|
%I #10 Mar 26 2019 15:55:46
%S 7,7,38,318,393,396,484,2076,2619,4099,5264,8556,18070,20732,27209,
%T 73351,356362
%N Greatest number m such that the fractional part of (Pi-2)^A153717(n) <= 1/m.
%F a(n) = floor(1/fract((Pi-2)^A153717(n))), where fract(x) = x-floor(x).
%e a(3)=38 since 1/39<fract((Pi-2)^A153717(3))=fract((Pi-2)^23)=0.02600...<=1/38.
%t A153717 = {1, 20, 23, 24, 523, 2811, 3465, 3776, 4567, 6145, 8507, 9353, 19790, 41136, 62097, 72506, 107346};
%t Table[fp = FractionalPart[(Pi - 2)^A153717[[n]]]; m = Floor[1/fp];
%t While[fp <= 1/m, m++]; m - 1, {n, 1, Length[A153717]}] (* _Robert Price_, Mar 26 2019 *)
%Y Cf. A081464, A153669, A153677, A153685, A153693, A153701, A137994, A153717, A154130.
%K nonn,more
%O 1,1
%A _Hieronymus Fischer_, Jan 06 2009
|
