A317907 Number of binary places to which n-th convergent of continued fraction expansion of Khintchine's constant matches the correct value.

0, -1, 5, 3, 9, 8, 12, 14, 16, 22, 25, 27, 30, 33, 39, 44, 42, 49, 52, 51, 56, 55, 64, 70, 73, 77, 81, 83, 82, 85, 88, 92, 93, 99, 101, 104, 109, 104, 111, 114, 117, 120, 122, 124, 126, 129, 131, 133, 136, 139, 138, 144, 138, 148, 151, 150, 153, 156, 158, 162

Offset

1,3

Comments

For number of correct decimal digits see A317908.

For the similar case of number of correct binary digits of Pi see A305879.

For the similar case of number of correct binary digits of log(2) see A317557.

The denominator of the k-th convergent obtained from a continued fraction satisfying the Gauss-Kuzmin distribution will tend to exp(k*A100199), A100199 being the inverse of Lévy's constant; the error between the k-th convergent and the constant itself tends to exp(-2*k*A100199), or in binary digits 2*k*A100199/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

A.H.M. Smeets, Table of n, a(n) for n = 1..19999

Formula

Lim_{n -> oo} a(n)/n = 2*log(A086702)/log(2) = 2*A100199/log(2) = 2*A305607.

Example

   n   convergent         binary expansion       a(n)
  ==  =============  ==========================  ====
   1     2 / 1       10.0...                       0
   2     3 / 1       11.0...                      -1
   3     8 / 3       10.101010...                  5
   4    43 / 16      10.1011...                    3
   5    51 / 19      10.1010111100...              9
  oo  lim = A317906  10.101011110111100111...     --

Prog

(Python)
i, cf = 0, []
while i <= 20100:
....c = A002211(i)
....cf, i = cf+[c], i+1
p0, p1, q0, q1, i, base = cf[0], 1, 1, 0, 1, 2
while i <= 20100:
....p0, p1, q0, q1, i = cf[i]*p0+p1, p0, cf[i]*q0+q1, q0, i+1
a0 = p0//q0
p0 = p0-a0*q0
i, p0, dd = 0, p0*base, [a0]
while i < 70000:
....d, p0, i = p0//q0, (p0%q0)*base, i+1
....dd = dd+[d]
n, pn0, pn1, qn0, qn1 = 1, a0, 1, 1, 0
while n <= 20000:
....p, q = pn0, qn0
....if p//q != a0:
........print(n, "- manual!")
....else:
........i, p, di = 0, (p%q)*base, a0
........while di == dd[i]:
............i, di, p = i+1, p//q, (p%q)*base
........print(n, i-1)
....n, pn0, pn1, qn0, qn1 = n+1, cf[n]*pn0+pn1, pn0, cf[n]*qn0+qn1, qn0

Crossrefs

Cf. A002210, A002211, A086702, A100199, A305607, A317906, A317908.

Sequence in context: A332528 A019955 A296345 * A075693 A198134 A082454

Adjacent sequences: A317904 A317905 A317906 * A317908 A317909 A317910

Keyword

sign,base

Author

A.H.M. Smeets, Aug 10 2018

Status

approved