%I #10 Mar 03 2020 06:56:47
%S 7,159,50,10,21,55,117,270,307,744,757,7804,13876,62099,70718,154755
%N Greatest number m such that the fractional part of Pi^A153710(n) <= 1/m.
%F a(n) = floor(1/fract(Pi^A153710(n))), where fract(x) = x-floor(x).
%e a(2)=159 since 1/160<fract(Pi^A153710(2))=fract(Pi^3)=0.0062766...<=1/159.
%t A153710 = {1, 3, 5, 9, 10, 11, 59, 81, 264, 281, 472, 3592, 10479,
%t 12128, 65875, 118885};
%t Table[fp = FractionalPart[Pi^A153710[[n]]]; m = Floor[1/fp];
%t While[fp <= 1/m, m++]; m - 1, {n, 1, Length[A153710]}] (* _Robert Price_, May 10 2019 *)
%Y Cf. A153662, A153670, A153678, A153686, A153694, A153702, A153710, A154130, A153722.
%Y Cf. A001672.
%K nonn,more
%O 1,1
%A _Hieronymus Fischer_, Jan 06 2009
%E a(16) from _Robert Price_, May 10 2019