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Irregular triangle read by rows, giving prime sequences (p-1|2n) appearing in the n-th von Staudt-Clausen sum.
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%I #32 Aug 30 2023 07:27:45

%S 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,

%T 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,

%U 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

%N Irregular triangle read by rows, giving prime sequences (p-1|2n) appearing in the n-th von Staudt-Clausen sum.

%C From _Gary W. Adamson_ & _Mats Granvik_, Aug 09 2008: (Start)

%C The von Staudt-Clausen 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.

%C A046886(n-1) = number of terms in row n.

%C The same terms in A143343 may be extracted from triangle A138239.

%C 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 Staudt-Clausen theorem.

%C The denominator of B_12 = 2730 = 2*3*5*7*13 = A027642(12) and A002445(6).

%C For example, B_12 = -691/2730 = (1 - 1/2 - 1/3 - 1/5 - 1/7 - 1/13).

%C The second operation is the von Staudt-Clausen 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)

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/vonStaudt-ClausenTheorem.html">von Staudt-Clausen Theorem</a>.

%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Von_Staudt-Clausen_theorem">Von Staudt-Clausen theorem</a>.

%e First few rows of the triangle:

%e 2;

%e 2, 3;

%e 2, 3, 5;

%e 2, 3, 7;

%e 2, 3, 5;

%e 2, 3, 11;

%e 2, 3, 5, 7, 13;

%e 2, 3;

%e ...

%e Sum for n=1 is 1/2 + 1/3, so terms are 2, 3;

%e sum for n=2 is 1/2 + 1/3 + 1/5, so terms are 2, 3, 5; etc.

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

%Y Cf. A000146, A002445, A027642, A138239, A143343.

%K nonn,easy,nice,tabf

%O 1,1

%A _Eric W. Weisstein_, Jan 27 2003

%E Edited by _N. J. A. Sloane_, Nov 01 2009 at the suggestion of _R. J. Mathar_