A081836 Let z(n) be the golden ratio (phi) truncated to n decimal digits; sequence gives maximum element in the continued fraction for z(n).

1, 2, 3, 5, 5, 6, 129, 15, 7, 10, 36, 65, 155, 70, 40, 20, 1122, 13, 15, 52, 52, 10, 19, 8, 87, 69, 42, 41, 30, 2036, 131, 86, 26, 41, 65, 231, 58, 161, 94, 94, 137, 137, 1323, 97, 14, 282, 15, 15, 122, 137, 80, 329, 164, 124, 748, 175, 7389, 2164, 101, 255, 201, 34, 17

Offset: 0

Author

Benoit Cloitre, Apr 11 2003

Keywords

• base
• nonn

Examples

			phi=(1+sqrt(5))/2=1.6180339887498948482045... so z(10)=1.6180339887 and the continued fraction for z(10) is [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 2, 1, 1, 4, 1, 10, 36, 2], hence a(10)=36.
		

Crossrefs

Cf. A001622, A081837 (for e), A373866 (for Pi).

Programs

Mathematica

A081836[n_] := Max[ContinuedFraction[Floor[GoldenRatio*10^n]/10^n]];
Array[A081836, 100, 0] (* Paolo Xausa, Jun 19 2024 *)