A080092 - OEIS

A080092

Irregular triangle read by rows, giving prime sequences (p-1|2n) appearing in the n-th von Staudt-Clausen sum.

2, 2, 3, 2, 3, 5, 2, 3, 7, 2, 3, 5, 2, 3, 11, 2, 3, 5, 7, 13, 2, 3, 2, 3, 5, 17, 2, 3, 7, 19, 2, 3, 5, 11, 2, 3, 23, 2, 3, 5, 7, 13, 2, 3, 2, 3, 5, 29, 2, 3, 7, 11, 31, 2, 3, 5, 17, 2, 3, 2, 3, 5, 7, 13, 19, 37, 2, 3, 2, 3, 5, 11, 41, 2, 3, 7, 43, 2, 3, 5, 23, 2, 3, 47, 2, 3, 5, 7, 13, 17, 2, 3

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

COMMENTS

From Gary W. Adamson & Mats Granvik, Aug 09 2008: (Start)

The von Staudt-Clausen theorem has two parts: generating denominators of the B_2n and the actual values. Both operations can be demonstrated in triangles A143343 and A080092 by following the procedures outlined in [Wikipedia - Bernoulli numbers] and summarized in A143343.

A046886(n-1) = number of terms in row n.

The same terms in A143343 may be extracted from triangle A138239.

Extract primes from even numbered rows of triangle A143343 but also include "2" as row 1. The rows are thus 1, 2, 4, 6, ..., generating denominators of B_1, B_2, B_4, ..., as well as B_1, B_2, B_4, ..., as two parts of the von Staudt-Clausen theorem.

The denominator of B_12 = 2730 = 2*3*5*7*13 = A027642(12) and A002445(6).

For example, B_12 = -691/2730 = (1 - 1/2 - 1/3 - 1/5 - 1/7 - 1/13).

The second operation is the von Staudt-Clausen representation of Bn, obtained by starting with "1" and then subtracting the reciprocals of terms in each row. (Cf. A143343 for a detailed explanation of the operations.) (End)

LINKS

Table of n, a(n) for n=1..92.

Eric Weisstein's World of Mathematics, von Staudt-Clausen Theorem.

Wikipedia, Von Staudt-Clausen theorem.

EXAMPLE

First few rows of the triangle:

2, 3;

2, 3, 5;

2, 3, 7;

2, 3, 5;

2, 3, 11;

2, 3, 5, 7, 13;