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%I #9 May 11 2019 01:44:36
%S 7,3,38,318,78,83,265,185,73351,356362
%N Greatest number m such that the fractional part of (Pi-2)^A153718(n) <= 1/m.
%F a(n) = floor(1/fract((Pi-2)^A153718(n))), where fract(x) = x-floor(x).
%e a(3) = 38 since 1/39 < fract((Pi-2)^A153718(3)) = fract((Pi-2)^23) = 0.02600... <= 1/38.
%t A153718 = {1, 2, 23, 24, 35, 41, 65, 182, 72506, 107346};
%t Table[Floor[1/FractionalPart[(Pi - 2)^A153718[[n]]]], {n, 1,
%t Length[A153718]}] (* _Robert Price_, May 10 2019 *)
%Y Cf. A153662, A153670, A153678, A153686, A153694, A153702, A153710, A153718, A154130.
%K nonn,more
%O 1,1
%A _Hieronymus Fischer_, Jan 06 2009
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