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a(n) = floor(b(n)), where b(n) = 1/cos(b(n-1)), b(0) = 1.
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%I #6 Jun 11 2020 07:46:53

%S 1,1,-4,-2,2,-2,9,-2,1,-4,-2,1,-4,-2,1,-4,-2,1,-3,-2,1,-3,-2,1,-4,-2,

%T 1,-4,-2,2,-2,12,1,1,-4,-2,1,-3,-2,2,-2,8,-3,-2,-6,1,16,-3,-2,7,-12,1,

%U -4,-2,2,-3,-3,-3,-2,-6,1,2,-2,2,-3,-2,-6,1,4,-3,-2,9,-2,2,-3,-2,-4,-2,1,-3,-2,1,-3,-2,2,-3,-2,-7,1,4,-2,70,4,-2,27

%N a(n) = floor(b(n)), where b(n) = 1/cos(b(n-1)), b(0) = 1.

%e a(3) = -2 because: b(0) = 1, b(1) = 1/cos(1) = 1.850815..., b(2) = 1/cos(1.850815) = -3.618291..., b(3) = 1/cos(-3.618291) = -1.125468... and a(3) = floor(b(3)) = -2

%o (PARI) default(realprecision,1000); b=1; for(n=0, 94, print1(floor(b)", "); b=1/cos(b))

%Y Cf. A000319.

%K sign

%O 0,3

%A Avi Peretz (njk(AT)netvision.net.il), Dec 26 2000

%E More terms from _Ralf Stephan_, Mar 23 2003