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%I #15 May 16 2020 15:26:18
%S 1,2,3,4,5,6,7,23,56,77,103,320,1477,1821,2992,15290,180168,410498,
%T 548816,672732,2601223
%N Minimal exponents m such that the fractional part of (11/10)^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 (11/10)^m is greater than the fractional part of (11/10)^k for all k, 1<=k<m.
%C The next such number must be greater than 2*10^5.
%C a(22) > 10^7. _Robert Price_, Mar 19 2019
%F Recursion: a(1):=1, a(k):=min{ m>1 | fract((11/10)^m) > fract((11/10)^a(k-1))}, where fract(x) = x-floor(x).
%e a(8)=23, since fract((11/10)^23)= 0.9543..., but fract((11/10)^k)<0.95 for 1<=k<=22;
%e thus fract((11/10)^23)>fract((11/10)^k) for 1<=k<23 and 23 is the minimal exponent > 7 with this property.
%t p = 0; Select[Range[1, 50000],
%t If[FractionalPart[(11/10)^#] > p, p = FractionalPart[(11/10)^#];
%t True] &] (* _Robert Price_, Mar 19 2019 *)
%o (Python)
%o A153687_list, m, n, k, q = [], 1, 11, 10, 0
%o while m < 10**4:
%o r = n % k
%o if r > q:
%o q = r
%o A153687_list.append(m)
%o m += 1
%o n *= 11
%o k *= 10
%o q *= 10 # _Chai Wah Wu_, May 16 2020
%Y Cf. A153663, A153671, A153683, A153679, A154130, A153695, A091560, A153711, A153719.
%K nonn,more
%O 1,2
%A _Hieronymus Fischer_, Jan 06 2009
%E a(18)-a(21) from _Robert Price_, Mar 19 2019
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