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Search: id:A065019
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| A065019 |
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Let phi be the golden number {1+sqrt(5)}/2 (A001622), let phi(n) be the number phi written in base 10 but truncated to n decimal digits. Sequence gives number of 1's at the beginning of the continued fraction expansion of phi(n). |
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3
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| 1, 3, 5, 11, 11, 13, 15, 17, 19, 21, 25, 27, 29, 31, 35, 35, 39, 41, 45, 49, 49, 51, 53, 55, 57, 61, 63, 65, 67, 69, 73, 75, 77, 81, 83, 83, 87, 91, 95, 95, 99, 99, 103, 103, 105, 107, 113, 113, 115, 117, 121, 123, 125, 129, 131, 133, 135, 137, 139, 141, 143, 147, 149
(list; graph; listen)
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OFFSET
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0,2
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COMMENT
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a(n) has the curious property of always being odd but is otherwise quite random. Nevertheless c = lim(n -> infinity) a(n)/n exists, about 2.3926 +/- 0.0004.
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FORMULA
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The value of lim n -> infinity a(n)/n is ln(10)/2/ln(phi)=2, 3924...
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EXAMPLE
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phi(6)=1.618033. The continued fraction expansion of phi(6) = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 129}. Hence a(6) = 15.
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MATHEMATICA
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gr = RealDigits[ N[ GoldenRatio, 250]] [[1]]; f[n_] := Block[ {k = 1}, While[ ContinuedFraction[ FromDigits[ {Take[ gr, n + 1 ], 1} ]] [[k]] == 1, k++ ]; k - 1]; Table[ f[n], {n, 0, 70} ]
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CROSSREFS
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Cf. A001622.
Sequence in context: A129738 A105603 A122133 this_sequence A071328 A006538 A066281
Adjacent sequences: A065016 A065017 A065018 this_sequence A065020 A065021 A065022
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KEYWORD
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nonn,base
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AUTHOR
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Benoit Cloitre (benoit7848c(AT)orange.fr) and Boris Gourevitch (boris(AT)314.net), Nov 02 2001
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EXTENSIONS
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Additional comments from Robert G. Wilson v (rgwv(AT)rgwv.com), Nov 02 2001
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