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%I #11 Mar 26 2019 11:47:32
%S 1,2,3,4,5,39,56,85,557,911,2919,2921,4491,11543,15724,98040,110932,
%T 126659
%N Minimal exponents m such that the fractional part of (Pi-2)^m obtains a maximum (when starting with m=1).
%C Recursive definition: a(1)=1, a(n) = least number m>a(n-1) such that the fractional part of (Pi-2)^m is greater than the fractional part of (Pi-2)^k for all k, 1<=k<m.
%C The next such number must be greater than 200000.
%C a(19) > 300000. - _Robert Price_, Mar 26 2019
%F Recursion: a(1)=1, a(k) = min{ m>1 | fract((Pi-2)^m) > fract((Pi-2)^a(k-1))}, where fract(x) = x-floor(x).
%e a(6)=39, since fract((Pi-2)^39)= 0.9586616565..., but fract((Pi-2)^k)<=0.9389018... for 1<=k<=38; thus fract((Pi-2)^39)>fract((Pi-2)^k) for 1<=k<39 and 39 is the minimal exponent > 5 with this property.
%t $MaxExtraPrecision = 100000;
%t p = 0; Select[Range[1, 10000],
%t If[FractionalPart[(Pi - 2)^#] > p, p = FractionalPart[(Pi - 2)^#];
%t True] &] (* _Robert Price_, Mar 26 2019 *)
%Y Cf. A153663, A153671, A153679, A153687, A153695, A153707, A153715, A153723, A154130.
%K nonn,more
%O 1,2
%A _Hieronymus Fischer_, Jan 06 2009
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