A131047 - OEIS

A131047

(1/2) * (A007318 - A007318^(-1)).

1, 0, 2, 1, 0, 3, 0, 4, 0, 4, 1, 0, 10, 0, 5, 0, 6, 0, 20, 0, 6, 1, 0, 21, 0, 35, 0, 7, 0, 8, 0, 56, 0, 56, 0, 8, 1, 0, 36, 0, 126, 0, 84, 0, 9

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OFFSET

1,3

COMMENTS

Row sums = (1, 2, 4, 8, ...). A131047 * (1,2,3, ...) = A087447 starting (1, 4, 10, 24, 56, ...). A generalized set of analogous triangles: (1/(Q+1)) * (P^Q - 1/P), Q an integer, generates triangles with row sums = powers of (Q+1). Cf. A131048, A131049, A131050, A131051 for triangles having Q = 2,3,4 and 5, respectively.

A007318, Pascal's triangle, = this triangle + A119467, since one triangle = the zeros or masks of the other. - Gary W. Adamson, Jun 12 2007

LINKS

Table of n, a(n) for n=1..45.

FORMULA

Let A007318 (Pascal's triangle) = P, then A131047 = (1/2) * (P - 1/P); deleting the right border of zeros.

EXAMPLE

First few rows of the triangle:

0, 2;

1, 0, 3;

0, 4, 0, 4;

1, 0, 10, 0, 5;

0, 6, 0, 20, 0, 6;