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%I #19 May 16 2020 15:40:35
%S 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,
%T 27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,
%U 50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,110,180,783,859,1803,7591,10763,19105,50172,355146,1101696,1452050,3047334,3933030
%N Minimal exponents m such that the fractional part of (101/100)^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 (101/100)^m is greater than the fractional part of (101/100)^k for all k, 1<=k<m.
%C The next such number must be greater than 10^6.
%C a(84) > 10^7. _Robert Price_, Mar 21 2019
%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(5)=5, since fract((101/100)^5)=0.05101005, but fract((101/100)^k)=0.01, 0.0201, 0.030301, 0.04060401 for 1<=k<=4; thus fract((101/100)^5)>fract((101/100)^k) for 1<=k<5.
%t p = 0; Select[Range[1, 20000],
%t If[FractionalPart[(101/100)^#] > p, p = FractionalPart[(101/100)^#];
%t True] &] (* _Robert Price_, Mar 21 2019 *)
%o (Python)
%o A153671_list, m, n, k, q = [], 1, 101, 100, 0
%o while m < 10**4:
%o r = n % k
%o if r > q:
%o q = r
%o A153671_list.append(m)
%o m += 1
%o n *= 101
%o k *= 100
%o q *= 100 # _Chai Wah Wu_, May 16 2020
%Y Cf. A153663, A154130, A153675, A153679, A153687, A153695, A091560, A153711, A153719.
%K nonn
%O 1,2
%A _Hieronymus Fischer_, Jan 06 2009
%E a(72)-a(83) from _Robert Price_, Mar 21 2019
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