A318058 a(n) is the number of decimal places to which the n-th convergent of the continued fraction expansion of the golden section matches the correct value.

0, -1, 0, 1, 1, 1, 2, 2, 2, 3, 2, 4, 4, 5, 5, 5, 6, 5, 7, 7, 8, 7, 9, 9, 9, 10, 10, 10, 11, 10, 12, 12, 13, 12, 13, 14, 15, 14, 15, 16, 16, 16, 17, 17, 17, 18, 18, 19, 18, 20, 20, 21, 21, 21, 22, 22, 23, 23, 24, 23, 24, 25, 25, 26, 26, 27, 27, 27, 28, 28, 29, 29, 29, 30, 30

Offset: 1

Author

A.H.M. Smeets, Aug 14 2018

Keywords

• sign
• base

Comments

The correct decimal value of the golden section is given in A001622; the continued fraction terms of the golden section is given in A000012.

For the number of correct decimal digits of the golden section see A318057.

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.

Examples

			   n   convergent         decimal expansion     a(n)
  ==  =============  =========================  ====
   1    1 / 1         1.0                         0
   2    2 / 1         2.0                        -1
   3    3 / 2         1.5                         0
   4    5 / 3         1.66                        1
   5    8 / 5         1.60                        1
   6   13 / 8         1.62                        1
   7   21 / 13        1.615                       2
   8   34 / 21        1.619                       2
   9   55 / 34        1.617                       2
  10   89 / 55        1.6181                      3
  oo  lim = A001622   1.6180339887498948482      --
		

Crossrefs

Cf. A000012, A001622, A002390, A097348, A202543, A318057.

Programs

Python

p, q, i, base = 1, 1, 0, 10
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)

Formula

Limit_{n -> oo} a(n)/n = 2*log(A001622)/log(10) = 2*A002390/log(10) = A202543/log(10) = 2*A097348.