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%I #23 Mar 27 2022 03:21:24
%S 1,2,3,4,5,6,13,14,15,21,22,24,25,30,31,32,33,38,41,43,48,49,50,53,57,
%T 59,60,61,65,67,69,70,72,76,77,78,79,83,84,85,87,88,89,94,95,96,99,
%U 101,104,105,106,107,111,112,114,116,119,121,122,123,124,130,131,134,135
%N Numbers m such that the absolute values of the real and imaginary part of zeta(1/2 + m*i) are both < 1.
%C Conjecture: lim_{n->infinity} a(n)/n = C exists, with 5 < C < 6. [The conjecture was based on erroneous terms; C is about 2.05 (see graph). - _Vaclav Kotesovec_, Feb 18 2021]
%H Vaclav Kotesovec, <a href="/A072564/b072564.txt">Table of n, a(n) for n = 1..10000</a>
%H Vaclav Kotesovec, <a href="/A072564/a072564.jpg">Plot of a(n)/n for n = 1..10000</a>
%e zeta(1/2 + 15*i) = (0.1471...) + (0.7047...)*i; 0.1471... < 1 and 0.7047... < 1, hence 15 is in the sequence.
%t Select[Range[100], Abs[Re[Zeta[1/2 + #*I]]] < 1 && Abs[Im[Zeta[1/2 + #*I]]] < 1 &] (* _Vaclav Kotesovec_, Feb 18 2021 *)
%o (PARI) isok(m) = my(x=zeta(1/2+m*I)); (abs(real(x)) < 1) && (abs(imag(x)) < 1); \\ _Michel Marcus_, Feb 18 2021
%K nonn
%O 1,2
%A _Benoit Cloitre_, Aug 06 2002
%E Corrected by _Michel Marcus_, Feb 18 2021