|
%I #13 Mar 19 2019 11:28:24
%S 1,17,37,237,599,615,6638,13885,1063942,9479731
%N Minimal exponents m such that the fractional part of (11/10)^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 (11/10)^m is less 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(11) > 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(2)=17, since fract((11/10)^17)=0.05447.., but fract((11/10)^k)>=0.1 for 1<=k<=16; thus fract((11/10)^17)<fract((11/10)^k) for 1<=k<17.
%t p = 1; Select[Range[1, 50000],
%t If[FractionalPart[(11/10)^#] < p, p = FractionalPart[(11/10)^#];
%t True] &] (* _Robert Price_, Mar 19 2019 *)
%Y Cf. A081464, A153669, A153689, A153677, A154130, A153693, A153701, A137994, A153717.
%K nonn,more
%O 1,2
%A _Hieronymus Fischer_, Jan 06 2009
%E a(9)-a(10) from _Robert Price_, Mar 19 2019
| 