

A080092


Irregular triangle read by rows, giving prime sequences (p12n) appearing in the nth von StaudtClausen sum.


8



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
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OFFSET

1,1


COMMENTS

The von StaudtClausen 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(n1) = number of terms in row n.
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 StaudtClausen 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 StaudtClausen 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



EXAMPLE

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.


MATHEMATICA

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


CROSSREFS



KEYWORD

nonn,easy,nice,tabf


AUTHOR



EXTENSIONS



STATUS

approved



