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%I #10 May 11 2019 01:44:41
%S 1,7,32,53,189,131,2665,10810,2693,1976,3697,4289,26577,483367
%N Greatest number m such that the fractional part of Pi^A153712(n) >= 1-(1/m).
%F a(n) = floor(1/(1-fract(Pi^A153712(n)))), where fract(x) = x-floor(x).
%e a(3) = 32, since 1-(1/33) = 0.9696... > fract(Pi^A153712(3)) = fract(Pi^15) = 0.96938... >= 0.96875 = 1-(1/32).
%t A153712 = {1, 2, 15, 22, 58, 109, 157, 1030, 1071, 1274, 2008, 2322,
%t 5269, 151710};
%t Table[Floor[1/(1 - FractionalPart[Pi^A153712[[n]]])], {n, 1,
%t Length[A153712]}] (* _Robert Price_, May 10 2019 *)
%Y Cf. A153664, A153672, A153680, A153688, A153696, A153704, A153712, A154130, A153724.
%Y Cf. A001672.
%K nonn,more
%O 1,2
%A _Hieronymus Fischer_, Jan 06 2009
%E a(14) from _Robert Price_, May 10 2019
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