%I #14 Mar 26 2019 11:46:59
%S 1,2,11,11,23,28,69,85,115,964,1153,1292,1296,1877,34015,156075,952945
%N Greatest number m such that the fractional part of e^A153701(n) <= 1/m.
%F a(n) = floor(1/fract(e^A153701(n))), where fract(x) = x-floor(x).
%e a(3)=11 since 1/12 < fract(e^A153701(3)) = fract(e^3) = 0.0855... <= 1/11.
%t A153701 = {1, 2, 3, 9, 29, 45, 75, 135, 219, 732, 1351, 3315, 4795,
%t 4920, 5469, 28414, 37373};
%t Table[fp = FractionalPart[E^A153701[[n]]]; m = Floor[1/fp];
%t While[fp <= 1/m, m++]; m - 1, {n, 1, Length[A153701]}] (* _Robert Price_, Mar 25 2019 *)
%Y Cf. A081464, A153669, A153677, A153685, A153693, A153701, A154130, A153713, A153721.
%Y Cf. A000149.
%K nonn,more
%O 1,2
%A _Hieronymus Fischer_, Jan 06 2009