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A153687
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Minimal exponents m such that the fractional part of (11/10)^m obtains a maximum (when starting with m=1).
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11
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1, 2, 3, 4, 5, 6, 7, 23, 56, 77, 103, 320, 1477, 1821, 2992, 15290, 180168
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| 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.
The next such number must be greater than 2*10^5.
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FORMULA
| 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).
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EXAMPLE
| a(8)=23, since fract((11/10)^23)= 0.9543..., but fract((11/10)^k)<0.95 for 1<=k<=22;
thus fract((11/10)^23)>fract((11/10)^k) for 1<=k<23 and 23 is the minimal exponent > 7 with this property.
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CROSSREFS
| Cf. A153663, A153671, A153683, A153679, A154130, A153695, A091560, A153711, A153719.
Sequence in context: A070759 A106275 A031054 * A142594 A010351 A183530
Adjacent sequences: A153684 A153685 A153686 * A153688 A153689 A153690
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KEYWORD
| nonn,more
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AUTHOR
| Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jan 06 2009
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