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%I #10 Mar 26 2019 19:20:11
%S 2,4,16,11,16799,11199,5536,92694,61796,41197,23242,55710,137921,
%T 257825,5271294,706641581,471094387,314062925
%N Greatest number m such that the fractional part of (3/2)^A153662(n) <= 1/m.
%F a(n):=floor(1/fract((3/2)^A153662(n))), where fract(x) = x-floor(x).
%e a(3)=16 since 1/17<fract((3/2)^A153662(3))=fract((3/2)^4)=0.0625=1/16.
%t A153662 = {1, 2, 4, 7, 3328, 3329, 4097, 12429, 12430, 12431, 18587, 44257, 112896, 129638, 4264691, 144941960, 144941961, 144941962};
%t Table[fp = FractionalPart[(3/2)^A153662[[n]]]; m = Floor[1/fp];
%t While[fp <= 1/m, m++]; m - 1, {n, 1, Length[A153662]}] (* _Robert Price_, Mar 26 2019 *)
%Y Cf. A002379, A081464, A153662, A153663, A153664, A153665, A153667, A153668.
%K nonn,more
%O 1,1
%A _Hieronymus Fischer_, Dec 31 2008
%E a(15)-a(18) from _Robert Price_, May 09 2012
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