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

Author

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

Keywords

• nonn
• base

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.

Examples

			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.
		

Crossrefs

Cf. A001622.

Programs

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} ]

Formula

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

Extensions

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