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

Offset: 1

Author

Eric W. Weisstein, Jan 27 2003

Keywords

• nonn
• easy
• nice
• tabf

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)

Examples

			First few rows of the triangle:
  2;
  2, 3;
  2, 3, 5;
  2, 3, 7;
  2, 3, 5;
  2, 3, 11;
  2, 3, 5, 7, 13;
  2, 3;
  ...
Sum for n=1 is 1/2 + 1/3, so terms are 2, 3;
sum for n=2 is 1/2 + 1/3 + 1/5, so terms are 2, 3, 5; etc.
		

Links

• Eric Weisstein's World of Mathematics, von Staudt-Clausen Theorem.
• Wikipedia, Von Staudt-Clausen theorem.

Crossrefs

Cf. A000146, A002445, A027642, A138239, A143343.

Programs

Mathematica

row[n_] := Select[ Prime /@ Range[n+1], Divisible[2n, # - 1] &]; Flatten[Table[row[n], {n, 0, 25}]] (* Jean-François Alcover, Oct 12 2011 *)

Extensions

Edited by N. J. A. Sloane, Nov 01 2009 at the suggestion of R. J. Mathar