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%I #29 Feb 22 2020 20:54:24
%S 0,2,4,8,1,3,2,531440,1,1,3,1,8,4,2,
%T 22528399544939174411840147874772640,1,1,4,8,1,3,1,1,531440,2,3,1,8,4,
%U 2
%N Continued fraction for ternary expansion of Liouville's number interpreted in base 3 (A012245).
%C The continued fraction of the number obtained by reading A012245 as a ternary fraction.
%C Except for the first term, the only values that occur in this sequence are 1,2,3,4, and values 3^((m-1)*m!)-1 for m > 1. The probability of occurrence P(a(n) = k) are given by:
%C P(a(n) = 1) = 3/8,
%C P(a(n) = 2) = 1/8,
%C P(a(n) = 3) = 1/8,
%C P(a(n) = 4) = 1/8 and
%C P(a(n) = 3^((m-1)*m!)-1) = 2^-(m+1) for m > 1.
%C More generally it seems that for any base > 2, P(a(n) <= base+1) = 3/4, P(a(n) > base+1) = 1/4, and P(a(n) = base^((m-1)*m!)-1) = 2^-(m+1) for m > 1.
%H A.H.M. Smeets, <a href="/A317414/b317414.txt">Table of n, a(n) for n = 0..62</a>
%F a(n) = 1 if and only if n in {floor(8*n/3) + A317627(n) | n > 0}.
%F a(n) = 2 if and only if n in {8*n - 10 + 3*A089013(n-1) | n > 0}.
%F a(n) = 3 if and only if n in {16*n - 11 | n > 0} union {16*n - 6 | n > 0}.
%F a(n) = 4 if and only if n in {16*n - 14 | n > 0} union {16*n - 3 | n > 0}.
%F a(n) = 3^((m-1)*m!)-1 iff n in {2^m*(1+k*4) - 1 | k >= 0} union {2^m*(3+k*4) | k >= 0} for m > 1.
%p with(numtheory): cfrac(add(1/3^factorial(n),n=1..7),30,'quotients'); # _Muniru A Asiru_, Aug 11 2018
%t ContinuedFraction[ FromDigits[ RealDigits[ Sum[1/10^n!, {n, 8}], 10, 10000], 3], 60] (* _Robert G. Wilson v_, Aug 09 2018 *)
%o (Python)
%o n,f,i,p,q,base = 1,1,0,0,1,3
%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, A089013, A317627.
%Y Cf. A058304 (in base 10), A317413 (in base 2), A317661 (in base 4).
%K nonn
%O 0,2
%A _A.H.M. Smeets_, Jul 27 2018
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