A318057 a(n) is the number of binary places to which n-th convergent of continued fraction expansion of the golden section matches the correct value.

0, -2, 3, 2, 5, 2, 6, 9, 10, 9, 13, 12, 15, 16, 19, 16, 20, 22, 24, 25, 27, 29, 28, 30, 33, 32, 36, 32, 38, 32, 41, 42, 44, 45, 46, 47, 50, 48, 52, 54, 53, 56, 53, 58, 59, 60, 64, 62, 66, 62, 67, 69, 71, 73, 75, 74, 77, 78, 80, 82, 81, 84, 81, 87, 81, 88, 90

Offset

1,2

Comments

The correct binary value of the golden section is given in A068432; the continued fraction terms of the golden section is given in A000012.

For the number of correct decimal digits of the golden section see A318058.

The denominator of the k-th convergent obtained from a continued fraction tend to k*A001622; the error between the k-th convergent and the constant itself tends to 1/(2*k*A001622), or in binary digits 2*k*log(A001622)/log(2) bits after the binary point.

The sequence for quaternary digits is obtained by floor(a(n)/2), the sequence for octal digits is obtained by floor(a(n)/3), and the sequence for hexadecimal digits is obtained by floor(a(n)/4).

Links

Table of n, a(n) for n=1..67.

Formula

Lim {n -> oo} a(n)/n = 2*log(A001622)/log(2) = 2*A002390/log(2) = A202543/log(2) = 2*A242208.

Example

   n   convergent         binary expansion       a(n)
  ==  =============  ==========================  ====
   1    1 / 1          1.0                         0
   2    2 / 1         10.0                        -2
   3    3 / 2          1.1000                      3
   4    5 / 3          1.101                       2
   5    8 / 5          1.100110                    5
   6   13 / 8          1.101                       2
   7   21 / 13         1.1001110                   6
   8   34 / 21         1.1001111001                9
   9   55 / 34         1.10011110000              10
  10   89 / 55         1.1001111001                9
  oo  lim = A068432    1.1001111000110111011110   --

Prog

(Python)
p, q, i, base = 1, 1, 0, 2
while i < 20200:
....p, q, i = p+q, p, i+1
a0, p, q = p//q, q, p
i, p, dd = 0, p*base, [0]
while i < 30000:
....d, p, i = p//q, (p%q)*base, i+1
....dd = dd+[d]
n, pn, qn = 0, 1, 0
while n < 20000:
....n, pn, qn = n+1, pn+qn, pn
....if pn//qn != a0:
........print(n, "- manual!")
....else:
........i, p, q, di = 0, (pn%qn)*base, qn, 0
........while di == dd[i]:
............i, di, p = i+1, p//q, (p%q)*base
........print(n, i-1)

Crossrefs

Cf. A000012, A001622, A002390, A068432, A202543, A242208, A318058.

Sequence in context: A099552 A252749 A076199 * A136096 A284694 A072591

Adjacent sequences: A318054 A318055 A318056 * A318058 A318059 A318060

Keyword

sign,base

Author

A.H.M. Smeets, Aug 14 2018

Status

approved