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%I #52 Jun 19 2021 09:09:06
%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 (A092874).
%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 occurrence 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.
%e 0.76562505... = 0 +1/( 1+ 1/(3+1/(3+1/(1+1/(2+...))))) - _R. J. Mathar_, Jun 19 2021
%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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