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).

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

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