0,5

Sum of n-th row terms = rightmost term of next row. Row sums = A014182: (1, 0, -1, 1, 2, -9, 9, 50, -267,...).

Right border = A014182 shifted: (1, 1, 0, -1, 1, 2, -9,...).

(A007318^(-1) * (A014182 * 0^(n-k))) 0<=k<=n

A007318^(-1) = the inverse of Pascal's triangle.

Given A014182: (1, 0, -1, 1, 2, -9, 9,...) = expansion of exp(1-x-exp(-x), we preface A014182 with a "1" getting (1, 1, 0, -1, 1, 2, -9,...).

Then diagonalize it as an infinite lower trianglular matrix R =

1;

0, 1;

0, 0, 0;

0, 0, 0, -1;

0, 0, 0, 0, 1;

...

Finally, take the inverse binomial transform of triangle R, getting A143987.

Given the inverse of Pascal's triangle by rows, we apply termwise products of

equal numbers of terms in the sequence: (1, 1, 0, -1, 1, 2, -9, 9,...).

First few rows of the triangle =

1;

-1, 1;

1, -2, 0;

-1, 3, 0, -1;

1, -4, 0, 4, 1;

-1, 5, 0, -10, -5, 2;

1, -6, 0, 20, 15, -12, -9;

-1, 7, 0, -35, -35, 42, 63, 9;

1, -8, 0, 56, 70, -112, -252, 72, 50;

...

Example: row 4 = (1, -4, 0, 4, 1) = termwise products of (1, -4, 6, -4, 1) and (1, 1, 0, -1, 1).= (1*1, -4*1, 6*0, -4*-1, 1*1).

A007318, Cf. A014182

Sequence in context: A096335 A129503 A144185 this_sequence A112760 A096087 A128138

Adjacent sequences: A143984 A143985 A143986 this_sequence A143988 A143989 A143990

tabl,sign

Gary W. Adamson (qntmpkt(AT)yahoo.com), Sep 07 2008