

A167146


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


1



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
(list;
graph;
refs;
listen;
history;
text;
internal format)



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 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

Cf. A002410, A275579.
Sequence in context: A087789 A071830 A276982 * A020891 A090967 A272475
Adjacent sequences: A167143 A167144 A167145 * A167147 A167148 A167149


KEYWORD

nonn


AUTHOR

Stephen Crowley, Oct 28 2009


STATUS

approved



