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%I #29 Mar 16 2019 15:22:00
%S 1,68,142,341,395,490,585,1164,1707,26366,41358,46074,120805,147332,
%T 184259,205661,385710,522271,3418770,3675376,9424094
%N Minimal exponents m such that the fractional part of (1024/1000)^m obtains a minimum (when starting with m=1).
%C Recursive definition: a(1)=1, a(n) is the least positive integer m such that the fractional part of (1024/1000)^m is less than the fractional part of (1024/1000)^k for all k, 1 <= k < m.
%C a(21) >= 4.5*10^6. - _David A. Corneth_, Mar 15 2019
%C a(22) > 10^7. _Robert Price_, Mar 16 2019
%F Recursion: a(1):=1, a(k):=min{ m>1 | fract((1024/1000)^m) < fract((1024/1000)^a(k-1))}, where fract(x) = x-floor(x).
%e a(2)=68, since fract((1024/1000)^68) = 0.016456..., but fract((1024/1000)^k) >= 0.024 for 1 <= k <= 67; thus fract((1024/1000)^68) < fract((1024/1000)^k) for 1 <= k < 68.
%t $MaxExtraPrecision = 10000;
%t p = .999;
%t Select[Range[1, 50000],
%t If[FractionalPart[(1024/1000)^#] < p,
%t p = FractionalPart[(1024/1000)^#]; True] &] (* _Robert Price_, Mar 15 2019 *)
%o (PARI) upto(n) = my(res = List(), r = 1, p = 1); for(i=1, n, c = frac(p *= 1.024); if(c<r, r=c; print1(i", "); listput(res,i))); res \\ _David A. Corneth_, Mar 15 2019
%Y Cf. A081464, A153669, A153681, A154130, A153685, A153693, A153701, A137994, A153717.
%K nonn,more
%O 1,2
%A _Hieronymus Fischer_, Jan 06 2009
%E a(18) from _Robert Price_, Mar 15 2019
%E a(19)-a(20) from _David A. Corneth_, Mar 15 2019
%E a(21) from _Robert Price_, Mar 16 2019
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