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Search: id:A153671
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| A153671 |
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Minimal exponents m such that the fractional part of (101/100)^m obtains a maximum (when starting with m=1). |
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+0
15
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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, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 110, 180
(list; graph; listen)
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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 (101/100)^m is greater than the
fractional part of (101/100)^k for all k, 1<=k<m.
The next such number must be greater than 10^6.
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FORMULA
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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).
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EXAMPLE
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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.
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CROSSREFS
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Cf. A153663, A154130, A153675, A153679, A153687, A153695, A153703, A153711, A153719.
Sequence in context: A130734 A090108 A090109 this_sequence A090107 A130696 A146297
Adjacent sequences: A153668 A153669 A153670 this_sequence A153672 A153673 A153674
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KEYWORD
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nonn
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AUTHOR
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Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jan 06 2009
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