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A153663
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Minimal exponents m such that the fractional part of (3/2)^m reaches a maximum (when starting with m=1).
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21
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1, 5, 8, 10, 12, 14, 46, 58, 105, 157, 163, 455, 1060, 1256, 2677, 8093, 28277, 33327, 49304, 158643, 164000, 835999, 2242294, 25380333, 92600006
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OFFSET
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1,2
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COMMENTS
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Recursive definition: a(1)=1, a(n) = least number m such that the fractional part of (3/2)^m is greater than the
fractional part of (3/2)^k for all k, 1<=k<m.
The next such number must be greater than 305000.
The fractional part of k=835999 is .999999 5 which is greater than (k-1)/k. The fractional part of k=2242294 is .999999 8 which is greater than (k-1)/k. The fractional part of k=25380333 is .999999 98 which is greater than (k-1)/k. The fractional part of k=92600006 is .999999 998 which is greater than (k-1)/k. So, all additional numbers in this sequence must be in A153664 and >3*10^8. - Robert Price, May 09 2012
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LINKS
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FORMULA
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Recursion: a(1):=1, a(k):=min{ m>1 | fract((3/2)^m) > fract((3/2)^a(k-1))}, where fract(x) = x-floor(x).
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EXAMPLE
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a(2)=5, since fract((3/2)^5)=0.59375, but fract((3/2)^k)=0.5, 0.25, 0.375, 0.0625 for 1<=k<=4; thus
fract((3/2)^5)>fract((3/2)^k) for 1<=k<5.
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MATHEMATICA
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a[1] = 1; a[n_] := a[n] = For[m = a[n-1]+1, True, m++, f = FractionalPart[(3/2)^m]; If[AllTrue[Range[m-1], f > FractionalPart[(3/2)^#]&], Print[n, " ", m]; Return[m]]];
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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