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%I #10 Apr 18 2019 11:41:28
%S 1,1,1,3,16,24,45,158,410,946,1182,8786,16159,20188,61392,78800,78959,
%T 217556
%N Greatest number m such that the fractional part of (Pi-2)^A153719(m) >= 1-(1/m).
%F a(n) = floor(1/(1-fract((Pi-2)^A153719(n)))), where fract(x) = x-floor(x).
%e a(5) = 16, since 1-(1/17) = 0.941176... > fract((Pi-2)^A153719(5)) = fract((Pi-2)^5) = 0.9389... >= 0.9375 = 1-(1/16).
%t $MaxExtraPrecision = 100000;
%t A153719 = {1, 2, 3, 4, 5, 39, 56, 85, 557, 911, 2919, 2921, 4491,
%t 11543, 15724, 98040, 110932, 126659};
%t Floor[1/(1 - FractionalPart[(Pi - 2)^A153719])] (* _Robert Price_, Apr 18 2019 *)
%Y Cf. A153663, A153671, A153679, A153687, A153695, A091560, A153711, A153719, A154130.
%K nonn,more
%O 1,4
%A _Hieronymus Fischer_, Jan 06 2009