A317413 Continued fraction for binary expansion of Liouville's number interpreted in base 2 (A092874).

0, 1, 3, 3, 1, 2, 1, 4095, 3, 1, 3, 3, 1, 4722366482869645213695, 4, 3, 1, 3, 4095, 1, 2, 1, 3, 3, 1

Offset

0,3

Comments

The continued fraction of the number obtained by reading A012245 as binary fraction.

Except for the first term, the only values that occur in this sequence are 1, 2, 3, 4 and values 2^((m-1)*m!) - 1 for m > 2. The probability of occurrence P(a(n) = k) are given by:

P(a(n) = 1) = 1/3,

P(a(n) = 2) = 1/12,

P(a(n) = 3) = 1/3,

P(a(n) = 4) = 1/12 and

P(a(n) = 2^((m-1)*m!)-1) = 1/(3*2^(m-1)) for m > 2.

The next term is roughly 3.12174855*10^144 (see b-file for precise value).

Links

A.H.M. Smeets, Table of n, a(n) for n = 0..48

Formula

a(n) = 1 if and only if n in A317538.

a(n) = 2 if and only if n in {24*m - 19 | m > 0} union {24*m - 4 | m > 0}.

a(n) = 3 if and only if n in A317539.

a(n) = 4 if and only if n in {12*m + A014710(m-1) - 2*(A014710(m-1) mod 2) | m > 0}

a(n) = 2^((m-1)*m!)-1 if and only if n in {3*2^(m-2)*(1+k*4) - 1 | k >= 0} union {3*2^m-2)*(3+k*4) | k >= 0} for m > 2.

Example

0.76562505... = 0 +1/( 1+ 1/(3+1/(3+1/(1+1/(2+...))))) - R. J. Mathar, Jun 19 2021

Maple

with(numtheory): cfrac(add(1/2^factorial(n), n=1..7), 24, 'quotients'); # Muniru A Asiru, Aug 11 2018

Mathematica

ContinuedFraction[ FromDigits[ RealDigits[ Sum[1/10^n!, {n, 8}], 10, 10000], 2], 60] (* Robert G. Wilson v, Aug 09 2018 *)

Prog

(Python)
n, f, i, p, q, base = 1, 1, 0, 0, 1, 2
while i < 100000:
....i, p, q = i+1, p*base, q*base
....if i == f:
........p, n = p+1, n+1
........f = f*n
n, a, j = 0, 0, 0
while p%q > 0:
....a, f, p, q = a+1, p//q, q, p%q
....print(a-1, f)

Crossrefs

Cf. A012245, A014710, A317538, A317539.

Cf. A058304 (in base 10), A317414 (in base 3).

Sequence in context: A244328 A073067 A003637 * A359838 A110628 A107292

Adjacent sequences: A317410 A317411 A317412 * A317414 A317415 A317416

Keyword

nonn,base,cofr

Author

A.H.M. Smeets, Jul 27 2018

Status

approved