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A317413 Continued fraction for binary expansion of Liouville's number interpreted in base 2 (A012245). 6

%I

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

%T 1,3,3,1

%N Continued fraction for binary expansion of Liouville's number interpreted in base 2 (A012245).

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

%C 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 occurence P(a(n) = k) are given by:

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

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

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

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

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

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

%H A.H.M. Smeets, <a href="/A317413/b317413.txt">Table of n, a(n) for n = 0..48</a>

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

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

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

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

%F 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.

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

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

%o (Python)

%o n,f,i,p,q,base = 1,1,0,0,1,2

%o while i < 100000:

%o ....i,p,q = i+1,p*base,q*base

%o ....if i == f:

%o ........p,n = p+1,n+1

%o ........f = f*n

%o n,a,j = 0,0,0

%o while p%q > 0:

%o ....a,f,p,q = a+1,p//q,q,p%q

%o ....print(a-1,f)

%Y Cf. A012245, A014710, A317538, A317539.

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

%K nonn,base,cofr

%O 0,3

%A _A.H.M. Smeets_, Jul 27 2018

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Last modified December 11 18:34 EST 2019. Contains 329925 sequences. (Running on oeis4.)