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%I #8 Oct 15 2013 22:32:23
%S 6,14,26,58,142,326,734,1713,3713,8057,17869,38985,84046,180010,
%T 385846,823687,1737474,3680099,7755978,16282918,34142786,71419857,
%U 148960009,310320958,645191390,1339363921,2777235410,5750237373,11891042257,24563702542,50684981730
%N Smallest integer at which the value of truncated Mertens function equals 2^n.
%C It appears that the ratio of a(j+1)/a(j) is a bit larger than 2 and perhaps tends to 2. Why?
%F Solutions to Min(x : A088004(x) = 2^n}, i.e. a(n) = Min(x: A002321(x) + A000720(x) = 2^n)
%t s = 0; k = 1; Do[ While[s = s + MoebiusMu[k]; s + PrimePi[k] < 2^n, k++ ]; Print[k]; k++, {n, 20}]
%Y Cf. A002321, A000720, A088004, A093772, A093773, A002110, A093774, A093775.
%K nonn
%O 1,1
%A _Labos Elemer_, May 03 2004
%E a(21) - a(24) from _Robert G. Wilson v_, May 06 2004
%E a(25)-a(31) from _Donovan Johnson_, Jun 21 2012