0,13

Row sums are A040000. Diagonal sums are A112469. Inverse is A112467. Row sums of k-th power are 1,k+1,k+1,k+1,.... Note that C(n,k)=sum{j=0..n-k, C(n-j-1,n-k-j)}.

Equals row reversal of triangle A112555 up to sign, where log(A112555) = A112555 - I. Unsigned row sums equals A052953 (Jacobsthal numbers + 1). 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. - Paul D. Hanna (pauldhanna(AT)juno.com), Jan 20 2006

Number triangle T(n, k)=sum{j=0..n-k, C(n-j-1, n-k-j)*(-1)^(n-k-j)}

G.f. of matrix power T^m: (1+(m-1)*x)*(1+m*x)/(1+m*x-x*y)/(1-x). G.f. of matrix log: x*(1-2*x*y+x^2*y)/(1-x*y)^2/(1-x). - Paul D. Hanna (pauldhanna(AT)juno.com), Jan 20 2006

Triangle starts

1;

1,1;

1,0,1;

1,1,-1,1;

1,0,2,-2,1;

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

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

Matrix log begins:

0;

1,0;

1,0,0;

1,1,-1,0;

1,1,1,-2,0;

1,1,1,1,-3,0; ...

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

Cf. A112465.

Cf. A112555 (reversed rows), A052953, A072547 (central terms), A112556.

Sequence in context: A031282 A085685 A112465 this_sequence A086275 A066855 A058914

Adjacent sequences: A112465 A112466 A112467 this_sequence A112469 A112470 A112471

easy,sign,tabl

Paul Barry (pbarry(AT)wit.ie), Sep 06 2005