%I #43 Jan 09 2021 02:55:35
%S 1,2,3,5,6,8,11,13,15,18,20,23,25,28,30,33,37,39,42,46,49,52,55,58,61,
%T 64,68,71,74,78,82,85,89,92,95,99,103,107,110,114,118,121,126,129,133,
%U 137,140,144,148,153,156,160,165,168,172,176,180,184,189,193,197
%N Positions of ones in A275737.
%C a(n) appears to be asymptotic to log((n+1)!). The question at MathOverflow discusses a related but more complicated sequence.
%H Jinyuan Wang, <a href="/A275341/b275341.txt">Table of n, a(n) for n = 1..10000</a> (terms 1..1000 from G. C. Greubel)
%H Mats Granvik, <a href="http://mathoverflow.net/questions/237796/what-explains-the-asymptotic-and-the-pattern-in-this-sequence-related-to-riemann">What explains the asymptotic and the pattern in this sequence related to Riemann zeta zeros?</a>.
%F a(n) is the positions of ones in round(im(zetazero(n + 1))/(2*Pi)) - round(im(zetazero(n))/(2*Pi)), where n starts at 1.
%e The sequence A275737: round(im(zetazero(n + 1))/(2*Pi)) - round(im(zetazero(n))/(2*Pi)) where n=1,2,3,4,5, starts:
%e 1, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1,...
%e The positions of ones in that sequence are a(n):
%e 1, 2, 3, 5, 6, 8, 11, 13, 15, 18, 20, 23, 25, 28, 30,...
%e Compare this to round(log((n+1)!)) A046654:
%e 1, 2, 3, 5, 7, 9, 11, 13, 15, 18, 20, 23, 25, 28, 31,...
%t Flatten[Position[Differences[Round[Im[ZetaZero[Range[135]]]/(2*Pi)]], 1]]
%Y Cf. A046654, A275579, A275737.
%K nonn
%O 1,2
%A _Mats Granvik_, Jul 28 2016