OFFSET
1,1
COMMENTS
I strongly suspect that lim_{n -> infinity} a(n)/n = 3/4. - Stephen Crowley, Oct 28 2009
LINKS
G. C. Greubel, Table of n, a(n) for n = 1..5000
FORMULA
From Mats Granvik, Jan 15 2018: (Start)
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.
(End)
MAPLE
[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
MATHEMATICA
Table[2*Round[Im[ZetaZero[n]]/(2*Pi)], {n, 1, 70}] (* Mats Granvik, Jan 15 2018 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Stephen Crowley, Oct 28 2009
STATUS
approved