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A317413
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Continued fraction for binary expansion of Liouville's number interpreted in base 2 (A092874).
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6
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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
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OFFSET
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0,3
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COMMENTS
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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).
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LINKS
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FORMULA
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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.
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EXAMPLE
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0.76562505... = 0 +1/( 1+ 1/(3+1/(3+1/(1+1/(2+...))))) - R. J. Mathar, Jun 19 2021
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MAPLE
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with(numtheory): cfrac(add(1/2^factorial(n), n=1..7), 24, 'quotients'); # Muniru A Asiru, Aug 11 2018
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MATHEMATICA
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ContinuedFraction[ FromDigits[ RealDigits[ Sum[1/10^n!, {n, 8}], 10, 10000], 2], 60] (* Robert G. Wilson v, Aug 09 2018 *)
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PROG
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(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)
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CROSSREFS
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KEYWORD
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nonn,base,cofr
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AUTHOR
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STATUS
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approved
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