

A167146


a(n) = (Im(rz(n))  Im(log(exp(rz(n)))))/Pi where rz(n) is the nth zero of Zeta(s).


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4, 6, 8, 10, 10, 12, 14, 14, 16, 16, 16, 18, 18, 20, 20, 22, 22, 22, 24, 24, 26, 26, 26, 28, 28, 30, 30, 30, 32, 32, 34, 34, 34, 36, 36, 36, 36, 38, 38, 40, 40, 40, 42, 42, 42, 42, 44, 44, 44, 46, 46, 46, 48, 48, 48, 50, 50, 50, 52, 52, 52, 54, 54, 54, 56, 56, 56, 56, 58, 58
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OFFSET

1,1


COMMENTS

I strongly suspect that lim_{n > infinity} a(n)/n = 3/4.  Stephen Crowley, Oct 28 2009


LINKS



FORMULA

a(n) = (Im(zetazero(n))  Im(log(exp(1/2  i*Im(zetazero(n))))))/Pi, where i = sqrt(1).
a(n) = 2*A275579(n) = 2*round(Im(zetazero(n))/(2*Pi)), verified for n=1..100000.
a(n) = (Im(zetazero(n))  arctan(cos(Im(zetazero(n))), sin(Im(zetazero(n)))))/Pi, verified for n=1..100000.
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MAPLE

[seq(round(evalf((Im(rzerof(n))Im(ln(exp(rzerof(n)))))/Pi)), n = 1 .. 100)] # where rzerof(n) is the nth zero of the Riemann zeta function, the rounding is simply for presentation purposes, the values are actually integers


MATHEMATICA

Table[2*Round[Im[ZetaZero[n]]/(2*Pi)], {n, 1, 70}] (* Mats Granvik, Jan 15 2018 *)


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



