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%I #10 Mar 03 2020 02:50:29
%S 1,2,11,11,964,34015,156075,952945,170942,247768,397506
%N Greatest number m such that the fractional part of e^A153702(n) <= 1/m.
%F a(n) = floor(1/fract(e^A153702(n))), where fract(x) = x - floor(x).
%e a(3) = 11 since 1/12 < fract(e^A153702(3)) = fract(e^3) = 0.0855... <= 1/11.
%t Floor[1/(#-Floor[#])]&/@Exp[Select[Range[1000],FractionalPart[E^#]<(1/#)&]] (* _Julien Kluge_, Sep 20 2016 *)
%Y Cf. A153662, A153670, A153678, A153686, A153694, A153702, A154130, A153714, A153722.
%Y Cf. A000149.
%K nonn,more
%O 1,2
%A _Hieronymus Fischer_, Jan 06 2009
%E a(10)-a(11) from _Jinyuan Wang_, Mar 03 2020
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