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%I #19 Mar 24 2019 14:37:47
%S 1,7,50,62,324,3566,66877,108201,123956,132891,182098,566593,3501843
%N Minimal exponents m such that the fractional part of (10/9)^m obtains a minimum (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 (10/9)^m is less than the fractional part of (10/9)^k for all k, 1 <= k < m.
%C The next such number must be greater than 2*10^5.
%C a(14) > 10^7. - _Robert Price_, Mar 24 2019
%F Recursion: a(1):=1, a(k):=min{ m>1 | fract((10/9)^m) < fract((10/9)^a(k-1))}, where fract(x) = x-floor(x).
%e a(2)=7, since fract((10/9)^7) = 0.09075.., but fract((10/9)^k) >= 0.11... for 1 <= k <= 6; thus fract((10/9)^7) < fract((10/9)^k) for 1 <= k < 7.
%t $MaxExtraPrecision = 100000;
%t p = 1; Select[Range[1, 10000],
%t If[FractionalPart[(10/9)^#] < p, p = FractionalPart[(10/9)^#];
%t True] &] (* _Robert Price_, Mar 24 2019 *)
%Y Cf. A081464, A153669, A153677, A153685, A153697, A154130, A153701, A137994, A153717.
%K nonn,more
%O 1,2
%A _Hieronymus Fischer_, Jan 06 2009
%E a(12)-a(13) from _Robert Price_, Mar 24 2019
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