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Search: id:A153679
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| A153679 |
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Minimal exponents m such that the fractional part of (1024/1000)^m obtains a maximum (when starting with m=1). |
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13
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| 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, 27, 28, 29, 82, 134, 1306, 2036, 6393, 34477, 145984
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OFFSET
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1,2
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COMMENT
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Recursive definition: a(1)=1, a(n) = least number m>a(n-1) such that the fractional part of (1024/1000)^m is greater than
the
fractional part of (1024/1000)^k for all k, 1<=k<m.
The next such number must be greater than 5*10^5.
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FORMULA
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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).
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EXAMPLE
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a(30)=82, since fract((1024/1000)^82)= 0.99191990..., but fract((1024/1000)^k)<0.9893 for 1<=k<=81;
thus fract((1024/1000)^82)>fract((1024/1000)^k) for 1<=k<82 and 82 is the minimal exponent >29 with this property.
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CROSSREFS
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Cf. A153663, A153671, A153675, A154130, A153687, A153695, A153703, A153711, A153719.
Sequence in context: A004441 A004438 A109425 this_sequence A160546 A087143 A080683
Adjacent sequences: A153676 A153677 A153678 this_sequence A153680 A153681 A153682
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KEYWORD
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nonn,more
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AUTHOR
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Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jan 06 2009
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