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EXAMPLE
| 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).
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