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%I A143343
%S A143343 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,
%T A143343 1,2,3,1,5,1,1,1,1,1,2,3,1,1,1,1,1,1,1,11,1,1,1,1,1,1,1,1,1,1,1,1,1,2,
               3,
%U A143343 1,5,1,7,1,1,1,1,1,13
%N A143343 Triangle read by rows, Bernoulli number generator based on the Von Staudt-Clausen 
               theorem by modifying A127093.
%C A143343 Triangle A138243 has a different arrangement of the same terms in each 
               row.
%D A143343 Wikipedia (Bernoulli numbers).
%F A143343 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.
%e A143343 Begin with Triangle A127093:
%e A143343 1;
%e A143343 1, 2;
%e A143343 1, 0, 3;
%e A143343 1, 2, 0, 4;
%e A143343 ..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:
%e A143343 1;
%e A143343 1, 2;
%e A143343 1, 2, 3;
%e A143343 1, 1, 1, 1;
%e A143343 1, 2, 3, 1, 5;
%e A143343 1, 1, 1, 1, 1, 1;
%e A143343 1, 2, 3, 1, 1, 1, 7;
%e A143343 1, 1, 1, 1, 1, 1, 1, 1;
%e A143343 1, 2, 3, 1, 5, 1, 1, 1, 1;
%e A143343 1, 1, 1, 1, 1, 1, 1, 1, 1, 1;
%e A143343 1, 2, 3, 1, 1, 1, 1, 1, 1, 1, 11;
%e A143343 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1;
%e A143343 1, 2, 3, 1, 5, 1, 7, 1, 1, 1, 1, 1, 13;
%e A143343 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1;
%e A143343 1, 2, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1;
%e A143343 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1;
%e A143343 ...
%e A143343 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.
%e A143343 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).
%Y A143343 Cf. A027642, A002445, A138239, A127093, A000012.
%Y A143343 Sequence in context: A167684 A094646 A124448 this_sequence A138243 A131796 
               A131797
%Y A143343 Adjacent sequences: A143340 A143341 A143342 this_sequence A143344 A143345 
               A143346
%K A143343 nonn,tabl
%O A143343 0,2
%A A143343 Gary W. Adamson & Mats O. Granvik (qntmpkt(AT)yahoo.com), Aug 09 2008

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