

A065019


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).


2



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
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OFFSET

0,2


COMMENTS

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.


LINKS

Table of n, a(n) for n=0..62.


FORMULA

The value of lim n > infinity a(n)/n is log(10)/2/log(phi)=2.3924...


EXAMPLE

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.


MATHEMATICA

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} ]


CROSSREFS

Cf. A001622.
Sequence in context: A170835 A122133 A273377 * A071328 A273037 A006538
Adjacent sequences: A065016 A065017 A065018 * A065020 A065021 A065022


KEYWORD

nonn,base


AUTHOR

Benoit Cloitre and Boris Gourevitch (boris(AT)314.net), Nov 02 2001


EXTENSIONS

Additional comments from Robert G. Wilson v, Nov 02 2001


STATUS

approved



