OFFSET
0,1
COMMENTS
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
B(0) = 1 = 2/2 (1 or 2/2 chosen arbitrarily)
B(1) = 1/2
B(2) = 1/6 = 1/2 - 2/6
B(3) = 0 = 1/2 - 3/6
B(4) = -1/30 = 1/2 - 4/6 + 2/15
B(5) = 0 = 1/2 - 5/6 + 5/15
B(6) = 1/42 = 1/2 - 6/6 + 9/15 - 8/105
B(7) = 0 = 1/2 - 7/6 + 14/15 - 28/105
B(8) = -1/30 = 1/2 - 8/6 + 20/15 - 64/105 + 8/105.
The constant values along the columns of denominators are A190339(n).
With B(0)=1, B(2) = 1/2 -1/3, (reduced case), the last fraction of the B(2*n) is
We can continue this method of sum of fractions yielding Bernoulli numbers.
Starting from 1/6 for B(2*n+2), we have:
B(2) = 1/6
B(4) = 1/6 - 3/15
B(6) = 1/6 - 5/15 + 20/105
B(8) = 1/6 - 7/15 + 56/105 - 28/105.
FORMULA
T(n,k) = A190339(k).
EXAMPLE
Triangle begins
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, 15, 105, 105;
2, 6, 15, 105, 105, 231;
2, 6, 15, 105, 105, 231;
2, 6, 15, 105, 105, 231, 15015;
2, 6, 15, 105, 105, 231, 15015;
MATHEMATICA
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 *)
CROSSREFS
KEYWORD
nonn,frac,tabf
AUTHOR
Paul Curtz, Apr 21 2013
STATUS
approved