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%I #31 Jan 06 2018 22:07:17
%S 2,2,2,6,2,6,2,6,15,2,6,15,2,6,15,105,2,6,15,105,2,6,15,105,105,2,6,
%T 15,105,105,2,6,15,105,105,231,2,6,15,105,105,231,2,6,15,105,105,231,
%U 15015,2,6,15,105,105,231,15015
%N Irregular triangle of the denominators of the unreduced fractions that lead to the second Bernoulli numbers.
%C The triangle of fractions A192456(n)/A191302(n) leading to the second Bernoulli numbers written in A191302(n) is the reduced case. The unreduced case is
%C B(0) = 1 = 2/2 (1 or 2/2 chosen arbitrarily)
%C B(1) = 1/2
%C B(2) = 1/6 = 1/2 - 2/6
%C B(3) = 0 = 1/2 - 3/6
%C B(4) = -1/30 = 1/2 - 4/6 + 2/15
%C B(5) = 0 = 1/2 - 5/6 + 5/15
%C B(6) = 1/42 = 1/2 - 6/6 + 9/15 - 8/105
%C B(7) = 0 = 1/2 - 7/6 + 14/15 - 28/105
%C B(8) = -1/30 = 1/2 - 8/6 + 20/15 - 64/105 + 8/105.
%C The constant values along the columns of denominators are A190339(n).
%C With B(0)=1, B(2) = 1/2 -1/3, (reduced case), the last fraction of the B(2*n) is
%C 1, -1/3, 2/15, -8/105, 8/105, ... = A212196(n)/A181131(n).
%C We can continue this method of sum of fractions yielding Bernoulli numbers.
%C Starting from 1/6 for B(2*n+2), we have:
%C B(2) = 1/6
%C B(4) = 1/6 - 3/15
%C B(6) = 1/6 - 5/15 + 20/105
%C B(8) = 1/6 - 7/15 + 56/105 - 28/105.
%C With the odd indices from 3, all these B(n) are the Bernoulli twin numbers -A051716(n+3)/A051717(n+3).
%F T(n,k) = A190339(k).
%e Triangle begins
%e 2;
%e 2;
%e 2, 6;
%e 2, 6;
%e 2, 6, 15;
%e 2, 6, 15;
%e 2, 6, 15, 105;
%e 2, 6, 15, 105;
%e 2, 6, 15, 105, 105;
%e 2, 6, 15, 105, 105;
%e 2, 6, 15, 105, 105, 231;
%e 2, 6, 15, 105, 105, 231;
%e 2, 6, 15, 105, 105, 231, 15015;
%e 2, 6, 15, 105, 105, 231, 15015;
%t nmax = 7; b[n_] := BernoulliB[n]; b[1] = 1/2; bb = Table[b[n], {n, 0, 2*nmax-1}]; diff = Table[ Differences[bb, n], {n, 1, nmax}]; A190339 = diff // Diagonal // Denominator; Table[ Table[ Take[ A190339, n], {2}], {n, 1, nmax}] // Flatten (* _Jean-François Alcover_, Apr 25 2013 *)
%Y Cf. A051716, A051717, A141044, A181131, A190339, A191302, A192456, A212196.
%K nonn,frac,tabf
%O 0,1
%A _Paul Curtz_, Apr 21 2013
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