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%I #13 Mar 26 2019 19:20:19
%S 2,2,2,2,3,14,31,33,69,137,222,318,901,1772,2747,12347,16540,18198,
%T 135794,222246,570361,2134829,6901329,75503109,814558605
%N Greatest number m such that the fractional part of (3/2)^A153663(n) >= 1-(1/m).
%F a(n) = floor(1/(1-fract((3/2)^A153663(n)))), where fract(x) = x-floor(x).
%e a(5)=3, since 1-(1/4)=0.75>fract((3/2)^A153663(5))=fract((3/2)^12)=0.746...>=1-(1/3).
%t A153663 = {1, 5, 8, 10, 12, 14, 46, 58, 105, 157, 163, 455, 1060, 1256, 2677, 8093, 28277, 33327, 49304, 158643, 164000, 835999, 2242294, 25380333, 92600006};
%t Table[fp = FractionalPart[(3/2)^A153663[[n]]]; m = Floor[1/(1-fp)];
%t While[fp >= 1 - (1/m), m++]; m - 1, {n, 1, Length[A153663]}] (* _Robert Price_, Mar 26 2019 *)
%Y Cf. A002379, A081464, A153662, A153663, A153664, A153665, A153666, A153668.
%K nonn,more
%O 1,1
%A _Hieronymus Fischer_, Dec 31 2008
%E a(22)-a(25) from _Robert Price_, May 10 2012