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a(n) = (-1)^(1+n+A088585(n)).
1

%I #10 Jul 13 2020 06:22:11

%S 1,1,1,-1,1,1,-1,-1,1,1,1,-1,-1,1,-1,-1,1,1,1,-1,1,1,-1,-1,-1,1,1,-1,

%T -1,1,-1,-1,1,1,1,-1,1,1,-1,-1,1,1,1,-1,-1,1,-1,-1,-1,1,1,-1,1

%N a(n) = (-1)^(1+n+A088585(n)).

%C Apparently the partial products of this sequence form the Hankel transform of A023359: 1, 1*1 = 1, 1*1*1 = 1, 1*1*1*-1 = -1, 1*1*1*-1*1 = -1, ... and 1, 1, 1, -1, -1, -1, 1, -1, -1, -1, -1, 1, -1, -1, 1, -1, -1, ... is the Hankel transform of A023359.

%H <a href="/index/Fo#fold">Index entries for sequences obtained by enumerating foldings</a>

%Y Cf. A023359, A088585.

%K sign

%O 0,1

%A _Philippe Deléham_, Aug 21 2006