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a(2n) = A164555(n). a(2n+1) = A027641(n).
3

%I #6 Mar 30 2012 18:52:05

%S 1,1,1,-1,1,1,0,0,-1,-1,0,0,1,1,0,0,-1,-1,0,0,5,5,0,0,-691,-691,0,0,7,

%T 7,0,0,-3617,-3617,0,0,43867,43867,0,0,-174611,-174611,0,0,854513,

%U 854513,0,0,-236364091,-236364091,0,0,8553103,8553103,0,0,-23749461029,-23749461029,0,0,8615841276005

%N a(2n) = A164555(n). a(2n+1) = A027641(n).

%C Formally, these are the numerators of a sequence of fractions defined by alternating A164555(n)/A027642(n) with A027641(n)/A027642(n),

%C which apart from the third term duplicates the Bernoulli numbers.

%C Essentially a duplication of the entries of A027641.

%Y Cf. A141056, A164020, A141056.

%K sign,frac,less

%O 0,21

%A _Paul Curtz_, Apr 10 2010

%E Edited by _R. J. Mathar_, Jun 07 2010