0,2

Triangle A138243 has a different arrangement of the same terms in each row.

Wikipedia (Bernoulli numbers).

To paraphrase Clausen, we first add unity to the divisors of n; equivalent to adding Triangle A000012 (an infinite lower triangular matrix with all 1's); to triangle A127093 (in which rows record the divisors of n). Extract primes from the result, change all nonprimes to 1 and append a column of 1's as the left border.

Begin with Triangle A127093:

1;

1, 2;

1, 0, 3;

1, 2, 0, 4;

..in which the divisors of n are recorded in ascending order. To this triangle 1 to each term, then change all nonprimes to 1. Finally, append a column of 1's as the left border; getting:

1;

1, 2;

1, 2, 3;

1, 1, 1, 1;

1, 2, 3, 1, 5;

1, 1, 1, 1, 1, 1;

1, 2, 3, 1, 1, 1, 7;

1, 1, 1, 1, 1, 1, 1, 1;

1, 2, 3, 1, 5, 1, 1, 1, 1;

1, 1, 1, 1, 1, 1, 1, 1, 1, 1;

1, 2, 3, 1, 1, 1, 1, 1, 1, 1, 11;

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1;

1, 2, 3, 1, 5, 1, 7, 1, 1, 1, 1, 1, 13;

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1;

1, 2, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1;

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1;

...

Using Clausen's algorithm to obtain the denominators of Bn, we take row products, matching the results to (all rows, A027642, denominators of Bernoulli numbers Bn; or to B_2n, cf. A002445).). Row 6 = (1, 2, 3, 1, 1, 1, 7) so the denominator of B6 = (1*2*3*1*1*1*7) = 42.

In the second part of the Von Staudt-Clausen theorem, we obtain the Bernoulli numbers for n-th row by starting with "1" then subtracting reciprocals of primes in each row. Thus B10 = 5/66 = (1 - 1/2 - 1/3 - 1/11), where the primes in row 10 are (2, 3, 11).

Cf. A027642, A002445, A138239, A127093, A000012.

Adjacent sequences: A143340 A143341 A143342 this_sequence A143344 A143345 A143346

Sequence in context: A132815 A094646 A124448 this_sequence A138243 A131796 A131797

nonn,tabl

Gary W. Adamson & Mats O. Granvik (qntmpkt(AT)yahoo.com), Aug 09 2008