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 A143343 Triangle read by rows, Bernoulli number generator based on the Von Staudt-Clausen theorem by modifying A127093. 4
 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, 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 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS Triangle A138243 has a different arrangement of the same terms in each row. LINKS Wikipedia, Bernoulli number FORMULA 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), change all nonprimes to 1 and append a column of 1's as the left border. EXAMPLE Begin with Triangle A127093: 1; 1, 2; 1, 0, 3; 1, 2, 0, 4; 1, 0, 0, 0, 5; 1, 2, 3, 0, 0, 6; in which the divisors of n are recorded in ascending order. To this triangle add 1 to each term: 2; 2, 3; 2, 1, 4; 2, 3, 1, 5; 2, 1, 1, 1, 6; 2, 3, 4, 1, 1, 7; then change all nonprimes to 1. 2; 2, 3; 2, 1, 1; 2, 3, 1, 5; 2, 1, 1, 1, 1; 2, 3, 1, 1, 1, 7; Finally, append a column of 1's as the left border; getting: 1; 1, 2; 1, 2, 3; 1, 2, 1, 1; 1, 2, 3, 1, 5; 1, 2, 1, 1, 1, 1; 1, 2, 3, 1, 1, 1, 7; Problem is we get a different triangle of data than originally published: 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). PROG (PARI) T(n, k) = if (! isprime(v = 1+k*0^(n % k)), 1, v); listT(nn) = {print(1); for (n=1, nn, print1(1, ", "); for (k=1, n, print1(T(n, k), ", "); ); print(); ); } \\ Michel Marcus, Apr 27 2014 CROSSREFS Cf. A027642, A002445, A138239, A127093, A000012. Sequence in context: A167684 A094646 A124448 * A138243 A273136 A273137 Adjacent sequences:  A143340 A143341 A143342 * A143344 A143345 A143346 KEYWORD nonn,tabl,uned AUTHOR Gary W. Adamson & Mats Granvik, Aug 09 2008 STATUS approved

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Last modified January 17 23:15 EST 2019. Contains 319251 sequences. (Running on oeis4.)