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%I #15 Jun 20 2017 23:31:23
%S 3,5,7,8,10,12,13,15,17,19,20,22,24,25,27,29,31,32,34,36,37,39,41,43,
%T 44,46,48,49,51,53,54,56,58,60,61,63,65,66,68,70,72,73,75,77,78,80,82,
%U 84,85,87,89,90,92,94,96,97,99,101,102,104,106,107,109,111,113
%N Least k such that G(k) > 3 - 1/2^n, where G(k) is the sum of the first k terms of the geometric series 1 + 2/3 + (2/3)^2 + ....
%C Many of terms in this sequence are that same as A186219(n+2) but not all.
%D Mohammad K. Azarian, Geometric Series, Problem 329, Mathematics and Computer Education, Vol. 30, No. 1, Winter 1996, p. 101. Solution published in Vol. 31, No. 2, Spring 1997, pp. 196-197.
%H G. C. Greubel, <a href="/A185050/b185050.txt">Table of n, a(n) for n = 0..5000</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/GeometricSeries.html">Geometric Series</a>
%e a(1) = 5 because 1 + 2/3 + (2/3)^2 + (2/3)^3 + (2/3)^4 > 3 - 1/2.
%t lst = {}; n = s = 0; Do[s = s + (2/3)^k; If[s > 3 - 1/2^n, AppendTo[lst, k + 1]; n++], {k, 0, 112}]; lst
%K nonn
%O 0,1
%A _Arkadiusz Wesolowski_, Dec 25 2012
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