A392060 Number of decimal places which are known after n terms of the continued fraction for the golden ratio.

0, 0, 1, 2, 2, 2, 3, 3, 3, 3, 3, 5, 5, 6, 6, 6, 6, 6, 8, 8, 8, 8, 10, 10, 10, 11, 11, 11, 11, 11, 13, 13, 13, 13, 14, 15, 15, 15, 16, 17, 17, 17, 18, 18, 18, 19, 19, 19, 19, 21, 21, 22, 22, 22, 23, 23, 24, 24, 24, 24, 25, 26, 26, 27, 27, 28, 28, 28, 29, 29, 30

Offset

1,4

Comments

n continued fraction terms is convergent c(n) = Fibonacci(n+1)/Fibonacci(n).

Further continued fraction terms are taken to be unknown and the worst case (maximum absolute difference) is a single additional 1 and so at worst c(n+1).

a(n) is therefore the length of agreement between c(n) and c(n+1).

It is expected n/a(n) -> 1/log_10(1+phi) = A392067 due to the magnitude of abs(c(n) - c(n+1)), but proving this depends on not having runs of consecutive equal terms long enough to push the ratio to some higher constant infinitely often.

Runs of consecutive equal terms occur when c(n) and c(n+1) span a run of 0 digits or 9 digits in the golden ratio and only several more continued fraction terms can narrow that interval enough to be sure whether 0's or 9's.

Any run of equal terms is finite since the golden ratio is not of the form x/10^y.

Links

Jwalin Bhatt, Table of n, a(n) for n = 1..10000

Example

For n=7, the continued fraction of 7 terms is c(n) = [1; 1,1,1,1,1,1] = 21/13 and the worst case continuation is c(n+1) = 34/21,
    21/13 = 1.61 53...
    34/21 = 1.61 90...
            ^ ^^ a(7) = 3 places agree

Mathematica

a[1]=0; a[n_]:=LengthWhile[RealDigits[N[FromContinuedFraction[Table[1, n]], n]][[1]]-RealDigits[N[FromContinuedFraction[Table[1, n+1]], n]][[1]], #==0&]; Array[a, 71] (* James C. McMahon, Feb 13 2026 *)

Prog

(Python)
from sympy import floor, fibonacci
from fractions import Fraction
from os.path import commonprefix
def reliable_digits_from_frac(bound1, bound2, prec):
    order = 10**prec
    trunc_bound1 = floor(bound1*order) / order
    trunc_bound2 = floor(bound2*order) / order
    return commonprefix([f'{trunc_bound1:.{prec}f}', f'{trunc_bound2:.{prec}f}'])
def a(n):
    bound1 = Fraction(fibonacci(n+1), fibonacci(n))
    bound2 = Fraction(fibonacci(n+2), fibonacci(n+1))
    rel_digits = reliable_digits_from_frac(bound1, bound2, n)
    return len(rel_digits) - 1 if rel_digits else 0

Crossrefs

Cf. A001622, A392067.

Cf. A369715 (between convergent and true value).

Sequence in context: A097561 A162345 A048689 * A069923 A095840 A131343

Adjacent sequences: A392057 A392058 A392059 * A392061 A392062 A392063

Keyword

nonn

Author

Jwalin Bhatt, Jan 27 2026

Status

approved