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%I #32 Mar 22 2015 20:21:41
%S 1,1,3,1,2,13,0,1,5,3,-1,-1,2,29,119,0,-1,-1,1,31,5,1,1,-1,-8,-1,43,
%T 253,0,1,1,4,-4,-1,41,7,-1,-1,-1,4,8,4,-1,29,239,0,-1,-1,-8,-4,4,8,1,
%U 31,9,5,5,7,-4,-116,-32,-116,-4,7,71,665,0
%N Ascending antidiagonal numerators of the table of repeated differences of A164558(n)/A027642(n).
%C The difference table of Bernoulli(n,2) or B(n,2) = A164558(n)/A027642(n) is defined by placing the fractions in the upper row and calculating further rows as the differences of their preceding row:
%C 1, 3/2, 13/6, 3, 119/30, ...
%C 1/2, 2/3, 5/6, 29/30, ...
%C 1/6, 1/6, 2/15, ...
%C 0, -1/30, ...
%C -1/30, ...
%C etc.
%C The first column is A164555(n)/A027642(n).
%C In particular, the sums of the antidiagonals
%C 1 = 1
%C 1/2 + 3/2 = 2
%C 1/6 + 2/3 + 13/6 = 3
%C 0 + 1/6 + 5/6 + 3 = 4
%C etc. are the positive natural numbers. (This is rewritten for Bernoulli(n,3) in A157809).
%C We also have for Bernoulli(.,2)
%C B(0,2) = 1
%C B(0,2) + 2*B(1,2) = 4
%C B(0,2) + 3*B(1,2) + 3*B(2,2) = 12
%C B(0,2) + 4*B(1,2) + 6*B(2,2) + 4*B(3,2) = 32
%C etc. with right hand sides provided by A001787.
%C More generally sum_{s=0..t-1} binomial(t,s)*Bernoulli(s,q) gives A027471(t) for q=3, A002697 for q=4 etc, by reading A104002 downwards the q-th column.
%t nmax = 11; A164558 = Table[BernoulliB[n,2], {n, 0, nmax}]; D164558 = Table[ Differences[A164558, n], {n, 0, nmax}]; Table[ D164558[[n-k+1, k+1]] // Numerator, {n, 0, nmax}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Feb 04 2015 *)
%Y Cf. A027641, A027642, A074909, A085737, A085738, A104002, A157809, A157920, A157930, A157945, A157946, A157965, A164555, A164558, A190339, A158302, A181131 (numerators and denominators of the main diagonal).
%K sign,tabl,frac,easy
%O 0,3
%A _Paul Curtz_, Feb 03 2015
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