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%I #28 Dec 26 2023 12:23:05
%S 4,6,8,10,10,12,14,14,16,16,16,18,18,20,20,22,22,22,24,24,26,26,26,28,
%T 28,30,30,30,32,32,34,34,34,36,36,36,36,38,38,40,40,40,42,42,42,42,44,
%U 44,44,46,46,46,48,48,48,50,50,50,52,52,52,54,54,54,56,56,56,56,58,58
%N a(n) = (Im(rz(n)) - Im(-log(exp(-rz(n)))))/Pi where rz(n) is the n-th zero of Zeta(s).
%C I strongly suspect that lim_{n -> infinity} a(n)/n = 3/4. - _Stephen Crowley_, Oct 28 2009
%H G. C. Greubel, <a href="/A167146/b167146.txt">Table of n, a(n) for n = 1..5000</a>
%F From _Mats Granvik_, Jan 15 2018: (Start)
%F a(n) = (Im(zetazero(n)) - Im(-log(exp(-1/2 - i*Im(zetazero(n))))))/Pi, where i = sqrt(-1).
%F a(n) = 2*A275579(n) = 2*round(Im(zetazero(n))/(2*Pi)), verified for n=1..100000.
%F a(n) = (Im(zetazero(n)) - arctan(cos(Im(zetazero(n))), sin(Im(zetazero(n)))))/Pi, verified for n=1..100000.
%F (End)
%p [seq(round(evalf((Im(rzerof(n))-Im(-ln(exp(-rzerof(n)))))/Pi)), n = 1 .. 100)] # where rzerof(n) is the n-th zero of the Riemann zeta function, the rounding is simply for presentation purposes, the values are actually integers
%t Table[2*Round[Im[ZetaZero[n]]/(2*Pi)], {n, 1, 70}] (* _Mats Granvik_, Jan 15 2018 *)
%Y Cf. A002410, A275579.
%K nonn
%O 1,1
%A _Stephen Crowley_, Oct 28 2009
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