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A157683
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Decimal expansion of Levy-Cantor distribution area constant.
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1
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6, 8, 8, 5, 4, 9, 5, 7, 8, 7, 0, 2, 7, 2, 6, 1, 8, 6, 9, 6, 3, 1, 9, 5, 5, 2, 1, 5, 5, 5, 7, 7, 6, 7, 6, 1, 1, 7, 4, 8, 1, 3, 9, 4, 6, 6, 5, 6, 8, 8, 0, 4, 4, 1, 7, 9, 4, 4, 9, 3, 0, 9, 9, 4, 6, 2, 7, 5, 4, 3, 8, 7, 8, 9, 0, 5, 3, 6, 7, 3, 9, 4, 2, 9, 5, 6, 8, 0, 3, 2, 1, 1, 3, 8, 4, 3, 2, 6, 1, 8
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OFFSET
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1,1
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COMMENTS
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The idea that the Levy distribution could be Bernoulli-like gave me this result:
m = mean, a = Fractal dimension as the Levy exponent,
f(x, a, m) = (1 + a) * log(x - m)/((x - m)^(1 + a) - 1)
On a {0,1} domain with mean = 1/2 you have to use the absolute value function.
It gives a cusp or Lambda point at 1/2.
Here is what it came from:
f(x, m, t) = t/((x-m)^(1+a) + t^(1+a)): Levy distribution;
as Bernoulli in t:
f(t) = t/(exp(t) - 1), so exp(t) = exp(log[(x-m)^(1+a) + t^(1+a) + 1)), and it gives t = (1+a)*log(x-m) + log(1 + (t^(1+a) + 1)/(x-m)^(1+a))
Limit for large (x-m)^(1+a) is: t = (1+a)*log(x-m).
In the Bernoulli f(t) that gives the distribution. Cantor dimension is a = log(2)/log(3). So taking it with m = 0, you get a finite constant area of a long tailed Levy-Cantor distribution that is Bernoulli-Boson like.
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LINKS
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EXAMPLE
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6.88549578702726186963195521555776761174813946656880441794493...
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MATHEMATICA
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a = Log[2]/Log[3]; m = 0; c = Integrate[(1 + a) * Log[x - m]/((x - m)^(1 + a) - 1), {x, 0, Infinity}] (* Bagula *)
RealDigits[Log[6, 3]*(PolyGamma[1, Log[6, 2]] + PolyGamma[1, Log[6, 3]]), 10, 100][[1]] (* Charles R Greathouse IV, Feb 03 2011 *)
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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