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%I #6 Jan 21 2019 19:01:55
%S 0,1,2,3,4,5,6,7,8,9,10,11,12,13,15,16,17,18,19,20,22,23,24,26,27,28,
%T 29,31,33,34,35,36,38,39,41,42,44,45,46,47,50,51,53,54,55,57,58,61,62,
%U 63,64,66,68
%N Construct the sequence given by the floor of the imaginary part of zeros of the Riemann zeta function; take complement.
%C Complement to A013629. Similar to A002410 and A122526, which use "round" instead of "floor".
%e The first zero is 14.13472.. so 0,1,2,3,4,5,6,7,8,9,10,11,12,13 are part of the sequence.
%e The second zero is 21.02203.. so 15,16,17,18,19,20 are in the sequence too.
%e The third zero is 25.0108.. so we get 22,23,24, etc.
%Y Cf. A002410, A122526.
%K fini,nonn
%O 0,3
%A _Jorge Coveiro_, Jul 02 2006
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