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%I #11 Apr 18 2019 11:42:05
%S 1,7,32,53,189,2665,10810,26577,128778,483367
%N Greatest number m such that the fractional part of Pi^A153711(m) >= 1-(1/m).
%F a(n) = floor(1/(1-fract(Pi^A153711(n)))), where fract(x) = x-floor(x).
%e a(3) = 32, since 1-(1/33) = 0.9696... > fract(Pi^A153711(3)) = fract(Pi^15) = 0.96938... >= 0.96875 = 1-(1/32).
%t $MaxExtraPrecision = 100000;
%t A153711 = {1, 2, 15, 22, 58, 157, 1030, 5269, 145048, 151710};
%t Floor[1/(1-FractionalPart[Pi^A153711])] (* _Robert Price_, Apr 18 2019 *)
%Y Cf. A153663, A153671, A153679, A153687, A153695, A091560, A153711, A154130, A153723.
%Y Cf. A001672.
%K nonn,more
%O 1,2
%A _Hieronymus Fischer_, Jan 06 2009
%E a(9)-a(10) from _Robert Price_, Apr 18 2019
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