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%I #23 Dec 09 2013 09:32:54
%S 0,1,1,1,0,-1,0,1,0,-3,0,5,0,-691,0,35,0,-3617,0,43867,0,-1222277,0,
%T 854513,0,-1181820455,0,76977927,0,-23749461029,0,8615841276005,0,
%U -84802531453387,0,90219075042845,0
%N Numerators of interleaved A063524(n) and A002427(n)/A006955(n).
%C Numerators of Br(n) = 0, 1, 1, 1/2, 0, -1/6, 0, 1/6, 0, -3/10, 0, 5/6, 0, -691/210,... complementary Bernoulli numbers.
%C A164555(n)/A027642(n) is an autosequence of second kind. Its inverse binomial transform is the signed sequence and its main diagonal is the double of the first upper diagonal.
%C Br(n) is an autosequence of first kind. Its inverse binomial transform is the signed sequence and its main diagonal is A000004=0's.
%C Br(n) difference table:
%C 0, 1, 1, 1/2, 0, -1/6,...
%C 1, 0, -1/2, -1/2, -1/6, 1/6,... =A140351(n)/A140219(n)
%C -1, -1/2, 0, 1/3, 1/3, 0,...
%C 1/2, 1/2, 1/3, 0, -1/3, -1/3,...
%C 0, -1/6, -1/3, -1/3, 0, 8/15,...
%C -1/6, -1/6, 0, 1/3, 8/15, 0,... etc.
%F a(2n)=A063524(n). a(2n+1)=A002427(n).
%F a(n) = numerators of n * b(n) with b(n)=0 followed by A164555(n)/A027642(n) = 0, 1, 1/2, 1/6, 0,... in A165142(n).
%F a(n+1) = numerators of Br(n+1) = Br(n) + A140351(n)/A140219(n), a(0)=Br(0)=0.
%t a[0] = 0; a[1] = a[2] = 1; a[n_] := 2*n*BernoulliB[n-1] // Numerator; Table[a[n], {n, 0, 36}] (* _Jean-François Alcover_, Nov 25 2013 *)
%Y Cf. A050925: a similar sequence, because 2*(n+1)*B(n) and (n+1)*B(n) have the same numerator.
%K sign,frac
%O 0,10
%A _Paul Curtz_, Oct 05 2013
%E Cross-ref. to A050925 by _Jean-François Alcover_, Dec 09 2013
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