-3,-4,-2,-1,1; ...

where adjacent sums in row n of T^-1 gives row n+1 of T.

(PARI) {T(n, k)=local(x=X+X*O(X^n), y=Y+Y*O(Y^k)); polcoeff(polcoeff((1+2*x+x*y)/((1-x*y)*(1+x+x*y)), n, X), k, Y)}

(PARI) {T(n, k)=local(m=1, x=X+X*O(X^n), y=Y+Y*O(Y^k)); polcoeff(polcoeff(1/(1-x*y) + m*x/((1-x*y)*(1+x+x*y)), n, X), k, Y)} (Hanna)

Cf. A108561, A084247, A001045, A072547, A112556.

Cf. A112468 (reversed rows).

Adjacent sequences: A112552 A112553 A112554 this_sequence A112556 A112557 A112558

Sequence in context: A137581 A113414 A112185 this_sequence A108561 A104579 A079531

sign,tabl

Paul D. Hanna (pauldhanna(AT)juno.com), Sep 21 2005