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

Author

A.H.M. Smeets, Jul 27 2018

Keywords

• nonn
• base
• cofr

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).

Examples

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

Links

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

Crossrefs

Cf. A012245, A014710, A317538, A317539.

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

Programs

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 *)

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)

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.