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%I #14 Mar 21 2019 17:15:53
%S 1,70,209,378,1653,2697,4806,13744,66919,67873,75666,81125,173389,
%T 529938,1572706,4751419,7159431,7840546,15896994,71074288,119325567
%N Minimal exponents m such that the fractional part of (101/100)^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 (101/100)^m is less than the fractional part of (101/100)^k for all k, 1<=k<m.
%C The next term is greater than 2*10^8.
%F Recursion: a(1):=1, a(k):=min{ m>1 | fract((101/100)^m) < fract((101/100)^a(k-1))}, where fract(x) = x-floor(x).
%e a(2)=70, since fract((101/100)^70)=0.00676..., but fract((101/100)^k)>=0.01 for 1<=k<=69; thus fract((101/100)^70)<fract((101/100)^k) for 1<=k<70.
%t p = 1; Select[Range[1, 5000],
%t If[FractionalPart[(101/100)^#] < p, p = FractionalPart[(101/100)^#];
%t True] &] (* _Robert Price_, Mar 21 2019 *)
%Y Cf. A081464, A154130, A153673, A153677, A153685, A153693, A153701, A137994, A153717.
%K nonn,more
%O 1,2
%A _Hieronymus Fischer_, Jan 06 2009
%E a(15)-a(21) from _Robert Gerbicz_, Nov 22 2010
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