A143987 Eigentriangle of (A007318)^(-1); row sums = A014182, exp(1-x-exp(-x)).

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, -1, 9, 0, -84, -126, 252, 756, 324, -450, -267, 1, -10, 0, 120, 210, -504, -1890, -1080, 2250, 2670, 413

Offset: 0

Author

Gary W. Adamson, Sep 07 2008

Keywords

• tabl
• sign

Comments

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,...).

Examples

			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).
		

Crossrefs

Cf. A007318, A014182.

Formula

(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 triangular 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,...).