0,2

Let the triangle = T. Row sums = 1. Row sums of n-th binomial transform

of T = powers of (n-1). Then multiply the results by the partial sum operator,

(1; 1,1; 1,1,1;...) to obtain Bernoulli's "Sums of Powers".

Inserting zeros to account for (n+1) terms per row, right border = Bernoulli numbers:

(A106458): (1, 1/2, 1/6, 0, -1/30, 0, 1/42, 0, -1/30, 0, 5/66,...).

Robert M. Young, "Excursions in Calculus", MAA, 1992 James Bernoulli, "Ars conjectandi", posthumously published in 1713; in which Bernoulli gives the table "Summae Potestatum (Sums of Powers) [Cf. Young, p. 86].

Let row 0 = 1; followed by the corrected table, giving denominators:

1;

2, 2;

3, 2, 6;

4, 2, 4;

5, 2, 3, -30;

6, 2, 12, -12;

7, 2, 2, -6, 42;

8, 2, 12, -24, 12;

9, 2, 3, -15, 9, -30;

10, 2, 4, -10, 2, -20;

11, 2, 6, -1, 1, -2, 66;

... The complete triangle with row 0 = 1, along with numerators:

1;

1/2, 1/2;

1/3, 1/2, 1/6;

1/4, 1/2, 1/4;

1/5, 1/2, 1/3, -1/30;

1/6, 1/2, 5/12, -1/12;

1/7, 1/2, 1/2, -1/6, 1/42;

1/8, 1/2, 7/12, -7/14, 1/12;

1/9, 1/2, 2/3, -7/15, 1/2, -3/20;

1/10, 1/2, 3/4, -7/10, 1/2, -3/20;

1/11, 1/2, 5/6, -1/1, 1/1, -1/2, 5/66;

...

A106458

Sequence in context: A108499 A107753 A078224 this_sequence A128710 A095757 A144368

Adjacent sequences: A159685 A159686 A159687 this_sequence A159689 A159690 A159691

tabl,sign

Gary W. Adamson (qntmpkt(AT)yahoo.com), Apr 19 2009