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%I #17 Sep 04 2015 08:19:31
%S 0,0,0,1,2,5,5,7,7,5,5,11,11,91,91,-9,-9,1207,1207,-10849,-10849,
%T 65879,65879,-783127,-783127,61098739,61098739,-2034290233,
%U -2034290233,72986324461,72986324461
%N Numerators of the autosequence preceding Br(n)=A229979(n)/(1 followed by A050932(n)).
%C Br(n)=0, 1, 1, 1/2, 0, -1/6, 0, 1/6, 0, -3/10, 0, 5/6, 0, -691/210, 0,.. .
%C a(n) is the numerators of Bp2(n)=0, 0, 0, 1, 2, 5/2, 5/2, 7/3, 7/3, 5/2, 5/2, 11/5, 11/5, 91/30, 91/30,... . Bp2(n) is an autosequence like Br(n).
%C With possible future sequences we can write the array PB
%C 1, 0, 0, 0, 0, 0, 0, 0, 0,
%C 1, 1, 0, 0, 0, 0, 0, 0, 0,
%C 1, 3/2, 1, 0, 0, 0, 0, 0, 0,
%C 1, 5/3, 2, 1, 0, 0, 0, 0, 0,
%C 1, 5/3, 5/2, 5/2, 1, 0, 0, 0, 0,
%C 1, 49/30, 5/2, 7/2, 3, 1, 0, 0, 0,
%C 1, 49/30, 7/3, 7/2, 14/3, 7/2, 1, 0, 0,
%C 1, 58/35, 7/3, 3, 14/3, 6, 4, 1, 0,
%C 1, 58/35, 5/2, 3, 7/2, 6, 15/2, 9/2, 1, etc.
%C The first column is A000012. The second A165142(n+1)/(1 followed by A100650(n)). The third is Bp2(n+1). The next others are built by the same way. From the second,every column is based on A164555(n)/A027642(n).
%C With negative (2*n+2)-th diagonals,the array without 0's is the triangle NPB. The sum of every row is
%C 1, 0, 1/2, -1/3, 1/3, -11/30, 11/30, -12/35, 12/35, -79/210, 79/210,... .
%C See A176250(n+2)/A100650(n).
%C The inverse of NPB is A193815(n)/(A003056(n) with 1 instead of 0).
%e a(0)=a(1)=0, a(i)=numerators of 0+Br(0)=0, 0+Br(1)=1, 1+Br(2)=2, 2+Br(3)=5/2, 5/2+Br(4)=5/2,... .
%t nmax = 30; Br[0] = 0; Br[1] = Br[2] = 1; Br[n_] := Numerator[2*n*BernoulliB[n-1]] / Denominator[n*BernoulliB[n-1]]; Bp2 = Join[{0, 0}, Table[Br[n], {n, 0, nmax-2}] // Accumulate]; a[n_] := Numerator[Bp2[[n+1]]]; Table[a[n], {n, 0, nmax}] (* _Jean-François Alcover_, Dec 18 2013 *)
%Y Cf. A233316.
%K sign
%O 0,5
%A _Paul Curtz_, Dec 13 2013
%E a(17)-a(30) from _Jean-François Alcover_, Dec 18 2013
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